Chemistry

Chemistry

Problem

Reaction 1: A+B→C+DΔG∘=+12 kJ mol−1 \text{A} + \text{B} \rightarrow \text{C} + \text{D} \quad \Delta G^{\circ}=+12\,\text{kJ mol}^{-1} Reaction 2: D→B+EΔG∘=−25 kJ mol−1 \text{D} \rightarrow \text{B} + \text{E} \quad \Delta G^{\circ}=-25\,\text{kJ mol}^{-1} The overall process is A→C+E \text{A} \rightarrow \text{C} + \text{E}. Which of the following statements best explains why the overall process is thermodynamically favorable? a) Species B acts as a catalyst, lowering the activation energy of Reaction 1 so the overall ΔG∘\Delta G^{\circ} becomes negative. b) The overall free-energy change is negative because species D is produced and then consumed, so its terms cancel when the individual ΔG∘\Delta G^{\circ} values are added. c) The overall ΔG∘\Delta G^{\circ} is the average of the two values, (−25+12)/2=−6.5  kJ mol−1(-25+12)/2 = -6.5\;\text{kJ mol}^{-1}, which is negative and therefore spontaneous. d) Because the magnitude of −25  kJ mol−1-25\;\text{kJ mol}^{-1} is larger than +12  kJ mol−1+12\;\text{kJ mol}^{-1}, the more exergonic reaction dominates, making the process favorable without adding the values.

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