## Chemistry: The Molecular Science (5th Edition)

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