Answer
$1.48\times10^{90}$
Work Step by Step
$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[2\Delta G_{f}^{\circ}(CO_{2},g)]-[2\Delta G_{f}^{\circ}(CO,g)+\Delta G_{f}^{\circ}(O_{2},g)]$
$=[2(-394.4\,kJ/mol)]-[2(-137.2\,kJ/mol)+(0)]$
$=-514.4\,kJ/mol$
$\Delta G^{\circ}=-RT\ln K\implies \ln K=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{-514.4\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}=207.6226$
Then, $K=e^{207.6226}=1.48\times10^{90}$
