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

To resolve this issue, an analysis of the thermodynamic parameters, specifically enthalpy change (\( \Delta H \)) and entropy change (\( \Delta S \)), for the provided chemical reaction is necessary. The spontaneity of a reaction is determined by the Gibbs free energy change (\( \Delta G \)), calculated as:

\(\Delta G = \Delta H - T \Delta S\)

• A reaction is spontaneous when \( \Delta G \lt 0 \).
• The given reaction is endothermic, meaning \( \Delta H \gt 0 \).

We will examine the conditions under which the reaction is non-spontaneous at water's freezing point (0°C or 273 K) and spontaneous at its boiling point (100°C or 373 K).

1. At 273 K (freezing point): The reaction is non-spontaneous, implying:
   • \( \Delta G \gt 0 \), which translates to \( \Delta H - 273 \Delta S \gt 0 \).
   • Given a positive enthalpy, this inequality indicates that \( \Delta H \) exceeds \( 273 \Delta S \).
2. At 373 K (boiling point): The reaction is spontaneous, implying:
   • \( \Delta G \lt 0 \), which translates to \( \Delta H - 373 \Delta S \lt 0 \).
   • This suggests that \( \Delta H \) is less than \( 373 \Delta S \).

The transition from non-spontaneous to spontaneous between these temperatures implies the following:

• \( \Delta S \gt 0 \): An increase in entropy facilitates spontaneity at elevated temperatures.

Considering that both \( \Delta H \) and \( \Delta S \) are positive, the conclusion is:

Both \( \Delta H \) and \( \Delta S \) are positive.