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CHAPTER 18

Thermodynamics and Equilibrium

CHAPTER TERMS AND DEFINITIONSNumbers in parentheses after definitions give the text section in which the terms are explained. Starred terms are italicized in the text. Where a term does not fall directly under a text section heading, additional information is given for you to locate it.

thermodynamics* study of the relationship between heat and other forms of energy involved in a chemical or physical process (chapter introduction)

internal energy (U) sum of the kinetic and potential energies of the particles making up a system (18.1)

state function property of a system, such as temperature and pressure, that depends only on its present condition and that is a defining factor of that condition (18.1)

work (w) energy exchange that results when a force F moves an object through a distance d; w = F d (18.1)

first law of thermodynamics the change in internal energy of a system ΔU equals the heat absorbed or lost by the system ±q plus the work done on or by the system ±w; ΔU=q+ w (18.1)

enthalpy (H) thermodynamic quantity defined by the equation H = U + PV (18.1)

spontaneous process physical or chemical change that occurs by itself (18.2, introductory section)

nonspontaneous* describes a physical or chemical change that occurs only with the aid of an outside force or agent (18.2, introductory section)

entropy (S) thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy (18.2)

second law of thermodynamics the total entropy of a system and its surroundings always increases for a spontaneous process; ΔS > q/T (18.2)

statistical thermodynamics* rigorous method of defining the entropy of a system in terms of a sum over the energy levels available to the system (18.2)

reversible* describes a process that can go in one direction or the other (18.2, marginal note)

third law of thermodynamics a substance that is perfectly crystalline at 0 K has an entropy of zero (18.3)

standard (absolute) entropy (S) entropy value for the standard state of a species (18.3)

free energy (G) thermodynamic quantity defined by the equation G = H – TS (18.4)

standard states* conditions under which to report the properties of matter: 1 atm pressure for pure substances, 1 M concentration for ions in solution (18.4)

standard free energy of formation ( ) free-energy change when one mole of substance is formed from its elements in their stablest states at one atmosphere of pressure and at a specified temperature (usually 25C) (18.4)

Copyright © Houghton Mifflin Company. All rights reserved.

393 Chapter 18: Thermodynamics and Equilibrium

thermodynamic equilibrium constant (K) equilibrium constant in which the concentrations of gases are expressed in partial pressures in atmospheres, whereas the concentrations of solutes in liquid solutions are expressed in molarities (18.6)

CHAPTER DIAGNOSTIC TEST1. Determine whether each of the following statements is true or false. If the statement is false,

change it so that it is true.

a. A spontaneous process is a physical or chemical change that occurs without warning. True/False: ________________________________________________________________

__________________________________________________________________________

b. Entropy is a measure of disorder in a physical or chemical system. True/False: __________

__________________________________________________________________________

c. Entropy increases going from solid to liquid to gas. True/False: _______________________

__________________________________________________________________________

d. A reaction is spontaneous as written if ΔG is a large positive number. True/False: __________________________________________________________________________

__________________________________________________________________________

2. Calculate the change in entropy ΔS for the following precipitation reaction:

Ag+(aq) + Cl–(aq) AgCl(s)

Use text Table 18.1 for standard entropy values.

3. Predict by inspection the sign of ΔS for each of the following reactions, and explain your answers.

a. H2(g) + C2H6(g) CH4(g)

b. 2SO2(g) + O2(g) 2SO3(g)

c. 2NO2(g) 2NO(g) + O2(g)

d. Pb(NO3)2(aq) + Na2SO4(aq) PbSO4(s) + 2NaNO3(aq)

4. The heat of vaporization of liquid bromine, Br2(l), at 25C is 30.7 kJ/mol:

Br2(l) Br2(g)

If liquid bromine at 25C has an entropy of 152.2 J/(K ∙ mol), what is the entropy of one mole of the vapor in equilibrium with the liquid at that temperature?

5. Calculate ΔG for the following precipitation reaction at 25C:

2Ag+(aq) + S2–(aq) Ag2S(s)

Use values of and S from text Appendix C.

6. Calculate the free-energy change ΔG for the following reaction using values of from text Appendix C:

2Li(s) + 2H2O(l) 2Li+(aq) + 2OH–(aq) + H2(g)

7. Give the expression for K for the following reactions.


Chapter 18: Thermodynamics and Equilibrium 394

a. 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

b. NH4OH(aq) NH3(g) + H2O(l)

c. 2H2O2(l) 2H2O(l) + O2(g)

8. Determine whether each of the reactions in Problem 3 is spontaneous or nonspontaneous as written. Use values in text Appendix C and give reasons for your answers.

9. Calculate Ka for the ionization of formic acid, HCOOH, in aqueous solution. Use values of in text Appendix C.

10. Calculate the value of the thermodynamic equilibrium constant at 648 K for the following reaction:

N2(g) + 3H2(g) 2NH3(g)

ΔG for this reaction at the given temperature is 43.2 kJ.

ANSWERS TO CHAPTER DIAGNOSTIC TESTIf you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1.

a. False. A spontaneous process is a physical or chemical change that can occur by itself, requiring no continuing outside agency to make it happen. (18.2, introductory section)

b. True. (18.2)

c. True. (18.2)

d. False. A reaction is spontaneous as written if ΔG is a large negative number. (18.4)

2. ΔS= –32.9 J/K (18.3, PS Sk. 3)

3.

a. ΔS 0; there is no change in the number of gas molecules.

b. ΔS < 0; there are three molecules of gas on the left but only two on the right.

c. ΔS > 0; there are two molecules of gas on the left and three on the right.

d. ΔS < 0; a solid is formed on the right. (18.3, PS Sk. 2)

4. ΔS = 255 J/(K ∙ mol) (18.2, PS Sk. 1)

5. ΔG = –278.3 kJ (18.4, PS Sk. 4)

6. ΔG = –427.8 kJ (18.4, PS Sk. 5)



7.

a. K =

b. K =

c. Water is not a solvent and must be included in the expression:

K = (18.6, PS Sk. 7)

8.

a. Spontaneous; ΔG < 0

b. Spontaneous; ΔG < 0

c. Nonspontaneous; ΔG > 0

d. Spontaneous; ΔG < 0 (18.4, PS Sk. 6)

9. Ka = K = 2.1 10–4 (18.6, PS Sk. 8)

10. K = 3.3 10–4 (18.6, PS Sk. 8)

SUMMARY OF CHAPTER TOPICS

18.1 First Law of Thermodynamics; Enthalpy

Learning Objectives Define internal energy, state function, work, and first law of thermodynamics.

Explain why the work done by the system as a result of expansion or contraction during a chemical reaction is –PV.

Relate the change of internal energy U and heat of reaction q.

Define enthalpy H.

Show how heat of reaction at constant pressure qp equals the change of enthalpy H.

Exercise 18.1A gas is enclosed in a system similar to that shown in text Figure 18.2. More weights are added to the piston, giving a total mass of 2.20 kg. As a result, the gas is compressed, and the weights are lowered 0.250 m. At the same time, 1.50 J of heat evolves from (leaves) the system. What is the change in internal energy of the system ΔU? The force of gravity on a mass m is mg, where g is the constant acceleration of gravity (g = 9.80 m/s2).

Wanted: ΔU, in joules

Given: Mass of weights = 2.20 kg, piston dropped 0.250 m, heat evolved (q) = –1.50 J; F = mg, g = 9.80 m/s2.

Known: q = –1.50 J, as heat is evolved; ΔU = q + w; w = F d, the sign of w is + because the work is done on the system; 1 J = 1 kg ∙ m2/s2.



Solution: Substitute algebraically, then with numbers, into

ΔU = q + w = q + (F d) = q + (mg d)

= 1.50 J + 2.20 kg 9.80 m/s2 0.250 m

ΔU = 3.89 J

Exercise 18.2Consider the combustion (burning) of methane, CH4, in oxygen:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

The heat of reaction at 25C and 1.00 atm is –890.2 kJ. What is the change in volume when 1.00 mol CH4 reacts with 2.00 mol O2? (You can ignore the volume of liquid water, which is insignificant compared with volumes of gases.) What is w for this change? Calculate ΔU for the change indicated by the chemical equation.

Wanted: ΔV, in L; w, in kJ; ΔU, in kJ

Given: equation; t = 25C; P = 1 atm; q = –890.2 kJ; 1 mol CH4; 2.00 mol O2.

Known: PΔV = ΔnRT; w = PΔV and is + because work is done on the system (the volume of the system decreases); ΔU = q + w, 1 L = 10–3 m3; 1 Pa ∙ m3 = 1 J; 1 atm = 1.01 105 Pa..

Solution: Since there are 3 mol of gas on the left of the equation and 1 mol on the right, Δn = 2 mol, and

ΔV = = = 48.9 L

w = +PΔV = 1.01 105 Pa 48.9 L =

= 4.94 103 J = 4.94 kJ

ΔU = q + w = –890.2 kJ + 4.94 kJ = –885.3 kJ

18.2 Entropy and the Second Law of Thermodynamics

Learning Objectives Define spontaneous process.

Define entropy.

Relate entropy to disorder in a molecular system (energy dispersal).

State the second law of thermodynamics in terms of system plus surroundings.

State the second law of thermodynamics in terms of the system only.

Calculate the entropy change for a phase transition. (Example 18.1)

Describe how H – TS functions as a criterion of a spontaneous reaction.

Problem-Solving Skill1. Calculating the entropy change for a phase transition. Given the heat of phase transition and

the temperature of the transition, calculate the entropy change ΔS of the system (Example 18.1).



Stop a minute and think of your room at home or in the dorm. If you are like most of us, you envision clothes hanging over a chair, books scattered, maybe a sock or two half hidden under the bed. Such a room gives a vivid example of the second law of thermodynamics—that entropy increases in a spontaneous process. It takes work to hang up the clothes, to arrange the books neatly, and to put the socks in the hamper.

It is a good idea to memorize the expression for entropy, S = q/T, and to remember that the entropy change in a system during a spontaneous change is ΔS > q/T (i.e.,nonreversible).

Exercise 18.3Liquid ethanol, C2H5OH(l), at 25C has an entropy of 161J/(K∙mol). If the heat of vaporization ΔHvap at 25C is 42.3 kJ/mol, what is the entropy of the vapor in equilibrium with the liquid at 25C?

Wanted: entropy of the vapor, in J/(K ∙ mol)

Given: S for liquid = 161 J/(K ∙ mol); ΔHvap = 42.3 kJ/mol; T = 25C.

Known: The entropy of the vapor = entropy of the liquid plus ΔS of vaporization; qp= ΔH; ΔSvap = ΔHvap/T; 25C = 298 K.

Solution: The calculation is

ΔSvap = = 1.420 102 J/(K ∙ mol)

Svap = (161 + 142) J/(K ∙ mol) = 303 J/(K ∙ mol)

18.3 Standard Entropies and the Third Law of Thermodynamics

Learning Objectives State the third law of thermodynamics.

Define standard entropy (absolute entropy).

State the situations in which the entropy usually increases.

Predict the sign of the entropy change of a reaction. (Example 18.2)

Express the standard change of entropy of a reaction in terms of standard entropies of products and reactants.

Calculate S for a reaction. (Example 18.3)

Problem-Solving Skills2. Predicting the sign of the entropy change of a reaction. Predict the sign of ΔS for a reaction to

which the rules listed in the text can be clearly applied (Example 18.2).

3. Calculating ΔS for a reaction. Given the standard entropies of reactants and products, calculate the entropy of reaction ΔS (Example 18.3).



It is important to note that the zero point of entropy is at 0 K and that we talk of absolute entropies S of substances rather than of entropies of formation. Recall that substances have an enthalpy of formation

and that the zero point for these values is the element in its most stable state at standard conditions.

Exercise 18.4Predict the sign of ΔS for each of the following reactions.

a. CaCO3(s) CaO(s) + CO2(g)

b. CS2(l) CS2(g)

c. 2Hg(l) + O2(g) 2HgO(s)

d. 2Na2O2(s) + 2H2O(l) 4NaOH(aq) + O2(g)

Known: Entropy is a measure of disorder; solids are more ordered than liquids, which are more ordered than gases.

Solution:

a. ΔS > 0; a molecule of gas is formed.

b. ΔS > 0; a molecule of gas is formed.

c. ΔS < 0; the solid product is more ordered than the reactants.

d. ΔS > 0; a molecule of gas is formed.

Exercise 18.5Calculate the change of entropy ΔS for the reaction given in Example 18.2(a). The standard entropy of glucose, C6H12O6(s), is 212 J/(K ∙ mol). See text Table 18.1 for other values.

Wanted: ΔS in J/K

Given: S for C6H12O6 = 212 J/(K ∙ mol)

Known: reaction from Example 18.2(a); text Table 18.1 for S values


C6H12O6(s) 2C2H5OH(l) + 2CO2(g)

S 212 2(161) 2(213.7) J/(K ∙ mol)

ΔS = [2(161) + 2(213.7) – 212] J/K

ΔS = 537 J/K

It is important to recall that in Section 14.2 you learned that the equilibrium-constant expression and its value depend on the coefficients used to write the reaction. It is customary to write the coefficients using the smallest ratio of whole numbers. But the reaction is still correct as long as you use any numbers in that ratio.

Thus the values for the entropy of reaction calculated in Example 18.3 and in Exercise18.5 both depend on the coefficients used to write the individual reactions. The entropies would be different if different coefficients were used.

This is also true of the enthalpies of reaction ΔH calculated in Chapter 6. Sometimes the term mole of reaction is used to describe this idea. This use of the word mole does not designate 6.02 1023



particles. It merely means “the reaction as written” or “the quantities specified in the reaction as written.” In Example 18.3, where the reaction is written as

2NH3(g) + CO2(g) NH2CONH2(aq) + H2O(l)

“one mole of reaction” means “2 6.02 1023 molecules of NH3 and 6.02 1023 molecules of CO2 react to form 6.02 1023 molecules of urea and 6.02 2023 molecules of water.” In Exercise 18.5, “one mole of reaction” means “6.02 1023 molecules of glucose yield 26.02 1023 molecules of ethyl alcohol and 2 6.02 1023 molecules of CO2.” So the units for the entropies of reaction could be written J/(K ∙ mol), mol meaning “mole of reaction.” We will refer to this concept again in the next section and in the following chapter.

18.4 Free Energy and Spontaneity

Learning Objectives Define free energy G.

Define the standard free-energy change.

Calculate G from H and S. (Example 18.4)

Define the standard free energy of formation G.

Calculate G from standard free energies of formation. (Example 18.5)

State the rules for using G as a criterion for spontaneity.

Interpret the sign of G. (Example 18.6)

Problem-Solving Skills4. Calculating ΔG from ΔH and ΔS. Given enthalpies of formation and standard entropies of

reactants and products, calculate the standard free-energy change ΔG for a reaction (Example 18.4).

5. Calculate ΔG from standard free energies of formation. Given the free energies of formation of reactants and products, calculate the standard free-energy change ΔG for a reaction (Example 18.5).

6. Interpreting the sign of ΔG. Use the standard free-energy change to determine the spontaneity of a reaction (Example 18.6).

Note that the zero point of free energy of formation is the element in its most stable state at 1 atm pressure and usually 25C, as with the enthalpy of formation. It is a good idea to memorize the expression for the free-energy change of a reaction: ΔG = ΔH – TΔS. You must know that ΔG for a spontaneous reaction is negative.

Exercise 18.6Calculate ΔG for the following reaction at 25C. Use data given in text Tables 6.2 and 18.1.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Known: ΔG = ΔH – TΔS; values are in text Table 6.2; S values are in text Table 18.1.

Solution: Calculations follow:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)



–74.9 2(0) –393.5 2(–241.8) kJ/mol

S° 186.1 2(205.0) 213.7 2(188.7) J/(K ∙ mol)

ΔH = [–393.5 + 2(–241.8) – (–74.9)] kJ

= –802.2 kJ

ΔS = {[213.7 + 2(188.7)] – [186.1 + 2(205.0)]} J/K

= –5.0 J/K

ΔG = ΔH – TΔS

ΔG = 802.2 kJ {298 K }

ΔG = –800.7 kJ

Let us again refer to the term mole of reaction. The free-energy change also could be thought of as per “mole of reaction” for the reaction with coefficients specified. This means that the units for the ΔH of the preceding reaction could be calculated as

ΔG = +

= 802.2 kJ / mol (of rxn)

We could do the same for the calculation of ΔS for the reaction.

In another text you might see ΔH or ΔG written as kJ/mol, where mol means “mole of reaction.”

Exercise 18.7Calculate ΔG at 25C for the following reaction using values of .

CaCO3(s) CaO(s) + CO2(g)

Known: values given in text Table 18.2



–1128.8 –603.5 –394.4 kJ/mol

ΔG = [–603.5 + (–394.4) – (–1128.8)] kJ

ΔG = 130.9 kJ



Exercise 18.8Which of the following reactions are spontaneous in the direction written? See text Table 18.2 for data.

a. C (graphite) + 2H2(g) CH4(g)

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

c. 4HCN(g) + 5O2(g) 2H2O(l) + 4CO2(g) + 2N2(g)

d. Ag+(aq) + I–(aq) AgI(s)

Wanted: spontaneity of reaction

Given: values in text Table 18.2

Known: If ΔG > 0, reaction is nonspontaneous; if ΔG < 0, reaction is spontaneous.

Solution: All the reactions are spontaneous, as the following calculations show:

a. C (graphite) + 2H2(g) CH4(g)

0 2(0) –50.8 kJ/mol

ΔG = –50.8 kJ

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

2(0) 2(0) 2(–237.1) kJ/mol

ΔG = [2(–237.1)] kJ

ΔG = –474.2 kJ

c. 4HCN(g) + 5O2(g) 2H2O(l) + 4CO2(g) + 2N2(g)

4(124.7) 5(0) 2(–237.1) 4(–394.4) 2(0) kJ/mol

ΔG = [2(–237.1) + 4(–394.4) – 4(124.7)] kJ

ΔG = –2551 kJ

d. Ag+(aq) + I–(aq) AgI(s)

77.12 –51.59 –66.19 kJ/mol

ΔG = [–66.19 – (–51.59 + 77.12)] kJ

ΔG = –91.72 kJ

18.5 Interpretation of Free Energy

Learning Objectives Relate the free-energy change to maximum useful work.

Describe how the free energy changes during a chemical reaction.



18.6 Relating ΔG° to the Equilibrium Constant

Learning Objectives Define the thermodynamic equilibrium constant K.

Write the expression for a thermodynamic equilibrium constant. (Example 18.7)

Indicate how the free-energy change of a reaction and the reaction quotient are related.

Relate the standard free-energy change to the thermodynamic equilibrium constant.

Calculate K from the standard free-energy change (molecular equation). (Example 18.8)

Calculate K from the standard free-energy change (net ionic equation). (Example 18.9)

Problem-Solving Skills7. Writing the expression for a thermodynamic equilibrium constant. For any balanced chemical

equation, write the expression for the thermodynamic equilibrium constant (Example18.7).

8. Calculating K from the standard free-energy change. Given the standard free-energy change for a reaction, calculate the thermodynamic equilibrium constant (Examples 18.8 and 18.9).

Memorize the relationship between ΔG and the thermodynamic equilibrium constant: ΔG = RT ln K. Remember also that K = Kc for reactions of solutes in liquid solution, K = Kp when the reaction involves only gases, and K includes both pressures and concentrations in heterogeneous systems.

You may well ask why R, the gas constant, shows up here when many reactions do not involve gases. The presence of R in the expression for ΔG is due to the intimate relationship between R and how the mole is defined. Recall from your study of the ideal gas law (text Section 5.3) that the amount of gas in a given volume (22.41 L) at a specified temperature (273.15 K) and pressure (1 atm) was called a mole. Also recall that R is the constant ratio of these standard conditions, PV/T. Thus R defines the mole. In this expression for ΔG, the word mole in the units of R, J/(K ∙ mol), means “mole of reaction,” explained previously in this chapter.

Exercise 18.9Give the expression for K for each of the following reactions.

a. CaCO3(s) CaO(s) + CO2(g)

b. PbI2(s) Pb2+(aq) + 2I–(aq)

c. H+(aq) + HCO3–(aq) H2O(l) + CO2(g)

Solution: The calculations are

a. K = = Kp

b. K = [Pb2+][I–]2 = Ksp

c. K =



Exercise 18.10Use data from text Table 18.2 to obtain the equilibrium constant Kp at 25C for the reaction


Note that values of are needed for CaCO3 and CaO, even though the substances do not appear in

Kp = .

Wanted: Kp (from ΔG)

Given: reaction; text Table 18.2 ( )

Known: ΔG = RT ln K; K for this reaction = Kp; R = 8.31 J/(K ∙ mol), so in order to have the units work out properly, we must report ΔG as kJ/mol; T = 298 K.



–1128.8 –603.5 –394.4 kJ/mol

ΔG = [–603.5 + (–394.4) – (–1128.8)] kJ

ΔG = 130.9 kJ/mol

Rearrange the expression for ΔG to solve for Kp:

ln Kp = = = 52.50

Kp = 2 1023

Exercise 18.11Calculate the solubility product constant for Mg(OH)2 at 25C. The values (in kJ/mol) are as follows: Mg2+(aq), –454.8; OH–(aq), –157.3; Mg(OH)2(s), –833.7.

Known: ΔG = –RT ln Ksp; R (see Exercise 18.10) and T given above


Mg(OH)2(s) Mg2+(aq) + 2OH–(aq)

–933.9 –456.0 2(–157.3) kJ/mol

ΔG = [–454.8 + 2(–157.3) – (–833.7)] kJ

ΔG = 64.3 kJ/mol

ln Ksp = = = 25.97

K = 5 1012



18.7 Change of Free Energy with Temperature

Learning Objectives Describe how G at a given temperature (GT) is approximately related to H and S at

that temperature.

Describe how the spontaneity or nonspontaneity of a reaction is related to each of the four possible combinations of signs of H and S.

Calculate G and K at various temperatures. (Example 18.10)

Problem-Solving Skill9. Calculating ΔG and K at various temperatures. Given ΔH and ΔS at 25C, calculate ΔG and

K for a reaction at a temperature other than 25C (Example 18.10).

Exercise 18.12The thermodynamic equilibrium constant for the vaporization of water,

H2O(l) H2O(g)

is Kp = . Use thermodynamic data to calculate the vapor pressure of water at 45C. Compare your answer with the value given in text Table 5.6.

Wanted: at 45C

Known: = Kp; ΔG = –RT ln Kp; need values from text Tables 6.2 and 18.1; R= 8.31 J/(K ∙ mol); T = 318 K


H2O(l) H2O(g)

–285.8 –241.8 kJ/mol

S 69.9 188.7 J/(K ∙ mol)

ΔH = [–241.8 – (–285.8)] kJ = 44.0 kJ

ΔS = (188.7 – 69.95) J/K = 118.75 J/K

ΔG = ΔH – TΔS

ΔG = 44.0 kJ 318 K = 6.237 103 J

which we will express as 6.237 103 J/mol.

ln Kp = = = 2.360

Kp = 9.4 102 atm

To compare with the value in text Table 5.6, convert to mmHg:

9.5 102 atm = 72 mmHg



The text table value is 71.9 mmHg. The calculated value is the text table value to two significant figures.

Exercise 18.13To what temperature must magnesium carbonate be heated to decompose it to MgO and CO2 at 1 atm? Is this higher or lower than the temperature required to decompose CaCO3? Values of (in kJ/mol) are as follows: MgO(s),601.2; MgCO3(s), –1111.7. Values of S(in J/K) are as follows: MgO(s),26.9; MgCO3(s), 65.9. Data for CO2 are given in text Tables 6.2 and 18.1.

Wanted: T at which MgCO3 will decompose; compare with T for CaCO3

Given: values; S values; see text Tables 6.2 and 18.1.

Known: ΔG = 0 at that T; ΔGo = ΔHo – TΔSo; assume ΔH and ΔS are constant, so we can use their values at 25C for the calculation.

Solution: If ΔG = 0, then ΔH = TΔS, and T = ΔHo/ΔSo. Calculations of ΔH and ΔS follow:

MgCO3(s) MgO(s) + CO2(g)

–1111.7 –601.2 –393.5 kJ/mol

ΔS 65.9 26.9 213.7 J/(K ∙ mol)

ΔH = [–601.2 + (–393.5) – (–1111.7)] kJ

ΔH = 117.0 kJ

ΔS = [(26.9 + 213.7) – 65.9] J/K

ΔS = 174.7 J/K

To solve for temperature:

T =

T = 6.70 102 K; T = 398C

This temperature is considerably lower than the 848C decomposition temperature of CaCO3 given in the text immediately preceding this exercise.

ADDITIONAL PROBLEMS1. As a sample of gas cools at 1 atm pressure, it loses 54 J of heat, and the volume changes by 0.200

L. What are the values of q, w, and ΔU for this process?

2. Gray tin, which exhibits the diamond structure, is transformed above 13.2C to white tin, which has a metallic nature. The S values at 25C are 44.8 and 51.5 J/(K ∙ mol), respectively. Predict ΔH for the phase transition.

3. The ΔH value for the transition of PCl3(l) to PCl3(g) is 32.7 kJ. The observed boiling point is 75C. Predict the entropy change on boiling 2 mol of the liquid.



4. Will the entropy change for each of the following processes probably be positive or negative? Explain why.

a. C (diamond) C (graphite)

b. NH3(g) + HCl(g) NH4Cl(s)

c. NaCl(s) Na+(aq) + Cl–(aq)

d. CaO(s) + CO2(g) CaCO3(s)

5. The enthalpy of formation of ICl(g) is 17.8 kJ/mol at 25C. Calculate the absolute entropy change for the reaction

I2(s) + Cl2(g) ICl(g)

S for I2(s) = 116.1 J/(K ∙ mol); S for Cl2(g) = 223.0 J/(K ∙ mol); S for ICl(g) = 247.4 J/(K ∙ mol).

6.

a. Using the following values for and S:

H2O(g): = –241.8 kJ/mol; S = 188.7 J/(K ∙ mol)

H2O(l): = –285.8 kJ/mol; S = 69.9 J/(K ∙ mol)

calculate ΔH and ΔS for the process

H2O(l) H2O(g) at 1 atm and 25C

b. Use your calculated values to determine ΔG for this process.

c. Is this process spontaneous at 25C?

d. If not, how could you make it spontaneous?

7. Calculate ΔG for the complete combustion of 1 mol methanol, CH3OH(l), to CO2(g) and H2O(g). The values are –166.2, –394.4, and –228.6 kJ/mol, respectively.

8. Use the following information to calculate ΔG at 325.37 K for the reaction given. What is important about this temperature?

2NO2(g) N2O4(g)

33.18 9.16 kJ/mol

S° 239.9 304.2 J/(K ∙ mol)



9. The reaction


is nonspontaneous at 25C and 1 atm. This does not surprise us when we know that marble, which is essentially CaCO3 with small amounts of impurities present, is an excellent building stone. The white marble Parthenon on the Acropolis in Athens has stood for 2500 years and has not decomposed to CaO and CO2. Limestone (CaCO3 with more impurities than occur in white marble) is used extensively for making quicklime, CaO, an important ingredient in building plaster. The limestone is mixed with fuel that burns and heats the stone, decomposing it according to the foregoing equation. How hot must we heat the limestone to make quicklime? Use the following values for :

CaCO3(s), –1206 kJ/mol; CaO(s), –635.5 kJ/mol; CO2(g), –393.5 kJ/mol

and for S:

CaCO3(s), 92.9 J/(K ∙ mol); CaO(s), 38.21 J/(K ∙ mol); CO2(g), 213.7 J/(K ∙ mol)

10. For each of the following systems, calculate the thermodynamic equilibrium constant at 25C and 1 atm. From this value of the constant, predict whether the reaction will produce only negligible amounts of the products, will yield moderate amounts of products, or will go essentially to completion toward the right.

a. 2HI(g) + Cl2(g) 2HCl(g) + I2(g) ΔG = –174.6 kJ

b. 2NO2(g) N2O4(g) ΔG = –4.78 kJ

c. N2(g) + 2O2(g) 2NO2(g) ΔG = 103 kJ

ANSWERS TO ADDITIONAL PROBLEMSIf you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1. q = –54 J, w = +20.2 J, ΔU = –34 J (18.1)

2. ΔH ≈ 1.9 kJ/mol (18.2, PS Sk. 1)

3. ΔS ≈ 188 J/K (18.1, 18.2, PS Sk. 1)

4.

a. ΔS should be positive because graphite is softer and the disorder greater than in diamond.

b. ΔS will be negative; the disorder in NH4Cl(s) crystals is much less than the disorder in NH3(g) and HCl(g). Furthermore, the reaction proceeds from 2 mol of a gas to 1 mol of a solid—with a decrease in the number of individual molecular units.

c. ΔS will be positive because NaCl in solution is dissociated into 2 mol of ions per mole of NaCl dissolved, and these ions, with their freedom of motion, are more disordered than the highly ordered Na+ and Cl– ions in crystalline NaCl.

d. ΔS will be negative because a gas with large entropy becomes incorporated into a solid with much less entropy, and the reaction proceeds with 2 mol of reactants forming 1 mol of product. (18.3, PS Sk. 2)

5. ΔS = 77.8 J/K (18.3, PS Sk. 3)

6.



a. ΔH = 44.0 kJ (6.8, 18.1)

ΔS = 118.8 J/K (18.3, PS Sk. 3)

b. ΔG = 8.60 kJ (18.4, PS Sk. 4)

c. Because ΔG is positive, the process is not spontaneous at 25C. (18.4, PS Sk. 6)

d. The process can be made spontaneous by increasing the temperature until –TΔS exceeds ΔH, giving a negative value of ΔG. We know that water boils at 100C. At this temperature, H2O(l) and H2O(g) are in equilibrium, and ΔG for the vaporization process must be zero. For the process to become spontaneous, the temperature must be raised an infinitesimal amount above 100C. (18.7, PS Sk. 9)

7. ΔG = –685.4 kJ (18.4, PS Sk. 5)

8. ΔG = 0. This is the temperature at which the stoichiometry of the balanced equation represents the relative amounts of the substances at equilibrium. (18.7, PS Sk. 9)

9. The temperature must be kept above 1100 K. (18.2, 18.3, 18.4, 18.7, PS Sk. 3, 4, 6, 9)

10.

a. K = Kp = 4 1030; the reaction goes essentially to completion.

b. K = Kp = 6.89; only moderate amounts of products form.

c. K = Kp = 9 10–19; negligible amounts of products form. (18.6, PS Sk. 8)

CHAPTER POST-TEST1. The enthalpy change when liquid dimethyl sulfide, (CH3)2S(l), vaporizes at 25C is 28.0 kJ/mol.

What is the entropy change when one mole of vapor in equilibrium with the liquid condenses to liquid at 25C if the entropy of this vapor at 25C is 285.7J/(K∙ mol)?

2. Predict by inspection the sign of ΔS for each of the following reactions, and explain your answers.

a. 2NOCl(g) 2NO(g) + Cl2(g)

b. Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

c. N2(g) + 3H2(g) 2NH3(g)

d. HCOOH(aq) + OH–(aq) HCOO–(aq) + H2O(l)

3. Calculate the change in entropy ΔS for the following neutralization reaction:

Ca2+(aq) + 2OH–(aq) + H2SO4(aq) CaSO4(s) + 2H2O(l)

Use text Appendix C for standard entropy values.

4. Calculate the ΔG for the following reaction at 25C:

2NO2(g) 2NO(g) + O2(g)

Is the reaction spontaneous as written? Use values of and S from text AppendixC.

5. Determine whether each of the following statements is true or false. If the statement is false, change it so that it is true.

a. A flow of entropy, a transfer of disorder, always follows a temperature change.

True/False: ________________________________________________________________



__________________________________________________________________________

b. The second law of thermodynamics states that the total entropy of a system and surroundings always increases for a spontaneous process. True/False:

__________________________________________________________________________

__________________________________________________________________________

c. The free-energy change of a reacting system is the amount of energy available to do work. True/False: ________________________________________________________________

__________________________________________________________________________

d. The zero point for free energy is the same as that for entropy. True/False:

__________________________________________________________________________

__________________________________________________________________________

6. Calculate the free-energy change ΔG for the following reaction using values of from text Appendix C:

Mg(s) + 2HCl(aq) Mg2+(aq) + 2Cl–(aq) + H2(g)

7. Determine whether each of the reactions in Problem 2 is spontaneous or nonspontaneous as written. Give reasons for your answers. [Use text Appendix C. (NOCl) = 66.36 kJ/mol.]

8. The work that a reacting system open to the atmosphere can do is PΔV work, where the atmosphere is pushed away. Using the ideal gas law (text Section 5.3), express this work as a more useful term, and calculate the maximum amount of work that could be done in the following reaction at 25C:

2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(g)

9. Use data from text Table 18.2 to calculate the thermodynamic equilibrium constant for the reaction

2H2O2(l) 2H2O(l) + O2(g)

at 25C. for H2O2(l) = –120.4 kJ/mol.

10. To what temperature must potassium chlorate be heated to decompose it to potassium chloride and oxygen gas? Use values of and S from text Appendix C. Values for KClO3 are

= –391.20 kJ/mol, S = 142.97 J/K ∙ mol.



ANSWERS TO CHAPTER POST-TESTIf you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1. ΔS = –94.0 J/K (18.2, PS Sk. 1)

2.

a. ΔS > 0; there are two molecules of gas on the left and three on the right.

b. ΔS > 0; there is one molecule of gas on the right.

c. ΔS < 0; there are four molecules of gas on the left and only two on the right.

d. ΔS 0. One would expect the molecular HCOOH to be associated with water molecules and the small OH– to be tightly surrounded by water molecules. In the products, the COOH– ion would not have quite the association with water the OH– does, but water associates with itself. (The calculated value is 8.08 J/K.) (18.3, PS Sk. 2)

3. ΔS = 305 J/K (18.3, PS Sk. 3)

4. ΔG = 70.5 kJ; reaction is nonspontaneous. (18.4, PS Sk. 4)

5.

a. True. (18.2)

b. True. (18.2)

c. True. (18.5)

d. False. The zero point for free energy is the element in its standard state at 1 atm pressure and a specified temperature (usually 25C). The zero point for entropy is a perfect crystalline substance at 0 K. (18.3, 18.4)

6. ΔG = –456.01 kJ (18.4, PS Sk. 5)

7.

a. Nonspontaneous; ΔG > 0

b. Spontaneous; ΔG < 0

c. Spontaneous; ΔG < 0

d. Spontaneous; ΔG < 0 (18.4, PS Sk. 5, 6)

8. wmax = ΔnRT = 22.3 kJ (18.1, 18.5)

9. K = Kp = 9 1040 (18.6, PS Sk. 8)

10. The reaction is spontaneous at any temperature. (18.7, PS Sk. 9)


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