At $60°\mathrm{c}$,one mole of dinitrogen tetroxide is fifty percent dissociated$\left({\mathrm{N}}_2{\mathrm{O}}_4\rightleftharpoons 2{\mathrm{NO}}_2\right)$. the standard free energy change at this temperature and at 1atm.

(log 1.33 = 0.1239)

$$\begin{gathered} {\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right) \\ 1\mathrm{mol} 2\mathrm{mol} \\ 1-\mathrm{x} 2\mathrm{x} \\ \mathrm{where} \mathrm{x}=50\%=0.5 \end{gathered}$$

$$\begin{gathered} {\mathrm{X}}_{{\mathrm{N}}_2{\mathrm{O}}_4}=\frac{0.5}{1.5} =\frac{1}{3} ; {\mathrm{X}}_{{\mathrm{NO}}_2}=\frac{1}{1.5}=\frac{2}{3} \\ {\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{O}}_4} =\frac{1}{3}\times 1 \mathrm{atm}. ; {\mathrm{P}}_{{\mathrm{NO}}_2}=\frac{2}{3}\times 1 \mathrm{atm}. \\ \mathrm{Equilibrium} \mathrm{constant} , {\mathrm{K}}_{\mathrm{P}}=\frac{{\left( {\mathrm{P}}_{{\mathrm{NO}}_2}\right)}^2}{{\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{O}}_4}}=\frac{\frac{2}{3}\times \frac{2}{3}}{\frac{1}{3}}=1.33 \mathrm{atm} \end{gathered}$$

$$\begin{gathered} ∆{\mathrm{G}}^{\mathrm{o}}=-2.303{\mathrm{RTlogK}}_{\mathrm{P}} \\ =-2.303(8.314)333(0.1239) \\ =-790 \mathrm{J} {\mathrm{mol}}^{-1} \end{gathered}$$