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The standard free energy change of a reaction is $\Delta \mathrm{G}^{\circ}=-115 \mathrm{~kJ}$ at $298 \mathrm{~K}$. Calculate the equilibrium constant $\mathrm{k}_{\mathrm{p}}$ in $\log \mathrm{k}_{\mathrm{p}}$ $\left(\mathrm{R}=8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right)$
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The correct answer is:
$20.16$
$\begin{aligned} \Delta \mathrm{G}^{\circ} &=-115 \mathrm{~kJ} \text { at } 298 \mathrm{~K} . \\ \mathrm{Now}, \Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\ \mathrm{R} &=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \& \mathrm{~T}=298 \mathrm{~K} . \\ \Delta \mathrm{G}^{\circ} &=-2.303 \times 8.314 \times 298 \times \log \mathrm{k}_{\mathrm{p}} \\ \log \mathrm{k}_{\mathrm{P}} &=\frac{-115 \times 10^{3}}{-2.303 \times 8.314 \times 298}=20.155 \\ &=20.16 \end{aligned}$
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