JEE Chemistry 2025 Question with Solution

Given: A reversible reaction has $\Delta H > 0$ and $\Delta S > 0$.

Find: The temperature condition for spontaneity in terms of $T_e$.

For spontaneity, Gibbs free energy must be negative:

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

A reaction is spontaneous when:

$$\Delta G < 0$$

So,

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

Rearranging,

$$T\Delta S > \Delta H$$

At equilibrium temperature $T_e$, we have:

$$\Delta G = 0$$

Therefore,

$$\Delta H = T_e\Delta S$$

Substituting this relation, the spontaneity condition becomes:

$$T > \frac{\Delta H}{\Delta S} = T_e$$

Therefore, the reaction becomes spontaneous when the temperature is greater than the equilibrium temperature. The correct option is C.

Given: $\Delta H$ and $\Delta S$ are both positive.

Find: When the reaction becomes spontaneous.

From

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

with $\Delta H > 0$ and $\Delta S > 0$, increasing $T$ makes the term $-T\Delta S$ more negative. Hence $\Delta G$ becomes negative only at sufficiently high temperature.

The boundary occurs at equilibrium:

$$\Delta G = 0 \Rightarrow T = T_e$$

So spontaneity occurs for:

$$T > T_e$$

Therefore, the correct option is C.