Answer
$K^{\circ}=6.75$
$K^{\circ}\ne K_{c}$ for this reaction.
$K_{p}=7.0$ is almost equal to $K^{\circ}$.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta_{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(N_{2}O_{4},g)]-[2\Delta_{f}G^{\circ}(NO_{2},g)]$
$=(97.89\,kJ/mol)-[2(51.31\,kJ/mol)]$
$=-4.73\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K^{\circ}$
$\implies K^{\circ}=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{-4.73\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$
$=6.75$
From table 12.1,
$K_{c}=1.7\times10^{2}$
Therefore, $K^{\circ}\ne K_{c}$ for this reaction.
But $K_{p}=7.0$ is almost equal to $K^{\circ}$.
