Is it necessary criteria that we draw Ellingham diagram for only one mole of oxygen consumed for a given reaction

In my textbook and many other reference text shows Ellingham diagram for only one mole oxygen consumed. Why is this done?




(Disclaimer: during composition of this answer I had many issues with typing and syntax, and I may have lost some battles. Improvements in these areas are much encouraged.)


Yes, a fair thermodynamic comparison of the oxidation processes requires using one mole of $\ce{O2}$ in the reaction.


For example, if we look up free energies of formation for magnesia and alumina at $25°\text{C}$, we get (from (<https://en.wikipedia.org/wiki/Standard_Gibbs_free_energy_of_formation>):


$\ce{MgO}: -569.3 \text{kJ/mol}$


$\ce{Al2O3}: −1582 \text{kJ/mol}$


It looks like the free energy release for alumina formation far exceeds that for magnesia formation. But now look at the balanced chemical reaction equations:


$\ce{Mg + (1/2)O2 -> MgO}$


$\ce{2Al + (3/2)O2 -> Al2O3}$


Thus the free energy data, each reported per mole of the compound, corresponds to one-half mole $\ce{O2}$ for magnesia versus 3/2 mole $\ce{O2}$ for alumina. So to match the oxygen consumption and get the fair thermodynamic comparison, we have to multiply the quoted bagnesia value by $2$ and the quoted alumina value by $2/3$


$\ce{MgO}: -1139 \text{kJ/mol } \ce{O2}$


$\ce{Al2O3}: -1055 \text{kJ/mol } \ce{O2}$


So when we match oxygen bases, we find that magnesia has the lower free energy of formation.


When we reduce free energy of firmation to one mole of $\ce{O2}$, this free energy becomes equal to $RT\ln(p\_{O\_2})$ or $RT\ln(f\_{O\_2})$, where thecargument of the logarithm is the equilibrium ideal-gas pressure or **fugacity** of $\ce{O2}$. The significance of the magnesium/magnesia couple being lower than the aluminum/alumina couple on the Ellingham diagram is that the former drives oxygen pressure or fugacity down to a lower level and thus magnesium is a thermodynamicalky more powerful reactant for aluminum than oxygen at $25°\text{C}$.


You'll also see that the magnesium/magnesia couple is actually lower than the aluminum/alumina couple only up to a temperature of about $1400°\text{C}$. At hotter temperature aluminum becomes more powerful thermodynamically than magnesium. Thus occurs because magnesium is rather volatile and boils off into a gas. The large entropy increase in the metal when this happens makes reaction with oxygen to form the oxide (which is still solid) less favorable. Aluminum does not boil off at temperatures commonly used in industry and so retains its fill reducing power. In contrast, with iron it is the oxide that boils off, so that the increased entropy of the gaseous oxide makes iron eventually catch up with the normally more reactive zinc.


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