The reaction, $\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al} \rightarrow \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}\left(\Delta \mathrm{G}^{\ominus}=-421 \mathrm{~kJ}\right)$ is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

Question

The reaction, $\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al} \rightarrow \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}\left(\Delta \mathrm{G}^{\ominus}=-421 \mathrm{~kJ}\right)$ is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?,

MARKS App · Accepted Answer

This is explained on the basis of $\mathrm{K}_{\text {eq }}$, the equilibrium constant. In the given redox reaction, all reactants and products are solids at room temperature, so, there is no equilibrium between the reactants and products and hence the reactions does not occur at RT. At high temperature, $\mathrm{Cr}$ melts and values of $\mathrm{T} \Delta \mathrm{S}$ increases. As a result, the value of $\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}$ becomes more negative and hence the reaction proceeds rapidly.

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