The standard Gibb's free energy change, $\Delta \mathrm{G}^{\circ}$ is related to equilibrium constant,…

The standard Gibb's free energy change, $\Delta \mathrm{G}^{\circ}$ is related to equilibrium constant, $\mathrm{K}_{\mathrm{P}}$ as
  1. $\quad \mathrm{K}_{\mathrm{P}}=-\mathrm{RT} \ln \Delta \mathrm{G}^{\circ}$
  2. $\quad \mathrm{K}_{\mathrm{P}}=\left[\frac{\mathrm{e}}{\mathrm{RT}}ight]^{\Delta \mathrm{G}^{\circ}}$
  3. $\mathrm{K}_{\mathrm{P}}=-\frac{\Delta \mathrm{G}}{\mathrm{RT}}$
  4. $\mathrm{K}_{\mathrm{P}}=\mathrm{e}^{-\Delta \mathrm{G}^{\circ} / \mathrm{RT}}$

Solution

$\Delta \mathrm{G}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{p}}$ or $\mathrm{K}_{\mathrm{p}}=\mathrm{e}^{\Delta \mathrm{G} / \mathrm{RT}}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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