1. For spontaneity, the Gibbs free energy change must be negative:

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

A reaction is feasible (spontaneous) when:

$\Delta G < 0$

2. The reaction is stated to be endothermic, so:

$\Delta H > 0$

Thus, options with negative $\Delta H$ are immediately rejected.

• Option A: $\Delta H < 0$ ❌
• Option C: $\Delta H < 0$ ❌

So only B or D may be possible.

3. The reaction is non-spontaneous at the freezing point of water ($0^\circ C = 273\,K$), but feasible at the boiling point of water ($100^\circ C = 373\,K$).

This means:

• At low temperature: $\Delta G > 0$
• At high temperature: $\Delta G < 0$

4. Since increasing temperature makes the reaction spontaneous, the term $-T\Delta S$ must become more negative as $T$ increases.

That can happen only if:

$\Delta S > 0$

Because then increasing $T$ increases the magnitude of $T\Delta S$, making:

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

change from positive to negative.

5. Therefore:

• $\Delta H > 0$
• $\Delta S > 0$

So the correct option is:

$\boxed{\text{B}}$

6. Check of all options:

• A: $\Delta H$ negative, $\Delta S$ positive → not endothermic ❌
• B: $\Delta H$ positive, $\Delta S$ positive → non-spontaneous at low $T$, spontaneous at high $T$ ✅
• C: both negative → not endothermic ❌
• D: $\Delta H$ positive, $\Delta S$ negative → becomes less spontaneous at higher $T$ ❌

Hence, the reaction must have both $\Delta H$ and $\Delta S$ positive.