JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Chemical Thermodynamics

Haber's synthesis $\mathrm{N}{}_2(\mathrm{g})+3\mathrm{H}{}_2(\mathrm{g})\to 2\mathrm{NH}{}_3(\mathrm{g})$ is exothermic, so $\Delta H_R^{\circ }$ is negative, and it proceeds with a decrease in the number of gaseous moles, so $\Delta S_R^{\circ }$ is negative; both are essentially independent of temperature and appear as constant lines below the axis. Since $\Delta H_R^{\circ }<0$, the quantity $-\Delta H_R^{\circ }/T$ is positive and falls towards zero as $T$ increases. From $-\Delta G_R^{\circ }/T=R\ln K=-\Delta H_R^{\circ }/T+\Delta S_R^{\circ }$, the curve for $-\Delta G_R^{\circ }/T$ lies below $-\Delta H_R^{\circ }/T$ by the constant $|\Delta S_R^{\circ }|$ and also decreases with temperature, crossing zero as $K$ drops below unity. The correct graph is the one showing $\Delta H_R^{\circ }$ and $\Delta S_R^{\circ }$ as constant negative lines, $-\Delta H_R^{\circ }/T$ positive and decreasing, and $-\Delta G_R^{\circ }/T$ decreasing beneath it.