# 18.6: Rotational Partition Functions of Diatomic Gases

The rotational energy levels of a diatomic molecule are given by

$E_{rot}(J) = \tilde{B} J (J + 1) \label{Eq0}$

where

$\tilde{B} = \dfrac{h}{8 π^2 I c}$

Here, \tilde{B} is the rotational constant expresses in cm-1. The rotational energy levels are given by

$E_j = \dfrac{J(J+1) h^2}{8 \pi I}$

where $$I$$ is the moment of inertia of the molecule given by $$μr^2$$ for a diatomic and $$μ$$ is the reduced mass and $$r$$ the bond length (assuming rigid rotor approximation). The energies can be also  expressed in terms of the rotational temperature, $$Θ_{rot}$$, which is defined as

$Θ_{rot} = \dfrac{r^2}{8 \pi^2 I k} \label{3.12}$

In the summation for the expression for rotational partition function ($$q_{rot}$$), Equation $$\ref{3.13}$$, we can do an explicit summation

$q_{rot} = \sum_{j=0} (2J+1) e^{-E_J/ k_B T} \label{3.13}$

if only a finite number of terms contribute. The factor $$(2J+1)$$ for each term in the expansion accounts for the degeneracy of a rotational state $$J$$. For each allowed energy $$E_J$$ from Equation $$\ref{Eq0}$$, then there are $$(2 J + 1)$$ states eigenstates it, then, the Boltzmann factor $$e^{ -E_J / k_B T}$$ has to be multiplied by $$(2J+ 1)$$ to properly account for all these states.

If the rotational energy levels are lying very close to one another, we can integrate similar to what we did for $$q_{trans}$$ previously to get

$q_{rot} = \int _0 ^{\infty} (2J+1) R^{-\tilde{B} J (J+1) / k_B T} dJ$

This the integration can be easily be done by substituting $$x = J ( J+1)$$ and $$dx = (2J + 1) dJ$$

$q_{rot} = \dfrac{k_BT}{\tilde{B}} \label{3.15}$

For a homonuclear diatomic molecule, rotating the molecule by 180° brings the molecule into a configuration which is indistinguishable from the original configuration. This leads to an overcounting of the accessible states. To correct for this, we divide the partition function by $$σ$$, which is called the symmetry number, which is equal to the distinct number of ways by which a molecule can be brought into identical configurations by rotations. The rotational partition function becomes,

$q_{rot}= \dfrac{kT}{\tilde{B} σ} \label{3.16}$

or commonly expressed in terms of $$Θ_{rot}$$

$q_{rot}= \dfrac{T}{ Θ_{rot} σ} \label{3.17}$

Example

What is the rotational partition function of $$H_2$$ at 300 K?

Solution:

The value of $$\tilde{B}$$ for $$H_2$$ is 60.864 cm-1. The value of $$k_B T} in cm-1 can be obtained by dividing it by \(hc$$, i.e., which is $$k_B T/hc = 209.7\; cm^{-1}$$ at 300 K. $$σ = 2$$ for a homonuclear molecule. Therefore from Equation $$\ref{3.16}$$,

$q_{rot}= \dfrac{k_BT}{\tilde{B} σ} = \dfrac{209.7 \;cm^{-1} }{(2) (60.864\; cm^{-1})} = 1.723$

Since the rotational frequency of $$H_2$$ is quite large, only the first few rotational states are accessible to at at 300 K