Question

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at the boiling point of water. Choose the correct option.

• A. Both \( \Delta H \) and \( \Delta S \) are (+ve)
• B. \( \Delta H \) is (-ve) but \( \Delta S \) is (+ve)
• C. \( \Delta H \) is (+ve) but \( \Delta S \) is (-ve)
• D. Both \( \Delta H \) and \( \Delta S \) are (-ve)

Hint:

For a reaction to become spontaneous at higher temperatures, the entropy change (\( \Delta S \)) must be positive. The reaction must absorb heat, indicating a positive enthalpy change (\( \Delta H \)).

Answer 1

The Correct Option is A

For a reaction to be spontaneous, the Gibbs free energy (\( \Delta G \)) must be negative: \[ \Delta G = \Delta H - T \Delta S \] where: - \( \Delta H \) is the enthalpy change, - \( \Delta S \) is the entropy change, - \( T \) is the temperature. For a reaction that is non-spontaneous at the freezing point of water and spontaneous at the boiling point of water, we can analyze the situation: - At the freezing point of water (273 K), the reaction is non-spontaneous, so: \[ \Delta G = \Delta H - T \Delta S>0 \] - At the boiling point of water (373 K), the reaction is spontaneous, so: \[ \Delta G = \Delta H - T \Delta S<0 \] From this, we can infer that: - \( \Delta H \) is positive: The reaction is endothermic. - \( \Delta S \) is positive: The reaction leads to an increase in disorder (entropy increases). Thus, both \( \Delta H \) and \( \Delta S \) are positive, which aligns with option (1).