Answer
7.83
Work Step by Step
$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[2\Delta G_{f}^{\circ}(BrCl,g)]-[\Delta G_{f}^{\circ}(Br_{2},g)+\Delta G_{f}^{\circ}(Cl_{2},g)]$
$=[2(-1.0\,kJ/mol)]-[(3.1\,kJ/mol)+(0\,kJ/mol)]$
$=-5.1\,kJ/mol$
$\Delta G^{\circ}=-RT\ln K\implies \ln K=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{-5.1\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}=2.058467$
Then, $K=e^{2.058467}=7.83$
