Question

A certain reaction is non spontaneous at 298 K. The entropy change during the reaction is 121 J K\(^{-1}\). Is the reaction endothermic or exothermic? The minimum value of \( \Delta H \) for the reaction is

• A. endothermic, \( \Delta H = 36.06 \, \text{kJ} \)
• B. exothermic, \( \Delta H = -36.06 \, \text{kJ} \)
• C. endothermic, \( \Delta H = 60.12 \, \text{kJ} \)
• D. exothermic, \( \Delta H = -60.12 \, \text{kJ} \)

Hint:

For non-spontaneous reactions, \( \Delta G>0 \), and the reaction is endothermic if \( \Delta H \) is positive.

Answer 1

The Correct Option is A

Step 1: Use the Gibbs free energy equation.
The Gibbs free energy equation is: \[ \Delta G = \Delta H - T \Delta S \] For non-spontaneous reactions, \( \Delta G>0 \), and we can use the given entropy and temperature to find \( \Delta H \).
Step 2: Conclusion.
Thus, the reaction is endothermic, and the minimum \( \Delta H = 36.06 \, \text{kJ} \).
Final Answer: \[ \boxed{\text{endothermic, } \Delta H = 36.06 \, \text{kJ}} \]