c
In the reaction \({A_2} \leftrightarrow 2A\)

Initially, Let \([{A_2}] = 1\,M\) and \([A] = 0\,M\)

After \(20\%\) dissociation , \(80\%\) of \(A_2\) remains.

\([{A_2}] = 1 \times \frac{{80}}{{100}} = 0.8\,M\)

\(20\%\) of \(1\,M\) is

 \(1 \times \frac{{20}}{{100}} = 0.2.\,[A] = 2 \times 0.2 = 0.4\,M\)

The equilibrium constant

\(K = \frac{{{{[A]}^2}}}{{[{A_2}]}};\) \(K = \frac{{{{[0.4]}^2}}}{{[0.8]}} = 0.2\) 

\(\Delta {G^o} =  - RT\,\ln \,K =  - 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}\)

                            \( \times 320\,K \times \ln \,0.2 = 4281\,J/mol\)