For the reaction 2NO(g) + O2(g) → 2NO2(g), the enthalpy change ΔH is negative and the entropy change ΔS is negative. The forward reaction is thermodynamically favored at which of the following temperatures?

• A. All temperatures
• B. Temperature has no effect
• C. Low temperatures
• D. High temperatures

Answer: (C)

Verified step by step guidance

1. Identify the thermodynamic parameters given: the enthalpy change (\( \Delta H \)) is negative, indicating the reaction is exothermic, and the entropy change (\( \Delta S \)) is negative, indicating a decrease in disorder during the reaction.
2. Recall the Gibbs free energy equation that determines reaction spontaneity: \[ \Delta G = \Delta H - T \Delta S \] where \( \Delta G \) is the Gibbs free energy change, \( \Delta H \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change.
3. Analyze the signs of \( \Delta H \) and \( \Delta S \) to understand how temperature affects \( \Delta G \): - Since \( \Delta H < 0 \), the reaction releases heat. - Since \( \Delta S < 0 \), the reaction leads to decreased entropy. Because \( \Delta S \) is negative, the term \( -T \Delta S \) becomes positive and increases with temperature.
4. Determine the temperature dependence of spontaneity: - At low temperatures, the \( -T \Delta S \) term is small, so \( \Delta G \approx \Delta H \), which is negative, favoring the forward reaction. - At high temperatures, the positive \( -T \Delta S \) term can outweigh the negative \( \Delta H \), making \( \Delta G \) positive and the reaction non-spontaneous.
5. Conclude that the forward reaction is thermodynamically favored at low temperatures because the negative enthalpy dominates, while at high temperatures the unfavorable entropy term reduces spontaneity.