07. EQUILIBRIUM (CHEMICAL) (HM)
Explanation:
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
\(K = \frac{{{{[A]}^2}}}{{[{A_2}]}};\) \(K = \frac{{{{[0.4]}^2}}}{{[0.8]}} = 0.2\)
\(\Delta {G^o} = - RT\,\ln \,K = - 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}\)
\( \times 320\,K \times \ln \,0.2 = 4281\,J/mol\)