||Computational Chemistry Comparison and Benchmark DataBase
Release 22 (May 2022) Standard Reference Database 101
National Institute of Standards and Technology
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How to get an enthalpy of formation from ab initio calculations
- Choose a reaction with experimentally known enthaplies of formation for all species except the one of interest.
- why do we need a whole reaction?
- what makes a good reaction?
- Calculate the energies for all species in the reaction at the same level of theory.
- What level of theory should I use?
- Optimize the geometries.
- Calculate vibrational frequencies. (Why do we need the vibrational frequencies?)
- (Optional) Calculate the electronic energy for all
optimized geometries at some higher level of theory.
- (Optional) Convert from 0K to 298.15K (or some other temperature).
- Combine the various energies to get a reaction energy.
- For each species we want the electronic energy plus the vibrational zero point energy.
- The reaction energy is the sum of the product
energies minus the sum of the reactant energies.
Use the enthalpy computation form, which has an example.
Using ab initio methods we can obtain ideal-gas enthalpies of formation,
though not usually directly.
The ab initio calculation provides us with an energy
for the process of separating the molecule into bare nuclei and electrons.
For example if we are interested in the enthalpy of formation of ethene
(C2H4 ) we can run a calculation, say
HF/6-31G* and get and energy for the optimized structure of -78.031718 hartrees. This
energy is for the (not very useful) reaction:
If we knew the experimental enthalpy of
formation of each species of the right side of this reaction we could obtain the
enthalpy of formation of difluoromethane. However it is better to use a reaction
that keeps the electrons and nuclei together. This helps cancel the systematic
errors in the calculations.
||C2H4 = 2C6+ + 4H+ + 16e-
Experimentally a reaction energy is measured.
From this reaction energy and known enthalpies of formation of
all but one of the species in the reaction, the enthalpy of formation of the unknown can be determined.
For example if we measured the enthalpy of reaction for
ethene reacting with oxygen to yield water and carbon dioxide.
We find a reaction enthalpy of -1,325.1 kJ/mol.
The relation between the reaction enthalpy and the individual enthalpies of formation is just
Σ ΔfH(product) -
Σ ΔfH(reactant). For
||C2H4 + 3O2 =
2H2O(gas) + 2CO2(gas)
Plugging in the known enthalpies of formation of H2O,
CO2 and O2 (The enthalpy of formation of O2 is
defined to be 0 kJ/mol):
||-1325.1 kJ/mol = 2 ΔfH(H2O) +
- 3 ΔfH(O2)
Rearranging this equation yields the enthalpy of formation of ethene
ΔfH(C2H4) = 61.1 kJ/mol.
||-1325.1 kJ/mol = 2 (-238.9 ) + 2 (-393.1) -
- 3 ΔfH(0)
We use a similar a similar procedure for determining the enthalpy of formation
of a species using ab initio calculations. We get the energy of the reaction by
calculating the ab initio energy of each species. Using HF/6-31G* calculated
energies we have:
||vibrational zpe (cm-1)
||total energy (hartrees)|
The reaction energy is 2 (-75.990175) + 2 (-187.622762) - (-77.982677) - 3
-0.401706 hartrees (-1054.66 kJ/mol).
Using this reaction enthalpy and the experimental values for the enthalpy of
formation of H2O, CO2 and O2 we have, similar to Eq. 2
This yields the enthalpy of formation of ethene
ΔfH(C2H4) = -209.34
kJ/mol. This poor result (we're off by 270 kJ/mol!) can be improved by the
choice of a better reaction. Ideally an isodesmic reaction (one that conserves
bond types) is best, but even an isogyric reaction
(one that conserves spin - O2 is a triplet species)
helps to cancel out systematic errors in the ab initio calculations.
||-1054.66 kJ/mol = 2 (-238.9 ) + 2 (-393.1) -
- 3 ΔfH(0)
An isodesmic reaction involving ethene could be
between ethene and propane to yield ethane and propene.
The number of each bond tpye is conserved: 1 C=C bond, 2 C-C bonds, 12 C-H bonds.
Using energies from an HF/6-31G* calculation we obtain a reaction energy of
-13.1 kJ/mol. Using experimental values for the heat of formation of
C3H8(-82.40), C2H6(-68.38), and
can obtain the enthalpy of formation of ethene of 61.82 kJ/mol. This value
is much closer to the reported experimental value of 60.99 kJ/mol.
||C2H4 + C3H8 = C2H6 + C3H6
In all the above examples we have used experimental enthalpies of formation
reported at 0 K (zero Kelvin). Usually it is more useful to predict enthalpies
of formation at room temperature (298.15 K) or some other temperature. To do
this we need to convert our 0 K enthalpy of formation to an enthalpy of
formation at our desired temperature.
This is described on the next page.