JEE Chemistry 2025 Question with Solution

Given: The reaction is endothermic, non-spontaneous at the freezing point of water, and spontaneous at the boiling point of water.

Find: The signs of $\Delta H$ and $\Delta S$.

For spontaneity, Gibbs free energy is given by

$$\Delta G = \Delta H - T\Delta S$$

A reaction is spontaneous when

$$\Delta G < 0$$

and non-spontaneous when

$$\Delta G > 0$$

Since the reaction is endothermic,

$$\Delta H > 0$$

At the freezing point of water, $T = 273 \, \text{K}$ and the reaction is non-spontaneous. Therefore,

$$\Delta H - 273\Delta S > 0$$

At the boiling point of water, $T = 373 \, \text{K}$ and the reaction is spontaneous. Therefore,

$$\Delta H - 373\Delta S < 0$$

As temperature increases, the term $T\Delta S$ must help make $\Delta G$ negative. This is possible only when

$$\Delta S > 0$$

Thus,

$$\Delta H > 0 \quad \text{and} \quad \Delta S > 0$$

Therefore, the correct option is A.

Given: The reaction is non-spontaneous at $273 \, \text{K}$ and spontaneous at $373 \, \text{K}$.

Find: Which combination of signs of $\Delta H$ and $\Delta S$ matches this behavior.

The Gibbs relation is

$$\Delta G = \Delta H - T\Delta S$$

The question states that the reaction is endothermic, so enthalpy change is positive.

$$\Delta H > 0$$

Now compare the effect of temperature. At lower temperature,

$$\Delta G > 0$$

while at higher temperature,

$$\Delta G < 0$$

This change means the subtractive term $T\Delta S$ becomes large enough at higher temperature to overcome the positive $\Delta H$. That can happen only if entropy change is positive.

$$\Delta S > 0$$

Hence both thermodynamic quantities are positive, and the reaction becomes spontaneous only at sufficiently high temperature.

Therefore, the correct option is A, that is, both $\Delta H$ and $\Delta S$ are (+ve).