Answer
46.2
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[2\Delta_{f}G^{\circ}(HI,g)]-[\Delta_{f}G^{\circ}(H_{2},g)+\Delta_{f}G^{\circ}(I_{2},g)]$
$=[2(1.70\,kJ/mol)]-[(0)+(19.327\,kJ/mol)]$
$=-15.93\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K^{\circ}$
$\implies K^{\circ}=e^{-\frac{\Delta _{r}G^{\circ}}{RT}}=e^{-\frac{-15.93\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(500.\,K)}}$
$=46.2$
