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?

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    $\begingroup$ Related: Rate Constant Units and Eyring Equation $\endgroup$
    – user7951
    Commented Dec 30, 2015 at 18:37
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    $\begingroup$ I don't believe this question is a duplicate. This question relates to the unit of the rate constant. The other SO question pertains to the order of the reaction and how the dimension of $k$ is accomodated for that. $\endgroup$
    – Ivo Filot
    Commented Dec 30, 2015 at 21:30
  • $\begingroup$ Concur, this is not a duplicate. The problem of $\mathrm{Hz}$ units carrying an implicit "cycles" in the numerator $\left(\mathrm{Hz}=\frac{\mathrm{cyc}}{\mathrm{s}}\neq\mathrm{s}^{-1} \equiv \frac{ \mathrm{rad} }{\mathrm{s}}\right)$ is rife throughout chemistry, spectroscopy in particular. It explains why we use both $h$ and $\hbar = \frac{h}{2 \pi}$, for example. $\endgroup$
    – hBy2Py
    Commented Jan 3, 2016 at 23:08
  • $\begingroup$ slaw, radians are dimensionless units: $\theta$ in radians is the arclength in meters traversed by a point per meter of distance away from the center of rotation about which $\theta$ is applied. So, $\mathrm{rad} = \frac{ \mathrm{m}}{\mathrm{m}} = 1$, and thus $\mathrm{rad \over s} = {1 \over \mathrm{s}}$. That said, the units collision in your question stands unaffected. $\endgroup$
    – hBy2Py
    Commented Jan 3, 2016 at 23:23
  • $\begingroup$ See my answer to chemistry.stackexchange.com/questions/10115/… which explains the units among other things. $\endgroup$
    – porphyrin
    Commented Sep 9, 2016 at 12:36

3 Answers 3


I have looked at the original article of Henry Eyring (J. Chem. Phys. 1935, 3, 107). 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.

  • $\begingroup$ Except your last sentence is not the case: the units of $h$ are $\mathrm{\frac{J}{Hz}} \equiv \mathrm{\frac{J~s}{cyc}}$, which should provide units of $\mathrm{\frac{cyc}{s}}$ to $k$. $\endgroup$
    – hBy2Py
    Commented Jan 4, 2016 at 12:42
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    $\begingroup$ In the original paper of Planck (Ann. d. Phys., 1901, 4, 553), the units for $h$ are given as $erg \cdot s$, which in S.I. would be $J \cdot s$. Since then, I believe we agree upon the dimensionality of this constant. For instance, IUPAC also defines it as $J \cdot s$. Where do these cycles originate from? $\endgroup$
    – Ivo Filot
    Commented Jan 4, 2016 at 14:24
  • $\begingroup$ It originates in the difference between angular and cyclic frequency. A wave with cyclic frequency $f = 1~\mathrm{\frac{cyc}{s}} = 1~\mathrm{Hz}$ has an angular frequency of $\nu = 2\pi~\mathrm{\frac{rad}{s}}$. 'Cycles' are generally taken as dimensionless, and radians are strictly dimensionless; thus we end up with two sets of units that, propagated incautiously, are both written as "$\mathrm{s}^{-1}$" but that differ numerically by a factor of $2\pi$. $\endgroup$
    – hBy2Py
    Commented Jan 4, 2016 at 14:37
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    $\begingroup$ Sure. That I understand. But neither Eyring nor Planck actually mention these in their publications. To be honest, I also never have heard of this during my own education and this (the cycles part) is also not mentioned in the course books that I used. I also looked in the books of Pauling (Introduction to Quantum Mechanics) and the one of Glasstone & Eyring (The Theory of Rate Processes); it is also not mentioned in these. What am I missing? $\endgroup$
    – Ivo Filot
    Commented Jan 4, 2016 at 14:43
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    $\begingroup$ For going to the effort of tracking down Eyring's original paper and working through a bunch of the math: BOUNTY! $\endgroup$
    – hBy2Py
    Commented Jan 11, 2016 at 16:28

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, 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:

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

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, 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, 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, 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, the general form of the Gaussian integrals of Eq. $\left(6\right)$ is:

$$ \int_0^\infty{e^{-ax^2}dx} = \frac{1}{2}\sqrt{\pi \over a} $$


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

The bracketed expression in Eq. $\left(7\right)$ matches exactly the expression given by Eyring 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 $2\pi\ \mathrm{rad}\over\mathrm{cyc}$.


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.

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    $\begingroup$ This unfortunately doesn't work numerically: $h\nu = \hbar\omega$; but $h\nu \neq \hbar\left(2\pi \omega\right)$. If one were to substitute $h=2\pi \hbar$, the final prefactor would be $k_\mathrm{B}T \over 2\pi \hbar$. My hunch is that there is a factor of $2\pi$ in the denominator that arises somewhere in the manipulation of the translational partition function, that was numerically (but not dimensionally) sensible to combine with $\hbar$ to yield the $h$ of the EE. My stat mech isn't strong enough to retrace the derivation, though. $\endgroup$
    – hBy2Py
    Commented Jan 9, 2016 at 14:27
  • $\begingroup$ Thanks for joining Chem.SE to post an answer, though! :-) $\endgroup$
    – hBy2Py
    Commented Jan 9, 2016 at 14:29
  • $\begingroup$ It's OK, @Brian. If in the standard EE $h$ is substituted by $2\pi\hbar$ then the rate constant in the l.h.s. will keep its units, i.e., Hz or cps, unchanged. Then, multiplying both sides of EE by $2/pi$ will remove the offending term $2\pi$ in the denominator of the prefactor while $k$ will be converted to $k' = 2\pi k$, where $k'$ will be in rad/s. $\endgroup$
    – user24239
    Commented Jan 9, 2016 at 19:03
  • $\begingroup$ I have converted your other answer to a comment here, please have a look at the help center for more information. $\endgroup$ Commented Jan 9, 2016 at 19:11
  • $\begingroup$ Thanks for giving it a shot, user24239, but this answer just notes what one might do to alter the units of Eyring's expression. It doesn't reach into Eyring's derivation, such as calculation of the partition function &c., to explain the presence of $h$ and how one might resolve the apparent units conflict. $\endgroup$
    – hBy2Py
    Commented Jan 10, 2016 at 18:21

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