Answer
$\Delta G^{\circ}_{rxn}=457.2\,kJ/mol$
$K_{p}=7.2\times10^{-81}$
Work Step by Step
According to the equation $\Delta G^{\circ}_{rxn}=\Sigma n\Delta G_{f}^{\circ}(products)-\Sigma m\Delta G_{f}^{\circ}(reactants)$,
$\Delta G^{\circ}_{rxn}=[2\Delta G_{f}^{\circ}(H_{2})+\Delta G_{f}^{\circ}(O_{2})]-[2\Delta G_{F}^{\circ}(H_{2}O(g))]$
$=[2(0)+0]-[2(-228.6\,kJ/mol)]$
$=457.2\,kJ/mol$
Recall that $\Delta G^{\circ}_{rxn}=-RT\ln K_{p}$
$\implies \ln K_{p}=-\frac{\Delta G^{\circ}_{rxn}}{RT}=-\frac{457.2\times1000\,J/mol}{(8.314\,J\,mol^{-1}K^{-1})(298\,K)}$
$=-184.5355$
$\implies K_{p}=e^{-184.5355}=7.2\times10^{-81}$
