JEE Main 2026 Apr 6 Shift 1, Chemistry Q25: Gibbs Energy Change and Equilibrium

Use $\Delta G^{\circ }=-RT\ln K=\Delta H^{\circ }-T\Delta S^{\circ }$, giving

$\Delta S^{\circ }=\frac{\Delta H^{\circ }+RT\ln K}{T}=\frac{\Delta H^{\circ }}{T}+R\ln K$.

Compute $\ln K=2.3\log (1.8\times 10^{-7})=2.3[\log 1.8-7]$. With $\log 1.8=\log (18/10)=\log 2+2\log 3-1=0.30+0.96-1=0.26$, so $\log K=-6.74$ and $\ln K=2.3\times (-6.74)=-15.50$.

$R\ln K=8.3\times (-15.50)=-128.65$ J K${}^{-1}$ mol${}^{-1}$.

$\frac{\Delta H^{\circ }}{T}=\frac{28400}{300}=94.67$ J K${}^{-1}$ mol${}^{-1}$.

$\Delta S^{\circ }=94.67-128.65=-33.98$ J K${}^{-1}$ mol${}^{-1}$.

Magnitude $\approx 34$ J K${}^{-1}$ mol${}^{-1}$.