For the reaction $2Mg + O_2 \rightarrow 2MgO$ the standard Gibbs free energy is $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ.$

Therefore, in a plot of $\Delta G^\circ$ vs $T$:

• the intercept is determined by $\Delta H^\circ$,
• the slope is $\frac{d(\Delta G^\circ)}{dT} = -\Delta S^\circ.$

Now determine the signs of $\Delta H^\circ$ and $\Delta S^\circ$ for formation of $MgO$.

(i) Sign of $\Delta H^\circ$

Formation of magnesium oxide is a highly exothermic reaction. Hence, $\Delta H^\circ < 0.$ So the graph starts at a negative value of $\Delta G^\circ$ at $T=0$.

(ii) Sign of $\Delta S^\circ$

Reactants contain a gas molecule: $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$ Gas is consumed and solid is formed, so disorder decreases. Hence, $\Delta S^\circ < 0.$

Since $\text{slope} = -\Delta S^\circ,$ and $\Delta S^\circ < 0$, we get $-\Delta S^\circ > 0.$ Thus the $\Delta G^\circ$ vs $T$ graph has a positive slope.

Combining both results:

• intercept is negative,
• slope is positive.

So the graph is a straight line starting below zero and rising with temperature.

This corresponds to Option A.