Answer
$K_{p}=7.92\times10^{-81}$
Work Step by Step
We find:
$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[2\Delta G_{f}^{\circ}(H_{2},g)+\Delta G_{f}^{\circ}(O_{2},g)]-[2\Delta G_{f}^{\circ}(H_{2}O,g)]$
$=[2(0)+(0)]-[2(-228.6\,kJ/mol)]=457.2\,kJ/mol$
$\Delta G^{\circ}=-RT\ln K_{p}\implies \ln K_{p}=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{457.2\times1000\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298.15\,K)}$
$=-184.44$
$K_{p}=e^{-184.44}=7.92\times10^{-81}$
