10-1 heat • 10-2 calorimetry • 10-3 enthalpy • 10-4 standard-state enthalpies...

OFB Chap. 10 12/17/2004

Chapter 10Thermochemistry

• 10-1 Heat• 10-2 Calorimetry• 10-3 Enthalpy• 10-4 Standard-State Enthalpies• 10-5 Bond Enthalpies• 10-6 The First Law of

Thermodynamics

OFB Chap. 10 22/17/2004

OFB Chap. 10 32/17/2004

Thermite Reaction2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)

OFB Chap. 10 42/17/2004


•Heat Capacity, Cp , is the amount of heat required to raise the temperature of a substance by one degree at constant pressure.

•Cp is always a positive number•Q and ∆T must be both negative or positive•If q is negative, then heat is evolved or given off and the temperature decreases•If q is positive, then heat is absorbed and the temperature increases

OFB Chap. 10 52/17/2004

Heat Capacity (continued)• Cp units are energy per temp. change• Cp units are Joule/oK or JK-1

• Units are J/K/mole or JK-1mol-1

• Molar heat capacity

OFB Chap. 10 62/17/2004

OFB Chap. 10 72/17/2004

Specific Heat Capacity• The word Specific before the name

of a physical quantity very often means divided by the mass

4.18H2O0.719O2

0.739SiO2

0.140Hg (l)cs

• Specific Heat Capacity is the amount of heat required to raise the temperature of one gram of material by one degree Kelvin (at constant pressure)

OFB Chap. 10 82/17/2004

OFB Chap. 10 92/17/2004

TmcqTmq

mCc

ipsrealtionshsome Equating

s

ps

∆∆

∆

=

===mTq /

sp

pps

ppsp

MccmC

Mcc

cMmncmcC

=

==

===

ipsrelationshsome Equating

OFB Chap. 10 102/17/2004


Example (not in book)a.) Calculate the molar heat capacity of quartz (SiO2) if cs

SiO2 = 0.739 J/K/gb) Calculate the amount of heat required to raise 45.0 Kg of rock (quartz) by 15°C.

Strategya) Use relationship cp = Mcs

b) Use q = mcs∆T

OFB Chap. 10 112/17/2004

Example (not in book)a.) Calculate the molar heat capacity of quartz (SiO2), if csSiO2=0.739 J/K/gb) Calculate the amount of heat required to raise 45.0 Kg of rock (quartz) by 15°C.

Solutiona) Find the molar mass of quartz

and Use relationship cp=Mcs

Kmol44.4J

Kg0.739J

mol60.08gMcc sp

°=

°

==

mcs∆cq =b) Use q= mcs∆T

OFB Chap. 10 122/17/2004

Exercise 10-1Exactly 500.0 kJ of heat is absorbed at a constant pressure by a sample of gaseous helium. The temperature increases by 15.0K.a.) Compute the heat capacity of the

sampleb.) The mass of the helium sample is

6.42 kg. Compute the specific heat capacity and molar heat capacity of helium.

Strategy– a.) Use the relationship Cp=q/∆T– B.) Use cs=Cp/m and use cp=Cp/n

OFB Chap. 10 132/17/2004

Exercise 10-1• Exactly 500.0 kJ of heat is absorbed at a constant

pressure by a sample of gaseous helium. The temperature increases by 15.0K.a.) Compute the heat capacity of the sampleb.) The mass of the helium sample is 6.42 kg.

Compute the specific heat capacity and molar heat capacity of helium.

Solution– a.) Use the relationship Cp=q/∆T

13

3

p

JK33.33x10K15

J500x10K15

500kJ∆TqC

−

=

=°

=°

=

113

13p

p

113

13p

s

mol20.8JK

4g/molg6.42x10KJ33.33x10

nC

c

g5.19JKg6.42x10KJ33.33x10

mC

c

−−−

−−−

=

°

==

=°

==

– b.) Use cs=Cp/m and use cp=Cp/n

OFB Chap. 10 142/17/2004

Equilibration Temperature

• When 2 bodies are placed in contact, heat is exchanged until they reach a common final temperature

• Heat gained by the cooler body equals the heat lost by the warmer body

• If q is positive, heat is gained• If q is negative, heat is lost

• Can also write similar equations for cs

2ifp221ifp11

2p221p11

21

TTcnTTcn∆Tcn∆Tcn

qq

−−=−

−=−=

OFB Chap. 10 152/17/2004


Example (not in book)A 43.9g piece of Copper (cs=0.385 J g-1 K - 1) at 135 °C is plunged into 254g of water (cs=4.18 J g-1 K - 1) at 39 °C. Assuming no heat is lost to the surroundings, what will be the final temperature?

a. 100.0 Cb. 87.0 Cc. 53.1 Cd. 40.5 Ce. None of these

OFB Chap. 10 162/17/2004

2s221s11

21

∆Tcm∆Tcmqq

−=−=

( )

( ) ( )

( ) ( )

danswer or 40.5CT43687.81078.6T

1061.7T41406.32281.516.9T39.0CT1061.7135CT16.90

39.0CTKg

4.18J254g

135CTKg

0.385J43.9g

f

f

ff

ff

f0H

fCu

Cu

2

==

−+=−−−=−

−

°

−

=−

°

Also Try 10-2 Example and Exercise

OFB Chap. 10 172/17/2004

• Heat is a means by which energy is transferred among systems

• For processes carried out at constant pressure, the heat absorbed equals a changein Enthalpy, H.

Enthalpy Change(note p denotes constant pressure)

• Enthalpy is a state property, which means it depends on its initial and final state, not the path to get there.

• Enthalpy is the heat content of a substance

Enthalpy

OFB Chap. 10 182/17/2004

OFB Chap. 10 192/17/2004

Enthalpy of Reaction• If a chemical reaction occurs at

constant pressure then• (Heat absorbed) = qp = ∆H• ∆H may be either positive or

negative– If ∆H positive = ∆H > 0 = q > 0 Means heat is absorbedAnd is called Endothermic– If ∆H negative = ∆H < 0 = q < 0 Means heat is given offAnd is called Exothermic

OFB Chap. 10 202/17/2004

CO (g) + ½ O2 (g) → CO2 (g)

∆H= -283kJ

An exothermic reaction

If Quantities doubled (x2), then double ∆H

2CO (g) + 1O2 (g) → 2CO2 (g) ∆H= -566kJ

If the original reaction is reversed

CO2 (g) → CO (g) + ½ O2 (g)

∆H= +283kJ

An endothermic reactionYou can Try Example 10-5

OFB Chap. 10 212/17/2004

Phase Changes• Phase changes are not chemical reactions

but involve enthalpy change

H2O(s)→ H2O(l) ∆Hfusion= + 6 kJ at 0C

∆Hfusion= + means endothermic (add heat)

if reversedH2O(l)→ H2O(s) ∆Hfreeze= - 6 kJ

• ∆Hfreeze= - ∆Hfusion at melting point

• ∆Hvaporization at Boiling point

= - ∆HCondensation

• Note that there are no T units, these occur at constant temperatures for freezing, fusion, vaporization or condensation

OFB Chap. 10 222/17/2004

OFB Chap. 10 232/17/2004

How much heat energy is needed to decompose 9.74g of HBr (g) (M =80.9 g/mol) into its elements?

H2 + Br2 → 2HBr (g)

∆H = -72.8 kJmol-1

Strategy:• Decomposition is actually

the reverse reaction• ∆H to decompose 2 moles of

HBr is +72.8kJ/mol

OFB Chap. 10 242/17/2004

How much heat energy is needed to decompose 9.74g of HBr (g) (M =80.9 g/mol) into its elements?

H2 + Br2 → 2HBr (g)

∆Hr = -72.8 kJmol-1

Solution:2HBr (g) → H2 + Br2

∆H = +72.8 kJmol-1

kJ 4.38moleHBr g 80.91

HBr g 9.76HBr moles 2

kJ 72.8∆Hq

=

==

OFB Chap. 10 252/17/2004

Hess’s Law• Sometimes its difficult or even

impossible to conduct some experiments in the lab.

• Because Enthalpy (H) is a State Property, it doesn’t matter what path is taken as long as the initial reactants lead to the final product.

• Hess’s Law is about adding or subtracting equations and enthalpies.

• Hess’s Law: If two or more chemical equations are added to give a new equation, then adding the enthalpies of the reactions that they represent gives the enthalpy of the new reaction.

OFB Chap. 10 262/17/2004

Standard State Enthalpies• Designated using a superscript °

(pronounced naught)• This is needed because Enthalpy varies

with conditions• Is written as ∆H°

• If no temperature is indicated in the sub-script, assume 25°C.

• OFB text appendix D lists standard enthalpies of formation for a variety of chemical species at 1 atm and 25°C.

• Standard State conditions– All concentrations are– All partial pressures are

OFB Chap. 10 272/17/2004

Standard State Enthalpies

• ∆Hr° is the sum of products minus the sum of the reactants

• For a general reactiona A + b B → c C + d D

• ∆Hr°= ΣH°products- ΣH°reactants

• from Hess’s Law

(B)Hb(A)Ha

(D)Hd(C)Hc∆Hof

of

of

of

or

∆∆

∆∆

−−

+=

OFB Chap. 10 282/17/2004


• Exercise 10-7 Suppose hydrazine and oxygen react to give dinitrogen pentaoxide and water vapor:

2N2H4 (l) + 7O2 (g) →2N2O5 (s) + 4H2O (g)

Calculate the ∆H of this reaction, given that the reaction:

N2 (g) + 5/2O2 (g)→N2O5 (s)

has a ∆H of -43 kJ.

OFB Chap. 10 292/17/2004

10-7 Exercise2N2H4 (l) + 7O2 (g) →

2N2O5 (s) + 4H2O (g)• ∆Hr° =

• What values to use?– ∆Hf°N2O5 (s) given ∆H°=-43.1 kJmol-1

– ∆Hf°H2O (g) look up in App. D– ∆Hf° N2H4 (l) look up in App. D– ∆Hf°O2 (g) look up in App. D

• ∆Hr° = (-1154.74) kJmol-1

OFB Chap. 10 302/17/2004

Typical Exam questionGiven the following thermo chemical

equations, calculate the standard enthalpy of formation for propane, C3H8 (g).

AnswersA. (-3131) kJmol-1

B. 129C. 4775D. (-1773)

1-r

22283

1-r

222

1-r

22

kJmol (-2452)∆H

O(l)4H(g)3CO(g)5O(g)HC 3.kJmol (-286)∆H

O(l)H(g)1/2O(g)H 2.kJmol (-393)∆H

(g)CO(g)OC(s) 1.

=

+→+

=

→+

=

→+

o

o

o

OFB Chap. 10 312/17/2004

BHbAHaDHdCHcHdDcCbBaA

General

ooooo ∆∆∆∆∆ −−+=

+→+

283

223

22

283

5143

)(4)(3)(5)(.3

OHHCHOHHCOHH

lOHgCOgOgHC

oo

ooo

∆∆

∆∆∆

−−

+=

+→+

(AnswerB) kJmol 129HC∆HX

5(0)X286)4(393)3(24520)(O∆H recall and s∆H' 3 are Given

183f

C3H8

2f

−==

−−−+−=−=

o

o

1-r

22283

1-2f

222

1-2f

22

kJmol (-2452)∆H

O(l)4H(g)3CO(g)5O(g)HC 3.kJmol (-286)O)(H∆H

O(l)H(g)1/2O(g)H 2.kJmol (-393))(CO∆H

(g)CO(g)OC(s) 1.

=

+→+=

→+=

→+

o

o

o

OFB Chap. 10 322/17/2004

Bond Enthalpy

• Bond Enthalpy (∆H Dissociation) is the enthalpy change in a reaction in which a chemical bond is broken in the gas phase.

• Bond Energy is the energy needed to break 1 mol of the particular bond.– Energy is released when bonds

are formed (exothermic)– Energy must be supplied when

bonds are broken (endothermic)

OFB Chap. 10 332/17/2004

Bond Enthalpies• The breaking of chemical bonds in

stable substances often generates highly reactive products (or intermediates)CH4 → ·CH3 + ·H ∆H°= + 439 kJmol-1

• Called Bond Enthalpy• OFB Table 10-3 p 462 gives Average

Bond Enthalpies

OFB Chap. 10 342/17/2004

• Average Bond Enthalpies

C2H6 → ·C2H5 + ·H ∆H°= + 410 kJmol-1

CHF3 → ·CF3 + ·H ∆H°= + 429 kJmol-1

CHCl3 → ·CCl3 + ·H ∆H°= + 380 kJmol-1

CHBr3 → ·CBr3 + ·H ∆H°= + 377 kJmol-1

average ∆H°C-H = + 412 kJmol-1

• Applications of Bond Enthalpy– Given a reaction– 1st Step is break all bonds to give free atoms in

the gas phase (Endothermic)– 2nd Step is form new bonds for the products.

(Exothermic)

OFB Chap. 10 352/17/2004

• Applications of Bond Enthalpy– Given a reaction– 1st Step is break all bonds to give free atoms

in the gas phase (Endothermic)– 2nd Step is form new bonds for the products.

(Exothermic)

OFB Chap. 10 362/17/2004

Exercise 10-10• Estimate the Standard Enthalpy of

reaction for the gas-phase reaction that forms methanol from methane and water and compare it with the ∆H°r obtained from the data in Appendix D.

CH4(g) + H2O(g)→CH3OH(g) + H2(g)

OFB Chap. 10 372/17/2004

H

C

H H

H

+O

H

H

H

HH

+ H HC

O H

- kJmol-1

Exothermic+ kJmol-1

Endothermic

3 C-H or 1 O-H or 1 C-O or 1 H-H or

4 C-H or 2 O-H or

FormedBroken

∆H= ∆H bonds broken + ∆H bonds formed

= (estimated)

Using Appendix D ∆H = 116 kJmol-1

CH4(g) + H2O(g)→CH3OH(g) + H2(g)

OFB Chap. 10 382/17/2004


• First Law of ThermodynamicsThe change in the internal energy of a system is equal to the work done on it plus the heat transferred to it.

• Second Law of ThermodynamicsIn a real spontaneous process the Entropyof the universe (meaning the system plus its surroundings) must increase.

• Third Law of ThermodynamicsIn any thermodynamic process involving only pure phases at equilibrium, the entropy change, ∆ S, approaches zero at absolute zero temperature; also the entropy of a crystalline substance approaches zero.

OFB Chap. 10 392/17/2004

First Law of Thermodynamics

∆E = q + w• q was previously defined as the Heat

Absorbed by a system• If q > 0 , heat is absorbed• If q < 0 , heat is given off• w is the work done on the body

Recall• w = F x d• w = ∆EKinetic=∆ (1/2mv2)• w = ∆Epotential= mg∆h

• In chemistry this kind of mechanical work is Pressure-Volume work (P-V)

OFB Chap. 10 402/17/2004

• Force exerted by heating = P1A– Where P1 is the pressure inside the vessel– Where A is the Area of the piston

• Pext = P1 if balanced• ∆V = A∆h•

– Units are atm·L or L atm– Where 1 L atm = 101.325 Joules

OFB Chap. 10 412/17/2004

• Expansion– ∆V > 0 therefore w < 0– The system does work on the

surroundings• Compression

– ∆V < 0 therefore w > 0– The surroundings have done work on

the system• Try Example and Exercise 10-11

– Calculate the work done on a gas and express it in Joules

OFB Chap. 10 422/17/2004

• Four P-V questions will be done as a PRS quiz in class.

• See lecture slides following

• Try to work them in advance of the lecture.

OFB Chap. 10 432/17/2004

Typical exam questions (1-4)A gas is compressed from 39.92L to 12.97L at a constant pressure of 5.00 atm. In the course of this compression 9.82 kJ of energy is released

Q1 The heat q for this process is1. 135 kJ2. -135 kJ3. -9.82 kJ4. 9.82 kJ

OFB Chap. 10 442/17/2004

A gas is compressed from 39.92L to 12.97L at a constant pressure of 5.00 atm. In the course of this compression 9.82 kJ of energy is released

Q2 The work w for this process is 1. 135 L atm2. -135 L atm3. -9.82 L atm4. 9.82 L atm

OFB Chap. 10 452/17/2004


Q3 ∆E for the process is1. -23.47 kJ2. 3.86 kJ3. 23.47 kJ4. 125 kJ

OFB Chap. 10 462/17/2004


Q4 ∆ H for the process isa -9.82 kJb -7.09 kJc 3.83 kJd 9.82 KJ

OFB Chap. 10 472/17/2004

Enthalpy and Energy• 1st Law of Thermodynamics

∆E = q + w – At constant volume ∆V =0– Thus w = -Pext ∆V = 0

∆E = qv (constant volume)• Recall as previously stated

∆H = qp (constant pressure)• Enthalpy is defined as

H = E + PV• or Expressed as a change in

Enthalpy∆H = ∆E + ∆(PV)

• Rearrange∆E = ∆H - ∆(PV)

• If PV = nRT is the ideal gas law

OFB Chap. 10 482/17/2004


• ∆ng is the change in the total chemical amount of gases in a reaction

• ∆ng = Total moles of product gases minus the Total moles of reactant gases

OFB Chap. 10 492/17/2004

kJ110.5∆H

1CO(g)(g)O21)C(graphite 2

−=

→+

Example• Calculate the internal energy @ 25°C

for the following reaction

Strategy– Step 1 calculate ∆(PV)– Step 2 calculate ∆E at 25°C

OFB Chap. 10 502/17/2004

Example• Calculate the internal energy @ 25°C

for the following reaction

kJ110.5∆H

1CO(g)(g)O21)C(graphite 2

−=

→+

• Solution– Step 1 calculate ∆(PV)

• ∆(PV) ≈ RT ∆ng • ∆ng = 1 – ½ = ½ mol• ∆(PV) ≈ RT ∆ng = (8.35 kJmol-1) (298 K) (1/2 mol)= 1.24 kJ

– Step 2 calculate ∆E at 25°C• ∆E = ∆H – ∆ (PV)• = – 110.5 – (1.24)• = – 111.74 kJ

wjb1

wjb1 William J Baron, 8/28/2002

OFB Chap. 10 512/17/2004

Chapter 10Thermochemistry Summary

if

pressureconstant at CapacityHeat denotes Cp where

p

T-T∆T

∆Tabsorbedheat

∆TqCCapacityHeat

=

===

11-pp molJK are units

nCc

CapacityHeat Molar

−= 1-1-ps gJK are units

mCc

CapacityHeat Specific

=

( ) ( )2if p221if p11

2 p221 p11

21

TTcnTTcn∆Tcn∆Tcn

qqeTemperaturionEquilibrat

−−=−−=−=

OFB Chap. 10 522/17/2004

– If ∆H positive = ∆H > 0 = q > 0 Means heat is absorbedAnd is called Endothermic– If ∆H negative = ∆H < 0 = q < 0 Means heat is given offAnd is called Exothermic

• Enthalpy, H.

qp= ∆H(note p denotes constant pressure)

• Enthalpy is a state property, which means it depends on its initial and final state, not the path to get there.

•Hess’s Law: If two or more chemical equations are added to give a new equation, then adding the enthalpies of the reactions that they represent gives the enthalpy of the new reaction.

OFB Chap. 10 532/17/2004

Standard State Enthalpies• ∆H° is the sum of products minus the

sum of the reactants• For a general reaction

a A + b B → c C + d D

(B)H(A)H

(D)H(C)H∆Hof

of

of

of

o

∆∆

∆∆

ba

dc

−−

+=

Bond Enthalpies∆H= ∆H bonds broken + ∆H bonds formed

First Law of ThermodynamicsThe change in the internal energy of a system is equal to the work done on it plus the heat transferred to it.

∆ E = q + w

OFB Chap. 10 542/17/2004

Enthalpy and Energy• Enthalpy is defined as

H = E + PV• or Expressed as a change in

Enthalpy∆H = ∆E + ∆(PV)

• For gases, If PV = nRT is the ideal gas law

∆(PV) ≈ ∆(nRT)= RT ∆ng

OFB Chap. 10 552/17/2004


• Examples / exercises10-1, 10-2, 10-5, 10-6, 10-7, 10-8, 10-9, 10-10, 10-11, 10-

12• HW Problems 11, 13, 19, 23, 33, 37, 40, 43,

49, 53, 59, 63

10-1 heat • 10-2 calorimetry • 10-3 enthalpy • 10-4 standard-state enthalpies...

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