Answer
$K=2\times10^{140}$
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(CO_{2},g)+2\Delta_{f}G^{\circ}(H_{2}O,g)]-[\Delta_{f}G^{\circ}(CH_{4},g)+2\Delta_{f}G^{\circ}(O_{2},g)]$
$=[(-394.359\,kJ/mol)+2(-228.572\,kJ/mol)]-[(-50.72\,kJ/mol)+2(0)]$
$=-800.783\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K$
$\implies K=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{-800.783\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298.15\,K)}}$
$=2\times10^{140}$
