Physical Chemistry , 1st ed.

In Chapter 14, we found that the rotational energy of a diatomic molecule is

Erot

J(J

2 I

1)^2

 (18.27)

where Jis the rotational quantum number,Iis the moment of inertia, and 
is Planck’s constant divided by 2. In addition, we found that the Jth rotational
state has a degeneracy of 2J 1, from the possible values of the zcomponent
of the total angular momentum. Using this in equation 18.26, the rotational
partition function becomes

qrot 



ifirst

(2J 1) exp

J(J

k

1

T

)^2 /2I



rotational level

which can be written as

qrot 



ifirst

(2J 1) expJ(J

2

IkT

1)^2



rotational level

where we have rearranged the exponent in the second expression. Like we did
with the expression for the vibrational partition function, we recognize that all of
the constants except T(and the rotational quantum number J) in the exponent
must collectively combine to yield a temperature unit, so that the overall expo-
nent is unitless. Therefore, we can define the following rotational temperaturer:

r 

2



I

2

k

 (18.28)

and replace the group of constants with r:

qrot 



ifirst

(2J 1) eJ(J+1)r/T

rotational level

(A diatomic molecule has only one defined rotation and so has only a single
r.) At high temperatures, the fraction r/Tis small and successive terms in the
summation are close to each other. We can again substitute an integral for the
summation:

qrot



J 0

(2J 1) eJ(J+1)r/TdJ (18.29)

where J, the rotational quantum number, is the variable.
This integral may seem problematic because of the presence ofJin the body
of the integral as well as the exponent. But note that in the exponent of the ex-
ponential, the expression J(J 1) can be written as J^2 J. The derivative of
J^2 Jis 2J 1. We can perform a classic substitution. Let

xJ^2 J

dx(2J 1) dJ

Now substitute: if we rewrite equation 18.29 as

qrot



J 0

eJ(J+1)r/T[(2J 1) dJ]

we get

qrot



x 0

exr/Tdx (18.30)

18.5 Diatomic Molecules: Rotations 629