The value of ${\log }_{10}$k for a reaction $\mathrm{A}\rightleftharpoons \mathrm{B}$ is [Given ${\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{H}}_{298\mathrm{K}}^0=-54.07{\mathrm{KJmol}}^{-1};{\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{S}}^0=10{\mathrm{JK}}^{-1}{ \mathrm{mol}}^{-1}$ and R=8.314 ${\mathrm{JK}}^{-1}$${\mathrm{mol}}^{-1}$; 2.303×8.314×298=5705]

$\begin{array}{l}\Delta G^0=-RT\ln K\\ {\Delta }_r{G}^0=-2.303RT{\log }_{10}K\\ \text{ But }{\Delta }_r{G}^0={\Delta }_r{H}^0-T{\Delta }_r{S}^0\\ =-54.07\times {10}^3-298\times 10\\ =-57050 \mathrm{J}\\ {\Delta }_r{G}^0=-2.303\times 8.314\times 298{\log }_{10}K\\ -57050=-5705{\log }_{10}K\\ \therefore {\log }_{10}k=10\end{array}$

Concept(s): Gibbs free energy