CHEM 160 Thermodynamics Chapter 19 Example Problems
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CHEM 160
Thermodynamics
Chapter 19
Example Problems
This list contains problems that will be worked as examples in class and additional problems for students to
try on their own.
1. Calculate ∆S°rxn for the following reaction at 25°C: 2SO2(g) + O2(g) → 2SO3(g)
Ans. –187.9 J/K
2. Calculate ∆S°rxn for the following reaction at 25°C: 2O(g) → O2(g)
Ans. –117 J/K
3. Calculate ∆S°rxn for the following reaction at 25°C: FeCl3(s) → Fe3+(aq) + 3Cl-(aq)
Ans. 320.5 J/K
4. Calculate ∆Suniv for the synthesis of ammonia at 25°C if ∆Ssys(∆S°rxn) = -199 J/K and
∆Hsys(∆H°rxn) = -92.6 kJ. Is this reaction spontaneous at 25°C? Ans. 112 J/K, yes
5. The synthesis of a particular reagent at 50°C leads to an increase in the system’s entropy
by 100 J/K and releases 85 kJ of heat energy. Calculate ∆Suniv for this process. Is this
reaction spontaneous at this temperature? Ans. 363 J/K, yes
6. For a particular reaction, ∆Hrxn = 53 kJ and ∆Srxn = 115 J/K. Calculate ∆G°rxn at (a) 25°C
and (b) 250°C? At which temperature is the reaction spontaneous? (c) At what
temperature does ∆G°rxn = 0? Ans. (a) 18.7 kJ, (b) –7.1 kJ, 250°C, 460.9K
7. Calculate the Gibbs free energy change for a reaction that has ∆Hrxn = 83 kJ and ∆Srxn =
188 J/K at 300°C. Is this reaction spontaneous at this temperature?
Ans. –24.7 kJ, yes
8. Calculate the entropy change, ∆Sfus (J/K) for the melting of 1 mole of a solid if it’s
melting point is 115°C and it’s heat of fusion, ∆Hfus, is 25 kJ/mol. Ans. 64.4 J/K
9. Calculate the standard free energy change for the following reaction at 25°C:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Ans. –818 kJ
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10. Calculate ∆Grxn for the following reaction if ∆G°rxn = 9.9 x 103 J/mol and (a) [A] = 0.8
M, [B] = 0.5 M, [C] = 0.05 M, and [D] = 0.05 M, and (b) [A] = 0.1 M, [B] = 1 M, [C] =
0.5 M, and [D] = 0.5 M. Is the reaction spontaneous?
2A + B --> C + D
Ans. (a) –2121 J, yes, (b) 17875 J, no
11. Calculate ∆G°rxn for the ionization of CH3COOH at 25°C. Ka = 1.8 x 10-5. Is the reaction
spontaneous under these conditions? Ans. 27 kJ, no
12. Calculate ∆G°rxn for the neutralization of a strong acid with a strong base. K = 1.0 x 1014
Is this reaction spontaneous under these conditions? Ans. –80 kJ, yes
13. The equilibrium constant for an exothermic reaction is 12.3 at 25°C. Calculate ∆G°rxn. If
∆H°rxn = -20 kJ, does the system become more ordered or disordered?
Ans. –6.22 kJ, more ordered since ∆S = -46.2 J/K
14. Calculate Keq for a reaction at 25°C if ∆H°rxn = 42.0 kJ and ∆S°rxn = 125 J/K. Is this
reaction reactant-favored or product-favored? Ans. 0.15, reactant-favored
15. Calculate Keq for a reaction at 25°C if ∆G°rxn = -12.0 kJ. Is this reaction reactant-favored
or product-favored? Ans. 127, product-favored
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