I'm doing a write-up for this for my students:
Source: Amirav Research Group (Facebook)
The energy of formation of isopropyl alcohol is $-318.2~\mathrm{kJ/mol}$ (kilojoules per mole), that of acetone is $-249.4~\mathrm{kJ/mol}$, so $68.8~\mathrm{kJ/mol}$ are gained on this end.
But I'm getting confused about the hydrogen end. I'm seeing that The energy of formation of $\ce{H+}$ is $0~\mathrm{J/mol}$ (it can be treated as a free proton), and that of $\ce{H2}$ is also $0~\mathrm{J/mol}$. Of course you would say that oxygen has $0~\mathrm{J/mol}$ as well, so since water has enthalpy of formation of $241.83~\mathrm{kJ/mol}$, I'd expect it to yield that much energy in the combustion.
But hydrogen gas has an “energy density” as a fuel of $71~\mathrm{kJ/mol}$. So the treatment that $\ce{H2}$ and $\ce{O2}$ have $0~\mathrm{J/mol}$ enthalpy of formation and water has enthalpy of formation of $242~\mathrm{kJ/mol}$ seems to not square with that. Where's the rest of the energy?
Particularly, in the diagram above you see the charge separation as the electron pair moves to the hydronium, producing hydrogen. But a charge separation (J = coulomb volt) is energy, so moving the electrons to the hydrogen has to be adding energy to them, which seems to indicate that the $\ce{H+} = 0~\mathrm{J}$ and $\ce{H2}$ also = $0~\mathrm{J}$ things is wrong. But I don't know where to go from here, does anyone feel confident helping with this?
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. Only use this environment for chemical formulae. I don't want to check every edit you make. $\endgroup$