1. For spontaneity at constant temperature and pressure, the criterion is:

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

A reaction is spontaneous when:

$\Delta G < 0$

2. Given in the question:

$\Delta H > 0, \quad \Delta S > 0$

So,

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

At low $T$, the term $T\Delta S$ is small, so $\Delta G$ tends to be positive. At high $T$, the term $T\Delta S$ becomes larger, so $\Delta G$ can become negative.

Thus, the reaction becomes spontaneous at sufficiently high temperature.

3. The equilibrium temperature $T_e$ is defined by:

$\Delta G = 0$

So,

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

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

4. For spontaneity:

$\Delta G < 0$

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

$T\Delta S > \Delta H$

Since $\Delta S > 0$, dividing by $\Delta S$ gives:

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

Hence, the reaction is spontaneous when:

$T > T_e$

5. Checking options:

• A: $T_e > T$ i.e. $T < T_e$ → not spontaneous
• B: $T > T_e$ → correct
• C: $T = T_e$ → equilibrium, not spontaneous driving force
• D: $T_e = 5T$ i.e. $T < T_e$ → not spontaneous in general

Therefore, the correct option is B.