B Derivation of the Exponential Energy Factor

In Lecture 5, we used the result that the fraction of collisions with relative kinetic 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 The Maxwell-Boltzmann Distribution

The kinetic theory of gases tells us that the probability density for finding a collision with relative kinetic energy \(E\) is given by the Maxwell-Boltzmann distribution:

\[f(E) = \frac{2\pi}{(\pi k_\mathrm{B}T)^{3/2}}\sqrt{E}\,\exp\left(-\frac{E}{k_\mathrm{B}T}\right)\]

where \(k_\mathrm{B}\) is the Boltzmann constant and \(T\) is the absolute temperature. This distribution is normalised such that:

\[\int_0^{\infty} f(E)\,\mathrm{d}E = 1\]

B.0.2 Calculating the Fraction with E ≥ E_a

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

\[P(E \geq E_\mathrm{a}) = \int_{E_\mathrm{a}}^{\infty} f(E)\,\mathrm{d}E\]

Substituting the Maxwell-Boltzmann distribution:

\[P(E \geq E_\mathrm{a}) = \int_{E_\mathrm{a}}^{\infty} \frac{2\pi}{(\pi k_\mathrm{B}T)^{3/2}}\sqrt{E}\,\exp\left(-\frac{E}{k_\mathrm{B}T}\right)\,\mathrm{d}E\]

To evaluate this integral, we make the substitution \(u = E/(k_\mathrm{B}T)\), so that \(E = uk_\mathrm{B}T\) and \(\mathrm{d}E = k_\mathrm{B}T\,\mathrm{d}u\). When \(E = E_\mathrm{a}\), we have \(u = E_\mathrm{a}/(k_\mathrm{B}T) \equiv \beta\). The integral becomes:

\[P(E \geq E_\mathrm{a}) = \frac{2\pi}{(\pi k_\mathrm{B}T)^{3/2}}\int_{\beta}^{\infty} \sqrt{uk_\mathrm{B}T}\,\mathrm{e}^{-u}\,k_\mathrm{B}T\,\mathrm{d}u\]

Simplifying:

\[P(E \geq E_\mathrm{a}) = \frac{2}{\sqrt{\pi}}\int_{\beta}^{\infty} \sqrt{u}\,\mathrm{e}^{-u}\,\mathrm{d}u\]

This integral can be evaluated using integration by parts. Let \(v = \sqrt{u}\) and \(\mathrm{d}w = \mathrm{e}^{-u}\,\mathrm{d}u\), so that \(\mathrm{d}v = \frac{1}{2\sqrt{u}}\,\mathrm{d}u\) and \(w = -\mathrm{e}^{-u}\):

\[\begin{align} \int_{\beta}^{\infty} \sqrt{u}\,\mathrm{e}^{-u}\,\mathrm{d}u &= \left[-\sqrt{u}\,\mathrm{e}^{-u}\right]_{\beta}^{\infty} + \int_{\beta}^{\infty} \frac{\mathrm{e}^{-u}}{2\sqrt{u}}\,\mathrm{d}u\\ &= \sqrt{\beta}\,\mathrm{e}^{-\beta} + \frac{1}{2}\int_{\beta}^{\infty} \frac{\mathrm{e}^{-u}}{\sqrt{u}}\,\mathrm{d}u \end{align}\]

For large \(\beta\) (which corresponds to \(E_\mathrm{a} \gg k_\mathrm{B}T\), a condition typically satisfied for chemical reactions), the second term becomes negligible compared to the first. Therefore:

\[\int_{\beta}^{\infty} \sqrt{u}\,\mathrm{e}^{-u}\,\mathrm{d}u \approx \sqrt{\beta}\,\mathrm{e}^{-\beta}\]

Substituting back:

\[P(E \geq E_\mathrm{a}) \approx \frac{2}{\sqrt{\pi}}\sqrt{\beta}\,\mathrm{e}^{-\beta} = \frac{2}{\sqrt{\pi}}\sqrt{\frac{E_\mathrm{a}}{k_\mathrm{B}T}}\exp\left(-\frac{E_\mathrm{a}}{k_\mathrm{B}T}\right)\]

For typical activation energies in chemical reactions, \(E_\mathrm{a}/(k_\mathrm{B}T) \gg 1\), which means the pre-exponential factor \(\sqrt{E_\mathrm{a}/(k_\mathrm{B}T)}\) is much larger than unity but changes only slowly with temperature compared to the exponential term. To a good approximation, we can therefore write:

\[P(E \geq E_\mathrm{a}) \approx \exp\left(-\frac{E_\mathrm{a}}{k_\mathrm{B}T}\right)\]

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:

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

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