A diagram plotting the standard Gibbs energy change of formation of oxides and similar compounds, normalized per mol of O₂, against temperature. The lower its line, the more stable the oxide a metal forms.
Japanese name: エリンガム図
Ellingham diagram appears in 4 practice questions across 1 topic, including Thermodynamics of materials.
Field: Thermodynamics of materials
An Ellingham diagram takes oxide formation reactions such as 2M + O₂ → 2MO, normalizes the standard Gibbs energy change ΔG° per mol of O₂, and plots it against temperature. Since ΔG° = ΔH° − TΔS° and ΔH° and ΔS° hardly depend on temperature, each reaction gives a straight line whose slope is −ΔS°. In metal oxidation gaseous O₂ is taken into a solid and ΔS° is negative, so almost all lines slope upward, and oxides become less stable at higher temperature. At the temperatures where a metal or oxide melts or boils, ΔS° changes and the line bends.
At a given temperature, the lower the line, the more negative ΔG° is and the more stable the oxide. The vertical distance between 2 lines equals the ΔG° of the reaction in which the metal of the lower line reduces the oxide of the upper line, so metals near the bottom, such as Al, Mg and Ca, can reduce many oxides. The line for carbon, 2C + O₂ → 2CO, slopes downward because the number of gas molecules increases and ΔS° is positive, and at higher temperatures it lies below the lines of more and more metal oxides. This is why iron ore can be reduced with coke in a blast furnace. Above the temperature where a line crosses ΔG° = 0, the oxide decomposes even in 1 bar of oxygen.
At equilibrium ΔG° = RT ln pO₂, so joining a point on a line at a given temperature to the point ΔG° = 0 at 0 K and extending the line lets you read the equilibrium oxygen partial pressure. Many Ellingham diagrams carry an oxygen partial pressure scale for this purpose, and some also have scales for H₂/H₂O and CO/CO₂ ratios. They can be used to estimate the conditions that prevent steel from oxidizing in a heat-treatment atmosphere, whether an oxide decomposes in vacuum, and the conditions under which Cr is selectively oxidized in stainless steel. However, the diagram shows equilibria between pure substances; reaction rates and differences in activity within alloys must be considered separately.
Answer: A. Gaseous O₂ is consumed to form a solid oxide, so the entropy change ΔS° is negative
An Ellingham diagram plots ΔG° = ΔH° − TΔS° for oxide formation per mol of O₂ against temperature, and if ΔH° and ΔS° are roughly constant, the slope of the line is −ΔS°. In metal oxidation, high-entropy gaseous O₂ is taken into a solid, so ΔS° is negative, around −0.2 kJ/(mol·K), and the slope is positive (upward). In other words, oxides become less stable at higher temperature. ΔH° depends little on temperature and only sets the height of the line. Diffusion and oxidation rates belong to kinetics and have nothing to do with the slope of the Ellingham diagram, which represents the equilibrium ΔG°. At the temperatures where a metal or oxide melts or boils, ΔS° changes and the line bends.
Answer: C. The gas increases from 1 mol to 2 mol, so ΔS° is positive, the line slopes downward, and carbon becomes a stronger reducing agent at higher temperature
In 2C + O₂ → 2CO, the gas increases from 1 mol of O₂ to 2 mol of CO, so ΔS° is positive (about +0.18 kJ/(mol·K)) and the ΔG° line slopes downward. As a result, at higher temperatures the CO line lies below the lines of many metal oxides, and carbon becomes able to reduce those oxides. This is why iron ore can be reduced with coke in a blast furnace. In C + O₂ → CO₂ the number of gas molecules does not change and ΔS° is nearly 0, so it is this reaction whose line is nearly horizontal. An upward slope is the case for metal oxidation, in which gaseous O₂ is taken into a solid.
Thermodynamically it can, since the ΔG° of the reaction in that direction is negative, but whether it actually proceeds depends on the reaction rate, the removal of products and the activities. Reactions between solids proceed poorly if contact is poor, and they can stop when the oxide formed covers the surface. There are also examples, such as the thermite reaction, that proceed all at once with a large release of heat.
Because when the amount of oxygen is the same, taking the difference between 2 formation reactions cancels the O₂, and the vertical distance between the lines is directly the ΔG° of the redox reaction. If it were drawn per mol of oxide, the number of oxygen atoms would differ with the formula, and the lines could not be compared directly. Diagrams for sulfides and nitrides are drawn per mol of S₂ or N₂ on the same idea.
No. An Ellingham diagram shows the equilibrium ΔG° and only indicates the thermodynamic tendency for an oxide to form. Aluminum corrodes little even though its ΔG° is strongly negative, because a dense oxide film blocks diffusion; the rate is set by kinetics such as diffusion through the film.
An Ellingham diagram plots ΔG° = ΔH° − TΔS° for oxide formation per mol of O₂ against temperature, and if ΔH° and ΔS° are roughly constant, the slope of the line is −ΔS°.
In an Ellingham diagram, what is the main reason why the ΔG…This corresponds to the Al₂O₃ line lying below the Cr₂O₃ line in the Ellingham diagram, and it is the principle of making metallic chromium by the aluminothermic process.
At 1000℃, take ΔG° for 4/3 Al + O₂ → 2/3 Al₂O₃ as −850 kJ/m…Activity Chemical potential Entropy of mixing Equilibrium oxygen partial pressure Gibbs energy Regular solution Spinodal decomposition