chemical thermodynamics 1

CHEMICAL THERMODYNAMICS and THERMOCHEMISTRY

THERMODYNAMICS• the study of the changes in

energy and transfers of energy that accompany chemical and physical processes.

• We need to address 3 fundamental questions.

1. Will two (or more) substances react when they are mixed under specified conditions?

2. If they do react, what energy changes and transfers are associated with their reaction?

3. If a reaction occurs, to what extent does it occur?

THERMODYNAMICS

SYSTEMS and SURROUNDINGS

Transfer of EnergyHeat - q ... occurs by conduction, convection, or radiation.... causes a substance's temperature to change... is positive if thermal energy flows into the system... is negative if thermal energy flows out of the system- joules------------------------------------------------------- improper: "heat flow" is redundant

WORK - wW is negative if work is done by system. Air does work on the environment: W < 0.

W is positive if work is done on the system.

Hand does work (squeezing) on the balloon: W > 0 (Alternative: system does negative work because force by air pressure on thumb is opposite to the direction of motion of the thumb.)

Heat > 0Work > 0

Heat < 0Work < 0

INTERNAL ENERGYInternal Energy U: (measured in joules) Sum of random translational, rotational, and vibrational kinetic energies ∆U: change in U ∆U > 0 is a gain of internal energy∆U < 0 is a loss of internal energy

Vibrational kinetic energy in solids The hotter the object, the larger the vibrational kinetic energy

Motions of a diatomic molecule in a fluid

Law of Conservation of ENERGY

• 1st Law of Thermodynamics"Energy can neither be created nor destroyed,

but only transferred from one system to another and transformed from one form to another."

or

"The internal energy of an isolated system is constant (even though that energy may be transformed from one type to another)."

For thermodynamic systems, the 1st Law is:

∆U = q + w

Changes in Internal energy

• Example: If 1200 joules of heat are added to a system and the system does 800 joules of work on the surroundings, what is the :

1. energy change for the system, Esys

2. energy change of the surroundings, Esurr

Thermodynamic state - set of conditions that specify all of the properties of the system

Thermodynamic states include: The number of moles and identity of each substance. The physical states of each substance. The temperature of the system. The pressure of the system.

Thermodynamic Terms

Thermodynamic Processes

Greek isos: equal baros: weight adiabatos: not passable-----------------------------------------------------

Isothermal: Same temperature Isobaric: Same pressure Isochoric: Same volume Adiabatic: Zero heat flow (Q = 0)

More Thermodynamic Terms

• State functions - properties of a system that depend only on the state (Initial and Final states)– State functions are always written using capital letters.

• State functions are independent of pathway.• An analog to a state function is the energy required to climb a

mountain taking two different paths.– E1 = energy at the bottom of the mountain

– E1 = mgh1

– E2 = energy at the top of the mountain

– E2 = mgh2

– E = E2-E1 = mgh2 – mgh1 = mg(h)

• Some examples of state functions are:

–T, P, V, E, H, G, and ∆S• Examples of non-state functions are:

– n, q, w

More Thermodynamic Terms

In thermodynamics we are often interested in changes in functions (state functions). We will define the change of any function X as: X = Xfinal – Xinitial

If X increases X > 0 If X decreases X < 0.

Energy Profiles Exothermic

reactions generate specific amounts of heat.

Energy of products < energy of reactants

Energy Profile• Endothermic

reactions require absorption of energy for complete formation into products

• Energy of products > energy of reactants

ENTHALPY, H

• The enthalpy change, H, is the change in heat content at constant pressure.– H = qp

• Most commonly employed energy measure for chemical reactions since most reactions analyzed occur at constant atmospheric pressure.

ENTHALPY of REACTIONS

• Hrxn is the heat of reaction.

– If Hrxn < 0 the reaction is exothermic.

– If Hrxn > 0 the reaction is endothermic.

• Hrxn = Hproducts - Hreactants

– Hrxn = Hsubstances produced - Hsubstances consumed

– Hrxn = Hfinal state – Hinitial state

Thermochemical Equations and Enthalpy of Reactions

• The stoichiometric coefficients in thermochemical equations must be interpreted as numbers of moles of each component.

• 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to as one mole of reactions.

moles 6 moles 5 moles 8 mole 1

kJ 3523 OH 6 CO 5O 8 HC )(22(g)2(g))12(5

This is an equivalent method of writing thermochemical equations.

kJ/mol 3523 - H OH 6 CO 5O 8 HC orxn)(22(g)2(g))12(5

Thermochemical Equations

N2(g) + 2O2(g) 2NO2(g) ∆Horxn = 15.6 kcal/mol

N2(g) + 2O2(g) + 15.6 kcal 2NO2(g)

Thermochemical Standard States Thermochemical standard state conditions

Standard T = 298.15 K. Standard P = 1.0000 atm.

Be careful not to confuse these values with STP.

Thermochemical standard states of matter Standard state of pure matter – the pure liquid or solid form. Gases –1.00 atm of pressure.

For gaseous mixtures - partial pressure =1.00 atm. Aqueous solutions – 1.00 M concentration.

∆Horxn

Standard Molar Enthalpies of Formation, Hf

o

• Standard molar enthalpy of formation - enthalpy for the reaction in which one mole of a substance is formed from its constituent elements.– Hf

o kJ/(mole of substance assigned)

• The standard molar enthalpy of formation for MgCl2(s) is:

Standard Molar Enthalpies of Formation, Hf

o

• Standard molar enthalpies of formation have been determined for many substances and are tabulated in Appendix K (Whitten et al, 8th ed).

• Standard molar enthalpy of elements in their most stable forms at 298.15 K and 1.000 atm is zero.– Reference state for all enthalpy calculations are the

constituent elements of the compound.

• Hfo of elements in their standard states: Zero

value

Elements Hfo kJ/mole

Ca(s) 0

C(s) (graphite) 0

H+(aq)

0

Br2(l) 0

I2(s) 0

S?(?) 0

Hg(?) 0

• Hfo of all compounds, ions and elements in

their non-standard states: Nonzero valueCompound Hf

o kJ/mole

CaF2(s) -1219.6

C(s) (diamond) +1.895

HNO3(aq) -207.36

HNO3(l) -205.0

H2O(g) -241.82

H2O(l) -285.83

Au(g) +366.1

Cu2+(aq) +64.77

Standard Molar Enthalpies of Formation, Hf

o

Example: The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for whichHo

rxn = -1281 kJ/mol.

P in standard state is P4

Phosphoric acid in standard state is H3PO4(s)

Using Enthalpies of Formation• For any process (can be reactions, physical

changes, or transformations)xA + yB nC + mD

∆Hprocess =

∑(coefficients•∆Hfproducts) – ∑(coefficients•∆Hfreactants)

∆Hprocess = (n•∆HfC + m•∆HfD) – (x•∆HfA + y•∆HfB)

• Set-up the equation for Enthalpy of Reaction in:2H2(g) + O2(g) 2H2O(l) at 25oC

∆Hrxno = (2 • ∆Hf, H2O(l)

o ) – (2 • ∆Hf, H2(g)o + ∆Hf, O2(g)

o )

Using Enthalpies of Formation

Set-up the equation for Enthalpy of Vaporization in:H2O(l) H2O(g) at 25oC

∆Hvapo = (1 • ∆Hf, H2O(g)

o ) – (1 • ∆Hf, H2O(l)o )

Set-up the equation for Enthalpy of Dissolution in:NaCl(s) Na+

(aq) + Cl-(aq)

∆Hdisso = (∆Hf, Na+(aq)

o + ∆Hf, Cl-(aq)o ) – (∆Hf, NaCl(s)

o )

• If the problem asks for the amount of heat involved (either released or absorbed) in the process:

heat of process = ∆Hprocess • moles of limiting reagent

Using Enthalpies

moles means

stoichiometry yeah! :D

mole)(n x Hq LRmolekJ

process

Example 1: Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere.

oHF(g) f

og)(H f

o)g(F f

oHF(g) frxn

H2

)H H( - )H2(H22

reaction of kJ/mole 542

HF mole

kJ 271

reaction of mole

HF mole 2H

HF kJ/mole 271H book, ofK Appendix From

rxn

oHF(g) f

• Example 2: Calculate the heat involved in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere.

Calculate ∆Hrxn = (2 • -1676 kJ/mol) – (0 + 0) = -3352 kJ/mole of reaction

Calculate # of moles = 15.0 g Al/ 26.98 g Al/mole Al = 0.555967383 moles Al

Calculate Heat = 0.555967383 moles Al x (-3352 kJ/ 4 moles Al)

= -465.9006672 kJ = -466 kJ or 466 kilojoules heat released

• Nitrogen oxide (NO) has been found to be a key component in many biological processes. It also can react with oxygen to give the brown gas NO2. When one mole of NO reacts with Oxygen, 57.0 kJ of heat is evolved.

a.) What is the ∆Hrxn?

b.) How many grams of nitrogen oxide must react with an excess of oxygen to liberate ten kilojoules of heat?

CALORIMETRY• Coffee-cup calorimeter -

used to measure the amount of heat produced (or absorbed) in a reaction at constant P– Method to measure qP for

reactions in solution.– Constant-Pressure

Calorimetry is used to measure enthalpy of reactions.

• Styrofoam-based calorimetry in general, measures ∆H

• EquationsFor heat transfer involving homogenous systems - PURE

substances (eg. water) q = m • c • ∆T

= mass • specific heat capacity • temperature change = g • ( J/g-oC) • oC

For heat transfer involving heterogenous systems (eg. entire calorimeter)

q = Ccal • ∆T = ( J/ oC) • oC

CALORIMETRY

Calculating Enthalpy of Reaction using Calorimetry

• Main Condition: No heat flows out of the calorimeter.

hheat

heat

surroundings

system

THEREFORE:qsystem = 0

The entire system is ADIABATIC

Because the entire system is adiabatic:qsystem = 0

The system is composed of the calorimeter, the main reactant molecules (in case of reaction), and the solvent:

qcalorimeter + qreaction + qwater = 0

Therefore:

(qcalorimeter+ qwater) = – qreaction Heat Surroundings = - Heat System

(qcalorimeter+ qwater) = – qreaction

qcalorimeter = Ccal • ∆T

qwater = mH2O • cH2O • ∆T

qreaction = ∆Horeaction x nLR

(Ccal∆T + mH2OcH2O∆T) = -(∆Horeaction x nLR)

reactionLR

OHOHcal Hn

)cmT(C22

REMEMBER:For reactions:

– ∆Horeaction is “negative”: EXOTHERMIC,

∴ ∆T is + (Tfinal > Tinitial ), Increase temperature

– ∆Horeaction is “positive”: ENDOTHERMIC,

∴ ∆T is - (Tfinal < Tinitial ), Decrease temperature

• Example: A coffee-cup calorimeter is used to determine the heat of reaction for the acid-base neutralization

CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O()

When we add 25.00 mL of 0.500 M NaOH at 23.000oC to 25.00 mL of 0.600 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947oC. The heat capacity of the empty calorimeter has previously been determined to be 27.8 J/0C. Assume that the specific heat of the mixture is the same as that of water, 4.18 J/g0C and that the density of the mixture is 1.02 g/mL. Determine the Heat of Reaction for this process.

Instead of reaction, a specific amount of heat is added (or removed) in the calorimeter (example, when a hot metal bar or ice cubes is placed in a pail of water):

(qcal + qwater) = - qinvolved

For direct heat transfer: heat is “negative” if it is ADDED to the calorimeter from a source heat is “positive” if it is REMOVED from the calorimeter by a source

• When 3.425 kJ of heat is added to a calorimeter containing 50.00 g of water the temperature rises from 24.00oC to 36.54oC. Calculate the heat capacity of the calorimeter in J/oC. The specific heat of water is 4.184 J/g oC.

Hess’s Law• Hess’s Law of Heat Summation

- if a particular reaction is carried out in a series of steps, the ∆H for the reaction is equal to the SUM of enthalpy changes of each step.

– Hess’s Law is true because H is a state function.

kJ 1648H OFe 2 O 3 Fe 4 2

kJ 454H FeO 2O Fe 2 1o

3(s)22(g)(s)

o(g)2(g)(s)

(s)322(g)(s) OFeO FeO

Given the following enthalpies of formation:

Determine the ∆Hrxn for :

– Balance the reaction first!

– Arrange reactions [1] and [2] so that they also have FeO and O2 as reactants and Fe2O3 as a product.

• Each reaction can be doubled, tripled, or multiplied by half, etc.• The Ho values are also doubled, tripled, halved, etc.• If a reaction is reversed the sign of the Ho is changed.

(s)322(g)(s) OFe2O 4FeO

For [1]: reverse, multiply by 2

For [2]: retain

kJ/mole) 8108H 2O 4Fe 4FeO 2[-1]

kJ/mole) 454(2H )O Fe 2 FeO 2( 2 2[-1]o

2(g)(g)(s)

o2(g)(g)(s)

kJ 1648H O2Fe 3O 4Fe 2 o3(s)22(g)(s)

kJ/mole -568H OFe2O 4FeO rxn(s)322(g)(s)

Hess’s Law

• Example: Given the following equations and Hovalues

calculate Ho for the reaction below. ?H NO 3 NO + ON o

gg2g2

Hess’s Law

• To produce silicon (used in semiconductors) from sand (SiO2), a reaction is used that can be broken down into three steps:SiO2(s) + 2C(s) Si(s) + 2CO(g) ∆H = 689.9 kJ/mole

Si(s) + 2Cl2(g) SiCl4(g) ∆H = -657 kJ/mole

SiCl4(g) + 2Mg(s) 2MgCl2(s) + Si(s) ∆H = -625.6 kJ/mole

a) Determine the thermochemical equation for the formation of silicon from silicon dioxide.b) What amount of heat is needed for the formation of one mole of silicon?

Bond Energy• Bond energy is the amount of energy required to

break the bond and separate the atoms in the gas phase.– To break a bond always requires an absorption of energy!

Bond Energies Table of average bond energiesMolecule Bond Energy (kJ/mol) F2 (F-F) 159

Cl2 (Cl-Cl) 243

Br2 (Br-Br) 192

O2 (O=O) 498

N2 (N≡N) 946

Hydrogen

Bond D(kJ/mol)

r(pm)

H-H 432 74H-B 389 119H-C 411 109H-Si 318 148H-Ge 288 153H-Sn 251 170H-N 386 101H-P 322 144H-As 247 152

Common Bond Energies (D) and Bond Lengths (r)

Bond D(kJ/mol)

r(pm)

H-O 459 96H-S 363 134H-Se 276 146H-Te 238 170H-F 565 92H-Cl 428 127H-Br 362 141H-I 295 161

Bond D(kJ/mol)

r(pm)

C-P 264 184C-O 358 143C=O 799 120C≡O 1072 113C-B 356C-S 272 182C=S 573 160C-F 485 135C-Cl 327 177C-Br 285 194C-I 213 214

Bond D(kJ/mol)

r(pm)

C-C 346 154C=C 602 134C≡C 835 120C-Si 318 185C-Ge 238 195C-Sn 192 216C-Pb 130 230C-N 305 147C=N 615 129C≡N 887 116

Bond Energies In gas phase reactions, Ho values may be related to

bond energies of all species in the reaction.

• Example: Use the bond energies to estimate the heat of reaction at 25oC for the reaction below.

(g)22(g)2(g)4(g) OH 2 CO O 2 CH

]BE 4 BE [2 - ]BE 2BE [4 H H-OOCOOH-Co298

You should know how to draw the structure :<

kJ/mole (464)] 4 741)( [2 - 498)]( 2(414) [4 Ho298

kJ/mole 686 - Ho298

Bond Energies

• Calculate the average N-H bond energy in ammonia, NH3.

H-No298ggg3 BE 3H H 3 + NNH

]H[]H 3H[H o

NH fo

H fo

N fo298 g3gg

3

o298

o298

NH kJ/mole 1173H

kJ/mole 11.46)218(3)7.472(H

bonds H-N molkJ

3

3H-N 391

NH moleH-N mole 3

NH kJ/mole 1173BE average

Relationship of H and E

• The total amount of heat energy that a system can provide to its surroundings at constant temperature and pressure is given by

H= E + PV – which is the relationship between H and E.

• H = change in enthalpy of system• E = change in internal energy of system• PV = work involved in the system

Internal Energy of Gas-phase reactions

• Gas-phase reactions may involve changes in volume, and as the volume of the container change (due to expansion or condensation of molecules), Work is involved in the system

∆ngas = (total moles of gaseous products) – (total moles of gaseous reactants)

Internal Energy of Gas-phase reactions

When Then Examples1. V2 = V1 PV = 0

ngas = 0 gas mol 2

2(g)(g)2

gas mol 2

(g)2(g) CO H OH CO

3. V2 < V1 PV < 0 ngas < 0

2. V2 > V1 PV > 0 ngas

> 0

gas mol 1

2(g)2(aq)

gas mol 0

(aq)(s) H ZnCl HCl 2 Zn

gas mol 2

3(g)

gas mol 4

2(g)2(g) NH 2 H 3 N

H= E + P V

For reactions with gases: H= E + (ngas)RT

But H = qP (Are these two definitions compatible?)

– Remember E = q + w.– We have also defined w = -PV .– Thus E = q + w = q -PV– Consequently, H = q- PV + PV = q

At constant pressure H = qP

Relationship of ∆H and ∆E

enthalpy, H: heat measured at constant pressure.

internal energy, E: heat measured at constant volume.

How much do the H and E for a reaction differ? The difference depends on the amount of work

performed by the system or the surroundings.

Relationship of ∆H and ∆E

Relationship of ∆H and ∆E

• For reactions in which the volume change is very small or equal to zero (for reactions in which there are no gaseous reactants or products).

• Determine the difference between ∆H and ∆U at 100oC for CO(g) + O2(g) CO2(g)

• Chlorine trifluoride is a toxic, intensely reactive gas. It was used in World War II to make incendiary bombs. It reacts with ammonia and forms nitrogen, chlorine, and hydrogen fluoride gases. When two moles of chlorine trifluoride reacts, 1196 kJ of heat is evolved.– Write the thermochemical equations for the reaction– What is the ∆Ho

f for ClF3?