Answer
3.1
Work Step by Step
$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[\Delta G_{f}^{\circ}(N_{2}O_{4},g)]-[2\Delta G_{f}^{\circ}(NO_{2},g)]$
$=(99.8\,kJ/mol)-[2(51.3\,kJ/mol)]$
$=-2.8\,kJ/mol$
$\Delta G^{\circ}=-RT\ln K\implies \ln K=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{-2.8\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}=1.13$
$K=e^{1.13}=3.1$
