Physical Chemistry , 1st ed.

difference between the dissociation limit and the very bottom of the electronic

potential well is labeled De. However, in reality a molecule will also be in its

(presumably) lowest vibrational energy state, which has some nonzero mini-

mum energy. That zero-point energy is ^12 h ,where is the classical vibrational

frequency of the diatomic molecule. The difference between the dissociation

limit and the ground-vibrational state of the electronic potential energy well is

labeled D 0. Thus, the relationship between Deand D 0 is

DeD 0   ^12 h (18.7)

We will be using the De-defined value for the dissociation energy. However,

many tables list D 0 , so it is important to keep track of which value is being re-

ported. Equation 18.7 can be used to go from one definition to the other. Both

values for dissociation energy are typically given as positive numbers, with D 0

having the slightly smaller magnitude. Values are also usually given in terms of

kilojoules per mole of molecules. Remember that in partition functions, val-

ues for individual moleculesmust be used.

For qelectof diatomic molecules, the benchmark for electronic energy is the

dissociation limit. This means that the dissociation limit is arbitrarily assigned

a value of zero energy. At the minimum of the potential energy surface, the

electronic energy of the molecule is therefore De. (It is negative because it is

going lowerin energy. Also, in this case Deis the energy needed for one mole-

cule,not one mole of molecules.) Using the explicit definition ofq, we have

qelectg 1 eDe/kT g 2 e^2 /kT g 3 e^3 /kT   

Because Deis typically large and is a positiveexponential, the first term in the

equation above typically dominates, and we can approximate the diatomic

molecule’s electronic partition function as

qelectg 1 eDe/kT (18.8)

However, if there are low-lying excited states, the explicit summation definition

for qelectmust be evaluated, keeping in mind that the zero point for electronic

energy is the dissociation limit of the molecule in its ground electronic state.

Example 18.3

The hydrogen molecule has a D 0 of 432 kJ/mol and a vibrational frequency

of 1.295 

1014 s^1. Calculate H 2 ’s electronic partition function at 298 K.

Assume that the ground electronic state is singly degenerate. Hydrogen’s first

excited electronic state lies 1.822 

10 ^19 J above the ground state and has

a degeneracy of 1.

Solution

Because we are given D 0 , we will have to calculate De. Using equation 18.7,

we have

De^4

m

32

o

k

l

J^1

1

00

k

0

J

J



1

2

(6.626 

10 ^34 Js)(1.295 

1014 s^1 )

In the first term, we have converted D 0 into number of joules per molecule.

Solving:

De7.61 

10 ^19 J

1 mol

6.02 

1023 molecules

622 CHAPTER 18 More Statistical Thermodynamics