ΔG° vs T plot for the formation of MgO, involving reaction
2Mg + O₂ → 2MgO, will look like :

Solution

Chemistry ma’am · 4 steps

The slope is -Δ S°. Once Mg boils, more gas is used up, so the line bends upward more steeply.

1. Slope means entropy. Write Δ G° = Δ H° - TΔ S°. Over a range where Δ H° and Δ S° stay almost constant, this is a straight line in T with slope -Δ S°.

   slope = d(Δ G°)/dT = -Δ S°

2. Below the boiling point. For 2Mg(s) + O₂(g) → 2MgO(s), one mole of gas disappears, so Δ S° is negative. The slope -Δ S° is therefore positive and the line rises gently.
3. Above the boiling point. Now magnesium is a gas: 2Mg(g) + O₂(g) → 2MgO(s). Three moles of gas disappear, so Δ S° becomes much more negative and the slope becomes much steeper. The line shows a kink at the boiling point of Mg.
4. So. The plot rises with a gentle slope, then bends upward to a steeper slope. That is graph (a).

The idea to keep. In an Ellingham plot the slope is -Δ S°. A kink appears at a phase change of the metal, and the line gets steeper when more gas is consumed.

Answer from the NTA final answer key. Solution by PrepWiser.