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In an Ellingham diagram, at the boiling point of a metal oxide and at the boiling point of a metal respectively, the slope of the curve changes as:
- Increasing slope and decreasing slope respectively.
- Decreasing slope and increasing slope respectively.
- Increasing slope and increasing slope respectively.
- Decreasing slope and decreasing slope respectively.
Correct answer: Decreasing slope and increasing slope respectively.
Solution
In the Ellingham diagram, the slope of the line (deltaG° vs T) equals -deltaS°. For oxidation reaction 2M + O2 -> 2MO: deltaS° = S°(2MO) - 2S°(M) - S°(O2). At the boiling point of the metal oxide: S°(MO) increases sharply (liquid to gas), making deltaS° more positive => slope (-deltaS°) decreases. At the boiling point of the metal: S°(M) increases sharply (solid/liquid to gas), making deltaS° more negative => slope (-deltaS°) increases.
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