The value of `Delta_f G^@` for the formation of `Cr_2 O_3` is -540 KJ `mol^(-1)` and that of `Al_2 O_3` is -827 KJ `mol^(-1)`. Is reduction of `Cr_2 O_3` possible with Al ?

Explanation

To determine if the reduction of \( \text{Cr}_2\text{O}_3 \) is possible with aluminum (Al), we can analyze the Gibbs free energy change (\( \Delta G \)) for the reaction. Here’s a step-by-step solution:

### Step 1: Write the formation reactions
We start by writing the formation reactions for both \( \text{Cr}_2\text{O}_3 \) and \( \text{Al}_2\text{O}_3 \):

1. For \( \text{Cr}_2\text{O}_3 \):
\[
2 \text{Cr} + \frac{3}{2} \text{O}_2 \rightarrow \text{Cr}_2\text{O}_3 \quad \Delta G_f = -540 \, \text{kJ/mol}
...