IndependentBoffin - 23-5-2011 at 00:10
Referring to:
https://engineering.purdue.edu/~propulsi/propulsion/comb/pro...
Can someone please explain to me why there is a switch in the rankings of heat of formation produced per gram of reactant when we compare Al/Mg
fluorides vs. Al/Mg oxides? I don't have a university chemistry background (just Year 12) so if the explanation is complicated you'll have to point me
in the direction of a reference available online for me to read 
So, for the flourides, Mg > Al hands down:
1) Mg + F<sub>2</sub> -> MgF<sub>2</sub> ΔH<sub>f</sub> = -2862 cal/g
2) 2Al + 3F<sub>2</sub> -> 2AlF<sub>3</sub> ΔH<sub>f</sub> = -844 cal/g
Why is the heat of formation of AlF<sub>3</sub> so poor? I'd have expected it to be much more comparable to MgF<sub>2</sub>.
For the oxides, Al > Mg:
1) 4Al + 3O<sub>2</sub> -> 2Al2O<sub>3</sub> ΔH<sub>f</sub> = -4000 cal/g
2) 2Mg + O<sub>2</sub> -> 2MgO ΔH<sub>f</sub> = -3610 cal/g
Oops corrected chemical formulae subscripts.
[Edited on 23-5-2011 by IndependentBoffin]
AndersHoveland - 23-5-2011 at 15:01
Remember, the heats of formation depend not only on the energy released, but also on the mass of the reactants.
In the case of aluminum and oxygen, the oxygen is relatively light. The answer to your question can mostly be explained by the weight of the reactants
and the reactant ratios in the equations.
However, another factor is that the formation of aluminum fluoride is not quite as favorable as that of magnesium fluoride because there are
three highly electronegative fluorine atoms withdrawing electric charge away from each aluminum atoms.
[Edited on 23-5-2011 by AndersHoveland]