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 solve this problem, we need to analyze the thermodynamic parameters of the given chemical reaction: the enthalpy change (\( \Delta H \)) and the entropy change (\( \Delta S \)). The Gibbs free energy change (\( \Delta G \)) determines the spontaneity of a reaction:

\(\Delta G = \Delta H - T \Delta S\) 

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

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

1. At the freezing point (273 K): The reaction is non-spontaneous, indicating:
   • \(\Delta G \gt 0\) which implies \(\Delta H - 273 \Delta S \gt 0\).
   • This inequality suggests that the \(\Delta H\) value is greater than \(273 \Delta S\), given that enthalpy is positive.
2. At the boiling point (373 K): The reaction is spontaneous, indicating:
   • \(\Delta G \lt 0\) which implies \(\Delta H - 373 \Delta S \lt 0\).
   • This indicates that \(\Delta H\) value is less than \(373 \Delta S\).

Between these two temperatures, as the reaction transitions from non-spontaneous to spontaneous, it suggests that:

• \(\Delta S \gt 0\): The increase in entropy helps the reaction become spontaneous at higher temperatures.

Given that both \(\Delta H\) and \(\Delta S\) are positive, the correct option is:

Both \(\Delta H\) and \(\Delta S\) are (+ve).

Answer 2

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).