Answer: C→CO reaction: solid+gas→gas, Delta_S > 0; most MO formations: solid+gas→solid, Delta_S < 0. Delta_G = Delta_H - T*Delta_S, so the C line decreases and MO lines increase with T.
- A C→CO reaction: solid+gas→gas, Delta_S > 0; most MO formations: solid+gas→solid, Delta_S < 0. Delta_G = Delta_H - T*Delta_S, so the C line decreases and MO lines increase with T
- B Carbon metal is generally assumed by convention to be far more chemically reactive than literally every other metal at every conceivable temperature in standard practice under most conditions encountered
- C Most of the various metal oxide lines are generally assumed by convention to remain permanently positioned above the carbon line at most temperature as frequently observed in practice
- D The observed slopes shown on the diagram for these lines are assumed by convention to have highly no underlying relationship to the entropy change of each reaction in many documented cases
Correct answer: A. C→CO reaction: solid+gas→gas, Delta_S > 0; most MO formations: solid+gas→solid, Delta_S < 0. Delta_G = Delta_H - T*Delta_S, so the C line decreases and MO lines increase with T
Explanation: For 2C + O<sub>2</sub> → 2CO: moles of gas increase (Delta_S > 0), so Delta_G decreases with T (more negative). For M + O<sub>2</sub> → MO<sub>2</sub>: moles of gas decrease (Delta_S < 0), so Delta_G increases with T.
An Ellingham diagram plots ΔG° of oxide formation against temperature for different metals; whichever line is LOWER (more negative ΔG°) at a given temperature reduces the oxide of any metal whose line sits above it - the basis of carbon reduction (Fe, Zn) vs electrolytic reduction (Al, Mg, Na) decisions.
Concept context
The science of extracting metals from ores and refining them for use. Covers concentration methods, reduction techniques, refining processes, and the thermodynamic principles that govern metal extraction.