At 60 ° C , dinitrogen tetroxide is fifty percent dissociated. Find its standard free energy change at…

The reaction showing the dissociation will be:

$\begin{array}{cccc}& {\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)& \rightleftharpoons & 2{\mathrm{NO}}_2\left(\mathrm{g}\right)\\ \mathrm{t}=0& 1 \mathrm{mol}& & 0 \mathrm{mol}\\ \mathrm{t}={\mathrm{t}}_{\mathrm{eq}}& 0.5 \mathrm{mol}& & 1 \mathrm{mol}\end{array}$

Clearly, from this equation we can say that the total number of moles is 

$0.5+1=1.5 \mathrm{mol}$

Hence, the partial pressure for the molecule will be

$$\begin{gathered} {\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{O}}_4} =\frac{ 0.5}{1.5}\times 1\mathrm{atm} = \frac{1}{3} \mathrm{atm} \\ {\mathrm{P}}_{{\mathrm{NO}}_2} = \frac{1}{1.5}\times 1\mathrm{atm} =\frac{2}{3} \mathrm{atm} \end{gathered}$$

Now, according to the law of chemical equilibrium we know that 

${\mathrm{K}}_{\mathrm{sp}} = \frac{{{\mathrm{P}}^2}_{{\mathrm{NO}}_2}}{{\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{O}}_4}} = \frac{{\left({\displaystyle \frac{2}{3}}\right)}^2}{{\displaystyle \frac{1}{3}}} = 1.33 \mathrm{atm}$

We know that at equilibrium the free Gibb's energy is: 

$$\begin{gathered} ∆\mathrm{G} = -2.303\times \mathrm{RT}\times \left({\log }_{10} {\mathrm{K}}_{\mathrm{P}}\right) \\ ∆\mathrm{G} = -2.303\times \left(8.314 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right)\times \left(333\mathrm{K}\right)\times \left[{\log }_{10} \left(1.33\right)\right] \\ ∆\mathrm{G} = -790 {\mathrm{Jmol}}^{-1} \end{gathered}$$