Physical Chemistry , 1st ed.

Example 18.4

For I 2 (g),vis 310 K. Calculate qvibfor I 2 at the following temperatures:

a.30 K

b.1000 K

Solution

a.At 30 K, we should probably be using the more explicit expression for qvib,

since we are well below the vibrational temperature. Therefore we would have

qvib

Evaluating, we have

qvib5.70 

10 ^3

b.At 1000 K, we are above the vibrational temperature, so we can use the ab-

breviated expression for qvib. Therefore we have

qvib^1

3

0

1

0

0

0

K

K3.23

Interestingly, if we use the more explicit definition ofqvib,we get qvib3.21.

(Try this yourself and see.) The abbreviated expression for qvibis thus a valid

approximation.

Most molecules have more than two atoms, so there are more vibrations

than one to consider. In Chapter 14, we found that a molecule that has N

atoms will have either 3N5 (for linear molecules; at least one of the vibra-

tions will be doubly degenerate) or 3N6 (for nonlinear molecules) normal

vibrations that are used to define its possible vibrational motions. (For sim-

plicity’s sake, we assume a nonlinear molecule so we don’t have to consider the

linear-verus-nonlinear issue at every turn, but you should recognize where the

differences will be.) The overall vibrational energy can therefore be separated

into 3N6 vibrational parts:

EvibE 1 E 2 E 3  E (^3) N 6
The subscripts 1 , 2 , and so on are the typical labels used to represent the in-
dividual normal modes of vibration. The vibrational partition function for a
polyatomic, nonlinear molecule is, by substituting into equation 18.9,
qvib 

ifirst
giexp (18.21)
vibrational level
Notice that there are two sums in this expression: a summation over the indi-
vidual vibrational modes(which is the summation in the exponent of the ex-
ponential) and a summation over the possible vibrational levels(which is the
summation indicated by the sign).
As we did before, we can separate the exponential into individual parts.
Equation 18.21 can be rewritten as
qvib

i 1
gieE^1 /kT

i 1
gieE^2 /kT

i 1
gieE^3 /kT
E 1 E 2 E 3  E (^3) N 6

kT
exp
2

3

(

1

3

0

0

K

K)





1 exp^3

3

1

0

0

K

K



626 CHAPTER 18 More Statistical Thermodynamics