Answer
$K^{\circ}=4.3\times10^{-7}=K_{c}$
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta_{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(HCO_{3}^{-},aq)+\Delta _{f}G^{\circ}(H^{+},aq)]-[\Delta_{f}G^{\circ}(H_{2}CO_{3},aq)]$
$=[(-586.77\,kJ/mol)+(0\,kJ/mol)]-(-623.08\,kJ/mol)$
$=36.31\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K^{\circ}$
$\implies K^{\circ}=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{36.31\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$
$=4.3\times10^{-7}$
From table 12.1,
$K_{c}=4.3\times10^{-7}=K^{\circ}$
