Let us consider a reversible reaction at temperature, T. In this reaction, both $$\Delta H$$ and $$\Delta S$$ were observed to have positive values. If the equilibrium temperature is $$T_e$$, then the reaction becomes spontaneous at:

For any process at a constant temperature and pressure, the criterion of spontaneity is based on the Gibbs free-energy change:

$$\Delta G = \Delta H - T\,\Delta S\quad -(1)$$

Key points to remember:

• If $$\Delta G \lt 0$$, the process is spontaneous.
• If $$\Delta G = 0$$, the system is at equilibrium.
• If $$\Delta G \gt 0$$, the process is non-spontaneous.

In the given reversible reaction, both enthalpy change and entropy change are positive, i.e.

$$\Delta H \gt 0,\qquad \Delta S \gt 0\quad -(2)$$

The equilibrium temperature $$T_e$$ is defined by the condition $$\Delta G = 0$$. Substituting $$\Delta G = 0$$ in equation $$(1)$$ gives

$$0 = \Delta H - T_e\,\Delta S$$

Rearranging, we obtain the expression for the equilibrium temperature:

$$T_e = \frac{\Delta H}{\Delta S}\quad -(3)$$

To find when the reaction becomes spontaneous, set $$\Delta G \lt 0$$ in equation $$(1)$$:

$$\Delta G \lt 0 \;\Longrightarrow\; \Delta H - T\,\Delta S \lt 0$$

Since both $$\Delta H$$ and $$\Delta S$$ are positive by $$(2)$$, divide by the positive quantity $$\Delta S$$ (which does not change the inequality direction):

$$\frac{\Delta H}{\Delta S} \lt T$$

Using equation $$(3)$$, $$\frac{\Delta H}{\Delta S}$$ equals $$T_e$$, so the condition becomes

$$T_e \lt T$$

Therefore, the reaction is spontaneous for temperatures greater than the equilibrium temperature.

Final result: The reaction becomes spontaneous when $$T \gt T_e$$, which corresponds to Option C.