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Question

(D) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K_c)$ is derived from the equation: $\Del

Vedclass · Accepted Answer

$-\Delta G^o = RT \ln K_c$

Vedclass · Answer

(D) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K_c)$ is derived from the equation: $\Delta G = \Delta G^o + RT \ln Q$. At equilibrium,the reaction quotient $Q = K_c$ and the change in Gibbs free energy $\Delta G = 0$. Substituting these values into the equation gives: $0 = \Delta G^o + RT \ln K_c$. Rearranging this,we get: $\Delta G^o = -RT \ln K_c$ or $-\Delta G^o = RT \ln K_c$.

The correct relationship between the standard free energy change in a reaction and the corresponding equilibrium constant $K_c$ is:

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