At 57°C, gaseous dinitrogen tetroxide is 50 % dissociated. Calculate the standard free energy change per mole of ${\mathrm{N}}_2{\mathrm{O}}_4$ (g) at this temperature and at 1 atm.

$(R=8.3{\mathrm{JK}}^{-1}{ \mathrm{mol}}^{-1}\text{ , In }10=2.3,\log 2=0.3,\log 3=0.48)$

Given Temperature $T_1={57}^{0}C+{273}^{0}C=330K$

Gaseous dinitrogen tetroxide=50 % dissociated

$R=8.3J{K}^{-1}mo{l}^{-1}$ ln 10=2.3, log2=0.3, log3=0.48

Standard free energy, $\Delta G^0=?$

The Reaction is: 

                                $N_2{O_4}_{\left(g\right)}\iff 2N{O}_{2\left(g\right)}$

Initial  (t = 1)                     1                    0 

At equilibrium (t = eq)    1-a                2a

Total moles   $=1-\alpha +2\alpha =1+\alpha$

a = 0.5  and P = 1 atm

$\therefore P_{N_2{O}_4}=\left(\frac{1-\alpha }{1+\alpha }\right)P{}_T$

$=\left(\frac{1-0.5}{1+0.5}\right)\times 1$

$=\frac{0.5}{1.5}atm$

$\therefore P{}_{N{O}_2}=\left(\frac{2\alpha }{1+\alpha }\right)P{}_T$

$\begin{array}{l}=\left(\frac{2\times 0.5}{1+0.5}\right)\times 1\\ =\frac{1}{1.5}atm\end{array}$

$\therefore K_P=\frac{P{}_{N{O}_2}^2}{P{}_{N_2{O}_4}}$

$=\frac{{\left(\frac{1}{1.5}\right)}^2}{\left(\frac{0.5}{1.5}\right)}$

$=\frac{4}{3}$

Gibbs energy formula, 

$\therefore \Delta G^0=-2.303nRT\log K_P$

$=-2.303\times 1\times 8.3\times 330\times \log \left(\frac{4}{3}\right)$

$=-2.303\times 8.3\times 330\left(0.6-0.48\right)$

$=-756Jmo{l}^{-1}$