Let's say we have standard chemical reaction given by $$A-B + C \leftrightharpoons [A\cdot\cdot B\cdot\cdot C]^{\dagger} \rightarrow A +B-C$$
I want to estimate the pre-exponential factor of this reaction using Transition State Theory.
From Fundamentals of Chemical Reaction Engineering by Davis and Davis, I see that $$\text{rate} = \nu _i C_{\dagger} = \nu _i \mathcal{K} C_{AB} C_{C}$$ Where $\mathcal{K} = C_{\dagger}/(C_{AB}C_C)$, and $C_i$ is the concentration of species $i$. Then we redefine $\mathcal{K}$ as $$\mathcal{K}_C = \frac{C_{\dagger}}{C_{AB}C_C} = \frac{Q'_{\dagger}}{Q'_{AB}Q'_C}$$ where $Q'_i$ is the partition function per unit volume. We redefine energies to lowest ground state, so $$\text{rate} = \nu _i C_{\dagger} = \nu _i \mathcal{K}e^{-E/k_BT}$$
We know that $Q = Q_tQ_rQ_{\nu}Q_{el}$, where $t$ is for translational, $r$ is for rotational, $\nu$ is for vibrational, $el$ is for electronic partition function. We also know that $Q'_{t} = \frac{(2\pi mk_B T)^{3/2}}{h^3}, Q_{el} \approx 1$.
Everything good so far.
My question lies with the rotational and vibrational partition function. $$Q_r = \frac{8\pi}{h^2 \sigma} (2\pi k_B T)^{3/2} (I_1I_2...)$$ where $\sigma$ is the rotational symmetry number. My question is, what is the rotational symmetry number? How do I find it? How do I find moment of inertia for molecules? More importantly, does a single atom like $C$ even have a rotational partition function? How do I calculate $Q_{v}$ for $A-B$? and $C$? The formula says $$Q_v = \prod_{s=1}^n (1-e^{-hv/k_BT})^{-1}$$
The end result says $$r = \frac{k_B T}{h} \cdot \frac{Q'_{\dagger}}{Q'_{A}Q'_{B}} e^{-E/k_BT}C_{AB}C_C$$
The final question would be, how do I simply something like this? With the $\frac{k_B T}{h}$, what exactly got simplified?
Edit: I figured out I can get moments of inertia from Calculated moments of inertia for molecules are listed in the NIST Computational Chemistry Comparison and Benchmark Database: http://cccbdb.nist.gov/).