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In A-level chemistry class today I was just told the following:

For a reaction with activation energy $E_A$, the proportion of reactant molecules with kinetic energy $\ge E_A$ is: $$e^{-E_A/RT}$$Where $R$ is the gas constant and $T$ is the temperature of the reactant(s?).

I find that very suspicious. We are told that the area under a Boltzmann distribution curve 'represents' the number of particles with energy greater than a given threshold, so the above statement should suggest: $$\int_{E_A}^\infty\text{Boltzmann}(t)\,\mathrm{d}t=Ke^{-E_A/RT}$$Up to some constant $K$ (we aren't taught quite what the distribution is nor any of the grisly thermodynamical details, so I threw in a constant $K$ just because I'm not sure whether it's computing proportions/probabilities or the total amount of substance).

This would imply that the Boltzmann distribution curve is exponential. However, if one sketches a Boltzmann distribution as we are taught to do in A-level it is clearly not exponential. It is vaguely skew-normal from what I can see. It does look exponential after a certain point (once you're over the central hump) but it isn't everywhere. I then suspect that what we are told in class is sort-of accurate and is applicable to most of the tame situations that could arise in A-level theory, but it is not always correct.

My question:

In terms as elementary as possible, could someone explain:

  • To what extent is the statement: "the proportion of molecules with sufficient energy is $e^{-E_A/RT}$" actually true?

Before someone says it, yes, I could look up precise definitions of the Boltzmann distribution and try to headache my way through this. But Googling mathematics definitions and trying to understand them is hard enough: in my experience, trying to do the same for any of the natural sciences is essentially impossible (everyone tells you a different thing up to a different level of expertise...) so I doubt I'd get anything out of that.

Many thanks for any response.

FShrike
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  • That's a good approximation. $E_A$ is always well above the central hump, otherwise most molecules would already have it and the reaction would just occur instantly. – Ivan Neretin Jan 16 '23 at 18:16
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    You are confusing two things. When you say the Boltzmann distribution is not exponential, you are correct. But the "number of molecules with energy greater than x" is not the distribution, it is the integration of the distribution from x to infinity. And that is exponential in most circumstances (essentially where the X is higher than the "hump" of the full distribution.) – matt_black Jan 16 '23 at 18:16
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    It seems to me that you are confusing the Maxwell-Boltzmann distribution with the Boltzmann distribution sometimes also referred to as Boltzmann ratio. – Paul Jan 16 '23 at 18:17
  • @matt_black When you say "in most circumstances", do you mean that there are other factors at play but the (Maxwell?)-Boltzmann distribution is approximately (locally) exponential most of the time? – FShrike Jan 16 '23 at 18:18
  • @Paul Perhaps. We don't learn the concept very precisely at A-level, we're just given a name for a distribution of number of molecules against kinetic energy and a rough outline of its shape and basic properties. – FShrike Jan 16 '23 at 18:19
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    @FShrike For most real reactions the proportion of molecules with higher than X activation energy is small. In other words the X is far away from the "hump" in the distribution. The resulting integration of the distribution will look very close to an exponential in X (same point as Ivan Neretin's comment). – matt_black Jan 16 '23 at 18:24
  • Ok. Thanks to all responders for the clarifications – FShrike Jan 16 '23 at 18:27
  • In an ideal gas the Boltzmann distribution tells you the proportion of molecules in given velocity range. This is purely exponential. The Maxwell-Boltzmann distribution does the same thing for the molecules' speeds. This is not a pure exponential, see https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution. The difference is important. – Ian Bush Jan 16 '23 at 18:30
  • @IanBush That's subtle. Does the difference arise from there being a significant likelihood of many particles having the same speed but at different velocities, thereby changing the probability distribution? I suppose what we draw in school (the kinetic energy, so a constant multiplied by the square of the speed) is a squared and scaled version of Maxwell-Boltzmann then – FShrike Jan 16 '23 at 18:33
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    Yes. Hit the nail on the head. A given speed corresponds to many different velocities due to the different directions a molecule can move in. – Ian Bush Jan 16 '23 at 18:37
  • For an extended discussion, see this possible duplicate: https://chemistry.stackexchange.com/questions/108854/boltzmann-factor-vs-graph-of-maxwell-boltzmann-distribution – Karsten Jan 16 '23 at 18:59
  • When $E_A/k_BT \gt 1$ then integrating the Boltzmann distribution from $E_A \to \infty$ does give a result $K\exp(-E_A/K_BT)$ where $K$ depends on $E_A/T$ so varies weakly compared to the exponential. (The Maxwell-Boltzmann distribution is $p(v)\sim \exp(-mv^2/(2k_BT)$ in velocity so in terms of energy $E=mv^2/2$ we get $p(E)\sim \exp(-E/k_BT)$.) – porphyrin Jan 17 '23 at 10:20

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