The equilibrium constant for a reaction is 10. What will be the value

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Question:

The equilibrium constant for a reaction is $10 .$ What will be the value of $\Delta G^{\theta} ? \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, T=300 \mathrm{~K}$.

Solution:

From the expression,

$\Delta G^{\theta}=-2.303 \mathrm{RT} \log K_{e q}$

$\Delta G^{\theta}$ for the reaction,

$=(2.303)\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K}) \log 10$

$=-5744.14 \mathrm{Jmol}^{-1}$

$=-5.744 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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