For a reaction,
4M(s) + nO₂(g) → 2M₂Oₙ(s)
the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which :

Solution

Chemistry ma’am · 4 steps

An oxide is stable only while forming it is spontaneous, so look for where Δ G crosses zero.

1. Stability means negative ΔG. The oxide is stable when its formation from the metal and oxygen is spontaneous, that is, when Δ G < 0. If Δ G > 0, the reverse reaction is favoured and the oxide breaks down into metal and oxygen.

   Δ G = Δ H - TΔ S

2. Why ΔG rises. This reaction uses up n moles of O₂ gas and makes only solids, so Δ S is negative. The slope of the Δ G against T line is -Δ S, which is positive. So Δ G starts negative and climbs as the temperature rises.
3. Read the plot. Below the temperature where the line crosses Δ G = 0, formation is spontaneous and the oxide is stable. Above it, the oxide decomposes. Changes in slope only mark phase changes such as melting or boiling of the metal. They say nothing about the sign of Δ G.
4. So. The required temperature is the point where the free energy change goes from negative to positive, option (a).

The idea to keep. On an Ellingham-type plot, the sign of Δ G decides stability. A kink only means a phase change. The oxide is stable only below the temperature where its line crosses zero.

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