Of the following reactions:(i) , A rightlefth | vedclass

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(C) The relationship between standard Gibbs free energy change $(\Delta G^{\circ})$ and the equilibrium constant $(K_{eq})$ is given by the equation:$

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$(iii)$

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(C) The relationship between standard Gibbs free energy change $(\Delta G^{\circ})$ and the equilibrium constant $(K_{eq})$ is given by the equation:$\Delta G^{\circ} = -RT \ln K_{eq}$ or $\log K_{eq} = -\Delta G^{\circ} / (2.303 \, RT)$From this relation,it is clear that $\log K_{eq}$ is inversely proportional to $\Delta G^{\circ}$.Therefore,the reaction with the most negative value of $\Delta G^{\circ}$ will have the largest positive value for $\log K_{eq}$,and consequently,the largest equilibrium constant $(K_{eq})$.Comparing the given values:$(i) \, 250 \, kJ \, mol^{-1}$$(ii) \, -100 \, kJ \, mol^{-1}$$(iii) \, -150 \, kJ \, mol^{-1}$$(iv) \, 150 \, kJ \, mol^{-1}$The most negative value is $-150 \, kJ \, mol^{-1}$,which corresponds to reaction $(iii)$.

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