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From the wikipedia for $KT$ (physics):

The rates and frequencies of many processes and phenomena depend not on their energy alone, but on the ratio of that energy and $kT$, that is, on $E / kT$ (see Arrhenius equation, Boltzmann factor). For a system in equilibrium in canonical ensemble, the probability of the system being in state with energy $E$ is proportional to $e^{−ΔE / kT}$.

I struggle to find a good explanation of the relationship between the two cases mentioned here, and why they have the same boltzmann exponential factor appearing in them. What is the missing link?

Another way of putting my question is: how do I relate the Boltzmann factor with the activation energy $e^{−E_a / kT}$ with the canonical (which I understand) $e^{−ΔE / kT}$?

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    $\begingroup$ Because for the process to happen, the system has to pass through a transition state at some $\Delta E$ higher than the average energy, which is part of the canonical ensemble. Thus, the probability of the process occurring is linked with the probability of the system being in the necessary configuration. $\endgroup$
    – Jon Custer
    Nov 11, 2016 at 15:37
  • $\begingroup$ I see. So what would be the $\Delta E=E_1- E_2$ if one is $E_a$? Say in the case of the Maxwell-Boltzmann distribution for gas molecules. $\endgroup$
    – usumdelphini
    Nov 11, 2016 at 15:38
  • $\begingroup$ That's an error on the Wiki page. Correction: "the probability of the system being in state with energy $E$ is proportional to $e^{-E/kT}$". $\endgroup$
    – lemon
    Nov 11, 2016 at 15:46
  • $\begingroup$ Thanks. Therefore, if I have a Maxwell-Boltzmann distribution, will the number of molecules that have energy higher than $E_a$ (call it $N_a$) be proportional to $N_0 e^{-E_a/KT}$, where $N_0$ is the total number of molecules? $\endgroup$
    – usumdelphini
    Nov 11, 2016 at 15:48

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