Chemistry: Atoms First 2e · Kinetics

Collision Theory

7 min read

On this page 9 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Molecules must meet before they can react, but meeting is not enough. makes this quantitative: a reaction occurs only when reactant particles collide and two further conditions hold. The colliding particles must carry enough kinetic energy to overcome the activation energy (E_a), the barrier separating reactants from products, and they must approach with the correct orientation, so that the atoms forming new bonds actually come into contact.

Because most collisions fail one or both tests, only a tiny fraction are effective — which is why many reactions that look as though they should be fast are actually slow. Collision theory explains in one framework why rates depend on temperature, concentration, surface area, and catalysts, and it gives rise to the linking temperature, activation energy, and rate.

Why this matters

Collision theory is the physical explanation behind every rate observation in this chapter: why food spoils faster in a warm kitchen than in a refrigerator, why powdered sugar ignites more readily than a sugar cube, and why a catalyst speeds a reaction by orders of magnitude. It also underlies enzyme function — enzymes lower the activation energy of biochemical reactions so life's chemistry proceeds rapidly at body temperature. For exams, it lets you predict — not memorize — how temperature, concentration, or surface area changes a rate.

The college version

Core Concepts

Collisions are necessary but not sufficient

Every reaction begins with a collision, but not every collision produces products. In the stratospheric reaction NO + O₃ → NO₂ + O₂, the molecules collide constantly, yet a given collision is effective only if it brings the nitrogen atom of NO into contact with a terminal oxygen of O₃ with enough energy. The fraction is the product of an energy factor and an .

Activation energy and the energy barrier

Even an energetically favorable reaction — one whose products are more stable than its reactants — requires an input of energy to get started, like pushing a boulder over a hilltop before it can roll down the other side. The activation energy is the height of that crest, measured from the average energy of the reactants, and only collisions with kinetic energy at least E_a can climb it. The fraction of molecules with energy at least E_a is the Boltzmann factor:

$f = e -E a /(RT)$

where R = 8.314 J mol^-1K^-1 is the gas constant and T is the absolute temperature. Raising the temperature both makes molecules collide more often and — far more importantly — increases the fraction carrying enough energy to react.

Orientation matters

Energy alone is not enough. In CO + NO₂ → CO₂ + NO, the carbon end of CO must strike the oxygen end of NO₂; a high-energy collision bringing the oxygen end of CO toward NO₂ simply bounces apart. The orientation factor (sometimes called the steric factor) is the fraction of collisions with the correct geometry, typically much less than 1.

The Arrhenius equation

Combining collision frequency, orientation, and the energy fraction gives the Arrhenius equation:

$k = A e -E a /(RT)$

Here A is the frequency factor, bundling how often collisions occur and what fraction are correctly oriented, and k is the rate constant. Taking the natural logarithm gives a linear form:

$ln k = ln A - E a R (1 T)$

so a plot of lnk versus 1/T is a straight line with slope -E_a/R — the standard experimental route to measuring an activation energy. For a reaction studied at two temperatures, the two-point form is:

$ln (k 2 k 1) = - E a R (1 T 2 - 1 T 1)$

How It Works / Step-by-Step Process

Estimate how often reactant particles collide — raised by concentration and surface area.
Find the fraction of collisions with energy ≥ E_a using the Boltzmann factor e^-E_a/(RT), then multiply by the orientation factor.
Combine: rate ∝ (collision frequency) × (orientation factor) × (energy fraction).
To quantify, use the Arrhenius equation: compare k at two temperatures, or read E_a from the slope of a lnk vs. 1/T plot.

Common Confusions

Do Not Confuse	With	Difference
Any collision	An effective collision	Most collisions lack the energy or orientation to react; only effective collisions contribute to the rate.
Raising temperature mainly increasing collision frequency	Raising temperature increasing the energy fraction	Frequency rises modestly ( ∝ T), but the fraction e^-E_a/(RT) rises exponentially — the dominant effect.
Activation energy	Overall energy change ΔH	E_a is the barrier height to the transition state; ΔH is the net energy difference between products and reactants. A reaction can be exothermic yet have a large E_a.
Frequency factor A	Rate constant k	A counts collisions and orientation; k also includes the energy fraction and is what experiments measure.
The 1/T plot slope	The activation energy itself	The slope of lnk vs. 1/T equals -E_a/R; multiply by -R to get E_a.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Molecules have to bump into each other to react, but most bumps do nothing. It is like two dancers in a crowded room: they only "react" if they bump hard enough and are facing the right way. Warming the room makes them move faster and bump harder, so more bumps work.

Worked example

Worked example 1: how much does a 10 K rise speed up a reaction?

A reaction has E_a = 50.0 kJ mol^-1. Compare its rate constant at 300 K and 310 K, assuming A stays constant.

Write the two-point Arrhenius form before substituting:

$ln (k 2 k 1) = - E a R (1 T 2 - 1 T 1)$

Convert the activation energy to joules so it cancels with R = 8.314 J mol^-1K^-1:

$E a = 50.0 kJ mol -1 \times 1000 J 1 kJ = 5.00 \times 104 J mol -1$

Substitute T₁ = 300 K, T₂ = 310 K:

$ln (k 2 k 1) = - 5.00 \times 104 8.314 (1310 - 1300)$

$ln (k 2 k 1) = -(6014)(0.003226 - 0.003333) = -(6014)(-1.075 \times 10 -4) = 0.647$

$k 2 k 1 = e 0.647 \approx 1.9$

The rate constant nearly doubles for a 10 K rise — a familiar rule of thumb for reactions with E_a near 50–60 kJ mol^-1. Dimensional analysis confirms the setup: converting kJ to J makes E_a/R come out in kelvin.

Worked example 2: what fraction of collisions can react?

At T = 300 K, a reaction has E_a = 75 kJ mol^-1. What fraction of collisions carries at least this energy?

Write the Boltzmann factor first:

$f = e -E a /(RT)$

Convert E_a to joules and substitute R = 8.314 J mol^-1K^-1:

$f = e -(7.5 \times 104)/(8.314 \times 300)$

$f = e -30.07 \approx 8.6 \times 10 -14$

Fewer than one collision in 10¹³ is energetic enough at room temperature — which is why such a reaction seems not to occur until the temperature is raised or a catalyst is added. Raising the temperature to 350 K gives f = e^{-(7.5 × 10⁴)/(8.314 × 350)} = e^-25.78 ≈ 6.4 × 10^-12, roughly 74 times larger.

Key takeaways

A collision is effective only if it has sufficient energy ( ≥ E_a) and correct orientation; most collisions fail one of these tests.
The fraction of molecules with energy ≥ E_a is e^-E_a/(RT); small increases in T can dramatically increase it.
The Arrhenius equation, k = A e^-E_a/(RT), links k to temperature; lnk vs. 1/T is linear with slope -E_a/R.
Larger E_a means a smaller rate constant; every 10 K rise roughly doubles many rates near room temperature.
Concentration and surface area change the frequency of collisions; temperature and catalysts change their effectiveness.
Collision theory is a model with limits: it assumes hard-sphere behavior and misses quantum effects and detailed bond-breaking dynamics, so it predicts trends, not exact rates.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

What two conditions must a collision satisfy to be effective?
Show answer
It must carry kinetic energy at least equal to the activation energy, and the particles must collide with the correct orientation.
A reaction has a very large activation energy. Fast or slow at room temperature, and why?
Show answer
Slow: the fraction of collisions with energy ≥ E_a, e^-E_a/(RT), is extremely small, so almost no collisions are effective.
How would you determine E_a from rate-constant data at several temperatures?
Show answer
Measure k at several temperatures, plot lnk versus 1/T, and multiply the slope by -R to obtain E_a.
Why does grinding a solid reactant into a powder speed up the reaction?
Show answer
Powdering increases the surface area, exposing far more particles to collisions, which raises the collision frequency and hence the rate.
If E_a = 0, what does the Arrhenius equation predict about the rate constant's temperature dependence?
Show answer
With E_a = 0, k = Ae⁰ = A: every collision is effective, and k is essentially independent of temperature.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Collision theory: The model that reactions occur when particles collide with enough energy and the right orientation.
Effective collision: A collision that actually produces products.
Activation energy Eₐ: The minimum energy a collision must have to break bonds and start the reaction.
Frequency factor A: The Arrhenius pre-exponential term combining collision frequency and orientation.
Orientation factor: The fraction of collisions with the correct geometry.
Arrhenius equation: k = A e^-E_a/(RT), the temperature dependence of the rate constant.

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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