Thermodynamic Principles of Metallurgy

Deciding how hot a reduction needs to run, and which reducing agent will actually do the job for a given metal oxide $M_xO_y$, isn't guesswork — it follows directly from Gibbs energy.

The basic criterion

For a thermal reduction to proceed at a given temperature, the Gibbs energy change of the reaction must be negative. Gibbs energy varies with temperature according to the familiar relation:

where $\Delta H$ is the enthalpy change and $\Delta S$ is the entropy change of the process. A reaction only goes forward when $\Delta G$ works out negative, and there are two distinct ways to engineer that:

1. Raise the temperature, when $\Delta S$ is positive. As $T$ increases, the $T\Delta S$ term grows, and once $T\Delta S$ overtakes $\Delta H$, $\Delta G$ swings negative.
2. Couple two reactions — an oxidation and a reduction — so that even if neither is favourable alone, their combined $\Delta G$ is negative. This is the trick underlying almost every real metallurgical reduction.

The Ellingham diagram

Plotting $\Delta_r G^\circ$ against $T$ for the formation of metal oxides (per mole of $O_2$ consumed) gives what's called an Ellingham diagram (Fig. 6.4), first devised by H.J.T. Ellingham. It is the single most useful graphical tool for choosing a reducing agent.

Why the lines slope upward. The plotted reaction is $2xM(s) + O_2(g) \rightarrow 2M_xO(s)$. A gas is consumed in forming the oxide, so molecular randomness decreases — $\Delta S$ is negative — which makes the $-T\Delta S$ term in equation 6.14 positive. So $\Delta_r G^\circ$ climbs upward (becomes less negative) as $T$ rises, giving most metal-oxide lines a positive slope.

Kinks mark phase changes. Each line is essentially straight except where the metal or its oxide changes phase (solid→liquid or liquid→gas) — at that temperature the slope visibly steepens on the positive side (the Zn/ZnO line, for instance, kinks where zinc melts).

The line crossing zero. As $T$ rises further, a line eventually crosses $\Delta_r G^\circ = 0$. Below that crossing, the oxide is thermodynamically stable ($\Delta_r G^\circ$ negative); above it, $\Delta_r G^\circ$ turns positive and the oxide will spontaneously decompose.

Similar diagrams can be drawn for sulphides and halides, and they make it obvious why reducing $M_xS$ compounds is generally much harder than reducing the corresponding oxide.

Two limitations worth remembering:

• The diagram only tells you whether reduction is thermodynamically possible, not how fast it happens — it says nothing about kinetics. It does, however, explain qualitatively why solid-state reactions are sluggish while reactions speed up once the ore melts (molecular randomness — and hence $\Delta S$ — jumps at a phase change).
• The $\Delta_r G^\circ$ values are derived via $\Delta G^\circ = -RT\ln K$, which presumes reactants and products are at equilibrium. In an actual furnace, solids and gases are often in contact for only a short time, so this assumption doesn't strictly hold.

Coupling reduction and oxidation

During reduction, the metal oxide decomposes and the reducing agent carries the released oxygen away. For the reducing agent to succeed, its own oxidation must have a $\Delta_r G^\circ$ negative enough that the sum with the oxide's decomposition is negative overall.

Write the oxide's decomposition as:

If carbon is the reducing agent, its oxidation can go two ways:

Coupling 6.15 with 6.16, or with 6.17, gives the two possible overall reduction reactions:

If carbon monoxide itself is the reducing agent, the relevant oxidation is:

giving the overall reaction:

Reading the diagram to pick a reducing agent …