Thermodynamics of materials
In an Ellingham diagram, what is the main reason why the ΔG° line for the oxidation of most metals, 2M(s) + O₂(g) → 2MO(s), slopes upward, with ΔG° becoming larger (closer to 0) at higher temperature?
- AGaseous O₂ is consumed to form a solid oxide, so the entropy change ΔS° is negative
- BAs temperature rises, the heat ΔH° released by oxidation grows
- CAs temperature rises, faster diffusion in the metal increases the oxidation rate
Show answer and explanation
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.