Essentials of Computational Chemistry

10.4 STANDARD-STATE HEATS AND FREE ENERGIES 367
is followed by the enthalpy change for combining them into the molecular structure, which
is the negative of the enthalpy of atomization. As an example, the 0 K heat of formation
of 2-butanone (methyl ethyl ketone, a widely used industrial solvent) is tabulated as
−51.9 kcal mol −1 . The molecule is composed of eight atoms of hydrogen, four of carbon,
and one of oxygen. The enthalpy cost to split 4 moles of hydrogen gas to create 8 moles
of hydrogen atoms at 0 K is 413.5 kcal, the cost to strip 4 moles of carbon atoms from an
infinite graphite block at 0 K is 1066.8 kcal, and the cost to split one-half mole of oxygen
gas to create 1 mole of oxygen atoms at 0 K is 59.0 kcal. The 0 K atomization enthalpy of
1 mole of 2-butanone is 1591.2 kcal. Thus, the tabulated 0 K heat of formation cited above
is determined as (413.5 + 1066.8 + 59.0 − 1591.2).
An important technical point that must be mentioned here is that some attention must be
paid to the states of the atoms to ensure that the difference between the molecular atomization
enthalpy and the enthalpies of formation of the atoms is carried out consistently. When one
atomizes a species like 2-butanone experimentally, each resulting atom will typically contain
a number of spin-unpaired electrons equal to its number of formal bonds in the molecule
(because each bond is being ruptured into two unpaired electrons, one on each atom formerly
involved in the bond). Thus, for instance, each carbon atom will have four unpaired electrons,
corresponding to the quintet S ( 5 S) term of the atom. However, this is not the ground state
of the carbon atom (the ground state is 3 P), so the value of 1066.8 kcal noted above for the
enthalpy change associated with stripping 4 moles of carbon atoms from a graphite block
may, under some experimental conditions, be measured as 680.6 kcal, the cost to generate
the 4 moles of C atoms in their 3 P ground state, plus 4 × 96.5 kcal mol −1 , where the latter
energy is the molar enthalpy cost to excite an atom of C from 3 Pto 5 S.
When one speaks of a computational atomization energy for a molecule, it should be
carefully specified whether the energies of the product atoms are being computed in their
ground states or in excited states that may be more convenient to work with for one reason
or another. This specification is also critical to determining a computed heat or free energy
of formation, as described next.
10.4.1 Direct Computation
Direct computation of a molecular heat or free energy of formation is something of a
misnomer, since it would imply computing the difference in H or G for some molecule
compared to the reference elemental standard states. Such a calculation might readily be
imagined for a molecule like HF, because the standard states of H and F are gaseous
diatomics. However, carrying out a high-level quantum mechanical calculation on an infinite
block of graphite is another matter altogether. As a result, almost all so-called direct
computations of heats of formation are carried out as illustrated in Figure 10.1. All quantities
in the large inset region are computed relative to the theoretical zero of energy (all
nuclei and electrons infinitely separated and at rest). To determine a standard-state molecular
thermodynamic quantity, the computed energy difference between the molecule and its
constituent atoms is added to the experimental thermodynamic value determined for the
identical atoms. For instance, if we wanted to predict the 298 K heat of formation above