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The standard Gibbs free energy change $\Delta G^{\prime}$ is related to equilibrium constant $K_p$ as
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Verified Answer
The correct answer is:
$K_p=e^{-\frac{\Delta G^T}{R T}}$
$K_p=e^{-\frac{\Delta C^T}{R T}}$
For equilibrium condition $\Delta \mathrm{G}=0$
So,
$0=\Delta G^{\mathrm{a}}+R T \ln K_p$
$\ln K_{\mathrm{p}}=-\Delta G^{\circ} / R T$
Now
$K_p=e^{-\frac{\Delta G^T}{R T}}$
For equilibrium condition $\Delta \mathrm{G}=0$
So,
$0=\Delta G^{\mathrm{a}}+R T \ln K_p$
$\ln K_{\mathrm{p}}=-\Delta G^{\circ} / R T$
Now
$K_p=e^{-\frac{\Delta G^T}{R T}}$
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