JEE Main 2025 Apr 2 Shift 2, Chemistry Q18: Gibbs Energy Change and Equilibrium

For Haber's process ${\mathrm{N}}_2(\mathrm{g})+3{\mathrm{H}}_2(\mathrm{g})\to 2N{\mathrm{H}}_3(\mathrm{g})$, the moles of gas decrease, so $\Delta S_R^{\circ }<0$. The forward reaction is exothermic, so $\Delta H_R^{\circ }<0$.

Both $\Delta H_R^{\circ }$ and $\Delta S_R^{\circ }$ are essentially temperature-independent over a moderate range, so they appear as nearly horizontal lines in the negative region.

Since $\Delta H_R^{\circ }$ is negative, $-\Delta H_R^{\circ }/T$ is positive and decreases with increasing $T$ (a $1/T$ curve approaching zero from above).

From $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$, dividing by $-T$ gives $-\Delta G^{\circ }/T=-\Delta H^{\circ }/T+\Delta S^{\circ }$. With $\Delta S^{\circ }<0$, this lies below the $-\Delta H^{\circ }/T$ curve and eventually becomes negative at high T.

Plot (1) shows exactly this behaviour.