The two equations are :(a) $\frac{4}{3} \mathrm{Al} + \mathrm{O}_2 \longrightarrow \frac{2}{3} \mathrm{Al}_2\mathrm{O}_3$(b) $2\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2\mathrm{MgO}$At the point of intersection of $\mathrm{Al}_2\mathrm{O}_3$ and MgO curves the ∆G° becomeszero for the reaction.For the reduction of MgO by Al consider the reaction : For this reaction the temperature should be chosen so that,$\Delta G^{\circ}(\mathrm{Al}_2\mathrm{O}_3) < \Delta_f G^{\circ}(\mathrm{MgO})$ Around 2000 K the line for $\mathrm{Al}_2\mathrm{O}_3$ lies below that of MgO, therefore, Al canreduce MgO. However, at a temperature below the intersection point oftwo lines MgO is more stable and cannot be reduced by Al.