Test Bank – Thermodynamics and Equilibrium: Complete Practice Questions, Detailed Solutions, and Verified Answers for Chemistry Exam Preparation (Updated 2025)

CHAPTER 18
Thermodynamics and Equilibrium


■ SOLUTIONS TO EXERCISES

Note on units and significant figures: The mole unit is omitted from all thermodynamic parameters such
as S, S, etc. If the final answer to a solution needs to be rounded off, it is given first with one
nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the
correct number of significant figures. In multistep problems, intermediate answers are given with at least
one nonsignificant figure; however, only the final answer has been rounded off.

18.1. When the liquid evaporates, it absorbs heat: Hvap = 42.6 kJ/mol (42.6  103 J/mol) at 25C, or
298 K. The entropy change, S, is
ΔH vap 42.6  103 J/mol
S = = = 142.9 J/(molK)
T 298 K
The entropy of one mole of the vapor equals the entropy of one mole of liquid (161 J/K) plus
142.9 J/K.
S = (161 + 142.9) J/(molK) = 303.9 = 304 J/(molK)

18.2. a. S is positive because there is an increase in moles of gas (ngas = +1) from a solid
reactant forming a mole of gas. (Entropy increases.)
b. S is positive because there is an increase in moles of gas (ngas = +1) from a liquid
reactant forming a mole of gas. (Entropy increases.)
c. S is negative because there is a decrease in moles of gas (ngas = −1) from liquid and
gaseous reactants forming two moles of solid. (Entropy decreases.)
d. S is positive because there is an increase in moles of gas (ngas = +1) from solid and
liquid reactants forming a mole of gas and four moles of an ionic compound. (Entropy
increases.)

18.3. The reaction and standard entropies are given below. Multiply the S values by their
stoichiometric coefficients, and subtract the entropy of the reactant from the sum of the product
entropies.




S = nS(products) − mS(reactants) = [(2  160.7 + 2  213.7) − 212] J/K = 536.8 = 537 J/K




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, Chapter 18: Thermodynamics and Equilibrium 711

18.4. The reaction, standard enthalpy changes, and standard entropies are as follows:




Calculate H and S for the reaction by taking the values for products and subtracting the
values for reactants.
H = nHf(products) − mHf(reactants) =
[(−393.5 + 2  −241.8) − (−74.87)] kJ = −802.23 kJ
S = nS(products) − mS(reactants) =
[(213.7 + 2  188.7) − (186.1 + 2  205.0)] J/K = −5.0 J/K
Now, substitute into the equation for G in terms of H and S ( = −5.0  10−3 kJ/K):
G = H − TS = −802.23 kJ − (298 K)(−5.0  10−3 kJ/K) = −800.74 = −800.7 kJ

18.5. Write the values of Gf multiplied by their stoichiometric coefficients below each formula:
CaCO3(s)  CaO(s)  CO2(g)
Gf: -1128.8 -603.5 -394.4 kJ

The calculation is
G = nGf(products) − mGf(reactants) =
[(−603.5) + (−394.4) − (−1128.8)] kJ = 130.9 kJ

18.6. a.
C(graphite)  2H2(g)  CH4(g)
Gf °: 0 0 -50.80 kJ
G° = [(-50.80) - (0)] kJ = -50.80 kJ (spontaneous reaction)
b.
2H2(g)  O2 (g)  2H2O(l)
Gf °: 0 0 2 x (-237.1) kJ
G° = [(2 x -237.1) - (0)] kJ = -474.2 kJ (spontaneous reaction)
c.
4HCN(g)  5O2(g)  2H2O(l)  4CO2(g)  2N2(g)
Gf°: 4 x 124.7 0 2 x (-237.1) 4 x (-394.4) 0 kJ
G° = [(2 x -237.1) + 4 x (-394.4) - (4 x 124.7)] kJ = -2550.6 kJ (spontaneous reaction)




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, 712 Chapter 18: Thermodynamics and Equilibrium


d.
Ag+(aq)  l-(aq)  Agl(s)
Gf °: 77.12 -51.59 -66.19 kJ
G° = [(-66.19) - (77.12 - 51.59)] kJ = -91.72 kJ (spontaneous reaction)

18.7. a. K = Kp = PCO 2



b. K = Ksp = [Pb2+][I−]2
PCO2
c. K= +
[ H ][ HCO3 ]

18.8. First, calculate G using the Gf values from Table 18.2.
CaCO3(s) CaO(s)  CO2(g)
Gf °: -1128.8 -603.5 -394.4 kJ

Subtract Gf of reactants from that of the products:
Gf = nGf(products) − mGf(reactants) =
[(−603.5) + (−394.4) − (−1128.8)] kJ = 130.9 kJ
Use the rearranged form of the equation, G = − RT ln K, to solve for ln K. To get compatible
units, express G in joules, and set R equal to 8.31 J/(molK). Substituting the numerical values
into the expression gives
ΔG° 130.9  103
ln K = = = −52.859
RT 8.31  298
K = Kp = e−52.859 = 1.10  10−23 = 1  10−23

18.9. First, calculate G using the Gf values in the exercise.

Mg(OH)2(s)  Mg2+(aq)  2OH-(aq)
Gf °: -833.7 -454.8 2 x (-157.3) kJ

Hence, G for the reaction is
G = [2  (−157.3) + (−454.8) − (−833.7)] kJ = 64.3 kJ
Now, substitute numerical values into the equation relating ln K and G.
ΔG° 64.3  103
ln K = = = −25.965
RT 8.31  298

K = Ksp = e−25.965 = 5.289  10−12 = 5  10−12




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, Chapter 18: Thermodynamics and Equilibrium 713

18.10. From Appendix C, you have
H2O(l) H2O(g)
H f °: -285.8 -241.8 kJ
S°: 69.95 188.7 J/K
Calculate H and S from these values.
H = [−241.8 − (−285.8)] kJ = 44.0 kJ
S = [188.7 − 69.95] J/K = 118.75 J/K
Substitute H, S ( = 0.11875 kJ/K), and T ( = 318 K) into the equation for GT:
GT = H − TS = 44.0 kJ − (318 K)(0.11875 kJ/K) = 6.23 kJ
Substitute the value of G (6.23  103 J) at 318 K into the equation relating ln K and G.
ΔG° 6.23  103
ln K = = = −2.360
RT 8.31  318
K = Kp = e−2.360 = 0.0943 = 0.09
Kp = PH O , so the vapor pressure of H2O is 0.09 atm (71.7 = 70. mmHg).
2



The value is 71.9 mmHg in Appendix B.

18.11. First, calculate H and S using the given Hf and S values.




H = [−601.2 + (−393.5) − (−1111.7)] kJ = 117.0 kJ
S = [(26.9 + 213.7) − 65.9] J/K = 174.7 J/K
Substitute these values into the expression relating T, H, and S ( = 0.1747 kJ/K).
ΔH ° 117.0 kJ
T= = = 669.72 K (lower than that for CaCO3)
ΔS ° 0.1747 kJ/K



■ ANSWERS TO CONCEPT CHECKS

18.1. The process is I2(s)  I2(g). The iodine atoms have gone from a state of some order (crystalline
iodine) to one that is more disordered (gas). The entropy will have increased.

18.2. The standard free energy of formation for NO(g) is 86.60 kJ/mol. This rather large positive value
means the equilibrium constant is small. At equilibrium, the NO concentration is low.




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