Temperature and activation energy

The Arrhenius equation came from the idea that molecules react only when they collide with enough energy to break bonds and reach the transition state. The plots show the fraction of molecules with at least that much energy, Ea, at two temperatures (shaded areas). The left panels show the distribution of kinetic energy: on a linear scale a realistic tail is often too thin to see, and the log scale below shows how small it really is and how strongly it grows with temperature. The textbook visualization uses the Maxwell–Boltzmann distribution of speeds, with the barrier marked at the speed whose kinetic energy equals Ea. If the pre-exponential factor A is taken to be constant, the ratio of areas equals the ratio of rate constants, k(T2)/k(T1).

Kinetic energy at two T values

T1T2

Textbook visualization: molecular speeds

T1T2

Kinetic energy, log scale

T1T2

Ratio of areas above Ea: T2 area / T1 area

How this is calculated; advanced options
Advanced options

The textbook plot is the Maxwell–Boltzmann distribution of molecular speeds. The barrier is drawn at the speed va whose kinetic energy equals Ea, so the shaded area is the same fraction as on the energy plots. The molar mass M only stretches the speed axis; it has no effect on that fraction.

f(v) dv = 4 pi (M/2 pi R T)^(3/2) v^2 exp(-M v^2/2RT) dv; v_a = sqrt(2 Ea / M)

For molecules moving in three dimensions, the fraction with molar kinetic energy between E and E + dE is

f(E) dE = (2/sqrt pi) (RT)^(-3/2) sqrt(E) exp(-E/RT) dE

The shaded area is the fraction of molecules with E above Ea. Integrating f(E) from Ea to infinity gives an exact formula, which is very nearly a simple exponential once Ea is a few times RT:

integral from Ea to infinity of f(E) dE = erfc(sqrt x) + 2 sqrt(x/pi) exp(-x), approximately 2 sqrt(x/pi) exp(-Ea/RT), where x = Ea/RT

This simplification, that the integral is approximately equal to the Boltzmann factor, is why the rate constant has the Arrhenius form, and why the ratio of rate constants at two temperatures depends only on Ea when A is constant:

k = A exp(-Ea/RT); k(T2)/k(T1) = exp[(Ea/R)(1/T1 - 1/T2)]

What about A? The pre-exponential factor collects everything else that matters: for a reaction between two molecules, how often they collide; and for any reaction, whether the molecules are oriented correctly and how their orbitals line up. It cannot be calculated from Ea. The ratio of areas plotted here is almost exactly the Arrhenius ratio above.

(For gases, simple collision theory multiplies this fraction by a collision frequency that grows as √T. That factor nearly cancels a 1/√T in front of the exponential in the fraction, which is one reason a temperature-independent A works so well.)

Liquids too. The Boltzmann distribution of energies is still observed in liquids, so the behavior shown here still holds, even though it may not be interpretable in terms of simple collision theory.

Jason D. Kahn, Dept. of Chemistry and Biochemistry, University of Maryland College Park, using Claude