Ellingham Diagram

Building the diagram. The change in Gibbs free energy for a reaction is given by ΔG = ΔH - TΔS ...... (1), where ΔH is the enthalpy change, T the temperature in kelvin, and ΔS the entropy change. For a reaction treated as an equilibrium process, the standard free energy change can also be obtained from the equilibrium constant: ΔG0 = -RT ln Kp.

Harold Ellingham used this relationship to calculate ΔG0 at a range of different temperatures for the reduction (equivalently, the formation) of many different metal oxides, treating each reduction as an equilibrium process. Plotting temperature on the x-axis against the standard free energy change for the FORMATION of each metal oxide on the y-axis gives, for each metal-oxide system, a nearly straight line -- with ΔS as the slope and ΔH as the y-intercept (matching equation (1) rewritten as a straight-line equation in T). This composite plot -- the graphical representation of how the standard Gibbs free energy of formation of various metal oxides varies with temperature -- is the ELLINGHAM DIAGRAM (Figure 1.4), and it summarises a very large amount of extractive-metallurgy information in a single picture: which oxide is more stable than another at a given temperature, and which element can therefore act as a reducing agent for which oxide.

Four observations read directly off the diagram.

1. For most metal-oxide formation lines, the SLOPE IS POSITIVE. This is because forming the oxide consumes gaseous O2 -- converting a mole of free-moving gas molecules into a solid oxide lattice -- which decreases the randomness (entropy) of the system, making ΔS negative; in the straight-line equation ΔG = ΔH - TΔS, a negative ΔS makes the -TΔS term positive and growing with T, so ΔG rises (becomes less negative) as temperature increases, i.e. a positive slope.

2. The line for the formation of CARBON MONOXIDE, 2C + O2 = 2CO, is the one clear exception: it has a NEGATIVE slope. Here ΔS is positive, because 2 moles of gaseous CO are produced from the consumption of only 1 mole of gaseous O2, i.e. moles of gas actually increase, raising randomness. This negative slope means CO becomes progressively MORE thermodynamically stable (its ΔG of formation becomes more negative) as temperature rises -- the opposite trend to almost every metal oxide.

3. As temperature increases, the ΔG for formation of a typical metal oxide becomes progressively LESS negative, reaching zero at some particular temperature; below that temperature ΔG is negative (oxide stable), and above it ΔG is positive (oxide unstable, i.e. decomposition is thermodynamically favoured). The general trend, therefore, is that metal oxides become less stable and easier to decompose as temperature rises. …