# Molecular Partition Functions¶

The base values

\begin{align} \beta&=\frac{1}{k_BT}\\ \Theta_{vib}&=\frac{h\nu}{k_B}\\ \Theta_{rot}&=\frac{\hbar^2}{2Ik_B}\\ \end{align}

## Building $$Q$$ (Monatomic)¶

### $$q_{trans}$$¶

Where the volume of the box that the energy is calculated in is $$V= 24.47 dm^3$$ by convention

$q_{trans}(T,V)=\bigg(\frac{2\pi mk_BT}{h^2}\bigg)^{3/2}V$

### $$q_{elec}$$¶

Where $$g_i$$ is the degeneracy of our level.

\begin{align} q_{elec}(T)&=\sum\limits_i^{levels}\mathrm{g}_ie^{-\beta\epsilon_i}\\ &or\\ q_{elec}(T)&=\mathrm{g}_1e^{\big(\frac{D_e}{k_BT}\big)}\\ \end{align}

Since we want don’t really care about absolute energy, we can calculate everything relative to the ground state energy

$q_{elec}(T)=\mathrm{g}_1+\mathrm{g}_2e^{-\beta\epsilon_2}+\mathrm{g}_3e^{-\beta\epsilon_3}+...:\hskip{1cm}$

The terms get rapidly smaller, so we typically only need one to three terms, depending on whether or not there’s a low lying excited state $$(\epsilon_0-\epsilon_1\leq12\kjmol)$$

Note

You can calculate the contribution of the state to the partition function by using $$\exp\big(-\epsilon_j/k_BT\big)$$
e.g.

$\exp\bigg(\frac{(-12\;KJ\cdot mol^{-1})(1000\;J^{-1})}{(1.380\e{-23}\; J\cdot K^{-1})(293\;K)(6.022\e{23}\:mol^{-1})}\bigg)=0.00724$
Personal note

It was at this point that I realised that Avogadro’s number has units of $$mol^{-1}$$, not $$mol$$

### Total Internal Energy (Monatomic)¶

We can now calculate the total internal energy of these two terms from the equation $$U=\color{purple}\frac{3}{2}N_Ak_BT\color{black}+\color{orange}N_A\mathrm{g}_2\epsilon_2exp\bigg(\frac{-\epsilon_2}{k_BT}\bigg)$$ Since the electronic term only involves the first excited state, it will have a very small contribution to $$U$$

Example

\begin{align} exp\bigg(\frac{-\epsilon_2}{k_BT}\bigg)&=\exp\bigg(\frac{(-178\;KJ\cdot mol^{-1})(1000\;J^{-1})}{(1.380\e{-23}\; J\cdot K^{-1})(298\;K)(6.022\e{23}\:mol^{-1})}\bigg)=6.9433\e{-32}\\ q_{elec}&=N_A\mathrm{g}_2\epsilon_2exp\bigg(\frac{-\epsilon_2}{k_BT}\bigg)\\ &=(6.022\e{23}\:mol^{-1})(2)(2.96053\e{−19}\:KJ)(6.9433\e{-32})\\ &=2.4757403\e{-29}\:\kjmol \end{align}

## Building $$Q$$ (Diatomic)¶

### $$q_{vib}$$¶

\begin{align} q_{vib}(T)&=\sum\limits_{n=0}^\infty e^{-\beta\epsilon_{vib}}=\frac{e^{\big(\frac{-h\nu}{2k_BT}\big)}}{1-e^{\big(\frac{-h\nu}{k_BT}\big)}}\\\\ &\hskip{2.5cm}or\\ q_{vib}(T)&=\frac{e^{\big(\frac{-\Theta_{vib}}{2T}\big)}}{1-e^{\big(\frac{-\Theta_{vib}}{T}\big)}}:\hskip{1cm} \Theta_{vib}=\frac{h\nu}{k_B} \end{align}

### $$q_{rot}$$¶

\begin{align} q_{rot}(T)&=\sum\limits_{J=0}^\infty(2J+1)e^{\big(\frac{-\Theta_{rot}J(J+1)}{T}\big)}: \hskip{1cm} \Theta_{rot}=\frac{\hbar^2}{2Ik_B}\\ &\hskip{3cm}or\\ q_{rot}(T)&=\frac{8\pi^2Ik_BT}{h^2}:\hskip{1cm} \Theta_{rot}<<T \end{align}

### Molecular Partition Function (Diatomic)¶

$\begin{gather} q(V,T)=\color{purple}q_\text{trans}\color{black}+\color{blue}q_\text{rot}\color{black}+\color{red}q_\text{vib}\color{black}+\color{orange}q_\text{elec}\\ q(V,T)=\color{purple}\bigg(\frac{2\pi mk_BT}{h^2}\bigg)^{3/2}V\color{black}\cdot\color{blue}\frac{T}{\sigma\Theta_{rot}}\color{black}\cdot\color{red}\frac{e^{\big(\frac{-\Theta_{vib}}{2T}\big)}}{1-e^{\big(\frac{-\Theta_{vib}}{T}\big)}}\color{black}\cdot\color{orange}\mathrm{g}_1e^{\big(\frac{D_e}{k_BT}\big)} \end{gather}$

### Ensemble Partition Function (Diatomic)¶

$Q(N,V,T)=\frac{q(V,T)^N}{N!}$

### Total Internal Energy (Diatomic)¶

We can now calculate the total internal energy of these two terms from the equation

$\begin{gather} \bar U=\color{purple}\bar U_{trans}\color{black}+\color{blue}\bar U_{rot}\color{black}+\color{green}U_{VPVE}\color{black}+\color{red}\bar U_{vib}\color{black}+\color{orange}\bar U_{elec}\\ \bar U=\color{purple}\frac{3}{2}RT\color{black}+\color{blue}RT\color{black}+\color{green}R\frac{\Theta_{vib}}{2}\color{black}+\color{red}R\frac{\Theta_{vib}}{e^{\big(\frac{\Theta_{vib}}{T}\big)}-1}\color{black}-\color{orange}N_AD_e \end{gather}$

## Building $$Q$$ (Polyatomic)¶

Translational and electronic energy are the same as with diatomic, but rotational and vibrational differ.

### $$q_{rot}$$¶

$$\sigma=$$ the symmetry number (similar to degenerate rotational states)

\begin{align} \text{For a spherical top:}\\ q_{rot}(T)&=\frac{\pi^{1/2}}{\sigma}\bigg(\frac{T}{\Theta_{rot}}\bigg)^{3/2}\\ \text{For a symmetric top:}\\ q_{rot}(T)&=\frac{\pi^{1/2}}{\sigma}\bigg(\frac{T}{\Theta_{rot,A}}\bigg)\bigg(\frac{T}{\Theta_{rot,B}}\bigg)^{1/2}\\ \text{For an asymmetric top:}\\ q_{rot}(T)&=\frac{\pi^{1/2}}{\sigma}\bigg(\frac{T^3}{\Theta_{rot,A}\Theta_{rot,B}\Theta_{rot,C}}\bigg)^{1/2} \end{align}

### $$\bar U_{rot}$$¶

$\bar U_{rot}=\frac{3}{2}RT$

### $$q_{vib}$$¶

$q_{vib}(T)=\prod\limits_{j=1}^\alpha\frac{e^{\big(\frac{-\Theta_{vib}}{2T}\big)}}{1-e^{\big(\frac{-\Theta_{vib}}{T}\big)}}$

### $$\bar U_{vib}$$¶

$E_{vib}=N_Ak_B\sum\limits_{j=1}^\alpha\Bigg(\frac{\Theta_{vib,j}}{2}+\frac{\Theta_{vib,j}}{e^{\big(\frac{\Theta_{vib}}{T}\big)}-1}\Bigg)$