Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

We need to determine the signs of $$\Delta H$$ and $$\Delta S$$ for a reaction that is endothermic, non-spontaneous at the freezing point of water (273 K), but spontaneous at the boiling point of water (373 K).

Because the reaction absorbs heat, it is endothermic, so $$\Delta H > 0\quad(\text{positive}).$$

Recall that the Gibbs free energy change is given by

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

and that a reaction is spontaneous when $$\Delta G < 0$$ and non-spontaneous when $$\Delta G > 0$$.

At the freezing point (T = 273 K), the reaction is non-spontaneous, so $$\Delta G > 0$$. Thus

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

which implies $$\Delta H > 273\,\Delta S$$.

At the boiling point (T = 373 K), the reaction becomes spontaneous, so $$\Delta G < 0$$ and hence

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

giving $$\Delta H < 373\,\Delta S$$, i.e., $$\Delta S > \frac{\Delta H}{373}$$.

Since $$\Delta H > 0$$, the inequality $$\Delta S > \frac{\Delta H}{373}$$ requires that $$\Delta S > 0$$ (positive).

Indeed, with both $$\Delta H > 0$$ and $$\Delta S > 0$$, at the lower temperature (273 K) the term $$T\Delta S$$ is relatively small so $$\Delta G = \Delta H - T\Delta S > 0$$ (non-spontaneous), while at the higher temperature (373 K) the term $$T\Delta S$$ becomes large enough to exceed $$\Delta H$$, making $$\Delta G < 0$$ (spontaneous).

The transition from non-spontaneous to spontaneous occurs at the critical temperature

$$T^* = \frac{\Delta H}{\Delta S}$$

which lies between 273 K and 373 K.

The correct answer is Option D: Both $$\Delta H$$ and $$\Delta S$$ are positive.