Bond Enthalpy and mean bond enthalpy


The standard molar enthalpy change of bond dissociation (DHd is the energy change when 1 mole of bonds is broken, the molecules and resulting fragments being in the gaseous state at 298K and a pressure of 100kPa.


  • This energy refers to a specific bond in a molecule, but if a molecule has 4 of the same bond (eg the C-H bonds in methane), then different dissociation energies can occur.

CH4(g) => CH3(g) + H(g)             DHd = +427 kJ mol-1 

CH3(g) => CH2(g) + H(g)              DHd = +371 kJ mol-1  

So it is much more useful to know the average amount of energy needed to break a particular bond. In this case, the process of breaking all the bonds in methane ending up with gaseous atoms.

So this process could be written as:

The enthalpy change for this reaction is +1646 kJ mol-1 , so the average bond enthalpy is +1646 / 4 = +412 kJ mol-1 . They can be looked up in data tables.

It is important to stress that these are mean or average bond enthalpies. If Average bond enthalpies are used to calculate an enthalpy change, the answer will be slightly out compared to a result obtained by other methods.

Why are they useful?

Their main use is in working out enthalpy changes for reactions. If we know the amount of energy needed to break a bond (endothermic), and the amount of energy we get back when a new bond forms (exothermic), then we can quite easily work out an approximate (because of the average nature of the bond enthalpies) value for the enthalpy change!

Remember: Breaking bonds requires energy (Endothermic) while making bonds releases energy (Exothermic).

Don't be scared by this equation below, it is just the correct terminology as used by chemists for working out enthalpy changes from bond energies.

This basically means that you add up all the energies of the broken bonds add up all the energies of the bonds that are reformed and subtract one from the other. Its another version of Hess' Law. Similar to the one that can be used when you know all the enthalpies of formation for the substances in a reaction.

An example.

The complete combustion of propane can be represented by the following equation:

or we could redraw it to represent the bonds present:

We now need to work out how many of each bond type we have broken.

  • 8xC-H

  • 2xC-C

  • 5xO=O

And then how many bonds have been formed!

  • 6xC=O

  • 8xH-O

So using data tables we can look up then average bond enthalpies from, and calculate the enthalpy change of the reaction.

Bond Type

Average bond enthalpy /kJ mol-1

C-H +413
C-C +347
O=O +498
C=O +805
H-O +464

Notice they are all endothermic.

So we can now do the sum, remember, sum of bonds broken - sum of bonds formed.

DHr= [(8x413)+(2x347)+(5x498)] - [(6x805)+(8x464)]

       = - 2054 kJ mol-1

The value for the enthalpy of combustion of propane from the data table is -2219kJ mol-1

This apparent error is due to fact that we use Average Bond Enthalpies in our calculations and the fact that the values above relate to the gaseous state, while the standard combustion state of water is liquid. If we allow for this, we get a value of -2226 kJ mol-1 which is pretty close to the value obtained above. The remaining difference must be down to the average bond enthalpy factor.

Now try the Energetics online test.