Answer
$K=4\times10^{-34}$
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(H_{2},g)+\Delta_{f}G^{\circ}(Cl_{2},g)]-[2\Delta_{f}G^{\circ}(HCl,g)]$
$=[(0)+(0)]-[2(-95.299\,kJ/mol)]$
$=190.598\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K$
$\implies K=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{190.598\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298.15\,K)}}$
$=4\times10^{-34}$
