Answer:
\[\Delta G=0\] because the reaction is in equilibrium.
\[\Delta G{}^\circ =-2.303\,RT\,\log \,K=-2.303\]\[\times 8.314\,J\,{{K}^{-1}}\,mo{{l}^{-1}}\,\times 300\,K\,\log \,{{10}^{2}}\]
\[=-11488\,J\,mo{{l}^{-1}}=-11.488\,kJ\,mo{{l}^{-1}}.\]
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