# Formation: So much heat, where does it all go?

I was comparing heat of formation numbers with specific heat values, and the numbers look really weird.

Take a typical example like magnesium fluoride. Its $\Delta H$ is -1124 kJ/mol and its heat capacity is almost certainly less than 0.1 kJ/mol K. Divide and you get at least 11000 Kelvins!

So the heat of formation doesn't seem to go into raising the temperature of the product. It can't be going into the surroundings either; otherwise the reaction would vaporize anything nearby. Where does it go, or what's wrong with my reasoning?

Edit: OK so let's not use a fluorine compound. Take aluminum nitride. -318 kJ/mol. Are we really going to suggest aluminum burns in nitrogen at over 3000 K?

Edit: Si3N4: -745 kJ/mol.

Edit: Regarding the article on adiabatic flame temperature: aluminum burns in oxygen at 4000 K. The $\Delta H$ of Al2O3 is -1680 kJ/mol. 1680/(4000-298) = 0.45 kJ/mol K. This would be a fantastic heat capacity whether we're talking constant pressure or constant volume.

• Well, I can certainly imagine that magnesium "burns" pretty extremly in a fluorine atmosphere. And magnesium burns with >3000°C in air. – DSVA Jul 2 '17 at 19:42
• en.wikipedia.org/wiki/Adiabatic_flame_temperature – Mithoron Jul 2 '17 at 20:41
• Magnesium burns vigorously in carbon dioxide with a very hot flame. I hesitate to think just how well it burns in fluorine. But I guess it might produce a lot of heat. – matt_black Jul 2 '17 at 20:53
• For these extremely exothermic reactions, a lot of energy is carried away in the form of photons, limiting how hot it can get. Recall that the power dissipated by a black body radiator is proportional to $T^4$, so it rises very quickly as the reaction becomes hotter. – Nicolau Saker Neto Jul 2 '17 at 22:53
• Thanks, but I think there is some more to the question than what I wrote. From what I understand, these adiabatic flame temperatures should in principle not be affected by emission of thermal radiation, but they are all significantly smaller than what your back-of-the-envelope calculation suggests – Nicolau Saker Neto Jul 3 '17 at 10:21