Calculating PM3 Heat of Formation D f H o

Testing Lewis Formulas and VSEPR Models with Quantum Theory
John Keller, Department of Chemistry and Biochemistry, Univ of Alaska Fairbanks
Calculating the PM3 Heat of Formation ∆fHo
The Heat of Formation ∆fHo (also called Enthalpy of Formation) is a property of a chemical
compound that reflects the stability of the atoms in the compound, compared to their stability in
the pure elemental form. More exactly, it equals ∆rHo for formation of the compound from the
elements, per mole of the compound. In other words, say n is the coefficient of the product in the
balanced chemical equation. ∆fHo for the compound then is ∆rH/n, and the units are kcal/mol or
kJ/mol.
For example ∆fHo for hydrogen fluoride is defined by the following reaction:
H2 (g) + F2 (g) → 2 HF (g) ∆rHo = -130.2 kcal/mol-rxn
Thus, ∆fHo(HF) = -65.1 kcal/mol.
Heats of Formation are really useful because they can be used to calculate ∆rHo for any reaction, if
the ∆fHo‘s are known for all the reactants and products.1 If it turns out that a tabulated ∆fHo value is
not available, it is usually possible to calculate it using quantum theory. In fact, the parameters
used in semi-empirical theories like PM3 are adjusted to maximize the accuracy of ∆fHo
calculations, among other molecular properties. So for most “normal” compounds, PM3 comes
remarkably close to the experimental ∆fHo. James Stewart’s paper on PM6 (a newer version of
PM3) and the Supporting Material, have several tables comparing calculated with experimental
∆fHo, geometry, and dipole moment values for many compounds.2, 3 The PM3 method is part of the
MOPAC2012 package available to the public at http://chem.uaf.edu/facilities/WebMO.
To better understand the “PM3 Heat of Formation” output, let’s consider ∆fHo for HF. This is best
done by arranging the chemicals along an enthalpy scale where the top of the diagram is the “high
energy” region (less stable), and the bottom of the diagram is the “low energy” region (more
stable); see the diagram on p. 2. This type of diagram is seen in many places in chemistry: it is very
important to get a good feel for it. One useful analogy says that the high energy, less stable state
corresponds to the top shelf of a bookcase with a book teetering on edge, and the low energy,
more stable state is the floor!
You can look at the Heat of Formation calculation in two ways: the “birds-eye” view in which ∆fHo
has essentially two components (p. 2), and a more detailed “computational” view, which shows
how the molecular part of the ∆fHo calculation is obtained (p. 3).
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The reference point, or “zero energy”, on this energy scale is the energy of the elemental forms of
hydrogen and fluorine, that is, H2 and F2. All the other energy states are relative to these. In step 1,
going up in energy, we break apart each molecule to give separate H and F atoms. In step 2,each H
combines with an F, which is quite exothermic.
1. The energy required to form the atoms from the element molecule is labeled ∆fHo(atom). These
values have been measured previously, or calculated by very accurate ab initio methods, and they
are stored in the program files. This does not require much storage space since PM3 covers only 40
main group elements and 6 transition metals. The ∆fHo(atom) values are “per atom”, therefore
forming 2 H’s requires 2∆fHo(H).
2. This is the bond forming step. The energy change equals the negative of the energy required to
split HF into separate H and F atoms, the latter being ∆Hatom(HF). Actually two HFs are formed in
the diagram, so the total energy change is -2∆Hatom(HF). ∆Hatom(HF) is a positive number that is
called the “molecular atomization energy”: it is calculated —behind the scenes—before PM3
∆fHo(HF) can be reported. The negative of this, -∆Hatom(HF), is called the “molecular binding
energy”. (In the HyperChem program used at UAF, this is the number shown on the status line as
“E”.)
The Heat of Formation of HF, ∆fHo(HF), is then equal to ½ the difference in the energy changes for
the two steps shown in the diagram.
The stored values of ∆fHo(H) and ∆fHo(F) are 52.10 kcal/mol and 18.89 kcal/mol, respectively, and
PM3 obtains a value of 133.76 kcal/mol for ∆Hatom(HF). Thus, ∆fHo(HF) is -62.76 kcal/mol. The
experimental value determined by directly reacting H2 (g) + F2 (g), is -65.1 kcal/mol. Pretty good!
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Now, how is the HF atomization energy ∆Hatom(HF) calculated? This is done in four separate steps
as shown below. The molecular geometry, in this case the H-F bond distance of 0.938 Å, is fixed.
1. All the valence electrons are removed from the molecule, leaving in place the nuclei. The energy
change for this step is equal to the negative of the energy change for adding the electrons back into
the orbitals, -Eelect(HF), which is the sum of the energy levels of the occupied molecular orbitals.
PM3 determines the energies (E) and shapes (Ψ) of the HF molecular orbitals as part of its
approximate solution to the Schrödinger equation HΨ=EΨ.
2. The nuclear repulsion energy Enuclear(HF) is obtained by an electrostatic repulsion calculation
using Coulombs Law: E = kQ1Q2/r2. Here k is a constant, Q is the charge on the ion, and r is the
distance between the nuclei.
3 and 4. HΨ=EΨ is solved for each type of atom. This is similar to step 1, except with the opposite
sign. Here it is the energy change for adding the electrons into the atomic orbitals of each atom,
Eelect(H) and Eelect(F). The latter are equal to the sum of the energies of the occupied atomic orbitals.
Finally, ∆Hatom(HF) is the difference between the endothermic step 1, and the three exothermic
steps 2, 3 and 4.
WebMO does not report any of these numbers from the MOPAC2012 calculation, except ∆fHo(HF).
But if one looks at the full MOPAC2012 output,4 the stepwise values are provided: Eelect(HF)
is -12,220.33 kcal/mol; Enuclear(HF) is 1695.52 kcal/mol; and Eelect(H) and Eelect(F) are
respectively, -301.48 and -10,089.58 kcal/mol. When these are combined with the correct signs,
they equal 133.75 kcal/mol.
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References and Notes
1. This works because enthalpy H is a “state function”, which means that changes in H depend only on the
initial and final states of the system. It does not matter if H2 and F2 are first split up into atoms, then form
two HF’s, or if they undergo a chemical reaction by some other mechanism: the enthalpy change is the same
both ways. Hess’s Law calculations are another application of this principle.
2. Stewart, J. J. P., Optimization of Parameters for Semiempirical Methods V: Modification of Nddo
Approximations and Application to 70 Elements. J. Mol. Model. 2007, 13, 1173-1213, DOI: 10.1007/s00894007-0233-4.
3. Stewart, J. J. P., Optimization of Parameters for Semiempirical Methods V: Modification of Nddo
Approximations and Application to 70 Elements. Supplemental Material. J. Mol. Model. 2007, 13, n/a.
4. The four numbers in the last paragraph can be obtained using MOPAC2012. Three separate PM3 jobs
must be run using the UAF WebMO site:
(1) Optimize the HF molecule, and open the "Raw output" link. Near the end of the page the
"ELECTRONIC ENERGY" of HF is given as -529.91339 EV (or 12,220.33 kcal/mol), and the "CORE-CORE
REPULSION" is reported as 73.5233 EV (or 1695.52 kcal/mol). In the parlance of the above discussion,
these are Eelect(HF) and Enuclear(HF), respectively.
(2) Energy calculation for H atom, which gives the ELECTRONIC ENERGY as -13.07332 EV (or -301.48
kcal/mol). This is Eelect(H).
(3) Energy calculation for F atom, which gives the ELECTRONIC ENERGY as -437.51717 EV (or -10,089.58
kcal/mol). This is Eelect(F).
Note: 1 EV = 1 electron volt = 23.061 kcal/mol
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