Collision Theory of Reaction Rates and the Steric Factor

Gas molecules colliding at precise orientations with kinetic energy vectors and activation energy barrier.

By Aniket Bhardwaj · 31 October 2026 · Updated 31 October 2026 · Chemical Kinetics: Theories of Reaction Rate

The Arrhenius equation describes how k changes with temperature, but it does not explain why molecules react. Collision theory gives the first physical picture. It treats molecules as hard spheres, counts how often they collide, and asks what fraction of collisions are energetic enough to react. This page covers the formula, the meaning of the steric factor P, and numericals of the type seen in IIT-JAM, GATE and CSIR NET.

Two bars compare rate constants of 0.63 L mol⁻¹ s⁻¹ at 500 K and 1.01 L mol⁻¹ s⁻¹ at 510 K, a 1.62-fold increase. — A 10 K rise increases the rate constant by about 62%, mainly through the exponential term.

The three conditions for a reaction

The molecules must collide.
The collision must carry at least the threshold energy E_a (along the line of centres, in the simplest model).
The molecules must be in a suitable orientation. This is the steric requirement.

The rate constant

k = P σ (8k_BT / πμ)^1/2 N_A exp(−E_a/RT)
Z_AB = σ (8k_BT / πμ)^1/2 N_A so k = P Z_AB exp(−E_a/RT)

σ = collision cross-section (m²), πd² for hard spheres of collision diameter d.
μ = reduced mass, m_Am_B/(m_A + m_B), in kg.
k_B = Boltzmann constant, 1.380649 × 10⁻²³ J/K; N_A = 6.022 × 10²³ mol⁻¹.
Z_AB here is the collision frequency factor per unit concentration, in m³ mol⁻¹ s⁻¹ (multiply by 1000 for L mol⁻¹ s⁻¹).
exp(−E_a/RT) = fraction of collisions with energy at least E_a.
P = steric factor, a number between 0 and 1 for simple reactions, representing the fraction of collisions with the correct orientation. Comparing Arrhenius’s A with P Z_AB, P = A_exp/Z_AB.

Note the temperature dependence: the prefactor goes as T^1/2, while the exponential changes much faster. So collision theory predicts that a plot of ln k against 1/T is very nearly, but not exactly, a straight line.

Worked example 1: collision frequency factor

Two identical molecules of mass 40 u collide with cross-section σ = 0.40 nm² at 500 K (practice values). Find Z_AB.

μ = 40 × 40 / 80 = 20 u = 20 × 1.66054 × 10⁻²⁷ kg = 3.3211 × 10⁻²⁶ kg.

(8k_BT / πμ)^1/2 = (8 × 1.380649 × 10⁻²³ × 500 / (π × 3.3211 × 10⁻²⁶))^1/2 = 727.5 m/s.

σ = 0.40 × 10⁻¹⁸ m². Z_AB = 0.40 × 10⁻¹⁸ × 727.5 × 6.022 × 10²³ = 1.75 × 10⁸ m³ mol⁻¹ s⁻¹.

In practical units: 1.75 × 10¹¹ L mol⁻¹ s⁻¹.

Worked example 2: rate constant with a steric factor

For the same reaction, E_a = 100 kJ/mol and P = 0.10. Find k at 500 K.

E_a/RT = 100000 / (8.314 × 500) = 24.054, so exp(−24.054) = 3.57 × 10⁻¹¹. Only about 3.6 collisions in 10¹¹ carry enough energy.

k = P Z_AB exp(−E_a/RT) = 0.10 × 1.75 × 10¹¹ × 3.57 × 10⁻¹¹ = 0.63 L mol⁻¹ s⁻¹.

Worked example 3: how k changes from 500 K to 510 K

Find k at 510 K for the same data and the ratio k(510)/k(500).

Z at 510 K goes as T^1/2: factor = (510/500)^1/2 = 1.0100, a rise of only about 1%.

Exponential: E_a/RT = 100000/(8.314 × 510) = 23.58, factor relative to 500 K = exp(24.054 − 23.58) ≈ 1.60.

Combined ratio = 1.62, so k(510) ≈ 1.01 L mol⁻¹ s⁻¹. A 10 K rise increases k by about 62% here, almost all through the exponential term.

Worked example 4: finding P from an experimental A

An experiment gives an Arrhenius pre-exponential factor A = 2.0 × 10⁹ L mol⁻¹ s⁻¹. Using Z_AB = 1.75 × 10¹¹ from Example 1, find P.

P = A / Z_AB = 2.0 × 10⁹ / 1.75 × 10¹¹ = 0.0114.

Only about 1% of collisions have the right geometry. A small P is typical for reactions between large molecules, where orientation matters. Simple collision theory cannot predict P; it must be taken from experiment.

Common mistakes

Using the mass instead of the reduced mass. μ must be used for two different molecules.
Using grams, not kilograms. Convert u to kg before applying the formula.
Mixing m³ and L. Convert at the end (× 1000).
Forgetting N_A when you want a molar rate constant.
Reading P > 1 as impossible. It is possible for some reactions (for example some ion reactions in solution), and it shows the simple model is incomplete.
Saying all collisions are effective. Only a tiny fraction carry the energy and orientation needed.

Exam relevance

Question type	Where to look
Calculate Z_AB or k	Compute μ, the mean speed term, then multiply
Interpret the steric factor	P = A_exp/Z_AB
Fraction of energetic collisions	exp(−E_a/RT)
Compare with transition state theory	Collision theory gives A from collisions; TST from partition functions

Practice values only. Check the current official syllabus of your exam.

Once you know E_a and the pre-exponential factor, check temperature effects with the Arrhenius calculator.

Open the Arrhenius Equation Calculator →

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