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The standard Gibb's free energy change, $\Delta \mathrm{G}^{\circ}$ is related to equilibrium constant, $\mathrm{K}_{\mathrm{p}}$ as
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Verified Answer
The correct answer is:
$K_{p}=e^{-\Delta G^{0} / R T}$
$\Delta G^{\circ}$ and $K_{p}$ are related as
$$
\begin{array}{l}
\Delta G^{\circ}=-R T \ln K_{p} \\
\Rightarrow \ln K_{p}=\frac{\Delta G^{\circ}}{R T} \Rightarrow K_{p}=e^{-\Delta G^{\circ} / R T}
\end{array}
$$
$$
\begin{array}{l}
\Delta G^{\circ}=-R T \ln K_{p} \\
\Rightarrow \ln K_{p}=\frac{\Delta G^{\circ}}{R T} \Rightarrow K_{p}=e^{-\Delta G^{\circ} / R T}
\end{array}
$$
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