As I am writing this I am at a conference and one of the participants just asked a question where he linked reaction barriers to durations for the reaction to complete. To paraphrase:

From our experience a reaction with an activation barrier of 15 kcal/mol should occur instantaneous at room temperature. An activation barrier of 20 kcal/mol takes about one to two minutes and an activation barrier of 25 kcal/mol needs about 10 hours.

I would like to rationalise this statement as it seems quite hand-weaving to me. How can I judge from (a possibly also computed) activation barrier how long a reaction will need to complete? For the the sake of the argument, let's only consider reactions that proceed in one step; generalisations may also be implied, but may be too complex.

In the comments (and the already existing answer) the Eyring equation is mentioned. If the connection between the energy values and the duration can be made with that, an illustrative example would be nice.

  • $\begingroup$ We do have the Eyring equation $k = \frac{k_{B}T}{h}e^{-\frac{\Delta G^{\ddagger}}{RT}}$. Have you tried plugging in some values? $\endgroup$
    – Zhe
    Commented Oct 26, 2016 at 12:28
  • $\begingroup$ @Zhe I'm aware of the equation and no, I have not tried plugging in some values. It was a statement I heard at the conference today and this was my way of making a mental note, so that I won't forget to investigate. I hope that this will provide some more insight further down the line. From this statement it seems that this is well established, although I never heard it, maybe established by experienced, and I wonder where did it come from and under what circumstances it will break down. I think the Eyring equation might give a first clue, but certainly won't be enough to justify it alone. $\endgroup$ Commented Oct 26, 2016 at 13:42
  • $\begingroup$ I plugged in some numbers for $T = 298\,\mathrm{K}$. For $15\,\mathrm{kcal/mol}$, the half-life of a first order reaction is 11 ms. For $20\,\mathrm{kcal/mol}$, 0.8 minutes. For $25\,\mathrm{kcal/mol}$, 64 hours. The top value is a bit off, but the others line up fairly well with "experience." $\endgroup$
    – Zhe
    Commented Oct 26, 2016 at 14:39
  • $\begingroup$ Because of the exponential relationship between activation energy and rate constant this becomes very small for relatively small changes in activation energy and can easily become years. Just as well as most organic material, i.e. us are exothermic wrt combustion with oxygen. Incidentally, the Eyring eqn as quoted above is only for first order reactions, need to multiply by ratio of partition functions for other reactions, but it is a good 'rule of thumb'. $\endgroup$
    – porphyrin
    Commented Oct 26, 2016 at 19:42
  • 1
    $\begingroup$ This earlier answer may provide some useful background. $\endgroup$
    – ron
    Commented Jan 4, 2017 at 16:09

3 Answers 3


Here are some back of the envelope numbers from using the Eyring equation:

$$k = \frac{k_{B}T}{h}e^{-\frac{\Delta G^{\ddagger}}{RT}}$$

Let's just assume we're at $298\ \mathrm{K}$ for the reaction, and the reaction is relatively simple:


I constructed the following table by plugging in values. $t_{1/2} = \frac{\ln 2}{k}$

\begin{array}{|c|c|} \hline \Delta G^{\ddagger}\ (\mathrm{kcal\,mol}^{-1})& k\ (\mathrm{s}^{-1}) & t_{1/2} \\ \hline\hline 15 & 63.4 & 10.9\ \mathrm{ms} \\ 20 & 0.0138 & 50.2\ \mathrm{s} \\ 25 & 2.98\cdot 10^{-6} & 64.6\ \mathrm{h} \\\hline \end{array}

The values for 15 and 20 $\mathrm{kcal\,mol}^{-1}$ seem pretty consistent with your rule. The top value is a bit off, but we're working with a very small numbers at this point, and there may be other sources of error that we're not accounting for in the model.

  • $\begingroup$ Thanks for doing this. You derived the value for the half-time from the first order rate law like $$\begin{align}c&=c_0\cdot e^{-kt}&\Leftrightarrow\ \ln c &=\ln c_0 -kt &\Rightarrow\ \ln\left(\frac{c_{1/2}}{c_0}\right) &= -kt_{1/2} &\Rightarrow\ t_{1/2} &=k^{-1}\ln2\end{align}$$ I am probably going to ask some follow up questions to this, especially why the last value is so far off. $\endgroup$ Commented Jan 5, 2017 at 3:12

If I understand your last statement correctly, what you would like to have is the reaction time $t$ as a function of the reaction barrier $\Delta G$. However, $t$ also depends on the conversion $c$ (for (pseudo)first order reactions, as is the assumption in the Eyring equation, conversion can never be 100%) and temperature $T$.

Although it has been mentioned, just for the sake of completion, here is the Eyring equation giving us the rate constant $k$:

$$k = \frac{k_B T}{h}e^{-\frac{\Delta G^\ddagger}{RT}}$$

We know that the half life $\lambda$ is:

$$\lambda = \frac{\ln(2)}{k}$$

The conversion $c(t)$ is related to this:

$$c = 1 - \frac{1}{2^{\frac{t}{\lambda}}}$$

If we solve this for $t$, we get:

$$t = \frac{\ln(\frac{1}{1-c})}{\ln(2)}\lambda = \frac{\ln(\frac{1}{1-c})}{k}$$

Where we can insert the Eyring equation for $k$, to get this final result:

$$t(\Delta G, c, T) = \frac{h \cdot \ln(\frac{1}{1-c})}{k_BT} \cdot e^{\frac{\Delta G^{\ddagger}}{RT}}$$

Here is a plot of the reaction time for some typical conversion rates at room temperature: tvG And another logarithmic plot, which makes it easier to get the involved time scales, from 10$^{-9}$ h (3.6 $\mu$s) to 100 h: enter image description here As you can see, reactions around 20 kcal/mol lie around the "typical" regime, from seconds to several hours, while reactions with $\Delta G^{\ddagger}$ < 15 kcal/mol proceed within milliseconds and reactions with $\Delta G^{\ddagger}$ > 25 kcal/mol may take days or weeks to complete.


The activation barrier you got, is it $\Delta G^‡$ or $E_a$ (the Arrhenius activation energy)? Depending on this you might either use the Eyring equation or the Arrhenius equation.

It is actually quite common to use the Eyring equation to calculate $\Delta H^‡$ and $\Delta S^‡$ and with that $\Delta G^‡$ from experimental rate constants, so doing it the other way round and using the equation for prediction is completly fine. But I would not say something like

An activation barrier of 20 kcal/mol takes about one to two minutes

but rather use half-lifes or 95% conversion or something similar. Just be careful if the reaction is not 1st order, since only there the half-life is independend on the concentration.

As an example see this paper from Joe Fox. At page two, bottom of the left column they predict one reaction to be 29-times faster than another one, purely based on calculated $\Delta G^‡$ values, using the eyring equation to calculate those rate constants. On page 21 of the supporting information you can see that they used the eyring equation to calculate $\Delta G^‡$ from experimental data.

As an extension to this, a similar approach should be possible to also judge the temperature of a reaction and linking it to the reaction duration and reaction barrier.

Yes it is, if you assume that your $\Delta G^‡$ is constant over that that temperature range.

  • 1
    $\begingroup$ This is not really what I was looking for. I know the Eyring equation and all its faults. Hence your answer is more or less at the same level of handwavyness as the original statement. I was looking for a more rigorous approach, a judgement of the degree of handwavyness, or at least a few calculated examples like in Zhe's comment to the question. The statement also was given as quoted, so there has to be some justification why it was said in that way. $\endgroup$ Commented Dec 28, 2016 at 13:12
  • $\begingroup$ Well, your question was quite different before the edit and imo my answer wasn't that bad to the questions present in the original post. I find it very questionable to alter the question in such an extreme way that the answers below seem to be quite offtopic. And as I said, people usually use the eyring equation in case of calculated activation energies to determine (relative) rates. If you want "authoritative uses", K.N. Houk is using this regularly in his papers, for example: pubs.acs.org/doi/abs/10.1021/ja505569a $\endgroup$
    – DSVA
    Commented Dec 28, 2016 at 13:50
  • 1
    $\begingroup$ I amended my question, I did not really change it. Your answer is as off-topic (if you really se it that way now) as it was before. It generally just boils down to: Use the Eyring equation, others do it, too. Which is not really a justification and exactly the same what Zhe said. When I amended my question I was careful enough not to invalidate what you have wrote (that would have been if I would have deleted the first paragraph). You can still edit your answer and obtain the bounty - I just found it not very helpful, sorry. $\endgroup$ Commented Dec 28, 2016 at 14:07
  • $\begingroup$ For example: I don't know what the kind of reaction energy the guy on the conference was going on about, probably Gibbs, what would be the difference in those approaches? You have not even stated the Eyring equation once and how do I get from the rate to a duration. That might be trivial for you, but the connection is not really that obvious to me. I don't understand what you mean with "but rather use half-lifes or 95% conversion or something similar." Links to papers are fine, but if they are behind a paywall it would be better to include the relevant portion in your own words or as quotes. $\endgroup$ Commented Dec 28, 2016 at 14:16
  • $\begingroup$ Before it was "how can I predict a rate from (calculated) energies?", now it's "if I use the eyring equation to predict rates, what are the problems, how to fix those and what are the alternatives ", those are quite different questions. My whole argument that you can use the eyring equation to do so seems quite stupid now that the eyring equation is mentioned in the question, and it also completly changes the "level" of discussion. Before it seems to me that we are talking on basically an undergrad level, leading to this simple answer. Now it's a much more complex and elaborate topic. $\endgroup$
    – DSVA
    Commented Dec 28, 2016 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.