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Explain the observations from the Ellingham diagram.

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1. For most of the metal oxide formation, the slope is positive. It can be explained as follows. Oxygen gas is consumed during the formation of metal oxides which results in the decrease in randomness. Hence, ∆S becomes negative and it makes the term, T∆S positive in the straight line equation.

2. The graph for the formation of carbon monoxide is a straight line with negative slope. In this case ∆S is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen gas. It indicates that CO is more stable at higher temperature.

3. As the temperature increases, generally ∆G value for the formation of the metal oxide become less negative and becomes zero at a particular temperature. Below this temperature, ∆G is negative and the oxide is stable and above this temperature ∆G is positive. This general trend suggests that metal oxides become less stable at higher temperature and their decomposition becomes easier.

4. There is a sudden change in the slope at a particular temperature for some metal oxides like MgO, HgO. This is due to the phase transition (melting or evaporation).

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