Animated Solution for Chemistry - General Principles and Processes of Isolation of Metals: For a reaction, $4M(s)+n{O}_2(g)⟶2M_2{O}_n(s)$ 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

Imagine you are a metallurgist trying to extract a pure metal from its ore. You need to know exactly at what temperature the metal oxide will willingly give up its oxygen. This is where the Ellingham Diagram becomes your best friend. It is a brilliant graphical representation that plots the standard Gibbs free energy of formation, ${\Delta }_f{G}^{\circ }$, against temperature, $T$.

Let's break down the reaction given to us:

Notice the physical states carefully. We are starting with a solid metal and a gaseous oxygen molecule, and we are ending up with a solid metal oxide.

In thermodynamics, entropy ($S$) is a measure of randomness or disorder. Gases are wild, free-roaming molecules with high entropy, while solids are highly ordered structures with low entropy. Because our reaction consumes a gas to form a solid, the overall randomness of the system takes a massive nosedive.

Mathematically, this means the change in entropy, $\Delta S$, is negative.

Now, let's bring in the master equation of thermodynamics:

Since $\Delta S^{\circ }$ is negative, the term $-T\Delta S^{\circ }$ becomes positive. As you crank up the temperature ($T$), this positive term grows larger and larger. Consequently, $\Delta G^{\circ }$ becomes less negative, moving upwards towards zero. This perfectly explains why the slope of the line in the Ellingham diagram is positive!

For any compound to be thermodynamically stable, its formation must be a spontaneous process. In the language of Gibbs, this means $\Delta G^{\circ }$ must be negative.

Looking at the Ellingham plot, as long as the line is below the horizontal axis (where $\Delta G^{\circ }=0$), the metal oxide is happy and stable. However, as temperature increases, the line marches upwards.

The exact temperature below which the oxide is stable is the precise point where the line intersects the $T$-axis. At this magical intersection, $\Delta G^{\circ }=0$. If you push the temperature even a fraction of a degree higher, $\Delta G^{\circ }$ becomes positive. The formation is no longer spontaneous, and the oxide will spontaneously decompose back into the metal and oxygen.

Therefore, the stability limit is inferred from the plot as the point at which the free energy change shows a transition from a negative to a positive value.