The equilibrium concentrations of the species in the reaction $\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$ are $2$, $3$, $10$ and $6{\text{molL}}^{-1}$, respectively, at $300\text{K}$. $\Delta G^{\circ }$ for the reaction is (Given $R=2{\text{calmol}}^{-1}{\text{K}}^{-1}$).

Solution & Explanation

\textbf{Answer:} (C) $-1381.80\text{cal}$ \textbf{Solution:} Equilibrium constant: $K=\frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}=\frac{10\times 6}{2\times 3}=10$. Then $\Delta G^{\circ }=-2.303RT\log K=-2.303\times 2\times 300\times \log (10)=-2.303\times 2\times 300\times 1=-1381.80\text{cal}$. Mapped: source lowercase answer 'c' → option C.