B Derivation of the Exponential Energy Factor

In Lecture 5, we used the result that the fraction of collisions with enough energy to react, i.e., with head-on collision energy greater than or equal to the activation energy \(E_\mathrm{a}\), is given by:

\[P(E \geq E_\mathrm{a}) = \exp\left(-\frac{E_\mathrm{a}}{RT}\right)\]

This expression can be derived from the kinetic theory of gases.

B.0.1 Head-On Collision Energy

Consider a molecule B approaching a molecule A. We can split the velocity of B relative to A into two parts: a component \(v\) along the line joining the centres of the two molecules when they collide, and a component perpendicular to this line. Only the component along the line of centres pushes the molecules together; the perpendicular component carries them past each other. The head-on collision energy is the kinetic energy associated with the component along the line of centres:

\[E = \frac{1}{2}\mu v^2\]

where \(\mu\) is the reduced mass of A and B.

B.0.2 The Distribution of Head-On Velocities

The kinetic theory of gases tells us that each component of the relative velocity of a pair of molecules follows a Gaussian (Maxwell-Boltzmann) distribution:

\[g(v) = \left(\frac{\mu}{2\pi k_\mathrm{B}T}\right)^{1/2}\exp\left(-\frac{\mu v^2}{2k_\mathrm{B}T}\right)\]

where \(k_\mathrm{B}\) is the Boltzmann constant and \(T\) is the absolute temperature. Because a gas has no preferred direction, this holds whichever direction the line of centres points in.

The distribution \(g(v)\) describes pairs of molecules, but the pairs that collide in a given time are not a random sample of all pairs. A pair approaching quickly closes the gap between the two molecules sooner than a pair approaching slowly, so the rate at which pairs with approach velocity \(v\) collide is proportional to \(v\) as well as to \(g(v)\). Only pairs that are approaching each other collide, so \(v > 0\). The distribution of head-on velocities across all collisions is therefore

\[h(v) \propto v\exp\left(-\frac{\mu v^2}{2k_\mathrm{B}T}\right) \qquad (v > 0)\]

B.0.3 From Velocities to Energies

To convert this into a distribution of head-on collision energies, we change variables from \(v\) to \(E = \frac{1}{2}\mu v^2\), so that \(\mathrm{d}E = \mu v\,\mathrm{d}v\). The factor of \(v\) in \(h(v)\) then cancels:

\[h(v)\,\mathrm{d}v \propto v\exp\left(-\frac{E}{k_\mathrm{B}T}\right)\frac{\mathrm{d}E}{\mu v} \propto \exp\left(-\frac{E}{k_\mathrm{B}T}\right)\mathrm{d}E\]

The head-on collision energies therefore follow a simple exponential distribution. Normalising, so that \(\int_0^{\infty} p(E)\,\mathrm{d}E = 1\), gives

\[p(E) = \frac{1}{k_\mathrm{B}T}\exp\left(-\frac{E}{k_\mathrm{B}T}\right)\]

which is the distribution plotted in Figure 5.2.

B.0.4 Calculating the Fraction with E ≥ E_a

The fraction of collisions with head-on energy greater than or equal to \(E_\mathrm{a}\) is obtained by integrating this distribution from \(E_\mathrm{a}\) to infinity:

\[P(E \geq E_\mathrm{a}) = \int_{E_\mathrm{a}}^{\infty} \frac{1}{k_\mathrm{B}T}\exp\left(-\frac{E}{k_\mathrm{B}T}\right)\mathrm{d}E = \exp\left(-\frac{E_\mathrm{a}}{k_\mathrm{B}T}\right)\]

Finally, using the relationship \(R = N_\mathrm{A}k_\mathrm{B}\) (where \(R\) is the gas constant and \(N_\mathrm{A}\) is Avogadro’s number), we can write this in the more familiar form, with \(E_\mathrm{a}\) now a molar energy:

\[P(E \geq E_\mathrm{a}) = \exp\left(-\frac{E_\mathrm{a}}{RT}\right)\]

This derivation shows that the exponential dependence of reaction rates on temperature arises naturally from the Maxwell-Boltzmann distribution of molecular speeds, and that the activation energy \(E_\mathrm{a}\) represents a threshold that determines which fraction of collisions have sufficient energy to react.