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The incorrect expression among the following is :
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Verified Answer
The correct answer is:
$\ln \mathrm{K}=\frac{\Delta \mathrm{H}^0-\mathrm{T} \Delta \mathrm{S}^0}{\mathrm{RT}}$
$\ln \mathrm{K}=\frac{\Delta \mathrm{H}^0-\mathrm{T} \Delta \mathrm{S}^0}{\mathrm{RT}}$
$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}$ and $\Delta \mathrm{G}^0=\Delta \mathrm{H}^0-\mathrm{T} \Delta \mathrm{S}^0$
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