# Boltzmann factor vs. graph of Maxwell–Boltzmann distribution

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

In the Arrhenius equation,

$$k = A \exp\left(-\frac{E_\mathrm A}{RT}\right)$$

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 $$\mathrm e^{-E_\mathrm{A}/RT}$$ 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?

• 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. – Ivan Neretin Jan 31 '19 at 19:56
• 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. – porphyrin Feb 1 '19 at 9:29
• @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. – Karsten Theis Feb 3 '19 at 21:17
• 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. – porphyrin Feb 4 '19 at 7:33
• Does the second half of this, or the appendix of porphyrin's answer there, address your question? The TL;DR is that porphyrin's first comment is pretty much correct, the area under the graph does generate an $\exp(-E/kT)$ form, although the maths is slightly more nuanced than simply doing the integral from $E_\mathrm A$ to $+\infty$. It also takes collision geometry into account, which seems to be one of your major points of contention. There are more references in my answer if that helps. – orthocresol Mar 20 '19 at 0:57

Rationale for posting an "answer" to my own question

[Comment from OP] I’m looking for an explanation and a graph that is better than what is currently in the first year college textbooks .

The comments to the question contain some excellent insight and references but there is no comprehensive answer to far, so I am collecting what was posted here. Maybe the depth of the comments and the lack of a highly rated answer reflects that it is impossible (at the first year undergraduate level at least) to make a meaningful quantitative connection between Maxwell-Boltzmann distribution of kinetic energy in a ideal gas on the one hand and the Arrhenius equation on the other hand.

Textbook treatment of the question

In first-year college textbooks, the Arrhenius equation is often rationalized for gas phase reactions via the collision theory. The quotes here are from OpenStax Chemistry as hosted by Libretext.org, but other textbooks have similar material:

Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term, $$e^{−E_a/RT}$$, is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.

This is great because it references collisions providing sufficient energy.

Then they go on to explain:

The Arrhenius equation (Equation 12.5.6) describes quantitatively much of what we have already discussed about reaction rates. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of $$E_a$$ results in a smaller value for $$e^{−E_a/RT}$$, reflecting the smaller fraction of molecules with enough energy to react. Alternatively, the reaction with the smaller $$E_a$$ has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of $$e^{−E_a/RT}$$, a larger rate constant, and a faster rate for the reaction.

This is pretty much correct if you don't over-interpret "fraction of molecules with enough energy to react". What is relevant is the energy made available by collisions, not the energy of the molecules, but if the molecules have higher kinetic energy, the collisions will be of higher energy. (There might be some energy associated with rotation and vibration, but since collision theory starts with a mono-atomic ideal gas and we are shooting for the simplest explanation, those will be ignored).

Then they show a figure (and this figure is fairly representative of the corresponding figures in other textbooks and those found online from various sources): This figure suggests that the fraction of molecules with kinetic energy higher than the activation energy will react. It also suggests that if you integrate over the shaded area, you get the Arrhenius equation.

If you take a close look at the curve, you will see that the slope at the origin is too small. It looks like the x-axis is the velocity rather than the energy, and the graph is accordingly distorted.

As an aside, there are two aspects of this figure that are not representative of what first-year students will experience and observe in the lab.

• The activation energy shown is on the order of a couple RT. This would be a reaction too fast to measure in an undergraduate lab, and it gives the impression that maybe 1 out of 10 collisions leads to a chemical reaction.
• The two temperatures differ by almost a factor two. If $$T_1$$ is room temperature, the second experiment would have to be done at 586 K, i.e. in an oven where most solvents would evaporate.

Relevant distribution for collision energy

The comments were very insightful. Ivan Neretin addressed the relevant degrees of freedom:

When the particles are about to react, in fact you are only interested in the distribution of one component of the velocity vector (say, vx). This distribution is a simple exponential decay.

I made a diagram to illustrate what happens in an inelastic collision: So you can't simply take the kinetic energy of one molecule (or add both kinetic energies together) but instead have to consider one degree of freedom each. The maximum energy provided by a collision is not the sum of the kinetic energies of the particle because the center of mass of the two particles combined continues on its path, so part of the kinetic energy is not available.

The wikipedia article on the Maxwell–Boltzmann distribution states:

Since the energy is proportional to the sum of the squares of the three normally distributed momentum components, this distribution is a gamma distribution; in particular, it is a chi-squared distribution with three degrees of freedom.

By the equipartition theorem, this energy is evenly distributed among all three degrees of freedom, so that the energy per degree of freedom is distributed as a chi-squared distribution with one degree of freedom.

So maybe the distribution of energies in a collision is a problem with two degrees of freedom ($$v_x$$ of either particle). The chi-square distribution with two degrees of freedom happens to be a pure exponential lacking a T-dependent pre-exponential factor, matching the Arrhenius equation.

Orthocresol gave a reference to the rigorous derivation of the Arrhenius equation from the Maxwell-Boltzmann distribution in one of his comments:

The TL;DR is that porphyrin's first comment is pretty much correct, the area under the graph does generate an exp(−E/kT) form, although the maths is slightly more nuanced than simply doing the integral from EA to +∞.

In this answer, orthocresol sketches the derivation, with two references that show the details.

Relevance to teaching at the first-year college level

Most reactions students will see and do will be in solution, where things are different than in the gas phase, but the Arrhenius equation still is a good description of rate vs. temperature. On the other hand, as porphyrin state in one of his comments, things are usually more complicated than a derivation of rates vs. temperature via the collision model suggests:

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.

So maybe it is best to treat the Arrhenius equation as a definition of the activation energy in an empirical sense.

A graph to illustrate collision energy distribution at multiple temperatures might be helpful.

Following the logic of deriving the Arrhenius equation from the collision model, this would simply be a graph of $$e^{−E_a/RT}$$ at different temperatures. For values of temperature relevant to first year experiments (eg. room temperature vs water bath), the two traces would be very close to each other, and almost zero at energies equal to a typical activation energy obtained in a first year lab (e.g. 80 kJ/mol for iodination of acetone). To see anything, you have to add an inset zooming in near the activation energy of interest. This graph is very different from the textbook graph. It shows the distribution of collision energies rather than the kinetic energy of individual particles, and the two temperatures chosen are explicitly given and more relevant to an undergraduate lab experience. It shows that the energy distribution is not so different between room temperature and body temperature (the average energy of a molecule is proportional to T, after all, and this is just a 6% difference in temperature). However, at typical activation energies (insets at 40 kJ/mol and 80 kJ/mol), this smaller difference results in a large difference in rate constants. It also shows that of the collisions of sufficient energy, most have an energy quite close to the activation energy (fast drop-off of probability with increasing energy).

• I haven't read your answer thoroughly but just wondered if you've looked at introductory pchem books, eg Atkins does a pretty good job presenting this. I suspect though that your approach may be more useful if you are aiming for intuition. – Buck Thorn Mar 26 '19 at 20:45
• @Night Writer My question is what is appropriate in General Chemistry. Once you reach PChem, there are more tools and skills you can rely on. – Karsten Theis Mar 26 '19 at 21:32
• Ok, but as soon as you move away from a reasonably mathematical explanation it suffices to wave your hands and say the tail of the Boltzmann distribution increases with T in such a way as to result in an exponential T dependence of collision activation rates. Anyway, I still need to look carefully at your answer! – Buck Thorn Mar 27 '19 at 7:34
• From a quick look I'd say that the figure 12.5.4a is simply wrong. When the activation barrier is lower more of the molecules in the tail (larger area of the distribution) have sufficient energy. – Buck Thorn Mar 27 '19 at 7:41
• @Night_Writer - The newest edition has this figure correct. I will edit my answer to include that. The libretext.org version I used was not updated yet. – Karsten Theis Mar 28 '19 at 16:18

Well , the relation between the Boltzmann factor and the Maxwell distribution curves is not that intuitive.

By looking at the Arrhenius equation, $$k=A\operatorname{e}^{\frac{-E_\text{a}}{RT}}$$

where the term $$\operatorname{e}^{\frac{-E_\text{a}}{RT}}$$ represents the fraction of collisions that are effective (fraction of molecules that have sufficient energy).

If we increase the temperature, the fraction of collisions that are effective ( fraction of molecules having sufficient energy) increases exponentially.

Now let's look at the Maxwell Boltzmann curve. Here the fraction of molecules is plotted against the kinetic energy at a fixed temperature $$T$$. Let's plot the graph at another temperature $$T+ 10$$ : As we can clearly see , the fraction of molecules having kinetic energy greater than a particular value increases with the increase of temperature that is the fraction of collisions that are effective increases which is what the Boltzmann factor in the Arrhenius equation tells us.

• Figure are from ncert.nic.in/ncerts/l/lech104.pdf – Karsten Theis Mar 25 '19 at 3:39
• IMHO, one should not consider speed nor energy distribution in 3D context, but from 1D context of energy and velocity component a single degree of freedom. The former has the Maxwell-Boltzmann distribution with maximum at the most probable speed, the latter has the Boltzmann distribution with the maximum at zero energy of velocity component. – Poutnik Mar 25 '19 at 9:50
• @Karsten Theis Does the answer address your question or you feel it is incomplete ? – Starboy Mar 25 '19 at 13:20
• @Starboy I was looking for a quantitative treatment (I now realize that for a first-year college course, a complete rigorous treatment is over-ambitious, but at least the quantitative parts shown should be correct). In your answer, I think that equating collisions of sufficient energy with molecules of sufficient energy is not correct. Also, I don't think the textbook images you are showing are actual graphs of the Maxwell Boltzmann distribution but rather artists renditions. The shape is odd, and the areas under the two curves seem different (should be one). – Karsten Theis Mar 26 '19 at 15:46