Answer
$2.09\times10^{-26}$
Work Step by Step
$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[2\Delta G_{f}^{\circ}(H_{2},g)+\Delta G_{f}^{\circ}(S_{2},g)]-[2\Delta G_{f}^{\circ}(H_{2}S,g)]$
$=[2(0\,kJ/mol)+(79.7\,kJ/mol)]-[2(-33.4\,kJ/mol)]$
$=146.5\,kJ/mol$
$\Delta G^{\circ}=-RT\ln K\implies \ln K=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{146.5\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}=-59.13$
Then, $K=e^{-59.13}=2.09\times10^{-26}$
