Answer
$7.07\times10^{29}$
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[2\Delta_{f}G^{\circ}(H_{2}O,g)]-[2\Delta_{f}G^{\circ}(H_{2},g)+\Delta_{f}G^{\circ}(O_{2},g)]$
$=[2(-228.572\,kJ/mol)]-[2(0)+(0)]$
$=-457.144\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K^{\circ}$
$\implies K^{\circ}=e^{-\frac{\Delta _{r}G^{\circ}}{RT}}=e^{-\frac{-457.144\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(800.\,K)}}$
$=7.07\times10^{29}$
