5.17: Statistical Mechnanical Transition State Theory
- Page ID
- 547112
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From transition state theory the rate constant is estimated as on a per molecule basis for a reaction of the form \(\ce{A + BC -> Products}\) is :
\[k = \kappa\frac{k_B T}{h} K^‡ [A][BC]\label{TST}\]
or
\[k = \kappa\frac{R T}{h} K^‡ [A][BC]\label{TSTmolar}\]
on a molar basis. Where \(K^‡\) is concentration equilibrium constant between the transition state and reactants. Thus, to formulate this in terms of statistical mechanics we need a statistical expression for the equilibrium constant.
Accounting for the changes upon forming the transition state
It seems straight forward to substitute in the expression for the statistical mechanical equilibrium constant:
\[K_{c}=\frac {q_{ABC^‡}/V}{(q_{A}/V)(q_{BC}/V)}.\label{Kc}\]
However, the complications become clear when we expand this out to the different degrees of freedom, using the more compact notation \(q^‡\) for the transition state species partition functions and subscripts e for electronic, v for vibrational, r for rotational and t for the translational degrees of freedom:
\[K^‡=V\frac {q^‡_{e}q^‡_{v}q^‡_{r}q^‡_{t}}{q_{Ae}q_{Av}q_{Ar}q_{At}q_{BCe}q_{BCv}q_{BCr}q_{BCt}}.\label{KwithDOF}\]
The ratio of the electronic partition functions can be separated out as the Boltzmann factor from the electronic contribution to the activation energy \(\Delta E_o\), the difference between the electronic energy of the transition state and the reactants:
\[K^‡=V\frac {q^‡_{v}q^‡_{r}q^‡_{t}}{q_{Av}q_{Ar}q_{At}q_{BCv}q_{BCr}q_{BCt}}exp\left(\frac{-\Delta E_o}{RT}\right).\label{Kwithelect}\]
The transition state has only 3 translational degrees of freedom, while there are 6 for the reactants.
\[q_\text{trans} = \left( \frac{2 π m kT} {h^2} \right)^{3/2} V = \frac{ V}{\Lambda^3} \label{parttransation}\]
Substituting this in and realizing that \((m_A + m_{BC})/(m_A m_{BC}) = 1/\mu\) we get:
\[K^‡=\frac {q^‡_{r}q^‡_{v}}{q_{Ar}q_{Av}q_{BCr}q_{BCv}}\left(\frac{h^2}{2\pi \mu k_B T}\right)^{3/2} exp\left(\frac{-\Delta E_o}{RT}\right).\label{Kwithelectandtrans}\]
With an estimate of the geometry of the transition state the rotational and vibrational partition functions can be estimated, taking care not to count motion along the reaction coordinate as a vibration. The geometries of the reactants are known so the rotational and vibrational partition functions can be calculated.
In the unimolecular case because the transition state and the reactant have the same mass the translational partition functions for the transition state and the reactant are the same and they cancel out. Thus for this calling the reactant A we have:
\[K^‡=\frac {q^‡_{r}q^‡_{v}}{q_{Ar}q_{Av}}exp\left(\frac{-\Delta E_o}{RT}\right).\label{Kunimol1}\]
If the transition state is substantially the same shape as the molecule, the rotational partitions functions may also cancel out. This would leave the vibrations that are not along the reaction coordinate.