Question
Which quantity out of $\Delta_\text{r}\text{G}$ and $\Delta_\text{r}\text{G}^\ominus$ will be zero at equilibrium?
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Answer
Gibbs energy for a reaction in which all reactants and products are in standard state. $\Delta_\text{r}\text{G}^\circ$ is related to the equilibrium constant of the reaction as follows:
$\Delta_\text{r}\text{G}=\text{A}_\text{r}\text{G}^\circ+\text{RT}\ \text{In}\ \text{K}$
At equalibrium, $0=\Delta_\text{r}\text{G}^\circ+\text{RT}\ \text{In}\ \text{A}^-(\Delta_\text{r}\text{G}=0)\ \text{or}\ \Delta_\text{r}\text{G}^\circ=-\text{RT}\ \text{In}\text{K}$$\Delta_\text{r}\text{G}^\circ=0$ when K = 1
For all other values of K, $\text{A}_\text{r}\text{G}^\circ$ will be non - zero.Need a full question paper?
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