The value of ${\log }_{10}K$ for a reaction $A\rightleftharpoons B$ is          

(Given:  $\Delta G^0=-57kJ/moleat298K$)

$\Delta G^0=\Delta H^0-T\Delta S^0=-2.303RT{\log }_{10}K$

$=-54.07\times {10}^3-298\times 10=-57050J{k}^{-1}mo{l}^{-1}$

$\Delta G^0=-2.303RT{\log }_{10}K$

$-57050=-2.303\times 8.314\times 298{\log }_{10}K$

$\therefore 10={\log }_{10}K$