JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Chemical Thermodynamics

For any process, spontaneity is governed by $\Delta G=\Delta H-T\Delta S$, with $\Delta G<0$ required for a spontaneous change.

The reaction is endothermic, so $\Delta H>0$.

At the freezing point of water (273 K) the reaction is non-spontaneous: $\Delta G=\Delta H-T\Delta S>0$, which means $\Delta H>T\Delta S$ at this temperature.

At the boiling point (373 K) it becomes spontaneous: $\Delta G=\Delta H-T\Delta S<0$, so $T\Delta S>\Delta H$ at this higher temperature.

For the $T\Delta S$ term to grow large enough with rising temperature to overtake the positive $\Delta H$, $\Delta S$ itself must be positive — a negative $\Delta S$ would make $-T\Delta S$ even more positive as T increases, never flipping the sign of $\Delta G$.

Both $\Delta H$ and $\Delta S$ are positive.