Computational Chemistry

wherenis a number large enough to represent thesubstancegraphite rather than
just some carbon oligomer. All the species in the figure will then be increased in
number by a factor ofn, but division by this common factor will still give us
Eq.5.190. Another point is that although hydrogen and oxygen are solids at 0 K, we
are considering isolated molecules being atomized.
To calculateDH-f0ðCH 3 OHÞwe need the 0 K heat of formation of C, H and O
atoms, i.e. the atomization energies of graphite, molecular hydrogen, and molecular
oxygen, and the 0 K atomization energy of methanol. The atomization energies of
hydrogen and oxygen can be calculated ab initio, but not that of graphite, which is a
very big “molecule”. For consistency we will use experimental values of all three
elemental atomization energies, as recommended [ 197 ]. From Eq.5.181, the 0 K
atomization energy of methanol is simply the ab initio energies of its constituent
atoms minus the ZPE-corrected ab initio of methanol:

DH-a0ðCH 3 OHÞ¼DEtotal0KðCð^3 PÞþ4Hð^2 SÞþOð^3 SÞÞ#DEtotal0KðCH 3 OHÞð 5 : 191 Þ

Experimental values ofDH-f0Cð^3 PÞ,DH-f0Hð^2 SÞ, andDH-f0Oð^3 SÞ(as well asDH-f0
for other atoms, and references to more extensive tabulations) are given in [ 197 ]; in
kJ mol#^1 :

C 711.2

H 216.035

O 246.8

To calculateDH-a0ðCH 3 OHÞwe need (Eq.5.191)DEtotal0K for C, H and O atoms in
the states shown and for methanol. First we use the old G2 method, just for
comparison with the value in [ 197 ]. As for [ 197 ], we’ll use Gaussian 94 [ 198 ].
We get these values (hartrees):

C #37.78430

H #0.50000

O #74.98203

CH 3 OH #115.53490

From Eq. (5.191)

DH-a0ðCH 3 OHÞ¼# 37 : 78430 þ 4 ð# 0 : 50000 Þ# 74 : 98203 #ð# 115 : 53490 Þh

¼# 114 : 76633 þ 115 :53490 h¼ 0 : 76857 ' 2625 :5 kJ mol#^1

¼ 2017 :88 kJ mol#^1

From Eq. (5.190)

DHf0-ðCH 3 OHÞ¼ 711 : 2 þ 4 ð 216 : 035 Þþ 246 : 8 # 2017 :88 kJ mol#^1

¼ 1822 : 1 # 2017 :88 kJ mol#^1 ¼# 195 :7 kJ mol#^1

316 5 Ab initio Calculations