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$$\begin{align} &\text{Aluminium oxide:} &\ce{4Al + 3O_2 &-> 2Al2O3} &(\Delta H_\mathrm f &= \pu{-1675 kJ/mol})\\[1.5em] &\text{Magnesium oxide:} &\ce{2Mg + O_2 &-> 2MgO} &(\Delta H_\mathrm f &= \pu{-602 kJ/mol})\\[1.5em] &\text{Iron (III) oxide:} &\ce{4Fe + 3O_2 &-> 2Fe2O3} &(\Delta H_\mathrm f &= \pu{-824 kJ/mol})\end{align}$$

Question:

  1. Why is the enthalpy of formation for aluminium oxide higher than that of magnesium oxide, even though magnesium is higher in the reactivity series?
  2. Why is enthalpy of formation for magnesium oxide lower than iron oxide even though its more reactive? Is this due to the number of bonds formed? Thus, magnesium is less exothermic as less bonds are formed?
  3. What determines reactivity? Why is magnesium more reactive than aluminium and why is aluminium more reactive than iron? Does it have anything to do with valency of the electrons?
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Because you are comparing apples with donuts. If I eat enough apples they will add more calories than that donut I gobbled for breakfast.

To compare food calories properly you need to match serving sizes. Similarly, to compare reaction energies fairly you need to use a common basis for the extent of reaction. For oxidation of metals the reactivity of the metals is usually compared by using one mole of oxygen as the common basis, as in Ellingham diagrams. The physical situation is if there is only a limited amount of oxygen available, which metal would that oxygen prefer to react with according to thermodynamics? (Not necessarily kinetics.)

So for the aluminum oxidation you should divide all the coefficients by three to get a reaction with one mole of oxygen, and then the enthalpy of formation is similarly divided by three. You would then get $-558\text{ kJ/mol }\ce{O2}$ versus $-602\text{ kJ/mol }\ce{O2}$ for magnesium oxidation.

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