For a particular reversible reaction at temperature $$T$$, $$\Delta H$$ and $$\Delta S$$ were found to be both $$+ve$$. If $$T_e$$ is the temperature at equilibrium, the reaction would be spontaneous when

For any process, the criterion of spontaneity at constant $$T$$ and $$P$$ is the sign of Gibbs free-energy change

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

At equilibrium the reaction is reversible and $$\Delta G = 0$$, so for that temperature $$T_e$$

$$0 = \Delta H - T_e \Delta S \;\;\Longrightarrow\;\; T_e = \frac{\Delta H}{\Delta S}$$

The data given are $$\Delta H \gt 0$$ and $$\Delta S \gt 0$$, hence $$T_e$$ is a positive quantity.

Now consider a temperature $$T$$ different from $$T_e$$.

Put $$T$$ in the Gibbs equation:

$$\Delta G = \Delta H - T\Delta S = \Delta H - \bigl(T_e + (T-T_e)\bigr)\Delta S = (\Delta H - T_e \Delta S) - (T-T_e)\Delta S$$

Since $$\Delta H - T_e \Delta S = 0$$ (by definition of $$T_e$$), this simplifies to

$$\Delta G = - (T - T_e)\Delta S$$

Because $$\Delta S \gt 0$$, the sign of $$\Delta G$$ is controlled solely by the factor $$(T - T_e)$$:

• If $$T \gt T_e$$, then $$(T - T_e) \gt 0$$, so $$\Delta G \lt 0$$ — the reaction is spontaneous.
• If $$T \lt T_e$$, then $$(T - T_e) \lt 0$$, so $$\Delta G \gt 0$$ — the reaction is non-spontaneous.

Therefore the reaction becomes spontaneous only at temperatures higher than the equilibrium temperature $$T_e$$.

Option B which is: $$T \gt T_e$$