\(\therefore \,\,\,\,0 = 0.591 - 0.0591\,\,\,\log \,{K_c}\)
\(\therefore \frac{{0.591}}{{0.0591}} = \log \,{K_c}\,\,\,\,\therefore \,\,\,10 = \log \,{K_c}\,\)
\(\,\therefore \,\,\,{K_c} = {10^{10}}\)
($F = 96,500\;C\;mo{l^{ - 1}}; \,\, R = 8.314\;J{K^{ - 1}}mo{l^{ - 1}})$