At $320 \mathrm{K}$, a gas ${\mathrm{A}}_2$ is 20 % dissociated to $\mathrm{A}\left(\mathrm{g}\right)$. The standard free energy change at $320K$} and $1 \mathrm{atm}$ in ${\mathrm{Jmol}}^{-1}$ is approximately : ($\mathrm{R}=8.314$ ${\mathrm{JK}}^{-1}{ \mathrm{mol}}^{-1}$; In 2=0.693; In 3=1.098 )

Initially, suppose $\left[{\mathrm{A}}_2\right]=1\mathrm{M}$ and $\left[\mathrm{A}\right]=0\mathrm{M}$

After $20\%$ dissociation, $80\%$ of ${\mathrm{A}}_2$ remains.

$\left[{\mathrm{A}}_2\right]=1\times \frac{80}{100}=0.8\mathrm{M}$

$20\%$ of $1\mathrm{M}$ is $1\times \frac{20}{100}=0.2.\left[\mathrm{A}\right]=2\times 0.2=0.4\mathrm{M}$

The equilibrium constant

$\mathrm{K}=\frac{[\mathrm{A}{]}^2}{\left[{ \mathrm{A}}_2\right]}$

$\mathrm{K}=\frac{[0.4{]}^2}{[0.8]}=0.2$

$\Delta {\mathrm{G}}^0=-\mathrm{RTlnK}=-8.314{\mathrm{JK}}^{-1}{ \mathrm{mol}}^{-1}\times 320 \mathrm{K}\times \ln 0.2=4281 \mathrm{J}/\mathrm{mol}$

Therefore, the correct option is (D).