enter image description here The Boltzmann distribution is a probability density function which expresses the probability of finding a particle in an energy state $\epsilon$ while in thermal equilibrium given a specific temperature. The Boltzmann distribution can be written as:

$$ p(\epsilon)=C\cdot\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)} $$

I have noticed that it is possible to derive the Arrhenius equation by slightly modifying the Boltzmann distribution. We simply have to express the Boltzmann Distribution in terms of the universal gas constant R by multiplying the Avogadro’s number.

$$ p\left(\epsilon\right)=C\cdot\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)}=C\cdot\exp{\left(\frac{-\epsilon\cdot N_\mathrm a}{k_\mathrm BT\cdot N_\mathrm a}\right)}\iff p\left(E\right)=C\cdot\exp{\left(\frac{-E}{RT}\right)} $$

Then normalize the function.

$$ C\int_0^\infty{\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)}\,\mathrm dE=1}\Longleftrightarrow C=\frac1{RT} $$

Finally, integrate the function from $E_a$ to $\infty$ to find the ratio of particles with sufficient energy.

$$ \frac1{RT}\int_{E_\mathrm a}^\infty\exp{\left(\frac{-E}{RT}\right)\,\mathrm dE}=\exp\left(\frac{-E_\mathrm a}{RT}\right) $$

This gives the Arrhenius exponential term.

However, I’m rather unsure about the validity of this demonstration, as the Boltzmann distribution fails to model gas particle behaviour conceptually. Since the Boltzmann distribution is decreasing over its entire domain, this suggests that given any gas particle, it has the highest probability to have energy equal to zero. In other words, any gas particle will most likely be immobile.

Given this inconsistency, is the above proof rigorous or is it just mathematical luck? If it is luck, then what is a more rigorous way of deriving the exponential term in the Arrhenius equation?

  • $\begingroup$ Your BD interpretation fails in 2 aspects: // (1) - it predicts highest probability of velocity and Ek for 1 defree of freedom, in 1D. For 3D, it directly switches to Maxwell-Boltzman D, with the nonzero highest probability of speed and energy. // (2) Highest probability of zero 1D velocity norm does not mean zero speed. similarly as highest probability density of electrons occurence does not mean the electron is supposed to be there. $\endgroup$
    – Poutnik
    Commented Apr 12, 2023 at 13:10
  • $\begingroup$ By other words, as 1D velocity and BD apply in Arrhenius eq. context, the highest probability of v_x = 0 does NOT mean the probability of |v_x| > 0 is zero. It just means v_x = 0 m/s is more probable than e.g. v_x = 300 m/s. $\endgroup$
    – Poutnik
    Commented Apr 12, 2023 at 15:01
  • $\begingroup$ Yes, I am aware that the BD. does not suggest that a given particle necessarily have velocity 0. However, it is still unlikely that a particle has highest possibility of not being in motion relative to a given axis. $\endgroup$ Commented Apr 12, 2023 at 17:05
  • 1
    $\begingroup$ You have to consider the integral probability. In fact, for BD, the integral probability says probability of v_x = 0 is zero, in spite of the fact the maximum of probability function is at v_x=0. As $\lim_{\Delta v_x \to 0}{(p(v_x=0) \mathrm{d}v_x)}=0$ $\endgroup$
    – Poutnik
    Commented Apr 12, 2023 at 18:10
  • 1
    $\begingroup$ Normally one would start from the Maxwell-Boltzmann distribution (i.e. the velocity distribution) and using the reaction cross section obtain an expression similar to the Arrhenius one. $\endgroup$
    – porphyrin
    Commented Apr 13, 2023 at 8:38

1 Answer 1


If the gas phase reaction is bimolecular $A + B \to C$ then we can observe the product formed by measuring in a certain direction at angles $\theta,\phi$ with the solid angle ($\Omega$) relative the the colliding pair. The rate is a product of terms

$$\displaystyle \frac{dN_C}{dt}=s(\theta,\phi,v)vf_Af_Bn_An_BVd\Omega$$

where $s$ is the reaction cross section at the angles given and relative velocity $v=|v_A-v_B|$. The terms $f_A,f_B$ are the probability densities of the species having a velocity in the range $v\to v+\delta v$ so are functions of the velocity alone. $V$ is the volume defined by the intersection of A and B and $n_A,n_B$ are their number densities.

The total rate is that value integrated over all angles $\theta,\phi$ and means that we can replace the cross section by its integrated value which now depends on the relative velocity only, and the volume also is extended to the whole reaction vessel's volume. This produces

$$\displaystyle \frac{dn_C}{dt}=\int_0^\infty\int_0^\infty v\sigma(v)\,f_Af_B\,n_An_B\,4\pi v_A^2 4\pi v_B^2 dv_Adv_B$$

The phenomenological rate from chemical kinetics is

$$\displaystyle \frac{dn_C}{dt}=kn_An_B$$

and so the rate constant $k$ is

$$\displaystyle k=\int\int_0^\infty v\sigma(v)\,f_Af_B4\pi v_A^2 4\pi v_B^2 dv_Adv_B$$

We will assume that the velocity distributions $f$ are given by the Maxwell-Boltzmann distribution

$$\displaystyle f_A= \left(\frac{m_A}{2\pi k_BT}\right)^{3/2}e^{-m_Av_A^2/2k_BT}$$

where $m_A$ is the mass of A etc. and $k_B$ the Boltzmann constant. Substituting for the $f_A,f_B$ gives, using the reduced mass $\mu$ and transforming to centre of mass coordinates, (v. tricky) gives the rate constant at temperature $T$ as

$$\displaystyle k(T)= \left(\frac{\mu}{2\pi k_BT}\right)^{3/2}\int_0^\infty v\sigma(v) e^{-\mu v^2/2k_BT}4\pi v^2dv$$

and as the collision energy is $E=(1/2)\mu v^2$ the rate constant is

$$\displaystyle k(T)=\frac{1}{k_BT}\left(\frac{8}{\pi\mu k_BT}\right)^{1/2}\int_0^\infty E\sigma(E) e^{-E/k_BT} dE$$

The final step is to make a model of the cross section for the collision. The hard sphere model in which only a fraction of the collision energy goes into reaction depending on the position of the collision, has a cross section of,

$$\displaystyle \sigma(E)=\pi d^2\left(1-\frac{E^*}{E}\right),\quad \text{if}\quad E\ge E^* $$

where $E^*$ is the minimum energy for reaction, and otherwise $\sigma (E)=0$ when $E\lt E^*$. Substituting and integrating gives

$$\displaystyle k(T)=\pi d^2\left(\frac{8k_BT}{\pi\mu }\right)^{1/2} e^{-E^*/k_BT} $$

This shows how the exponential term initially from the velocity of the gas atoms /molecules when transformed into relative collision energy produces an Arrhenius type rate constant.

There are other ways of working out the rate constant, such as starting with the speed distribution rather than the velocity distribution, but here I have followed closely that in 'Chemical Kinetics and Dynamics' by Steinfeld, Francisco and Hase, (publ. Prentice Hall) where a much fuller description can be found.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.