Why are there different values for enthalpy of combustion, depending on the calculation method? Take for example the combustion of ethanol: $$\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O (l)}$$

If I calculate this using enthalpy of formation:

\begin{array}{c|c} \text{Substance} & \Delta H^\circ /\pu{(kJ/mol)} \\ \hline \ce{H2O(l)} & -285.8 \\ \ce{CO2(g)} & -393.5 \\ \ce{C2H5OH(l)} & -277.0 \end{array}

\begin{align} \Delta H &= \left[2\cdot(\pu{-393.5 kJ/mol}) + 3\cdot(\pu{-285.8 kJ/mol})\right] -\left[\pu{-277.0 kJ/mol} + 3\cdot(\pu{0 kJ/mol})\right]\\ &= \pu{-1367.4 kJ/mol} \end{align}

If I calculate this using bond energy:

\begin{array}{c|c} \text{Bonds} & \Delta H^\circ /\pu{(kJ/mol)} \\ \hline \ce{C-C} & 347 \\ \ce{C-H} & 414 \\ \ce{O-H} & 464 \\ \ce{C-O} & 351 \\ \ce{C=O(CO2)} & 803 \\ \ce{O=O} & 498 \end{array}

\begin{align} \Delta H &= \pu{347 kJ/mol} + \pu{351kJ/mol} + \pu{464 kJ/mol} + 5\cdot(\pu{414 kJ/mol}) \\ &\quad + 3\cdot(\pu{498 kJ/mol}) + 4\cdot(\pu{-803 kJ/mol}) + 6\cdot(\pu{-464 kJ/mol})\\ &= \pu{-1270 kJ/mol} \end{align}

Why is there such a big difference between them?


3 Answers 3


Average bond enthalpies (note the word "average") are calculated in a different way from formation enthalpies. Formation enthalpies are well-defined and precise, meaning that two different people could measure them the same way and get the same answer. Average bond enthalpies are averages over many different types of bonds.

For example, there is only one way to make $\ce{CH4}$, but there are many different molecules with $\ce{C-H}$ bonds. The enthalpy of dissociation of a $\ce{C-H}$ bond in $\ce{CH4}$ is different from that in $\ce{C2H6}$, and in $\ce{C6H6}$, and really in any other molecule you can think of. To find average bond enthalpies, a selection of a large number of those bonds in many different molecules is taken, and the enthalpy of dissociation for each is measured. Then, the average (mean) is recorded. If you look at some different charts of bond enthalpies, you will notice two things: First, there are some numbers marked "exact" - $\ce{H-H}$, $\ce{Cl-Cl}$, $\ce{F-F}$, for example. This is because there is only one molecule with each of those types of bonds. The other thing you will notice is that depending on which chart you are using, many of the numbers will be slightly different. This is because there is no universal, unchanging standard describing which molecules are used to determine each bond - it depends upon what the people making the chart decided to use.

Because of this difference, when making predictions, average bond enthalpies are less accurate than formation enthalpies. In practice, they are used for quick estimations, or as a last resort when formation enthalpies are not available.

When I taught chemistry, I would have my students measure reaction enthalpies in lab, then calculate them using formation enthalpy tables and average bond dissociation enthalpy tables. The goal was to show them that average bond enthalpies can be reasonably close for rough estimates, but that formation enthalpies are much better in practice. It also showed them that accurately measuring reaction enthalpy in the lab is difficult!


Because calculation with bond energies are notoriously imprecise.

Those bond energies that you use do not reflect the exact C-C bond energy in ethanol, for example. It is merely an "average" C-C bond energy taken from a wide variety of organic compounds. Obviously if you use it to calculate the enthalpy change of a reaction specifically involving ethanol, your answer is not going to be correct.

Using bond energies is often only good enough for a ballpark estimate. You should never place much trust in those numbers - certainly not as much as you seem to have in your question.


It's because bond enthalpy calculations apply to molecules in their gas phase!

That is, you would have to account for enthalpy change of vaporization of ethanol.

Enthalpy change of formation: $$\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O (l)}$$

Gas phase reactio for bond energy calculations: $$\ce{C2H5OH(g) + 3O2(g) -> 2CO2(g) + 3H2O (g)}$$

If we take the enthalpy change of vaporization of water and ethanol as $41 \ \mathrm{kJ\ mol^{-1}}$ and $42 \ \mathrm{kJ\ mol{-1}}$, we get $\Delta H_f \approx -1270 - 41 - 42 \approx -1350 \ \mathrm{kJ\ mol^{-1}}$

Close enough!

I made a similar mistake here:

Discrepancy between calculated and experimental enthalpy change of formation? Hydrogen Iodide

  • 1
    $\begingroup$ Note that the bond enthalpy change is the major but not the only part of the reaction enthalpy, what can be easily seen in reaction enthalpy temperature dependence. $\endgroup$
    – Poutnik
    Commented Aug 15, 2023 at 11:44

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