For the gas-phase equilibrium N2O4(g)rightleftharpoons 2NO2(g) the standard Gibbs energy change at 298 K is Δ G^°=+5.4 kJ mol^-1 (standard pressure 1…

The relation $\Delta G^\circ=-RT\ln K_p$ gives $K_p=\exp(-5400/(8.314\times298))=0.113$. Start with 1 mol $\mathrm{N_2O_4}$ and let $\alpha$ dissociate, so $n(\mathrm{N_2O_4})=1-\alpha$, $n(\mathrm{NO_2})=2\alpha$ and the total is $1+\alpha$. The partial pressures are $P(1-\alpha)/(1+\alpha)$ and $2\alpha P/(1+\alpha)$, so $K_p=\dfrac{4\alpha^2P}{1-\alpha^2}$ with $P$ in bar. Hence $\alpha=\sqrt{K_p/(K_p+4P)}=\sqrt{0.113/(0.113+2.0)}=0.231$. Using $P=1$ bar by mistake gives $0.17$, which is the value at standard pressure and is therefore not the required answer. Lowering the pressure favours dissociation, so $\alpha$ must be larger than $0.17$, and the value $0.23$ is consistent with Le Chatelier's principle.