In my syllabus we only study about rate laws under the topic "Rates of reactions", but as I was interested in it, I started to search about that topic. And then I found the Arrhenius equation

$$\ln k = \ln A - \frac{E}{RT}.$$

But according to this equation rate constant $k$ doesn't have units because we are taking the natural logarithm of it. According to rate laws, rate constants have units and they depend on the reaction.

Are we talking about two different rate constants here? I searched this on internet and all the books I could refer but I couldn't find an answer.

  • $\begingroup$ Rewrite as $$\ln \frac{k}{A} = - \frac{E_A}{RT}$$ and $k$ is allowed to have the same units as $A$. $\endgroup$
    – Karsten
    Sep 24, 2019 at 10:58
  • $\begingroup$ I saw this equation in this format too but my problem occurred when I saw the above equation was used to draw a graph between lnK and T. $\endgroup$ Sep 24, 2019 at 13:07
  • $\begingroup$ There are two problems: The y-axis will be ln k, and you are supposed to read off ln A from the y-intercept. You circumvent the problem by plotting ln (k / units), where "units" are the appropriate units for k. The y-intercept will be ln(A / units). $\endgroup$
    – Karsten
    Sep 24, 2019 at 15:11
  • $\begingroup$ The graph would be ln k vs. 1/T to get a linear relationship (and -Ea / R as slope) $\endgroup$
    – Karsten
    Sep 24, 2019 at 15:12

2 Answers 2


The reason is because of sloppy notation. This sort of logarithmic plot is widespread in chemistry, though, and isn't limited to the Arrhenius equation, so do not expect it to change very much in the near future. You will have to get used to it.

With the Arrhenius equation

$$k = A \exp\left(-\frac{E_\mathrm a}{RT}\right)$$

you cannot directly take logarithms on both sides, because of the inherent units in $k$ and $A$. First you would have to divide through by a constant which we could call $k^\circ$:

$$\frac{k}{k^\circ} = \frac{A}{k^\circ} \, \exp\left(-\frac{E_\mathrm a}{RT}\right)$$

Now we can write

$$\ln \left( \frac{k}{k^\circ} \right) = \ln \left( \frac{A}{k^\circ} \right) - \frac{E_\mathrm a}{RT}$$

and as long as $k$, $k^\circ$, and $A$ are expressed in the same units, the resulting arguments of the logarithms will not have units. Note that the actual value of $k^\circ$ does not matter: you will of course get a different y-intercept, but the final values of $A$ and $E_\mathrm a$ that you get out of the Arrhenius plot will not be affected.

The division by $k^\circ$ is often implicit. When people write $\ln k$, what they usually mean is $\ln(k/k^\circ)$ where $k^\circ$ has the numerical value of $1$ and the units of whatever $k$ is in. Once you have the y-intercept, you take the exponential of that and tack the units back on to get $A$.

Although it is not rigorous, from a practical standpoint, this generally does not lead to much confusion as long as the author is consistent with their usage of units. Another reason for omission is that the actual value of $k^\circ$ has no physical effect, as previously discussed, and an overly pedantic focus on writing $k/k^\circ$ and $A/k^\circ$ can distract one from the chemical principles behind these equations, especially for more complicated equations which have many more symbols.

  • 1
    $\begingroup$ The above mentioned approach is described in detail in a nice article "Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions" in the Journal of Chemical Education. I think that this is rather a new invention, or to justify the "dimensionless-ness" of transcendental functions. The original authors never bothered to define their functions that way. They were only interested in the numerical values. DImensional analysis is rather a later invention, much later than the work of Arrhenius and Sorenson. $\endgroup$
    – AChem
    Sep 24, 2019 at 15:38
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    $\begingroup$ This is the same thing what confuses some people with the equilibrium constant $K$. A lot of undergraduate textbooks define $K$ as having units depending in the stoichiometry of the reaction in question. Then they move to Gibbs free energy of the reaction through $ \Delta G = - RT ln(K)$ and just kind of forget that they spend an awful amount of time on the units of $K$ seven lectures ago. $\endgroup$
    – user23638
    Sep 25, 2019 at 14:09
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    $\begingroup$ @Ezze, it’s not exactly the same situation. K is supposed to be dimensionless. The rate constant k is not supposed to be dimensionless. $\endgroup$ Sep 25, 2019 at 16:19

$$ \begin{align} k &= A \exp\left(-\frac{E_\mathrm a}{RT}\right) \label{eq:1}\tag{1}\\ \ln k &= \ln A -\frac{E_\mathrm a}{RT} \tag{2} \end{align} $$

If you use the equation \eqref{eq:1} to get the units for $A,$ you can get the units for $A$ as the exponential term is unitless. The units of $A$ are the same as those of the rate constant since $A$ carries the units for $k.$ The units for $A$ vary according to the order of the reaction^

  • The first order: $\pu{s^-1}$
  • The second order: $\pu{M^-1 s^-1}$

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