Answer:
\[\Delta {{G}^{{}^\circ
}}=-2.303RT\,{{\log }_{10}}{{K}_{p}}\]
\[\log
{{K}_{p}}=\frac{-\Delta {{G}^{{}^\circ }}}{2.303RT}\]
\[=-\frac{(-13.6\times
{{10}^{3}})}{2.303\times 8.314\times 298}\]
\[=2.38\]
\[{{K}_{p}}=Anti\log
(2.38)=2.4\times {{10}^{2}}\]
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