At $60^{\circ} C$,dinitrogen tetroxide is $50$ per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

(N/A) The dissociation reaction is $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.
If $N_{2}O_{4}$ is $50 \%$ dissociated,the mole fractions at equilibrium are:
$x_{N_{2}O_{4}} = \frac{1-0.5}{1+0.5} = \frac{0.5}{1.5} = \frac{1}{3}$
$x_{NO_{2}} = \frac{2 \times 0.5}{1+0.5} = \frac{1}{1.5} = \frac{2}{3}$
At $P = 1 \, atm$,the partial pressures are:
$p_{N_{2}O_{4}} = \frac{1}{3} \times 1 \, atm = 0.333 \, atm$
$p_{NO_{2}} = \frac{2}{3} \times 1 \, atm = 0.667 \, atm$
The equilibrium constant $K_{p}$ is:
$K_{p} = \frac{(p_{NO_{2}})^{2}}{p_{N_{2}O_{4}}} = \frac{(2/3)^{2}}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} \approx 1.333 \, atm$
Using the relation $\Delta_{r} G^{\ominus} = -RT \ln K_{p}$ at $T = 333 \, K$:
$\Delta_{r} G^{\ominus} = -8.314 \, J K^{-1} mol^{-1} \times 333 \, K \times \ln(1.333)$
$\Delta_{r} G^{\ominus} = -8.314 \times 333 \times 0.2877 \approx -796.5 \, J mol^{-1}$