Generally alkali metals form their oxides by an average increment of 6 kJ `mol^(-1) K^(-1)` in their entropies. If at `25^(@)C, Delta G_(f)^(@)` of `Al_(2)O_(3)(s)=-1582 kJ mol^(-1) , Delta H_(f)^(@)` of `Li_(2)O(s)=1244 kJ` and `Delta H_(f)^(@)` of `Na_(2)O(s)=1411 kJ`. The aluminium oxide can be reduced to Al metal by

Options

A. Sodium
B. Lithium
C. Both Na and Li
D. Neither Na nor Li

Explanation

To determine how aluminum oxide (Al₂O₃) can be reduced to aluminum metal (Al), we will analyze the thermodynamic feasibility of the reduction reaction using the given data.

### Step 1: Write the Reduction Reaction
The reduction of aluminum oxide can be represented as:
\[ \text{Al}_2\text{O}_3(s) + 6 \text{M}(s) \rightarrow 2 \text{Al}(s) + 3 \text{MO}(s) \]
where M can be either lithium (Li) or sodium (Na).

### Step 2: Calculate the Gibbs Free Energy Change for the Reaction
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