Thermodynamics

Terms/ Definitions• Thermodynamics

– Deals with the interconversion of heat an other forms of energy

• Thermochemistry– Deals with heat change in chemical reactions

• State Function– Function that depends only on the conditions (state) not on how the state was obtained

• Energy(E)– Internal energy = kinetic + potential energy

– Kinetic energy comes from molecular motion, electron motion etc.

– Potential energy comes from attractive and repulsive forces in nuclei, and interactions between molecules

• Enthalpy(H)– H = E + PV– Extensive property– State function– Can only measure the difference in enthalpy between two states

• Heat (Q,q)– Transfer of thermal energy between two bodies at different temperatures

• Work (W,w)– Form of energy, can be mechanical or non mechanical

– Mechanical work is normally pressure - volume work

– W = -PexV– Work done by the system on the surrounding = negative

– Work done by the surroundings on the system = positive

• System– Specific thing we are looking at

• Surroundings– Everything outside the system

• Universe– System + surroundings

• Open system– Can exchange both matter and energy with the surroundings

• Closed system– Can exchange energy but not matter with the surroundings

• Isolated system – No exchange of matter or energy with the surroundings

– Q = ?– Q = 0

• Thermochemical Equation– Chemical equation that includes the enthalpy change

– e.g. H2O(s) --> H2O(l) H = 6.01 kJ

• Heat flow– Endothermic, Q = +, heat absorbed by the system

– Exothermic, Q = -, heat given off by the system

• Units of energy– Joules(J), kilojoules(kJ)– Calories(cal), kilocalories(kcal)– 1 cal = 4.184 J– (Liter)(atmosphere) 1 L atm = 101.3 J

• Enthalpy– State function– Heat energy of state– Look at change between states H

• Extensive Property - depends on the amount

1 mole of the reaction as written

12

N2(g) 12

O2(g) NO(g) H90.4 kJ/mol

N2(g) O2(g) 2NO(g)

H?

2(90.4 kJ/mol)=180.8 kJ/mol

• Changes sign as reverse reaction:

• Follows Hess’ Law

12

N2(g) 12

O2(g) NO(g) H90.4 kJ/mol

NO(g) 12

N2(g) 12

O2(g) H?

90.4 kJ/mol

Hess’ LawIf a reaction can be considered to go by a series of steps, H of the reaction is the sum of H of the steps.

Reaction kJ/mol

1/2 N2(g) + 1/2 O2(g) NO(g) +90.4

NO2(g) NO(g) + 1/2 O2(g) +56.5

What is the value of H for the reaction:1/2 N2(g) + O2(g) NO2(g)

NO(g) + 1/2 O2(g) NO2(g) -56.5

1/2 N2(g) + 1/2 O2(g) NO(g) +90.4

1/2 N2(g) + O2(g) NO2(g)+33.9

Standard StatesMost stable form at 1 atmosphere and usually at 25oC

Indicate standard state by Ho

AllotropesMore than one stable form at 1 atmosphere

O2 and O3C(s) graphite and C(s) diamond

Standard heat of formation (Standard enthalpy of formation)

Heat change that occurs when 1 mole of a single product is formed from elements in their standard states.

C(s) graph + O2(g) CO2(g) -393.5 kJ

Hfo

HfoCO2 = -393.5 kJ/mol

e.g. C(s)graphiteH2(g) N2(g) Br2(l)

Fe(s) S(s) Hg(l) I2(s)

Write the formation reaction for the following:

NH4NO3(s)

benzene

aniline

N2(g) + H2(g) + O2(g)NH4NO3(s)2 3/2

C6H6(l)

C6H6(l)C(s)graphite + H2(g)6 3

C6H5NH2(l)

C6H5NH2(l) C(s)graphite + H2(g) + N2(g)6 7/2 1/2

Value of standard state of elements:

By definition, Ho formation of an element in its standard state = 0.

Which of the following will NOT have H formation = 0

H2(g) Ne(g) Cl2(g) I2(l)

Hg(s) Br2(l) Ca(s) Fe(l)

Calculate Heat of a reaction:

Hrxno

2. Heat of reaction can be calculated using Hess’ Law

1. Heat of reaction can be calculated from Heat of Formation data

Using Heat of formation dataIf data for all the reactants and products is given as , then is equal to the sum of of the products minus the sum of of the reactants.

Hfo

Hfo

Hfo

Hrxno

Hrxno Hf

o(products) Hfo(reactants)

1. Compound Ho

f kcal/mol C2H6 (g) - 20.2

CO2 (g) - 90.1 H2O (l) - 68.3

Calculate Horxn for the reaction:

C2H6(g) + O2(g) CO2(g) + H2O(l) 2 37/2

Hrxno Hf

o(products) Hfo(reactants)

Hrxno {2Hf

o(CO2(g))3Hfo(H2O(l))} {Hf

o(C2H6(g))+(72Hf

o(O2(g) )}

= {2(-90.1) + 3(-68.3)} - {1(-20.2) + 7/2 (?)}= {2(-90.1) + 3(-68.3)} - {1(-20.2) + 7/2 (0)}

= -364.9 kcal/ mol

Using Hess’ LawIf all information is not given as Hf, then need to go a different way

Use Hess’ LawGiven: Reaction Ho

(kJ/mol)

C(s)graph O2(g) CO2(g) 393.5

H2(g) 12

O2(g) H2O(l) 285.8

H2(g) 12

O2(g) H2O(g) 241.8

2C2H2(g) 5O2(g) 4CO2(g)+2H2O(l) 2598.8

Find H for the formation of C2H2 1. Write a target reaction: --> C2H2(g)2C(s) graph + H2(g)

2. Rearrange reactions so that when added together you get the target

4CO2(g)+2H2O(l) 2C2H2(g) 5O2(g) 2598.8

4C(s)graph 4O2(g) 4CO2(g) 4( 393.5)

2H2(g) O2(g) 2H2O(l) 2( 285.8)

--> 2C2H2(g)4C(s) graph + 2H2(g) +453.2

--> C2H2(g)2C(s) graph + H2(g) +453.2/2 = 226.6 kJ

Which of the following would have = 0?

H fo

A) C(s) diamondB) Hg(g)C) Xe(g)D) Br2(g)E) Cl - (aq)

Energy(E) - Internal Energy (KE + PE)Kinetic energy: comes from the

molecular motion and the electronic motion

Potential Energy: comes from attractive and repulsive forces in nuclei and from interactions between

molecules.First Law:Energy can be converted from one form to another, it can not be created or destroyed. i.e. The energy of the universe is constant.

Euniverse = Esystem + Esurroundingss = 0

Esystem = - Esurroundingss

All energies do not have to be the same form

Since we are primarily interested in what happens to a chemical system, we use the form:

E = q + w Q = heat

W = work

Sign convention

- +Heat(q) from system to from surroundings

surroundings to system (exothermic) (endothermic)

Work(w) by the system on by surroundingsthe surroundings on the system

Work includes all kinds of work, both mechanical and non-mechanical. We limit to mechanical work at this time.

Mechanical work:W = -PexV = - PV at constant P A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work is done?

A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work(in J) is done?

A) 10.0 JB) - 10.0 JC) 1.01 x 103 JD) 20.0 JE) - 1.01 x 103 J

A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work is done?

= - (5.0 atm)(4.0 L - 2.0 L) = - 10.0 L atm

1 L atm = 101.3 J

= (- 10.0 L atm) (101.3 J/ Latm) = - 1013 J = - 1.01 x 103 J

W = - PV

Relationship between E and q

E = q + w

w = - PexV E = q - PexV At constant volume DV = ?= 0

E = qv

Relationship between E and H

By definition:H = E + PVSo: H = E + (PV)At constant P: (PV) = P V

H = E + P VWhen you are looking at a reaction dealing with gases:

PV = nRT So: P V = nRT

And: H = E + nRT

Relationship between H and q

H = E + P V E = q + wH = q + w + P VW = - PexVAt constant P: Pex = P

W = - PVH = q - PV + P V

H = qP

A system at 1 atm and 11.0 L absorbs 300 J of heat and has 700 J of work performed on it. What is E for the process?

A) + 1000 JB) + 400 JC) - 400 JD) - 1000 J

A system at 1 atm and 11.0 L absorbs 300 J of heat and has 700 J of work performed on it. What is E for the process?

E = q + w q = + 300 J w = + 700 J

E = 300J + 700 J = 1000 J

CalorimetryDeals with the transfer of heat energy

Heat Capacity - Capacity of a system to store heat- Heat needed to raise the temperature of a system

by 1oC ( 1K) = J/Kq = (capacity of system)(T)

Specific Heat (c) -Heat needed to raise the temperature of 1 gramof material by 1oC (1K) = J/g K

Molar Heat Capacity (C) -

Heat needed to raise the temperature of 1 moleof material by 1oC (1K) = J/ mol K

q = msT

q = nCT

Constant volume CalorimetryBomb CalorimeterQ = E

Isolated system

qsystem = ? = 0

qsystem = qsample + qcalorimeter

qsample = - qcalorimeter

qsample not directly measurable

qcalorimeter can be measured

qcalorimeter = heat capacity of bomb(T) = - qsample

Since reaction carried out at constant volume qsample = E

Constant Pressure Calorimetry

Coffee Cup CalorimeterQ = H Isolated systemqsystem = ? = 0

Generally look at two differentkinds of systems1. Reaction in aqueous solution2. Metal solid in water

qsystem = qreaction + qsolution qreaction = - qsolution

qreaction not directly measurable qsolution can be measured

1.

qsolution = mcT = -qreaction

2.qsystem = qmetal + qwater

- qmetal = qwater

-(m)(c)T = (m)(c)T metal water

Since at constant pressureqreaction = Hreaction

0.500 L of 1.0 M Ba(NO3)2 at 25oC was mixed with 0.500 L of 1.0 MNa2SO4 at 25oC. A white precipitate forms (BaSO4) and the Temperature rises to 28.1oC. What is H/ mol of BaSO4 formed?Specific heat mix = 4.184 J/g K d of mix = 1.0 g/ mL

Assume a coffee cup calorimeter, P = constant

q transfer from system to surroundings = ? = 0Why?Coffee cup calorimeter assumes an isolated system

qsys = qrxn + qmix = 0

qrxn = - qmix Can’t measure heat of the ppt’n directly but canmeasure heat of mixing

qmix = msT = (1000mL)(1.0g/mL)(4.184J/g K)(28.1-25.0)

= 12970J = 12.970kJ

qppt = - qmix = - 12.970kJBut this is only for the amount ofBaSO4 made in this reaction and wewant per mole

0.500 L of 1.0 M Ba(NO3)2 at 25oC was mixed with 0.500 L of 1.0 MNa2SO4 at 25oC. A white precipitate forms (BaSO4) and the Temperature rises to 28.1oC. What is H/ mol of BaSO4 formed?Specific heat mix = 4.184 J/g K d of mix = 1.0 g/ mL

Need to determine the amount of BaSO4 that was formed in the reaction

Rxn: Ba(NO3)2 + Na2SO4 --> BaSO4 + 2NaNO3

Mole Ba(NO3)2 = VM = (0.500L)(1.0M) = 0.50 mol Mole Na2SO4 = VM = (0.500L)(1.0M) = 0.50 mol

Ba(NO3)2 + Na2SO4 --> BaSO4 + 2NaNO3 0.50 mol 0.50 mol

--> 0.50 molSo, the reaction produced 0.50 mol BaSO4 and that released - 12.97kJ

Therefore the production of 1 mol of BaSO4 will have H = ?

H = 2(- 12.97) = - 25.94 kJ = - 25.9 kJ

28.2 g Ni metal at 99.8 oC was placed in 150. g water at 23.50 oC. The final temperature at equilibrium was 25.00 oC. Find the specificheat of Ni in cal/ g K.

A) 0.107 cal/ g KB) 0.1066 cal/ g KC) - 0.1066 cal/ g KD) - 0.107 cal/g KE) 0.1267 cal/g K

28.2 g Ni metal at 99.8 oC was placed in 150. g water at 23.50 oC. The final temperature at equilibrium was 25.00 oC. Find the specificheat of Ni in cal/ g K.

qsystem = qmetal + qsolution qmetal = mcT for the Ni

qsolution = mcT for the water

qmetal = (28.2 g)coC

qsolution = (150 g)(1 cal/g oC)(25.0-23.5oC)

- qmetal = qsolution

- (28.2 g)coC = (150 g)(1 cal/g oC)(25.0-23.5oC)

cNi = 0.1066 cal/goC = 0.107 cal/ g K