Collision Model: Study Guide (Chemistry 5.5)

No collision, no reaction

In Topic 5.4 you saw that the rate law of an elementary reaction comes straight from its equation. The collision model explains why. Bonds can only break and form when the reacting particles are close enough for their electron clouds to interact. In practice, that means they must collide.

Think about one step, such as a chlorine atom reacting with an ozone molecule:

Cl + O₃ → ClO + O₂

For this to happen, a Cl atom and an O₃ molecule must meet. During the collision one O–O bond in ozone breaks and a new Cl–O bond forms. Everything this topic says follows from that picture.

But collisions are extremely common. In a gas at room conditions, each molecule collides with others billions of times every second. If every collision produced products, almost every gas reaction would be over in a fraction of a second. Most are not, so most collisions must fail.

What makes a collision successful

A collision leads to products only when two conditions are met at the same time.

1. Enough energy. Breaking bonds costs energy. The colliding particles must bring at least a minimum amount of kinetic energy into the collision, called the activation energy, Eₐ. In a weaker collision the particles slow down, push apart and separate unchanged.
2. A suitable orientation. The particles must meet the right way round, so that the atoms that need to bond actually touch. A collision on the “wrong” end does not rearrange the bonds, however hard it is.

For OH⁻ + CH₃Br, the oxygen must attack the carbon atom from the side opposite the bromine, so that a C–O bond can form as the C–Br bond breaks. A hit on the bromine end, or on one of the hydrogen atoms, cannot do this. Small, round particles (such as single atoms) have few orientation limits. Larger, less symmetrical molecules have more, so a smaller share of their collisions is correctly lined up.

Put together: only a small fraction of collisions are successful in most reactions. The rate of an elementary reaction depends on three things:

Factor	Question it answers	What changes it
Collision frequency	How many collisions happen each second?	Concentration (or gas pressure); temperature, slightly
Collision energy	What fraction of those collisions reach Eₐ?	Temperature (strongly); the size of Eₐ
Collision orientation	What fraction are lined up correctly?	The shapes of the particles (fixed for a given reaction)

Collision frequency and concentration

If you double the concentration of one reactant, each particle of the other reactant meets it twice as often. Collision frequency doubles, so the rate doubles. The fraction of collisions that succeed does not change, because the temperature (energy spread) and the particle shapes (orientation) are the same.

This is the particle-level reason behind the rate law for an elementary step. For A + B → products, collisions need one A and one B, so the collision frequency is proportional to [A] × [B], and rate = k[A][B]. The rate constant k contains the parts that do not depend on concentration: the energy fraction and the orientation fraction.

Worked example 1: concentration and collision frequency

Question. The step Cl + O₃ → ClO + O₂ is elementary. In a gas-phase experiment at constant temperature, its rate is 3.0 × 10⁻⁵ M s⁻¹ (invented data). Predict the rate (a) when [Cl] is doubled, (b) when [Cl] is doubled and [O₃] is tripled, and (c) when only [O₃] is halved. Explain (a) in terms of collisions.

1. The step is elementary and bimolecular, so rate = k[Cl][O₃]. The rate is first order in each reactant.
2. (a) Doubling [Cl] multiplies the rate by 2: 2 × 3.0 × 10⁻⁵ = 6.0 × 10⁻⁵ M s⁻¹.
3. (b) Multiply by 2 × 3 = 6: 6 × 3.0 × 10⁻⁵ = 1.8 × 10⁻⁴ M s⁻¹.
4. (c) Multiply by ½: 1.5 × 10⁻⁵ M s⁻¹.

Explanation for (a). With twice as many Cl atoms in the same volume, each O₃ molecule is hit by Cl atoms twice as often, so there are twice as many Cl–O₃ collisions per second. The temperature has not changed, so the same fraction of those collisions has enough energy and the right orientation. Twice as many successful collisions per second means twice the rate.

Check. The answers use only ratios, so k is never needed. A common slip in (b) is to add the factors (2 + 3 = 5) instead of multiplying them.

The Maxwell–Boltzmann distribution

In any sample, particles do not all have the same kinetic energy. Collisions constantly pass energy around, so at any moment a few particles are almost still, most have a middling energy, and a few have a very large energy. The Maxwell–Boltzmann distribution is the graph of this spread: number of particles (y-axis) against kinetic energy (x-axis).

Read three features from the curve:

• It starts at zero (no particle has negative energy), rises to a peak (the most common energy), then has a long tail to the right that approaches zero but never quite reaches it.
• The total area under the curve stands for the total number of particles.
• If you mark Eₐ on the energy axis, the area under the curve to the right of Eₐ stands for the particles that have at least the activation energy. Its share of the total area is a qualitative estimate of the fraction of collisions energetic enough to react.

How temperature changes the curve

Heating the sample raises the average kinetic energy. On the graph:

• The peak moves right (the most common energy is higher) and gets lower.
• The curve becomes wider and flatter, with a much thicker tail.
• The total area stays the same, because the number of particles has not changed.

The key result is in the tail. Because Eₐ usually lies far out in the tail, a modest temperature rise can make the area beyond Eₐ several times larger. In Figure 2 it grows about threefold.

Temperature also raises the collision frequency a little, because faster particles meet each other more often. But that effect is small. The large increase in rate comes from the much larger fraction of collisions with enough energy.

Worked example 2: why a small temperature rise matters so much

Question. An elementary reaction is run at 300 K and then at 320 K, with the same concentrations. Using the collision model, explain why the rate increases, and identify which effect is responsible for most of the increase.

1. Collision frequency. At 320 K the particles move faster on average, so they collide slightly more often.
2. Collision energy. At 320 K the Maxwell–Boltzmann curve is flatter and shifted to higher energy. A larger area lies beyond Eₐ, so a larger fraction of collisions have at least Eₐ.
3. Orientation. Unchanged: the particle shapes are the same.
4. Conclusion. More collisions per second, and a larger fraction of them succeed, so the rate increases. The second effect is far larger, because Eₐ lies in the tail of the curve, where the area changes fastest.

Background: the size of the two effects (not assessed). For a reaction with Eₐ = 60 kJ mol⁻¹, a standard model gives these values:

Quantity	300 K	320 K	Change
Fraction of collisions with energy ≥ Eₐ	about 1 in 2.8 × 10¹⁰	about 1 in 6.2 × 10⁹	× 4.5
Collision frequency (relative)	1	1.033	+ 3.3%

You do not need these numbers. They show why exam answers should give the energy fraction as the main reason for the faster rate. They also show how small the fraction of energetic collisions can be.

Linking the particle scale to what you measure

In the lab you never see a collision. You see a colour fade, a gas volume grow or a pH change, and you turn that into a rate. A strong answer joins the two scales in one chain:

observed change (rate doubles, reaction speeds up) ← number of successful collisions per second ← frequency × fraction with enough energy × fraction correctly oriented.

Then say which link changed and which did not. For a concentration change, only the frequency changes. For a temperature change, the energy fraction changes most, and the frequency a little. For a change of reactant (a different molecule with a different shape), the orientation fraction and Eₐ can both change. An answer that names only one link, such as “more collisions”, is usually incomplete.

Common misconceptions

• “Every collision causes a reaction.” No. Most collisions fail because they lack enough energy, the right orientation, or both.
• “A faster rate at high temperature is mainly due to more collisions.” Collision frequency rises only slightly. The main cause is the larger fraction of collisions with energy at least Eₐ.
• “Raising the temperature lowers the activation energy.” Eₐ does not change with temperature. More particles reach it.
• “Increasing concentration gives the particles more energy.” Concentration changes how often particles collide, not the energy of each collision. The fraction that succeeds is unchanged.
• “At a higher temperature the peak gets taller.” It gets lower and moves right. The area must stay the same, so a wider curve must be lower.
• “All particles at a given temperature have the same energy.” Temperature measures the average kinetic energy. The Maxwell–Boltzmann curve shows a wide spread at every temperature.
• “A hard enough collision always works.” Not if the orientation is wrong (Figure 1, panel 2). Energy and orientation are both needed.

Where this leads

Next, Topic 5.6 follows the energy of the colliding particles through a single collision: the reaction energy profile shows Eₐ as the climb from reactants to the transition state, and the Arrhenius equation turns this topic’s temperature argument into an equation. Later in the unit, catalysts change the picture by offering a route with a lower Eₐ. Now try the practice questions, then use the revision notes and the checklist.