JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Chemical Thermodynamics

For any reaction, $\Delta G=\Delta H-T\Delta S$. Row (A) has $\Delta H>0$ and $\Delta S<0$, so $\Delta G=(+)-T(-)$ is a sum of two positive terms at every temperature — the reaction is non-spontaneous at all T, exactly as the table states. Row (D) has $\Delta H<0$ and $\Delta S>0$, so $\Delta G=(-)-T(+)$ stays negative at every temperature — the reaction is spontaneous at all T, again exactly as the table states. Row (B) has $\Delta H>0$ and $\Delta S>0$: at low T, the positive $\Delta H$ term dominates over the small $T\Delta S$, so $\Delta G>0$ and the reaction is non-spontaneous at low T, becoming spontaneous only once T is high enough that $T\Delta S$ exceeds $\Delta H$; the table's claim of 'spontaneous at low T' reverses this. Row (C) has $\Delta H<0$ and $\Delta S<0$: at low T, the negative $\Delta H$ term dominates over the small $T|\Delta S|$, so $\Delta G<0$ and the reaction is spontaneous at low T, becoming non-spontaneous only at high T; the table's claim of 'non-spontaneous at low T' likewise reverses this. Only rows (A) and (D) state the spontaneity correctly.