For the chemical reaction X Y, the standard reaction Gibbs energy depends on temperature T (in K) as r G\,(in kJ mol-1)…

🧠 When $\Delta G^\circ > 0$: $K < 1$, reactant (X) is major; when $\Delta G^\circ < 0$: $K > 1$, product (Y) is major Evaluate $\Delta_r G^\circ = 120 - \frac{3}{8}T$ at each temperature.

🗺️ Checking each option Equilibrium when $\Delta G^\circ = 0$: $T = 120 \times 8/3 = 320\text{ K}$.

Below 320 K: $\Delta G^\circ > 0$ → $K < 1$ → X is major. Above 320 K: $\Delta G^\circ < 0$ → $K > 1$ → Y is major.

(a) Y at 300 K: $\Delta G^\circ = 120 - 112.5 = +7.5 > 0$ → X is major, not Y. FALSE ✗ (b) Y at 280 K: $\Delta G^\circ = 120 - 105 = +15 > 0$ → X is major. FALSE ✗ (c) X at 350 K: $\Delta G^\circ = 120 - 131.25 = -11.25 < 0$ → Y is major. FALSE ✗ (d) X at 315 K: $\Delta G^\circ = 120 - 118.1 = +1.9 > 0$ → X is major. TRUE ✓

⚠️ Trap: Many students calculate $\Delta G^\circ$ correctly but then misread the direction — positive $\Delta G^\circ$ means reactant ($X$) is favoured, not product.

$\boxed{\text{Answer: (d)}}$