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I am currently working on an investigation where I am burning isomers of butanol to test for the heat of combustion values to see which fuel is most efficient. The fuels I am testing are 1-butanol, 2-butanol, 2-methylpropan-1-ol (isobutanol) and 2-methylpopan-2-ol (tert-butyl alcohol).

I have researched that the heat of combustion values for these fuels are as follows:

  • 1-butanol: −2671 kJ/mol
  • 2-butanol: −2661.1 kJ/mol
  • Isobutanol: −2662.6 kJ/mol
  • Tert-butyl alcohol: −2644.8 kJ/mol

So evidently, my experimental results should also show similarity in the ordering of greatest to lowest heat of combustion values.

However, I am wondering why isomers of butanol have different heat of combustions. I know it has to do with the shape of the molecules and whatnot, but how does this affect the energy released? What about bond angles?

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The heat of combustion (or any reaction) is the energy that is set free when you make the products ($\ce{CO2}$ and $\ce{H2O}$ here, in their respective stochiomeric amounts) out of single atoms (this is the same for all isomers), minus the energy needed to break all bonds in the educts. The latter is different for the different isomers, because the energy in a chemical bond not only depends on the two partners but also on their neighbours (the "shape" of the molecules).

The usual procedure for this calculation is to not go over single atoms but the elements (and compounds) in their most stable form at standard conditions. These energies a called standard enthalpies of formation. There are long tables listing them for all possible compounds.

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