Answer
$K^{\circ}=4.9\times10^{-9}$
$K^{\circ}$ is close to $K_{c}$.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta_{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(Ca^{2+},aq)+\Delta _{f}G^{\circ}(CO_{3}^{2-},aq)]-[\Delta_{f}G^{\circ}(CaCO_{3},s)]$
$=[(-553.58\,kJ/mol)+(-527.81\,kJ/mol)]-(-1128.79\,kJ/mol)$
$=47.4\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K^{\circ}$
$\implies K^{\circ}=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{47.4\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$
$=4.9\times10^{-9}$
From table 12.1,
$K_{c}=2.8\times10^{-9}$
$K^{\circ}$ is close to $K_{c}$.
