Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetraoxide at 25 °C.

\({\rm{\Delta G = \Delta H - T\Delta S}}\)where,\({\rm{\Delta G }}\)is the change in Gibbs free energy,\({\rm{\Delta H}}\)is the change in enthalpy, T is the absolute temperature in Kelvin, and\({\rm{\Delta S}}\)is the change in entropy.

\(\Delta S = \frac{{\Delta H}}{T}\)

\(\Delta G_{f\left( {N{O_2}} \right)}^^\circ = 51.30\;{\rm{kJ}}/{\rm{mol}}\)

\({\rm{\Delta }}{{\rm{G}}_{{\rm{f}}\left( {{\rm{\;}}{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}} \right)}}{\rm{ = 99}}{\rm{.8\;kJ/mol}}\)

\(\Delta {G^^\circ } = \sum \Delta G_f^^\circ \)(products) \( - \sum \Delta G_f^^\circ \) (reactants)

\({\rm{ = 2\Delta G}}_{{\rm{N}}{{\rm{O}}_{\rm{2}}}}^{\rm{^\circ }}{\rm{ - \Delta }}{{\rm{G}}^{\rm{^\circ }}}{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}\)

\({\rm{ = 2 \times 51}}{\rm{.3 - 99}}{\rm{.8}}\)

\({\rm{ = 2}}{\rm{.8\;kJ/mop}}\quad {\rm{(1kJ = 1000\;J)}}\)

Or \(2800\;{\rm{J}}/\) mol

\({\rm{\Delta }}{{\rm{G}}^{\rm{^\circ }}}{\rm{ = - RTln}}{{\rm{K}}_{{\rm{eq}}}}\)

\( = \frac{{ - 2800}}{{8.314 \times 298}}\)

\({\rm{ln}}{{\rm{k}}_{{\rm{eq }}}}{\rm{ = - 1}}{\rm{.130}}\)

or\({{\rm{K}}_{{\rm{eq}}}} = {e^{ - 1.130}}\)

\({K_{{\rm{eq }}}} = 0.30\)