How to relate a reaction barrier to the time the reaction needs to proceed? 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.
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:

$$\ce{A-&gt;B}$$

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})&amp; k\ (\mathrm{s}^{-1}) &amp; t_{1/2} \\ \hline\hline
15 &amp; 63.4 &amp; 10.9\ \mathrm{ms} \\
20 &amp; 0.0138 &amp; 50.2\ \mathrm{s} \\
25 &amp; 2.98\cdot 10^{-6} &amp; 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.