Thermodynamics of materials
Which statement correctly describes the line for the carbon reaction 2C(s) + O₂(g) → 2CO(g) in an Ellingham diagram?
- ALike C + O₂ → CO₂, ΔS° is nearly 0, so the line is horizontal
- BLike metal oxidation, ΔS° is negative, so the line slopes upward and carbon becomes a weaker reducing agent at higher temperature
- CThe 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
Show answer and explanation
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.