Thermo Examples

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Chapter 6 Thermodynamics:

The First Law

Systems, States, and Energy (Sections 6.16.8)

Key Concepts

thermodynamics, statistical thermodynamics, system, surroundings, open system,closed system, isolated system, work, internal energy, translational kinetic energy,rotational kinetic energy, vibrational kinetic energy, mode of motion (degree offreedom), equipartition theorem, internal energy of an ideal gas, heat, calorie, adiabaticwall, diathermic (nonadiabatic) wall, first law of thermodynamics, state function,expansion work, free expansion, nonexpansion work, reversible isothermal expansion,reversible process, irreversible process, exothermic process, endothermic process,calorimetry, calorimeter, heat capacity, specific heat capacity, molar heat capacity

6.1 Systems

Example 6.1a Suggest a system, boundary, and surroundings for 10 moles of propane gas in a rigid metal

cylinder.

Answer One choice is that the 10 moles of propane gas constitute the system, the metal cylinder and

everything outside of it the surroundings, and the inside walls of the cylinder a real boundary.

This is an appropriate way to study the properties of the gas alone.

Example 6.1b Suggest a system, boundary, and surroundings for 500 mL of water in an open beaker.

Answer To study the properties of water alone, it should constitute the system. The walls of the beaker

constitute a real, physical boundary and the phase boundary between water and air constitutes

an imaginary boundary. The surroundings consist of everything outside the boundariesincluding the beaker, the platform on which it rests, and the atmosphere.

6.2 Work and Energy

Example 6.2a Describe three commonplace examples of how work is done on or by a system.

Answer Compression of a spring: If the spring is the system, work is done on the system by the

surroundings to compress the spring.

Compression of a gas mixture as in an automobile engine: If the gas mixture is the system,

work is done on the system by the surroundings (piston) to compress the gases.

Muscle contraction: As the muscle contracts, work is done on the surroundings.

Both mechanical and electrical work are involved as well as chemical processes.

Example 6.2b A plumber of mass 65 kg carries a toolbox of mass 15 kg to a fifth floor walk-up apartment

15 m above ground level. Calculate the work required for this process. Recall that

(difference in potential energy) = mghfrom Table 6.1 in the text.

Answer The difference in potential energy between the fifth and ground floors is equal to the work

done by the plumber. The force is mgand the distance is h. Both the plumbers mass andthat of the toolbox must be elevated.

m = (65 +15) kg = 80 kg

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Chapter 6

g= 9.81 ms2h= 15 mw= (80 kg)(9.81 ms2)(15 m) = 1.18 104J = 11.8 kJ

Example 6.2c Describe the internal energy change and work performed when a spring is compressed or

expanded.

Answer Work is done on the system by the surroundings to compress the spring. The compressed

spring has a greater capacity to do work than the uncompressed spring. In each case, the

internal energy of the system is increased by the amount of work done on it.

U> 0 and U= UfinalUinitial= w

Therefore, w> 0 when work is done onthe system.

Work is done on the surroundings by the system to expand the spring. The expanded spring

has a lesser capacity to do work than the unexpanded spring. The internal energy of the

system is decreased by the amount of work done by it.

U 0 when electrical work is done onthe system.

6.3 Expansion Work

Example 6.3a The pressure exerted on an ideal gas at 2.00 atm and 300 K is reduced suddenly to 1.00 atm

while heat is transferred to maintain the initial temperature of 300 K. Calculate q, w, and

Uinjoulesfor this process.

Solution For an ideal gas undergoing an isothermal process, U= 0. The work done at constantexternal pressure is w= Pex V, where V= VfinalVinitial .

Vinitial = (nRT)/Pinitial= (1 mol)(0.082

06 LatmK1mol1)(300 K)/(2.00 atm) = 12.3 L

Vfinal = (nRT)/Pfinal= (1 mol)(0.082

06 LatmK1mol1)(300 K)/(1.00 atm) = 24.6 L

V= Vfinal Vinitial= 24.6 L 12.3 L = 12.3 L (system expands)

w = Pex V= (1.00 atm)(12.3 L) = 12.3 Latm= (12.3 Latm)(101.325 JL1atm1) = 1.25 103J

Work is done by the system on the surroundings.

q= Uw= 0 (1.25 103J) = +1.25 103JHeat is absorbed by the system from the surroundings.

Example 6.3b Suppose the pressure change in Example 6.3a is carried out reversiblyas well as

isothermally. Calculate q, w, and Uinjoulesfor this process and compare the answersto those in the irreversibleexpansion in one step.

Answer The internal energy of an ideal gas depends only on temperature, and U= 0.Check the two formulas for isothermal, reversible work.

1 1final

initial

24.6 Lln (1.00 mol)(8.31447 J K mol )(300 K) ln

12.3 L

Vw nRT

V

= =

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Thermodynamics: The First Law

or 1 1initial

final

2.00 atmln (1.00 mol)(8.314 47 J K mol )(300 K) ln

1.00 atm

Pw nRT

P

= =

w= 1.73 103J and q= 1.73 103J

Note: The absolute value of the work done by the system in a reversibleexpansion is greater than the

value obtained in an irreversibleexpansion. It is possible to show that the maximum amount

of work is always done by the system in a reversible expansion.

Example 6.3c Suppose the pressure change in Example 6.3a is carried out in the opposite direction

(compression). Calculate q, w, and Uinjoulesfor this process in both a reversible andirreversibleway.

Answer The labels for initialandfinalstates are interchanged. The calculation for reversible

work results simply in a change in sign.

1 1final

initial

12.3 Lln (1.00 mol)(8.314 47 J K mol )(300 K) ln

24.6 L

Vw nRT

V

= =

= +1.73 103J and q= 1.73 103J

The calculation for irreversible work results in a quite different value.

w = Pex V= (2.00 atm)(12.3 L) = +24.6 Latm= (24.6 Latm)(101.325 JL1atm1) = +2.50 103J and q= 2.50 103J

Note: The value of the work done on the system in a reversiblecompression is smaller than the value

obtained in an irreversiblecompression. It is possible to show that the minimumamount of

work is always done onthe system in a reversiblecompression.

6.4 Heat

Example 6.4a Calculate the heat flow and change in internal energy if the temperature of 1 mol of Ar is

increased from 300 K to 900 K. Assume the atoms behave as an ideal gas and no work is

done.

Solution From Example 6.8a, Um= 3.74 kJmol1at 300 K. Tripling the temperature leads toUm= 11.22 kJmol

1at 900 K.

q= Um= Um(900 K) Um(300 K) = (11.22 3.74) kJmol1= 7.48 kJmol1

Example 6.4b Calculate the heat flow and change in internal energy if the temperature of 1 mol of H2is

increased from 300 K to 900 K. Assume the molecules behave as an ideal gas and no work is

done.

Solution In Example 6.8b, Um= 6.24 kJmol1at 300 K. Tripling the temperature leads to

Um= 18.72 kJmol1at 900 K.


Example 6.4c Calculate the heat flow and change in internal energy if the temperature of 1 mol of H2O is

increased from 300 K to 900 K. Assume the molecules behave as an ideal gas and no work isdone.

Solution In Example 6.8c, Um= 7.48 kJmol1at 300 K. Tripling the temperature leads to

Um= 22.44 kJmol1at 900 K.


Note: In all three examples, qand the change in internal energy are positive values, because heat

flows from the surroundings into the system. The amount of heat required to increase the

temperature (and internal energy) is smaller for atoms than for molecules, and is smaller for

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Chapter 6

linear molecules than for nonlinear molecules. The greater the number of modes of motion

(degrees of freedom), the greater is the capacityof a species to store energy.

6.6 The First Law

Example 6.6 An electric stirrer performs 40 kJ of work on a beaker of water and water transfers 1.5 103cal of heat to the surroundings. Calculate the change in internal energy of water in kJ.

Solution Convert calories to the units of kilojoules.

(1.5 103cal)(4.184 Jcal1)(103kJJ1) = 6.3 kJWork is done on the system (water), so w= 40 kJ > 0.

Heat is transferred from the system to the surroundings, so q= 6.3 kJ < 0.

U= q+w= (6.3 + 40) kJ = 34 kJ

U> 0 and the internal energy of the system has increased.

6.7 State Functions

Example 6.7a Describe the state of a system composed of 1 mol of Ar(g) at 1 bar and 300 K.

Solution The pressure and temperature are uniform throughout the gas. The ideal gas equation uniquely

determines the volume of the gas, V= nRT/P. All other properties applicable to a noble gas

under these conditions, such as density, refractive index, and heat capacity, are fixed as well.

Example 6.7b Describe the state of the following system: a partition that separates a sample of liquid water

at 10C from a sample of water at 30C is removed.

Solution The resulting system is not in any given state, because its properties are not fixed. The

temperature and density are not uniform throughout. The system will eventually approach a

uniform temperature and density as a result of diffusion. The state of the system can then be

defined.

Example 6.7c For spring break, two students decide to fly from Philadelphia (initial state) to Denver (final

state). Two of their classmates choose to drive from Philadelphia to Denver by way of New

Orleans (intermediate state). Describe one variable whose change is the same for the two

groups (state function) and another that is not.

Solution The destination (final state) and origin (initial state) are the same for both groups. The

distances traveled depend on the path taken, and, presumably, are not the same for the plane

and car.

Example 6.7d A laboratory technician assigned to stir an insulated beaker of water inadvertently leaves the

stirrer on overnight. In the morning, the technician notes that the temperature of the water has

risen 3C. The same temperature rise can also be accomplished by heating the water using aheating mantle. Compare the two processes in terms of the change in internal energy, and the

heat and work done on the system.

Solution For the two processes, the initial and final states are the same, so Uis also the same. In thefirst process, mechanical work is done on the system, q= 0, and U = w. In the second

process, w= 0, and U= q. Both qand ware clearly path dependent.

Note: Because the changes in state functions are path independent, the most convenient path may be

chosen for a given change in state both in the laboratory and in calculations.

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6.8 A Molecular Interlude: The Origin of Internal Energy

Example 6.8a Calculate the internal energy of 1 mol of Ar at 300 K. Assume the atoms behave as an ideal

gas.

Solution An atom has only three degrees of freedom in translational motion. The internal energy per

mole is then

Um = Um(translation) =3

2RT= (1.5)(8.31447 JK1mol1)(300 K)

= 3.74 103Jmol1

= 3.74 kJmol1

Example 6.8b Calculate the internal energy of 1 mol of H2at 300 K. Assume the molecules behave as an

ideal gas.

Solution All diatomic molecules are linear. The H2molecule has three degrees of freedom in

translational motion and two degrees of freedom in rotational motion. The internal energy per

mole is then

Um = Um(translation) + Um(rotation, linear) =3

2RT+RT= 5

2RT

= (2.5)(8.31447 JK1mol1)(300 K) = 6.24 103Jmol1= 6.24 kJmol1

Example 6.8c Calculate the internal energy of 1 mol of H2O vapor at 300 K. Assume the molecules behave

as an ideal gas.

Solution If we use the VSEPR model (Section 3.2), the H2O molecule is predicted to be nonlinear.

Thus, it has three degrees of freedom in translational motion and three degrees of freedom in

rotational motion. The internal energy per mole is then

Um = Um(translation) + Um(rotation, nonlinear) =3

2RT+ 3

2RT= 3RT

= (3)(8.31447 JK1mol1)(300 K) = 7.48 103Jmol1= 7.48 kJmol1

Notes: In all three examples, the internal energy is linearly proportional to the temperature of the

ideal gas. The vibrational contribution to the internal energy of an ideal gas is small at room

temperature and can usually be ignored. Its temperature dependence is, however, decidedly

nonlinear at normal temperatures. At veryhigh temperature, each vibrational degree offreedom of a molecule has a contribution of of 12kTfrom kinetic energy and 1

2kTfrom

potential energy if vibration is treated as a harmonic oscillator. The total contribution to the

internal energy is then kTper degree of freedom for vibrational motion.

Enthalpy (Sections 6.96.13)

Key Concepts

enthalpy, heat transfer at constant pressure, heat capacity of gases, heat capacity atconstant volume, heat capacity at constant pressure, molecular origin of heat capacity,physical change (phase transition), enthalpy of vaporization, enthalpy of fusion,enthalpy of freezing, enthalpy of sublimation, heating curves, temperature of asubstance at its melting or boiling point

Note: Study Guidenotation (notused in the text) qVand qP(heat transfer at

constant volume and constant pressure, respectively)

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Chapter 6

6.9 Heat Transfers at Constant Pressure

Example 6.9 A gas expands against a constant external pressure and does 25 kJ of expansion work on the

surroundings. During the process, 60 kJ of heat is absorbed by the system. Determine the

values of Hand U.

Solution Because this is a constant pressure process, H= qP= 60 kJ.U= HPV= H+ww= 25 kJ (work done by the system is negative)U= 60 kJ 25 kJ = 35 kJ

6.10 Heat Capacities at Constant Volume and Constant Pressure

Example 6.10 The heat capacity, CP , of 1.500 moles of F2is 46.95 JK1at 25C. Assuming the gas is

ideal, calculate the molar heat capacity at constant volume, CV,m, for F2at 25C.

Solution CP,m= CP/n= (46.95 JK1)/(1.500 mol) = 31.30 JK1mol1

CV,m= CP,mR= (31.30 8.31451) JK1mol1= 22.99 JK1mol1

6.11 A Molecular Interlude: The Origin of the Heat Capacities of Gases

Example 6.11a Calculate the molar heat capacity at constant volume, CV,m, for F2at 25C. Compare the

calculated value to the experimental one determined in Example 6.10.

Solution Diatomic molecules are linear.

CV,m=5

2R= 20.78628 JK1mol1

The experimental value of 22.99 JK1mol1is slightly larger than the calculated one. The

difference is associated with vibrational motion of the F2molecule.

Example 6.11b A 100-L vessel containing 6.00 mol of H2(g) at 2.00 atm is cooled to 203.0 K. Assuming the

gas behaves ideally, calculate Uand Hfor this process.

Solution Calculate the initial temperature, Tinitial .

Tinitial=PV/nR= [(2.00 atm)(100 L)]/[(6.00 mol)(0.08206 LatmK1mol1)]

Tinitial= 406.2 K

T= TfinalTinitial= (203.0 406.2) K = 203.2 K

CV,m=5

2R= 20.78628 JK1mol1= 0.02079 kJK1mol1

CP,m= CV,m+R= 29.10079 JK1mol1= 0.02910 kJK1mol1

From Sections 6.5 and 6.10

U = nCV,mT= (6.00 mol)( 0.02079 kJK1mol1)(203.2 K) = 25.3 kJ

H = nCP,mT= (6.00 mol)( 0.02910 kJK1mol1)(203.2 K) = 35.5 kJ

Note: This equation applies even though pressure is not constant. For an ideal gas, enthalpy depends

only on temperature.

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6.12 The Enthalpy of Physical Change

Example 6.12 Considering the relative strength of intermolecular forces in the liquid phase, arrange the

following substances in the order of increasing enthalpy of vaporization at the boiling

point of the liquid: CH3OH, C6H6 , Hg, CH4 .

Solution The intermolecular interactions of significance for the substances listed are London forces forC6H6and CH4, metallic bonding for Hg, and hydrogen bonding for CH3OH. The relative

strength of intermolecular forces increases in the order London forces, hydrogen bonding, and

metallic interactions. The strength of London interactions increases with the number of

electrons in the molecule (polarizability); thus, C6H6has stronger interactions than CH4 .

The expected order of increasing enthalpy of vaporization is CH4 , C6H6 , CH3OH, and Hg.

The values of standard enthalpies of vaporization in Table 6.3 in the text are 8.2, 30.8, 35.3,

and 59.3 kJmol1, respectively.

The Enthalpy of Chemical Change (Sections 6.146.22)

Key Concepts

reaction enthalpy, thermochemical equation, combustion, standard state, standardreaction enthalpy, Hesss law, reaction sequence, overall reaction, standard enthalpy ofcombustion, standard enthalpy of formation, lattice enthalpy, Born-Haber cycle, bondenthalpy, mean (average) bond enthalpy, temperature dependence of reaction enthalpy,Kirchhoffs law

6.14 Reaction Enthalpies

Example 6.14a Reaction with oxygen is a combustion reaction. Write the thermochemical equation for the

combustion of hydrogen gas at 25C. For the reaction of 1 mol H2(g), qP= 285.83 kJ.Determine the reaction enthalpy, Hr .

Solution The chemical equation H2(g) + O2(g) H2O(l) is balanced by inspection. Note that water

at 25C is a liquid. The balanced equation with the smallest whole integer coefficients is

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

For a constant pressure reaction, qP= H. The heat change is given for 1 mol H2(g),but the balanced equation requires 2 mol.

Thus, H= 2(285.83 kJ) = 571.66 kJ. The thermochemical equation is

2H2(g) +O2(g) 2H2O(l) H= 571.66 kJ

The reaction enthalpy is Hr= 571.66 kJmol1. Here mol refers to the enthalpy change

for 2 mol H2(g), 1 mol O2(g), and 2 mol H2O(l).

More explicitly, Hr= 571.66 kJ(2 mol H2)1, Hr= 571.66 kJ(1 mol O2)

1, and

Hr= 571.66 kJ(2 mol H2O)1.

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Chapter 6

Example 6.14b Use the thermochemical equation given below to determine how much heat is released, qsurr ,

when 350 g of propane, C3H8(g), is burned completely in a backyard barbecue at 25C.

C3H8(g) + 5 O2(g) 3CO2(g) +4

H2O(l) H= 2220 kJ

Solution The reaction enthalpy is Hr= 2220 kJmol1. Here mol refers to the enthalpy change for

1 mol C3H8(g), 5 mol O2(g), 3 mol CO2(g), and 4 mol H2O(l). More explicitly,

Hr= 2220 kJ(1 mol C3H8)1. The molar mass of propane is3(12.01) +8(1.0079) = 44.09 gmol1. The number of moles of propane is then

(350 g)/(44.09 gmol1) = 7.938 mol. The heat change for the system is

qP= nHr= (7.938 mol)(2220 kJmol1) = 1.76 104kJ.

The heat released to the surroundings is qsurr= qP= 1.76 104kJ = 17.6 MJ.

6.15 The Relation Between Hand U

Example 6.15a Calculate the change in internal energy, U, for the combustion reaction in Example 6.14b.

olution C3H8(g) + 5O2(g) 3

CO2(g) +4H2O(l) H= 2220 kJS

HU+(ngas)RT and UH(ngas)RT

ngas= n(CO2) [n(C3H8) +n(O2)] = 3 (1 +5) = 3 6 = 3

Be careful to exclude liquid water.

(ngas)RT= (3 mol)(8.31447 JK1mol1)(298.15 K) = 7437 J = 7.437 kJ

UH(ngas)RT = 2220 kJ (7.437 kJ) = 2213 kJ

Note: Use consistent energy units, either J or kJ. The internal energy change is a more positive value

than the enthalpy change for a reaction that consumesgas. The difference between HandUis small, only about 0.3%.

Example 6.15b Calculate the change in internal energy, U, for the dissociation of hydrogen gas into atoms at25C, given the following thermochemical equation:

H2(g) 2

H(g) H= +435.94 kJ

Solution ngas= n(H) n(H2 ) = 2 1 = +1

(ngas)RT= (+1 mol)(8.31447 JK1mol1)(298.15 K) = +2478.97 J = +2.479 kJ

UH(ngas)RT = +435.94 kJ (+2.479 kJ) = +433.46 kJ

Note: The internal energy change is a more negative value than the enthalpy change for a reaction

thatproducesgas. The difference between Hand Uis small, only about 0.6%.

6.16 Standard Reaction Enthalpies

Example 6.16 Compare the following two thermochemical reactions at 25C. Comment on the nature of the

difference in standard reaction enthalpy between reaction (2) and reaction (1).

(1) C3H8(g) + 5O2(g) 3

CO2(g) +4H2O(l) H

= 2220 kJ

(2) C3H8(g) + 5O2(g) 3

CO2(g) +4H2O(g) H

= 2044 kJ

Solution Reaction (2) is less exothermic than reaction (1). Reaction (2) produces water vapor and

reaction (1) produces liquid water. For reaction (1), Hr= 2220 kJmol1. For reaction (2),

Hr= 2044 kJmol1. Recall that mol refers to 1 mol C3H8 , 5 mol O2 , 3 mol CO2 , and

4 mol H2O. The enthalpy of vaporization of water is 44.01 kJmol1at 25C. The difference

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in standard reaction enthalpies between reaction (2) and reaction (1) is 176 kJ(4 mol H2O)1

or 44 kJmol1, which is the enthalpy of vaporization of water at 25C. Thus, the differencecorresponds to the vaporization of 4 mol of liquid water at 25C. The thermochemicalequations may be combined by subtracting reaction (1) from reaction (2). The result is

4H2O(l) 4H2O(g) H

= +176 kJ

Changing the stoichiometric coefficients to the simplest whole integers yields

H2O(l) H2O(g) Hr= +44 kJmol1= Hvap

Combining thermochemical equations is an application of Hesss law, which is discussed in the

next section.

6.17 Combining Reaction Enthalpies: Hesss Law

Example 6.17 Determine the standard reaction enthalpy for

(3) 2SO2(g) + O2(g) 2SO3(g) H

= ?

Use Hesss law to combine combustion reactions (1) and (2) given below.

(1) 2S(s) + 3O2(g) 2SO3(g) H

= 792 kJ

(2) S(g) + O2(g) SO2(g) H= 297 kJ

Solution In reaction (1), the proper number of moles of product SO3is present, and the reaction may be

combined as written. In reaction (2), the number of moles of SO2needs to be doubled and the

reactants and products interchanged. The thermochemical equations that combine to give

reaction (3) are then

(1) 2S(s) + 3O2(g) 2SO3(g) H

= 792 kJ

2(2) 2 SO2(g) 2 S(s) + 2 O2(g) H= +594 kJ____(3) 2SO2(g) + O2(g) 2

SO3(g) H= 198 kJ

6.18 The Heat Output of Reactions

Example 6.18a Write a balanced thermochemical equation for the combustion of one mole of octane (use

Table 6.4 in the text). Suppose an average automobile travels 11,000 miles in a year and

averages 21.0 miles per gallon of gasoline. Approximating gasoline as octane and assuming a

gallon of gasoline has a mass of 2.66 kg, calculate the enthalpy produced from the combustion

of fossil fuel and used by an average automobile per year. Assume the complete combustion

of octane in the engine.

Solution The thermochemical equation for one mole of octane is (Table 6.4 in the text)

C8H18(l) +25

2O2(g) 8

CO2(g) +9H2O(l) Hc= 5471 kJmol

1

(11000 miles / 21.0 milesgallon1) = 524 gallons of gasoline or octane

(524 gallons 2.66 kggallon1) = 1.39 103kg = 1.39 106g

molar mass of octane = [8(12.01) +18(1.0079)] = 114.22 gmol1(1.39 106g / 114.22 gmol1) = 1.22 104mol octane per year

(5471 kJmol11.22 104mol) = 6.67 107kJ = 66.7 GJ per year

Example 6.18b How many grams of CO2(g) are added to the atmosphere by the average automobile in one

year? Use the results of Example 6.18a.

nswer 4.29 106g = 4.29 Mg of CO2(g)A

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Chapter 6

6.19 Standard Enthalpies of Formation

Example 6.19a Using enthalpy of formation values in Appendix 2A, calculate the enthalpy of combustion of

two moles of octane.

Solution The thermochemical equation for the combustion of two moles of octane is

2C8H18(l) + 25O2(g) 16

CO2(g) +18H2O(l) Hc= ?

The standard enthalpy of formation of an element in its most stable form is 0.

Therefore, Hf[O2(g)] = 0.

Hr= nHf(products) nHf(reactants) = Hc

= 16Hf[CO2(g)] +18Hf[H2O(l)] 2Hf[C8H18(l)]

= 16(393.51 kJmol1) +18(285.83 kJmol1) 2(249.9 kJmol1)

= 6296.15 5144.94 +499.8 kJmol1

= 10941.3 kJmol1 at 298.15 K

Recall that mol refers to 2 mol of C8H18 , 25 mol O2 , 16 mol CO2 , and 18 mol H2O.

Example 6.19b Using enthalpy of formation values in Appendix 2A, calculate the enthalpy of dissociation ofone mole of hydrogen molecules into hydrogen atoms.

Answer H2(g) 2H(g) Hr= ?

Hr= 2Hf[H(g)] = 2(217.97 kJmol1) = 435.94 kJmol1 at 298.15 K

This enthalpy change is also known as the bond enthalpy, HB(HH), defined in Section 6.21.

Note: It is possible to obtain an unknown enthalpy of formation if a reaction enthalpy and the other

enthalpies of formation in that reaction are known. See Example 6.10 in the text.

6.21 Bond Enthalpies

Example 6.21a Write a thermochemical equation for breaking one mol of CH bonds in methane, CH4.

The enthalpy of formation of CH3(g) is 138.9 kJmol1. Use Appendix 2A for other data.

Solution CH4(g) CH3(g) +H(g) Hr= HB(CH)

Hr= Hf[CH3(g)] +Hf[H(g)] Hf[CH4(g)]

= 138.9 +217.97 (74.81) kJmol1= 431.7 kJmol1= HB(CH)

The CH bond has an averagebond enthalpy of 412 kJmol1in polyatomic molecules (seeTable 6.8 in the text). Note that mol refers to the amount of CH bonds. Breaking the firstCH bond in methane requires more heat than subsequent ones.

Example 6.21b Write a thermochemical equation for breaking one mol of carbonyl CH bonds in ethanal(acetaldehyde), CH3COH. The enthalpy of formation of CH3CO(g) is 18.8 kJmol

1.

Use Appendix 2A for other data.

Solution CH3COH(g) CH3CO(g) +H(g) Hr= HB(CH)

Hr= Hf[CH3CO(g)] +Hf[H(g)] Hf[CH3COH(g)]

= 18.8 +217.97 (192.30) kJmol1= 391.5 kJmol1= HB(CH)

The carbonyl CH bond in ethanal is considerably weaker than the averageCH bondenthalpy in polyatomic molecules.

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E

xample 6.21c Calculate the average CH bond enthalpy in methane, CH4. Use the data in Appendix 2A.

Solution CH4(g) C(g) +4H(g) Hr= 4HB(CH)

Hr= Hf[C(g)] +4Hf[H(g)] Hf[CH4(g)]

= 716.68 +4(217.97) (74.81) kJmol1= 1663.37 kJmol1

HB(CH) = Hr/4 = 415.84 kJmol1

The average CH bond enthalpy in methane is slightly larger than the averageCHbond enthalpy in polyatomic molecules.

Example 6.21d Calculate the average C=O bond enthalpy in carbon dioxide, CO2. The enthalpy of

formation of O(g) is 249.17 kJmol1. Use Appendix 2A for other data.

Solution CO2(g) C(g) +2O(g) Hr= 2HB(C=O)

Hr= Hf[C(g)] +2Hf[O(g)] Hf[CO2(g)]

= 716.68 +2(249.17) (393.51) kJmol1= 1608.53 kJmol1

HB(C=O) = Hr/2 = 804.27 kJmol1

The C=O double bond in carbon dioxide is considerably stronger than the averageC=O bond

enthalpy of 743 kJmol1in polyatomic molecules.

6.22 The Variation of Reaction Enthalpy with Temperature

Example 6.22a Calculate a value for the enthalpy of vaporization of liquid water at 298.15 K. Use

Kirchhoffs law to estimate the enthalpy of vaporization of liquid water at the normal

boiling point of 373.15 K. Compare the value to the one in Table 6.3 in the text. Use the

data in Appendix 2A.

Solution H2O(l) H2O(g) Hr= Hvap

Hr= Hf[H2O(g)] Hf[H2O(l)]

= 241.82 (285.83) kJmol1= +44.01 kJmol1= Hvapat 298.15 K

CP = CP,m[H2O(g)] CP,m[H2O(l)] = 33.58 75.29 JK1mol1= 41.71 JK1mol1= 0.04171 kJK1mol1

Hvap(373.15 K) = Hvap(298.15 K) +CP(373.15 298.15 K)

= 44.01 +(0.04171)(75.00) = 44.01 3.13 kJmol1

= 40.88 kJmol1

The value in Table 6.3 in the text is 40.7 kJmol1. The approximation is an excellent onein this case.

Example 6.22b The bond enthalpy of HH is 436 kJmol1at 298.15 K. Use Kirchhoffs law to estimate thebond enthalpy at 0 K. Use the data in Appendix 2A.

Solution H2(g) 2H(g) Hr= HB(HH) = 436 kJmol

1at 298.15 K

CP = 2CP,m[H(g)] CP,m[H2(g)] = 2(20.78) 28.82 JK1mol1

= +12.74 JK1mol1= +0.01274 kJK1mol1

HB(0 K) = HB(298.15 K) +CP(0 298.15 K)

= 436 +(0.01274)(298.15) = 436 3.8 kJmol1

= 432 kJmol1

The spectroscopic dissociation energy of H2(g) is 432.07 kJmol1, which corresponds to a

thermodynamic temperature of absolute zero. Any physical change of state that occurs at

low temperature is deliberatelyignored.

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Example 6.22c If H= 571.66 kJ for reaction (1) at 25C, calculate Hfor reaction (2) at 0C.

(1) 2H2(g) +O2(g) 2H2O(l)

(2) 2H (g) +O2 2(g) 2H2O(s)

Solution Solve the problem in two parts. First use Kirchoffs law to calculate Hfor Reaction (1)at 0C. Then consider the phase change from liquid water to solid water (Hesss law).The following information is taken from Table 6.3 in the text and Appendix 2A.

Hfus[H2O] = 6.01 kJmol1 CP,m[H2O(l)] = 75.29 JK

1mol1CP,m[H2(g)] = 28.82 JK

1mol1 CP,m[O2(g)] = 29.36 JK1mol1

Answer H= 585.27 kJ for reaction (2) at 0C

Note: This example illustrates how to modify Kirchoffs law when phase changes occur.

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