JEE Main 2026 — Thermodynamics (C) Question with Solution

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^o=-RT\ln K$.
We also know that $\Delta G^o=\Delta H^o-T\Delta S^o$.
Equating the two expressions: $-RT\ln K=\Delta H^o-T\Delta S^o$.
Dividing by $-RT$, we get: $\ln K=-\frac{\Delta H^o}{RT}+\frac{\Delta S^o}{R}$.
To convert natural log to base 10, use $\ln K=2.303{\log }_{10}K$:
$2.303{\log }_{10}K=-\frac{\Delta H^o}{RT}+\frac{\Delta S^o}{R}$.
Dividing by $2.303$: ${\log }_{10}K=-\frac{\Delta H^o}{2.303RT}+\frac{\Delta S^o}{2.303R}$.
Rearranging in the form of a straight line equation $y=mx+c$ where $y={\log }_{10}K$ and $x=\frac{1}{T}$:
${\log }_{10}K=\left(-\frac{\Delta H^o}{2.303R}\right)\frac{1}{T}+\frac{\Delta S^o}{2.303R}$.
Comparing the terms:
Intercept ($c$) = $\frac{\Delta S^o}{2.303R}$
Slope ($m$) = $-\frac{\Delta H^o}{2.303R}$.