I couldn't decide whether to ask this to Chemistry, Physics or Statistics stack exchange. Hopefully I made the right choice.

According to this Wikipedia page, while the speed of particles in an ideal gas follows a Maxwell-Boltzmann distribution, the energy distribution in a collection of ideal gas molecules follows a Chi-Square distribution with however many degrees of freedom as the molecule. i.e. 3 for monatomic, 5 for diatomic (at moderate temperatures).

As an example computation to try this out, I wanted to find the activation energy of a reaction which would have 10% of the monatomic gas molecules eligible for reaction at temperature $ 300 \text{ K} $.

From the Chi-Square distribution, we should have $ \frac{E}{RT} \sim \chi^2_3 $. From the inverse Chi-Square table, at a significance level (i.e. right-tail proportion of the distribution) of 0.1, the critical value is 6.251, which implies that we need $ \frac{E_a}{RT} = 6.251 $ so the activation energy should be no more than $ E_a = 8.314 \times 300 \times 6.251 = 15.5912 \text{ kJ mol}^{-1} $.

This seems like a reasonable number but then I tried checking the answer by working it out directly from the distribution of energies (also given on the Wikipedia page). I worked out that

$$ P(E > E_a) = \frac{2}{\sqrt{\pi } (R T)^{3/2}} \int_{E_a}^{\infty } \sqrt{E} \exp \left ( -\frac{E}{R T} \right ) \ \text{d}E $$

Evaluating this numerically for $ E_a = 15591.2 $ and $ T = 300 $ gave $ P(E > E_a) = 0.005847 $, or about 0.5% of the particles, which is very different to the 10% I specified when solving it the first way, showing that something went wrong when I was working it out.

A third method is to use the commonly-cited (but it seems nowhere derived) result from the Arrhenius equation that $ \exp(\frac{-E_a}{RT}) $ is the fraction of particles with $ E > E_a $. Using this gives a proportion of 0.00193 = 0.2%, which differs yet again from both previous answers. I am very skeptical of this result though because neither that integral nor the Chi-Square inverse pdf have a nice closed form, and this is a suspiciously clean simple formula to say it's supposed to do the same thing.

Does anyone know the correct way to go about finding particle energy proportions? It would be nice to be able use Chi-Square correctly here as I find it interesting how different areas of science and statistics fit together.


1 Answer 1


I believe I figured it out - I was off by a factor of 2 in my scaling.

The variable $ \frac{2E}{RT} $ has a Chi-Square distribution with 3 degrees of freedom (on further research this appears to be unrelated to the physical degrees of freedom of the molecule and is always just 3 - could be wrong here though).

So once we have our critical value from the Chi-square table, rearrange to get $ E_a = \frac{\chi^2_c RT}{2} $ as the activation energy for which (significance level)% of the molecules have above this energy.

Then, I got the distribution formula wrong, the correct equation should be

$$ P(E > E_a) = \int_{E_a}^{\infty} f_E(E) \ \text{d}E = 1 - \frac{2}{\sqrt{\pi}(RT)^{3/2}}\int_{0}^{E_a} \sqrt{E} \exp \left ( -\frac{E}{RT} \right ) \ \text{d}E. $$

In the example problem I gave, the corresponding activation energy is 7795.8 kJ / mol and this does produce the correct answer of 10% of the molecules.

The $ e^\frac{-E_a}{RT} $ factor representing this fraction of molecules with the activation energy does not work. I can only assume this is some commonly-taught thing in chemistry that nobody bothers to check the maths for because to me, it looks like it is completely wrong!


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