I am working with a bimolecular reaction for which I have computed the energy profile. I have output files for TS and reactants which I intend to use for rate constant calculations. From TST we have:

$$k=\frac{k_{\mathrm B}T}{h}\frac{Q^{\ddagger}}{Q_A Q_B}\exp\left(\frac{-E^\ddagger}{RT}\right)$$

But how do I calculate the rate constant from the Gaussian output? I know how to find $E^\ddagger$, but the partition functions are teasing me. In the output file values for $Q$ are shown both for the vibrational motion, the electronic, the translational, and the rotational:

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I have found that $Q=q'_\text{trans}q_\text{rot}q_\text{vib}q_e$

I am not too sure about what the prime in $q'_\text{trans}$ means other than it might have something to do with the translational function being per unit volume.

I have stumbled upon a book where it is explained that you have to convert the translational function to being per unit volume to get the correct units of $\pu{m3 mol-1 s-1}$ for bimolecular reactions. I do not quite understand this nor which of the vibrational partition functions to use (Total Bot or Total V=0).

Furthermore, I am uncertain if the unit of $\pu{mol-1}$ is just obtained by multiplication with the Avogadro constant and what the unit of the translational partition function exactly is. I thought the unit was $\pu{m-3}$ based on:

$$\operatorname{dim}q_\text{trans}=\frac{\operatorname{dim} q^\ddagger}{(\operatorname{dim} q_\ce{A})(\operatorname{dim} q_\ce{B})}= \frac{\mathsf{L}^{-3}}{(\mathsf{L}^{-3})(\mathsf{L}^{-3})} = \mathsf{L}^{-3}$$

But with introduction of the relative translational motion per unit volume as explained in the link, I am confused of where the unit comes into play.

  • $\begingroup$ Actually, I would prefer to have the rate constant in cm^3*molecules^-1*s^-1. To obtain this, I would need to multiply with 10^6 and the Avogadro constant. However, I don't see why I am able to multiply with the Avogadro constant to get molecules^-1 as there would need to be a mol unit in the expression. Otherwise, I would obtain mol*molecules^-1, which is undesired. $\endgroup$
    – Lea
    Dec 30 '18 at 9:35

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