In discussions of the Arrhenius equation describing how rates change with temperature, there is often a reference to collision theory.

In the Arrhenius equation,

$$ k = A e^{-\frac{E_\mathrm{A}}{R T}}$$

there is a Boltzmann factor interpreted as the fraction of collisions of sufficient energy to overcome the activation barrier. If you plot the Boltzmann factor $e^{-\frac{E_\mathrm{A}}{R T}}$ against temperature, it looks like a simple exponential decay (all quantities in the exponent are positive). However, most textbooks use a graph of the Maxwell–Boltzmann distribution (i.e. distribution of speeds of gas molecules), which is not a simple exponential decay but has a maximum near the average speed.

So how is a graph of the Maxwell-Boltzmann distribution relevant to the Arrhenius equation?

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    $\begingroup$ When the particles are about to react, in fact you are only interested in the distribution of one component of the velocity vector (say, $v_x$). This distribution is a simple exponential decay. The same is true for $v_y$ and $v_z$. Now if you try to convert that into the distribution of $|\vec v|=\sqrt{v_x^2+v_y^2+v_z^2}$, you'll get the Maxwell–Boltzmann thing. $\endgroup$ – Ivan Neretin Jan 31 at 19:56
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    $\begingroup$ Draw a vertical line on the the Maxwell-Boltzmann distribution. To the right of this is the fraction of molecules with sufficient energy to equal or exceed the reaction barrier and hence may react. $\endgroup$ – porphyrin Feb 1 at 9:29
  • $\begingroup$ @porphyrin I question whether all of the molecules to the right of your line would have collisions of sufficient energy to exceed the activation energy. As Ivan Neretin pointed out, only the component perpendicular to the collision plane is relevant. If two fast particles don't have a head-on collision but something analogous to a side-by-side car collision, there will be no reaction. Also, I don't think the fraction of molecules corresponds to the Boltzmann factor. $\endgroup$ – Karsten Theis Feb 3 at 21:17
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    $\begingroup$ The Arrhenius eqn is only a v crude model, sure molecules have to collide in the correct way, even if this is inline the wrong part may still collide depending on the molecule's size. Also vibrational energy can aide or impede collisions even if every thing else is ok. Have a look at RRKM, Tolman's stat mech model or better still Kramer's model of reactions. $\endgroup$ – porphyrin Feb 4 at 7:33

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