Units of rate constant (cycles per second vs radians per second) in Eyring equation

In the Eyring equation (EE),


$$k = \frac{k\_\mathrm B T}{h} \exp\left(\frac{-\Delta G\_{\mathrm f}}{RT}\right),$$


the units of $k$ are $\mathrm{s^{-1}}$. However, in general rate constants are usually expressed in $\frac{\mathrm{rad}}{\mathrm{s}}$. For instance, in expressions for damped oscillations of the form $\exp[(\mathrm{i} \omega - k)t]$, where $\omega$ is by definition in $\frac{\mathrm{rad}}{\mathrm{s}}$ and $k$ has to bear units providing consistent dimensions.


By expressing in the EE the energy $k\_\mathrm B T$ as $\nu h$, where frequency $\nu$ is in cycles per second $\left(\mathrm{Hz}=\frac{\mathrm{cyc}}{\mathrm{s}} = \mathrm{cps}\right)$, it appears that $k$ is also given in $\mathrm{cps}$ in this equation. The desired units of $\frac{\mathrm{rad}}{\mathrm{s}}$ would be obtained if $h$ were replaced by $\hbar$. However, I have never seen the EE formulated that way. Has anybody an idea in this matter?




I have looked at the original article of Henry Eyring ([*J. Chem. Phys.* **1935,** *3,* 107](http://dx.doi.org/10.1063/1.1749604)). In that article, Eyring explains the derivation of his famous equation. It basically boils down to the following:


You assume an equilibrium between the initial and transition state and consider that once a species crosses the barrier, it goes to the product state:


$$R \leftrightarrows R^{\dagger} \rightarrow P$$


The rate expression is then simply:


$$k = \nu K$$


where $k$ is the reaction rate constant, $\nu$ the crossing frequency and $K$ the equilibrium constant between the IS and TS.


I will not go into detail about how to calculate $K$, but just remember that $K$ can be obtained by calculating the quotient of the partition functions of the system in the transition and initial state, respectively. How $\nu$ is calculated, will hopefully answer your question. 


Let us assume that in the transition state, the bonds to be formed or broken are weak. In that case, we can model the reaction coordinate (the direction of the reaction) as a loose vibration. The partition function of a vibration is


$$
f = \frac{1}{1 - \exp \left(\frac{-h \nu}{k\_{b}T} \right)}
$$


Herein, $\nu$ is the frequency of the loose vibration. We assumed that this vibration was very weak (loose). In that case we can apply a Taylor expansion to the exponent and cut off the series after the first term. This gives us:


$$
f = \frac{1}{1 - 1 + \frac{h \nu}{k\_{b}T}} = \frac{k\_{b}T}{h \nu}
$$


Plugging this result back into our original rate expression gives


$$
k = \nu \frac{k\_{b}T}{h \nu} K^{\dagger} = \frac{k\_{b}T}{h} K^{\dagger}
$$


Here, I have put a $\dagger$ after the equilibrium constant to make clear that I have extracted one partition function (that of the loose vibration) out of the equilibrium constant.


From the dimensional analysis of $\frac{k\_{b}T}{h}$, you can see that $k$ should have units of $s^{-1}$ and not something else.


 6

**The Eyring equation is numerically correct, despite the apparent units problem.**


To understand the origin of the problem, one must go all the way back to the underlying statistical and quantum mechanics, since Eyring treated the motion across the transition state as being effectively a translation ([*J Chem Phys* **3**: 107, 1935](http://dx.doi.org/10.1063/1.1749604), p109, emphasis added):



> 
> The activated state is because of its definition always a saddle point with positive curvature in all degrees of freedom except the one which corresponds to crossing the barrier for which it is of course negative. ... A configuration of atoms corresponding to the activated state thus has all the properties of a stable compound except in the normal mode corresponding to decomposition ***and this mode because of the small curvature can be treated statistically as a translational degree of freedom.***
> 
> 
> 




---


The starting point is the Hamiltonian for a [particle in a box](http://quantummechanics.ucsd.edu/ph130a/130_notes/node136.html):


$$
H = - {\hbar^2 \over 2m}{d^2 \over dx^2} \tag{1}
$$


Additional to Eq. $\left(1\right)$, of course, is that the particle is confined by an infinite potential to the domain $x=\left(0,L\right)$. After a standard undergraduate physical chemistry [derivation](http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html#c5), the wavefunctions $\Psi\_n\!\left(x\right)$ of the time-independent Schrödinger equation $H\Psi\_n=E\_n\Psi\_n$ are:


$$
\Psi\_n\!\left(x\right) = \sqrt{2\over L}\sin{\left(n\pi {x\over L}\right)}\tag{2}
$$


Differentiating Eq. $\left(2\right)$ twice, substituting into the Schrödinger equation, and comparing the result to $E\_n\Psi\_n$ yields the following energy levels for the particle:


$$
E\_n = {\hbar^2 n^2 \pi^2 \over 2mL^2} \tag{3}
$$


Eq. $\left(3\right)$ can be converted to the more commonly used form, that of Eq. $\left(53\right)$ of [Salzman](http://cbc.arizona.edu/~salzmanr/480b/statt02/statt02.html), by a simple substitution of $\hbar = {h \over 2\pi}$:


$$
E\_n = {\hbar^2 n^2 \pi^2 \over 2mL^2} = {h^2 \over 4\pi^2}{n^2\pi^2 \over 2mL^2}
 = {h^2n^2 \over 8mL^2}\tag{4}
$$


***This is where the units problem in the Eyring equation originates.*** The factor of $\pi^2$ in the numerator derives from the form of $\Psi\_n$ and $H$, where the Hamiltonian requires twice-differentiating a sine function with the factor $n\pi$ in the argument. This $n\pi$ is a *fully unitless* scaling factor for the non-dimensional position $x/L$ that is needed for $\Psi\_n\!\left(0\right) = \Psi\_n\!\left(L\right) = 0$ to hold, as required by the particle-in-a-box problem definition and the mathematical properties of the sine and cosine functions. I assume a key motivation for performing the transformation of Eq. $\left(4\right)$ is cosmetic, as it removes an apparently superfluous factor of $\pi^2$. But, it admixes into the overall expression the $4\pi^2 \rightarrow \left({2\pi\ \mathrm{rad} / \mathrm{cyc}}\right)^2$ factor in the denominator that is required to maintain the correct units downstream, obfuscating the dimensionality.


The next step is to obtain the translational partition function $q\_\mathrm{t}$ which, per Eqs. $\left(54\right)$ and $\left(55\right)$ of [Salzman](http://cbc.arizona.edu/~salzmanr/480b/statt02/statt02.html), is:


$$
q\_\mathrm{t} = \sum\_{n\,=\,1}^\infty{\exp\!\left[{-{1\over k\_\mathrm{B}T}{\hbar^2n^2\pi^2 \over 2mL^2}}\right]} \left\{= \sum\_{n\,=\,1}^\infty{\exp\!\left[{-{1\over k\_\mathrm{B}T}{h^2n^2\over 8mL^2}}\right]}\right\} \tag{5}
$$


$$
q\_\mathrm{t} \approx \int\_0^\infty{\exp\!\left[-{1\over k\_\mathrm{B}T}{\hbar^2n^2\pi^2 \over 2mL^2}\right] dn} \left\{= \int\_0^\infty{\exp\!\left[-{1\over k\_\mathrm{B}T}{h^2n^2\over 8mL^2}\right] dn}\right\} \tag{6}
$$


In the above equations, I have provided the results for $E\_n$ of Eq. $\left(3\right)$ first, with the final results using the $E\_n$ of Eq. $\left(4\right)$ following it in curly brackets. I will continue with this convention below as needed. 


Per, e.g., [Wolfram Alpha](http://www.wolframalpha.com/input/?i=int%28exp%28-ax%5E2%29dx%2Cx%3D0..infty%29), the general form of the Gaussian integrals of Eq. $\left(6\right)$ is:


$$
\int\_0^\infty{e^{-ax^2}dx} = \frac{1}{2}\sqrt{\pi \over a}
$$


Thus:


$$
q\_\mathrm{t} \approx {L\over \hbar} \sqrt{m k\_\mathrm{B}T \over 2\pi}
\left\{= {L\over h}\sqrt{2\pi m k\_\mathrm{B}T}\right\} \tag{7}
$$


The bracketed expression in Eq. $\left(7\right)$ matches exactly the expression given by [Eyring](http://dx.doi.org/10.1063/1.1749604) on p108. Using manipulations I have not taken the time to retrace in detail, Eyring on p110 asserts the following expression for the prefactor of his now-eponymous equation:


$$
\left({\sqrt{2\pi m^\* k\_\mathrm{B}T} \over h}\right)\cdot {\overline{p}\over m^\*} = {k\_\mathrm{B}T \over h} \tag{8}
$$


This is made possible by derivation of the following expression (p110 and preceding):


$$
{\overline{p} \over m^\*} = {k\_\mathrm{B}T \over \sqrt{2\pi m^\* k\_\mathrm{B}T}} \tag{9}
$$


The expression in parentheses on the LHS of Eq. $\left(8\right)$ was apparently obtained by substituting $m^\*$ for $m$ and setting $L=1$ (p108) in the bracketed expression of Eq. $\left(7\right)$:



> 
> If we set the length $l\_i\!=\!1$ we have the number of unit cells per cm of length, a quantity frequently used in what follows.
> 
> 
> 


Using instead the unbracketed expression of Eq. $\left(7\right)$, transformed as in Eyring, with Eqs. $\left(8\right)$ and $\left(9\right)$ gives:


$$
{1\over\hbar}\sqrt{m^\*k\_\mathrm{B}T\over 2\pi}\cdot{\overline{p}\over m^\*} = 
{1\over\hbar}\sqrt{m^\*k\_\mathrm{B}T\over 2\pi}\cdot{k\_\mathrm{B}T \over \sqrt{2\pi m^\* k\_\mathrm{B}T}} = {k\_\mathrm{B}T \over 2\pi\hbar} \tag{10}
$$


Thus, **the result of Eq. $\left(10\right)$ is numerically equal to the prefactor reported by Eyring.** So, happily (and unsurprisingly!), none of the work performed with it in the last eight decades needs to be revisited. However, the cycles units discrepancy in Eyring's version arises because, as noted above, the factor of $2\pi$ appearing in the denominator of Eq. $\left(10\right)$ is a consequence of differentiation of the trigonometric translational wavefunction $\Psi\_n$ underpinning the prefactor derivation, and ***not*** present as a conversion factor of $2\pi\ \mathrm{rad}\over\mathrm{cyc}$.


 2

In the derivation of the Eyring equation (EE) quoted above by Ivo Filot, substitute consistently $\nu$ by $\omega = 2\pi\nu$ and $h$ by $\hbar$ wherever these two quantities appear. This will do no harm to the entire reasoning, but in the final equation the prefactor would be $$\frac{k\_{b}T}{\hbar}$$ and $k$ would be expressed in $rad/s^{-1}$, as it should be. I dare to claim that this would be the correct answer to the problem. I do not have enough courage to submit such a note to a peer-reviewed chemical journal indexed by ISI.


 0