# 18.6: Rotational Partition Functions Contain a Symmetry Number

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The rotational energy levels of a diatomic molecule are given by:

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

where:

$\tilde{B} = \dfrac{h}{8 π^2 I c} \nonumber$

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

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

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, $$Θ_\text{rot}$$, defined as:

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

The interpretation of $$θ_\text{rot}$$ is as an estimate of the temperature at which thermal energy ($$\approx kT$$) is comparable to the spacing between rotational energy levels. At about this temperature the population of excited rotational levels becomes important. See Table 1.

Table 1: Select Rotational Temperatures. In each case the value refers to the most common isotopic species.
Molecule $$H_2$$ $$N_2$$ $$O_2$$ $$F_2$$ $$HF$$ $$HCl$$ $$CO_2$$ $$HBr$$ $$CO$$
$$\Theta_\text{rot}$$ 87.6 2.88 2.08 1.27 30.2 15.2 0.561 12.2 2.78

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

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

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 $$\ref{Eq0}$$ there are $$(2 J + 1)$$ eigenstates. The Boltzmann factor must be multiplied by $$(2J+ 1)$$ to properly account for the degeneracy these states:

$(2J+ 1)e^{ -E_J / k T}$

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:

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

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

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

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 and 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:

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

or commonly expressed in terms of $$Θ_\text{rot}$$:

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

##### Example 18.6.1

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 $$k T$$ in cm-1 can be obtained by dividing it by $$hc$$, i.e., which is $$kT/hc = 209.7\; cm^{-1}$$ at 300 K. $$σ = 2$$ for a homonuclear molecule. Therefore from Equation $$\ref{3.16}$$,

\begin{align*} q_\text{rot} &= \dfrac{kT}{\tilde{B} σ} \\[4pt] &= \dfrac{209.7 \;cm^{-1} }{(2) (60.864\; cm^{-1})} \\[4pt] &= 1.723 \end{align*} \nonumber

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