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Lesson 3 · 25 min

Collision theory and activation energy

Warming by 10 °C adds under 2 % more collisions, yet many reactions go twice as fast. Build an Arrhenius plot from 1921 measurements on N₂O₅ and extract the activation energy.

In this lesson

  • Explain the temperature effect with collision theory: collision frequency, energy factor and orientation factor
  • Write the Arrhenius equation and obtain the activation energy from the slope of ln k against 1/T
  • Use the two-point form to calculate an activation energy or predict k at another temperature
  • Say when the rule of thumb “10 °C warmer, twice as fast” holds

Milk keeps for a week in the fridge but sours in a day in a summer kitchen; fireflies flash faster on warm evenings. Many reactions go twice as fast for every 10 °C of warming. Yet by kinetic theory, going from 25 °C to 35 °C raises the average molecular speed by only about 1.7 %, and the number of collisions per second by the same small amount. Where does the rest of the doubling come from?

Predict first

For a reaction with an activation energy of about 50 kJ/mol, warming from 25 °C to 35 °C roughly doubles the rate. What roughly doubles?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Collision theory

For a gas-phase reaction, molecules must collide — but the vast majority of collisions simply bounce. A collision is “effective” only if two conditions are met at once:

  1. Enough energy: the kinetic energy along the line of collision must reach at least the activation energy Ea, enough to stretch and break old bonds. By the Boltzmann distribution, about e−Ea/RTe^{-E_a/RT} of collisions qualify.
  2. The right orientation: the molecules must meet with the right parts facing. This is captured by an orientation (steric) factor p, often much smaller than 1.

So the rate constant ≈ collision frequency × p × e−Ea/RTe^{-E_a/RT}. The first two change slowly with temperature; the exponential is exquisitely sensitive — nearly all of temperature’s magic lives there. How the high-energy tail of the distribution swells on heating is told vividly in Molecules over the mountain.

k=A e−Ea/RTln⁡k=ln⁡A−EaR⋅1Tln⁡k2k1=−EaR(1T2−1T1)k = A\,e^{-E_\mathrm{a}/RT}\qquad \ln k = \ln A - \frac{E_\mathrm{a}}{R}\cdot\frac{1}{T}\qquad \ln\frac{k_2}{k_1} = -\frac{E_\mathrm{a}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)
The Arrhenius equation (1889). A plot of ln k against 1/T is a straight line with slope −Ea/R and intercept ln A.
AA
pre-exponential factor: collision frequency and orientation
EaE_\mathrm{a}
activation energy, J/mol
TT
absolute temperature, K

Worked example

The activation energy of N₂O₅ from two temperatures

For NX2OX5\ce{N2O5} decomposition, k=3.46×10−5 s−1k = 3.46\times10^{-5}\ \mathrm{s^{-1}} at 298 K and 4.87×10−3 s−14.87\times10^{-3}\ \mathrm{s^{-1}} at 338 K. Find Ea and predict k at 318 K.

ln k against 1/T · slope = −Ea/R

Arrhenius plot: activation energy from data

3.13.23.33.43.53.63.71000 / T (K⁻¹)−8−7−6−5−4−3−2ln k

Data (tick two points for the two-point method)

T / Kk / L mol⁻¹ s⁻¹
2780.00179
2880.00393
2980.00793
3080.0157
3180.0305
ln⁡k=ln⁡A−EaR⋅1T\ln k = \ln A - \dfrac{E_\mathrm{a}}{R}\cdot\dfrac{1}{T}
Least-squares fit Ea = 51.9 ± 0.3 kJ/mol A = 9.90 × 10⁶ L mol⁻¹ s⁻¹ · R² = 0.99986
Two-point method Ea = 52.1 kJ/mol

Predicted k: 0.0113 L mol⁻¹ s⁻¹

Fraction of collisions with E ≥ Ea, e^(−Ea/RT): 1.14 × 10⁻⁹

Heating by 10 K multiplies k by 1.93

The N₂O₅ values are Daniels and Johnston’s 1921 measurements; the iodine-clock values come from this site’s virtual-bench rate model (k₂₅ = 8.0 × 10⁻³ L mol⁻¹ s⁻¹, Ea = 52 kJ/mol) with about 3 % seeded measurement noise. The fit is ordinary least squares on ln k against 1/T; the Ea uncertainty is from the slope’s standard error. The collision fraction is e^(−Ea/RT), the share of collisions with energy above Ea in collision theory.
These are rate constants of the iodine clock from the virtual bench at 5–45 °C. Switch between the “k vs T” and “ln k vs 1/T” views, tick any two points to compare the two-point result with the fit, and drag the prediction temperature to watch k and the fraction of effective collisions change.

Concept check

A plot of ln k against 1/T gives a straight line. Its slope equals:

Try the calculation

For a reaction with Ea = 50 kJ/mol starting at 25.0 °C, how many kelvin of warming exactly double the rate constant?

K

Try the calculation

A reaction’s rate constant exactly doubles from 25 °C to 35 °C. What is its activation energy in kJ/mol?

kJ/mol

Key takeaways

  • An effective collision needs enough energy (a fraction of about e^(−Ea/RT)) and the right orientation.
  • Arrhenius: k = A e^(−Ea/RT); ln k against 1/T is a straight line of slope −Ea/R.
  • Warming speeds reactions mainly through the exponential; collision frequency rises by only a few per cent.
  • “Double per 10 °C” roughly holds only for Ea ≈ 50 kJ/mol near room temperature.
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Data and references

  1. S. Arrhenius, “Über die Reaktionsgeschwindigkeit bei der Inversion von Rohrzucker durch Säuren”, Z. Phys. Chem. 4, 226–248 (1889).
  2. F. Daniels, E. H. Johnston, J. Am. Chem. Soc. 43, 53–71 (1921) (N₂O₅ rate constants, 273–338 K).
  3. P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topics 17D and 18A (Arrhenius equation, collision theory).