Chapter 16 - Thermodynamics: Directionality of Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 737d: 68b

$K=6.75$

$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$ $=[\Delta_{f}G^{\circ}(N_{2}O_{4},g)]-[2\Delta_{f}G^{\circ}(NO_{2},g)]$ $=(97.89\,kJ/mol)-[2(51.31\,kJ/mol)]$ $=-4.73\,kJ/mol$ $\Delta _{r}G^{\circ}=-RT\ln K$ $\implies K=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{-4.73\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$ $=6.75$