U9.5 · Free Energy and Equilibrium (ΔG° = −RT ln K)Practice · 自由能与平衡（ΔG° = −RT ln K）
Lessons covered:U9.3 Gibbs Free Energy and Thermodynamic FavorabilityU9.5 Free Energy and Equilibrium (ΔG° = −RT ln K)
- 1Numeric
ΔG°f for NH₃(g) is −16.5 kJ/mol. Find K at 25 °C (298 K) for N₂(g) + 3H₂(g) → 2NH₃(g). R = 8.314 J/(mol·K).
× 10ExplanationΔG° = 2 × (−16.5) = −33.0 kJ. ln K = −ΔG° ÷ RT = 33000 ÷ (8.314 × 298) = 13.32, so K = 6.1 × 105. K is large, yet the reaction is extremely slow at room temperature, which is why industry uses a catalyst and higher temperatures.
- 2Numeric
A reaction has K = 2.0 × 10−3 at 25 °C (298 K). What is its ΔG° in kJ/mol? R = 8.314 J/(mol·K).
kJ/molExplanationΔG° = −RT ln K = −8.314 × 298 × ln(2.0 × 10−3) = −2478 × (−6.215) = +15397 J/mol ≈ +15.4 kJ/mol. K < 1, so ΔG° > 0.
- 3Multiple choice
A reaction has ΔG° = 0 at some temperature. What is its equilibrium constant K then?
ExplanationΔG° = −RT ln K = 0, so ln K = 0 and K = 1: at equilibrium products and reactants are comparable (by the expression). ΔG° = 0 doesn’t mean the reaction stops; it means a standard-state mixture happens to be at equilibrium.
- 4Multiple choice
At 25 °C, if a reaction’s ΔG° becomes 11.4 kJ/mol more negative, what happens to K?
Explanationlog K = −ΔG° ÷ (2.303RT). At 25 °C, 2.303RT ≈ 5.7 kJ/mol, so each 5.7 kJ/mol drop in ΔG° makes K 10 times larger. 11.4 ≈ 2 × 5.7, so K grows by about 10² = 100 times.
- 5Numeric
For C(s) + CO₂(g) → 2CO(g), ΔH° = +172.47 kJ and ΔS° = +175.86 J/K. Assuming these don’t change with temperature, estimate K at 1200 K. R = 8.314 J/(mol·K).
ExplanationAt 1200 K, ΔG° = 172.47 − 1200 × 0.17586 = −38.56 kJ. ln K = 38562 ÷ (8.314 × 1200) = 3.865, so K ≈ 48. At 25 °C K is tiny, but at high temperature K > 1 and CO is favored. Use T = 1200 K, not 298 K.
Review: U9.5 Free Energy and Equilibrium (ΔG° = −RT ln K)U9.3 Gibbs Free Energy and Thermodynamic Favorability
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