Calculating Molar Heat of Combustion

 
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Using Heat of Formation

A few pages away from the place where I looked up 1409.4 kJ/mol as the heat of combustion of ethanol, I can find the heats of formation of ethanol (-235.1 kJ/mol as a gas) of carbon dioxide (-393.5 kJ/mol) and of water (-285.8 kJ/mol as a liquid, -241.8 kJ/mol as a gas). These numbers are all negative, indicating that heat is released when all of these compounds are formed from their elements. Oxygen has a heat of formation of zero, since O2 (gas) is defined as the standard state for oxygen as an element.

To find the heat of combustion of ethanol, therefore, we add up all the heats of formation on the product side of the reaction:

2 × CO2 = 2 × -393.5 kJ/mol = -787.0 kJ/mol
3 × H2O = 3 × -285.8 kJ/mol = -857.4 kJ/mol

-1644.4 kJ/mol

And subtract all the heats of formation on the reactant side (just ethanol, -235.1 kJ/mol). This gives -1409.3 kJ/mol, indicating that 1409.3 kJ/mol of heat is generated in this reaction.

The basic principle we are using here is that energy cannot be created or destroyed in a process. To do the calculation, we pretend that broke all the reactants up into their standard states (i.e. O2(g) H2(g) etc) and then reformed them from these standard states into the products of the reaction. Since energy must be conserved during this process, the total energy released by taking the "scenic" route via the standard states must be the same as the total energy released by going directly from reactants to products.

Using Bond Energies

It is also possible to make very good guesses of the magnitude of the heat of combustion by considering which bonds are broken and which are formed. This would be useful if (as sometimes happens) you are considering one of the approximately twenty million organic compounds not featured in handbooks of thermodynamic data.

Bond Heat of Bond Formation (kJ/mol)
C-H -414
C-C -346
C-O -358
C=O (double bond) -804
O-H -463
O=O (double bond) -498

These heats of bond formation have been derived by measuring the heat of formation of many thousands of compounds. They are the "average" energy released when the bond a "typical" bond is formed. Since the bonds in every molecule are slightly different due to the presence of other groups nearby in the molecule, these values are not exact for every molecule, but they do allow us to construct quite useful estimates of the heat of a reaction (e.g. a combustion reaction).

Computational chemists are also on the point of being able to predict them from first principles, and have been for many years. Since these are average values (the heat of formation of a compound is not exactly divisible into unique bond contributions) they should not give an answer as exact as we obtained using the heat of formation approach, but let's try:

Once again, we are using conservation of energy to pretend we are taking a "scenic route" from the reactants to the products. This time, however, we go via individual atoms rather than "standard states". First, we work out how much energy it would take to break all the bonds in the reactants, then we work out how much energy we would get back if we formed all the bonds on the products side. Since energy is conserved, this is the same as going directly from the products to the reactants. (But we don't know the exact values for the bond energies, just averages, so the answer will not be exact.)

On the product side, we have

4 × C=O double bonds = 4 × -804 kJ/mol = -3216 kJ/mol
6 × O-H single bonds = 6 × -463 kJ/mol = -2778 kJ/mol

-5994 kJ/mol

On the reactant side,

5 × C-H single bonds = 5 × -414 kJ/mol = -2070 kJ/mol
1 × C-C single bond = 1 × -346 kJ/mol = -346 kJ/mol
1 × C-O single bond = 1 × -358 kJ/mol = -358 kJ/mol
1 × O-H single bond = 1 × -463 kJ/mol = -463 kJ/mol
3 × O=O double bonds = 3 × -498 kJ/mol = -1494 kJ/mol

-4731 kJ/mol

Taking the reactants away from the products leaves -1263 kJ/mol; pretty poor, you might think, even for government work. However, we have not yet taken into account the additional heat change involved in the condensation of water from gas to liquid. This is -44 kJ/mol, and as we formed three molecules of water, the true heat of combustion ought to be -1263 + (3 × -44) kJ/mol, or -1395 kJ/mol. And that's not too bad. (Indeed, I reckon it's pretty good going, considering we were using average data for a typical C-C bond etc... whatever that is!)

Now all we need is a convenient way to estimate heats of vaporisation/condensation...