27 Oct 2010

Take out Homework Objective SWBAT calculate work

done on a system and energy change for a heat transfer

Do now Describe how you would make 100 L of a 08 M LiCl solution from solid LiCl

Agenda

I Do nowII Homework answersIII Thermochemistry notes and

examples Part 1 Homework p 263 13 15 17 19

(ELS) Read p 239-245

Thermochemistry

Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy

can be converted from one form to another but can not be created or destroyed

We can not accurately measure total energy of a system

Instead we measure changes in energy ΔE

ΔEsys + ΔEsurr = 0 ΔEsys=- ΔEsurr

energy gained in one place must be lost somewhere else

In chem we are usually interested in just the ΔE to the system we are studying ΔE=q+w q=heat exchange between systems w=work done on or by the system

Sign Conventions heat and work

Process SignHeat absorbed by the system (endothermic)

+q

Heat absorbed by the surroundings (exothermic)

-q

Work done by the system on surroundings

-w

Work done on the system by the surroundings

+w

Exothermic process (-q) is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process (+q) is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Work Pressure and VolumeVPw

Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Agenda

I Do nowII Homework answersIII Thermochemistry notes and

examples Part 1 Homework p 263 13 15 17 19

(ELS) Read p 239-245

Thermochemistry

Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy

can be converted from one form to another but can not be created or destroyed

We can not accurately measure total energy of a system

Instead we measure changes in energy ΔE

ΔEsys + ΔEsurr = 0 ΔEsys=- ΔEsurr

energy gained in one place must be lost somewhere else

In chem we are usually interested in just the ΔE to the system we are studying ΔE=q+w q=heat exchange between systems w=work done on or by the system

Sign Conventions heat and work

Process SignHeat absorbed by the system (endothermic)

+q

Heat absorbed by the surroundings (exothermic)

-q

Work done by the system on surroundings

-w

Work done on the system by the surroundings

+w

Exothermic process (-q) is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process (+q) is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Work Pressure and VolumeVPw

Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Thermochemistry

Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy

can be converted from one form to another but can not be created or destroyed

We can not accurately measure total energy of a system

Instead we measure changes in energy ΔE

ΔEsys + ΔEsurr = 0 ΔEsys=- ΔEsurr

energy gained in one place must be lost somewhere else

In chem we are usually interested in just the ΔE to the system we are studying ΔE=q+w q=heat exchange between systems w=work done on or by the system

Sign Conventions heat and work

Process SignHeat absorbed by the system (endothermic)

+q

Heat absorbed by the surroundings (exothermic)

-q

Work done by the system on surroundings

-w

Work done on the system by the surroundings

+w

Exothermic process (-q) is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process (+q) is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Work Pressure and VolumeVPw

Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy

can be converted from one form to another but can not be created or destroyed

We can not accurately measure total energy of a system

Instead we measure changes in energy ΔE

ΔEsys + ΔEsurr = 0 ΔEsys=- ΔEsurr

energy gained in one place must be lost somewhere else

In chem we are usually interested in just the ΔE to the system we are studying ΔE=q+w q=heat exchange between systems w=work done on or by the system

Sign Conventions heat and work

Process SignHeat absorbed by the system (endothermic)

+q

Heat absorbed by the surroundings (exothermic)

-q

Work done by the system on surroundings

-w

Work done on the system by the surroundings

+w

Exothermic process (-q) is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process (+q) is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Work Pressure and VolumeVPw

Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

ΔEsys + ΔEsurr = 0 ΔEsys=- ΔEsurr

energy gained in one place must be lost somewhere else

In chem we are usually interested in just the ΔE to the system we are studying ΔE=q+w q=heat exchange between systems w=work done on or by the system

Sign Conventions heat and work

Process SignHeat absorbed by the system (endothermic)

+q

Heat absorbed by the surroundings (exothermic)

-q

Work done by the system on surroundings

-w

Work done on the system by the surroundings

+w

Exothermic process (-q) is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process (+q) is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Work Pressure and VolumeVPw

Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Sign Conventions heat and work

Process SignHeat absorbed by the system (endothermic)

+q

Heat absorbed by the surroundings (exothermic)

-q

Work done by the system on surroundings

-w

Work done on the system by the surroundings

+w

Exothermic process (-q) is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process (+q) is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Work Pressure and VolumeVPw

Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Exothermic process (-q) is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process (+q) is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Work Pressure and VolumeVPw

Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Work Pressure and VolumeVPw

Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Ex 1

A gas expands in volume from 20 L to 60 L at a constant temperature Calculate the work done by the gas if it expandsa) against a vacuumb) against a constant pressure of 12 atm

1 Lmiddotatm = 1013 JJoule = unit of heat

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Ex 2

A gas expands from 264 mL to 971 mL at constant temperature Calculate the work done (in J) by the gas if it expands

a) against a vacuumb) against a constant pressure of 400

atm

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Ex 3

The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Homework

p 263 13 15 17 19 (ELS) Read p 239-245

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

28 Oct 2010

Objective SWBAT calculate heat change to a system given a thermochemical equation

Do now The work done when gas is compressed in a cylinder is 462 J During this process there is a heat transfer of 128 J from the gas to the surroundings Determine the signs for q and w Calculate the energy change for this process

ΔE=q+w

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Agenda

I Do nowII Homework checkIII Thermochemistry part 2Homework p 263 24 25 27 (JMS)Read p 245-252 read lab handout

and do pre-lab questions

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

A gas expands and does P-V work on the surroundings equal to 279 J At the same time it absorbs 216 J of heat from the surroundings What is the change in energy of the system

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

H = H (products) ndash H (reactants)H = heat given off or absorbed during a reaction at constant pressure

Hproducts lt HreactantsH lt 0

Hproducts gt HreactantsH gt 0 64

H = E + PV

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Thermochemical Equations

H2O (s) H2O (l) H = 601 kJ

Is H negative or positive

System absorbs heat

Endothermic

H gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)H = -8904 kJ

Is H negative or positive

System gives off heat

Exothermic

H lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

H2O (s) H2O (l)H = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of H changesH2O (l) H2O (s)H = -601 kJ

bull If you multiply both sides of the equation by a factor n then H must change by the same factor n2H2O (s) 2H2O (l)

H = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) H 1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) H 3013kJmol

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Homework

p 263 24 25 26 Read LAB DO prelab read in

bookhellip

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

1 Nov 2010

Take out pre-lab questions Objective SWBAT calculate the

amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer

Do now For the chemical hand warmer predict the sign convention for heat (q) and work (w)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

3 Lab notebooks collected tomorrow 2 lab grades

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

2 Nov 2010

Objective SWBAT relate ΔE to ΔH and complete calorimetric calculations

Do now Type your second trial results into the spreadsheet

Homework Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Comparing ΔH and ΔE ΔE = ΔH ndash RTΔn

R = 8314JKmol T = temperature in Kelvin n = number of moles

For gas reactions the change in volume can be very large because hot gases expand

For reactions not involving gas ΔH and ΔE are often very similar

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Ex 1

Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC

2CO(g) + O2(g) 2CO2(g) ΔH = -5660 kJmol

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Ex 2

What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC

C(graphite) + frac12O2(g) CO(g) ΔH = -1105 kJmol

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Calorimetry

the measurement of heat changes q=CΔt

C = heat capacity of a substance the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius (=ms)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Homework

Lab notebooks due tomorrow ndash 2 lab grades

p 263 27 28 31 32 34 36 (JR)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

3 Nov 2010

Take out Homework and Lab notebook

Objective SWBAT calculate the standard enthalpy of formation and reaction for compounds

Do now An iron bar of mass 869 g cools from 94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Agenda

I Do nowII Homework checkIII Standard Enthalpy of FormationReaction

notesproblemsHomework p 264 42 43 46 51 53 58 63

(SR)Read p 258-261 Heats of solutiondilutionTomorrow Practice problemsreviewBring your book to classTest Monday (ch 6)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Announcement

Tutor needed for 10th grade chem student

You will earn community service hours

Interested See Ms Kern

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Standard Enthalpy of Formation and Reaction

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm

f

The standard enthalpy of formation of any element in its most stable form is zeroH0 (O2) = 0f

H0 (O3) = 142 kJmolfH0 (C graphite) = 0f

H0 (C diamond) = 190 kJmolf

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atmrxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn H0 (products)f=S H0 (reactants)fS-

Hessrsquos Law When reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn H0 (products)f= S H0 (reactants)fS-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Practice Problem

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

What if the reaction is not a single step

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) H0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) H0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) H0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)H0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935

kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

H0= -25988 kJmol

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Homework

p 264 42 43 46 51 53 58 63 Bring your book to class

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) H0=-3935 kJmolH2(g) + 12O2(g) H2O(l) H0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) H0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

63

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

66

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

77

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Gibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQ

R is the gas constant (8314 JKbullmol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

186

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

G0 = RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = mst

q = Ct

t = tfinal - tinitial

65

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

How much heat is given off when an 869 g iron bar cools from 940C to 50C

s of Fe = 0444 Jg bull 0C

t = tfinal ndash tinitial = 50C ndash 940C = -890C

q = mst = 869 g x 0444 Jg bull 0C x ndash890C = -34000 J

65

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Constant-Pressure Calorimetry

No heat enters or leaves

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = mst

qcal = Ccalt

65

Reaction at Constant PH = qrxn

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid

Hsub = Hfus + Hvap

( Hessrsquos Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 27 Oct 2010
• Agenda
• Thermochemistry
• Slide 4
• Slide 5
• Sign Conventions heat and work
• Slide 7
• Slide 8
• Ex 1
• Ex 2
• Ex 3
• Homework
• 28 Oct 2010
• Agenda (2)
• Slide 15
• Enthalpy of Chemical Reactions
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Example 1
• Example 2
• Homework (2)
• 1 Nov 2010
• Lab
• 2 Nov 2010
• Comparing ΔH and ΔE
• Ex 1 (2)
• Ex 2 (2)
• Calorimetry
• Ex 2
• Ex 3
• Homework (3)
• 3 Nov 2010
• Agenda (3)
• Announcement
• Standard Enthalpy of Formation and Reaction
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Practice Problem
• What if the reaction is not a single step
• Slide 45
• Slide 46
• Slide 47
• Homework (4)
• 4 Nov 2010
• Agenda (4)
• Chapter 6 Problem Set
• Slide 52
• 8 Nov 2010
• Agenda (5)
• Thermodynamics Part 2
• Slide 56
• Spontaneous Processes
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63
• Slide 64
• Slide 65
• Slide 66
• Slide 67
• Slide 68
• Practice
• Slide 70
• Slide 71
• Slide 72
• Practice (2)
• Slide 74
• Practice (3)
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Slide 80
• Slide 81
• Slide 82
• Practice (4)
• Slide 84
• Recap Signs of Thermodynamic Values
• Slide 86
• Slide 87
• Slide 88
• Slide 89
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Wherersquos Waldo
• Slide 95
• Slide 96
• Slide 97
• Slide 98
• Sample Problem
• Sample Problem (2)