Question at position 13 Consider the following reaction: 3NO2(g) + H2O(l) ⇌ 2HNO3(l) + NO(g) ∆G° = 9.4 kJ Calculate ∆G for this reaction at 25 °C under the following conditions: partial pressure of NO2 = 0.115 atm and partial pressure of NO = 9.70 atm. 20.4 kJ43.2 kJ11.2 kJ8.2 kJ31.1 kJ单项选择题
A
20.4 kJ
B
43.2 kJ
C
11.2 kJ
D
8.2 kJ
E
31.1 kJ

Question

Question at position 13 Consider the following reaction: 3NO2(g) + H2O(l) ⇌ 2HNO3(l) + NO(g) ∆G° = 9.4 kJ Calculate ∆G for this reaction at 25 °C under the following conditions: partial pressure of NO2 = 0.115 atm and partial pressure of NO = 9.70 atm. 20.4 kJ43.2 kJ11.2 kJ8.2 kJ31.1 kJ单项选择题

A

20.4 kJ

B

43.2 kJ

C

11.2 kJ

D

8.2 kJ

E

31.1 kJ

BlackTom AI · Accepted Answer

ΔG = ΔG° + RT ln Q. For this heterogeneous equilibrium, only gaseous (or aqueous) species appear in Q; pure liquids are omitted.

Relevant concepts:
- Reaction quotient Q uses activities; for gases use partial pressures (in atm) if ΔG° given for standard states in those units.
- Pure liquids (H2O, HNO3) are excluded from Q.
- Use R = 8.314 J·mol−1·K−1 (0.008314 kJ·mol−1·K−1) and T = 25.0 °C = 298.15 K.

Step-by-step calculation:
1) Write Q for 3NO2(g) + H2O(l) ⇌ 2HNO3(l) + NO(g):
   Q = P_NO / (P_NO2)^3
2) Insert pressures: P_NO = 9.70 atm, P_NO2 = 0.115 atm:
   (0.115)^2 = 0.013225
   (0.115)^3 = 0.013225 × 0.115 = 0.001520875
   Q = 9.70 / 0.001520875 = 6,377.91 (approx)
3) Compute ln Q:
   ln(6377.91) ≈ 8.7598
4) Compute RT in kJ:
   R = 0.008314 kJ·mol−1·K−1
   RT = 0.008314 × 298.15 = 2.47896 kJ·mol−1
5) Compute RT ln Q:
   RT ln Q = 2.47896 × 8.7598 ≈ 21.72 kJ·mol−1
6) Finally, ΔG = ΔG° + RT ln Q = 9.4 kJ + 21.72 kJ = 31.12 kJ ≈ 31.1 kJ

Why this is correct:
- The calculation uses the correct reaction quotient (excluding pure liquids), the correct form ΔG = ΔG° + RT ln Q, and consistent units (kJ).
- The positive ΔG (≈31.1 kJ) indicates the reaction is non-spontaneous under the given conditions.

Why the other options are incorrect:
- 20.4 kJ, 43.2 kJ, 11.2 kJ, 8.2 kJ do not match the result from correctly computing Q and RT ln Q; they likely stem from arithmetic errors (e.g., wrong exponent for (0.115)^3), incorrect ln(Q), wrong sign, or unit inconsistency.

Conclusion: The calculated ΔG ≈ 31.1 kJ, which matches the provided correct answer (31.1 kJ).

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