# Statistical Thermodynamics and Rate Theories/Rotational partition function of a linear molecule

## Derivation

The rotational partition function, $Q_{rot}$  is a sum over state calculation of all rotational energy levels in a system, used to calculate the probability of a system occupying a particular energy level. The open form of the partition function is an infinite sum, as shown below. By making a few substitutions and replacing the sum with an integral, an algebraic expression for the rotational partition function can be derived. $I=\mu {r_{e}}^{2}$  $q=\sum _{j}^{\infty }g_{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)$

The degeneracy, g, of a rotational energy level, j, is the number of different measurable states that have the same energy. For rotational energy levels, this is given by:

$g=2J+1$

The rotational energy of a molecule is:

$E_{J}={\frac {\hbar ^{2}}{2I}}J(J+1)$

Substituting these values into the open form of the partition function, we get

$q_{rot}=\sum _{j}^{\infty }(2J+1)\exp \left({\frac {-\hbar ^{2}}{2k_{B}TI}}J(J+1)\right)$

Since the spacings of the rotational energy levels is small, the sum can be approximated as an integral over J,

$q_{rot}=\int _{0}^{\infty }(2J+1)\exp \left({\frac {-\hbar ^{2}}{2k_{B}TI}}J(J+1)\right){\textrm {d}}J$

From a table of integrals:

$\int (2x+1)\exp(-ax(x+1)){\textrm {d}}x={\frac {\exp(-ax(x+1))}{a}}$

Letting x = J and $a={\frac {-\hbar ^{2}}{2k_{B}TI}}$  we get

$q_{rot}={-{\frac {-\exp(-ax(x+1))}{a}}}{\bigg |}_{0}^{\infty }$

$=0-{\frac {1}{a}}=-{\frac {1}{a}}$
$={\frac {2k_{B}T^{2}I}{\hbar ^{2}}}$

A symmetry factor $\sigma$  is introduced to account for the nuclear spin states of homonuclear diatomic molecules. $\sigma$  has a value of 2 for homonuclear diatomics and 1 for other linear molecules.

$q_{rot}={\frac {2k_{B}T^{2}I}{\hbar ^{2}\sigma }}$

The rotational characteristic temperature $\theta _{rot}$  is introduced to simplify the rotational partition function expression.

$\theta _{rot}={\frac {\hbar ^{2}}{2k_{B}I}}$

The physical meaning of the characteristic rotational temperature is an estimate of which thermal energy is comparable to energy level spacing. Substituting this into the partition function gives us

$q_{rot}={\frac {T}{\sigma \theta _{rot}}}$