# 18.6: Rotational Partition Functions Contain a Symmetry Number

- Page ID
- 204063

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 expressed 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 \(\PageIndex{1}\)

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 \(k_B T\) 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 \(\ref{3.16}\),

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

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

## Contributors and Attributions

- www.chem.iitb.ac.in/~bltembe/pdfs/ch_3.pdf