Answer
Requires work to be done.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[2\Delta_{f}G^{\circ}(Al,s)+\frac{3}{2}\Delta_{f}G^{\circ}(O_{2},g)]-[\Delta_{f}G^{\circ}(Al_{2}O_{3},s)]$
$=[2(0\,kJ/mol)+\frac{3}{2}(0\,kJ/mol)]-(-1582.3\,kJ/mol)$
$=1582.3\,kJ/mol$
As $\Delta _{r}G^{\circ}$ is positive, work needs to be done for this reaction to occur.
At least 1582.3 kJ/mol work is required.
