Three conditions in the collision model
Collision theory in 2.11.1 said that reactant particles must collide, and the collisions must be effective. AP sums this up as the collision model碰撞模型: for a collision to cause a reaction, all three conditions must hold:
- The particles collide. The number of collisions per unit time, the collision frequency, depends on reactant concentrations.
- The energy is enough. The collision’s energy must be at least the activation energy (Ea)活化能 so old bonds can break and new ones form. AP treats Ea as the minimum energy a collision needs (2.11.1 compared this with the definition in Chinese textbooks).
- The orientation is right. The particles must collide in a suitable direction, so the atoms that will bond can get close.
So the rate can be thought of as the product of the three:
Energy distribution: the Maxwell–Boltzmann distribution
At a given temperature, particles have a range of kinetic energies. The curve showing how many particles have each energy is the Maxwell–Boltzmann distribution麦克斯韦–玻尔兹曼分布 (U3.5 used it for the speeds of gas molecules). The area under the curve to the right of Ea represents the fraction of particles with energy at least Ea.
When the temperature rises:
- the curve gets lower and wider, and its peak moves right;
- the total area under the curve stays the same (the number of particles doesn’t change);
- the area to the right of Ea grows noticeably: a larger fraction has enough energy, so the rate increases and k increases.
Change the temperature and watch the shaded area to the right of Ea:
Schematic: the activation energies are far smaller than in real reactions so the shaded areas are visible. The fractions are calculated exactly from the energy distribution.
Why does heating speed reactions up so much?
Heating also makes particles move faster and collide more often, but that effect is small: collision frequency is roughly proportional to √T, so going from 300 K to 310 K adds less than 2%. The main effect is on the fraction of collisions with enough energy: for a reaction with Ea = 50 kJ/mol, the same 10 K rise nearly doubles that fraction.
Under “Temperature and energy” below, change T and Ea and compare the two bars; then open “Orientation” to see why enough energy isn’t always enough:
Cl + NOCl → Cl₂ + NO
Schematic model: collision frequency is calculated as proportional to √T; the fraction with enough energy uses collision theory’s e^(−Ea/RT), with R = 8.314 J/(mol·K).
Explaining the factors with the collision model
| Change | Collision frequency | Fraction with enough energy | k | Rate |
|---|---|---|---|---|
| Higher reactant concentration (or compressing a gas) | Increases | No change | No change | Increases |
| Greater surface area of a solid | Increases | No change | — | Increases |
| Higher temperature | Increases slightly | Increases greatly | Increases | Increases greatly |
| Adding a catalyst (U5.11) | No change | Increases (lower Ea) | Increases | Increases |
Concentration changes the rate, not k; temperature and catalysts change k.
A reaction that barely happens at room temperature often has strong bonds to break, so its activation energy is high and only a tiny fraction of collisions has enough energy. The N≡N bond in N₂, for example, is very strong (average bond enthalpy 941 kJ/mol), and N₂ and H₂ barely react at room temperature.
Example 1: Reading a distribution graph
Two Maxwell–Boltzmann curves for the same gas at T₁ and T₂: curve 1 has a taller peak farther left; curve 2 is lower and wider, with its peak farther right. Ea is marked on the graph. Which temperature is higher? At which temperature is the reaction faster? Use the graph to justify your answers.
- Identify: what the peak means. The peak sits at the most probable kinetic energy; the higher the temperature, the greater the average kinetic energy, so the peak moves right and the curve gets lower and wider. Curve 2’s peak is farther right, so .
- Compare total areas. The total area under both curves is the same, because the number of particles is the same; curve 2 is lower because the same particles are spread over a wider range of energies.
- Compare the area right of Ea. The area to the right of Ea is the fraction of particles with energy at least Ea. Curve 2 has much more area there than curve 1.
- Connect to rate. At T₂ a larger fraction of particles has enough energy, so more collisions per unit time are energetic enough to react: the reaction is faster at T₂.
- What graders look for. Base the reason on the area to the right of Ea (the fraction of particles with energy ≥ Ea), not just “higher temperature, faster molecules”.
Example 2: Which factor is at work?
A gas-phase reaction takes place in a rigid container. For each change, describe the effect on collision frequency, on the fraction of collisions with enough energy, and on k: (1) adding more reactant; (2) raising the temperature.
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Identify: the two factors in rate. One is collision frequency (depends on concentration); the other is the fraction of collisions with enough energy (depends on temperature and Ea). k reflects only the second, so it doesn’t change with concentration.
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(1) Adding reactant: collision frequency. The rigid container’s volume is fixed, so the reactant concentration rises; with more particles per unit volume, collisions are more frequent.
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(1) Adding reactant: energy fraction and k. The temperature is unchanged, so the energy distribution is unchanged: the fraction with enough energy and k are unchanged. The rate rises only because collisions are more frequent.
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(2) Raising temperature: energy fraction. The distribution shifts right and widens, so the area to the right of Ea grows a lot: the fraction with enough energy rises sharply. The particles also move faster, so collision frequency rises slightly.
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(2) Raising temperature: k. k increases. The rate rises mainly because a larger fraction of collisions has enough energy; the extra collisions contribute only a little.
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Summary.
- Add reactant: frequency ↑, fraction same, k same
- Raise T: frequency slightly ↑, fraction ↑↑, k ↑
- Raising the temperature
A larger fraction of collisions has enough energy.
- Adding more reactant
Changing concentration changes the rate, not k.
- Compressing gaseous reactants into a smaller volume
Concentrations rise; k stays the same.
- Adding a catalyst
The activation energy is lower (U5.11).
- Lowering the temperature
A smaller fraction of collisions has enough energy.
- Diluting a reactant solution with water
Concentrations fall; k stays the same.
- Adding an unreactive gas at constant volume
Neither the reactant concentrations nor the temperature change.
- Letting the reaction run longer
At constant temperature, k doesn't change over time.
- Moving the reaction from an ice bath to a 50 °C water bath
The temperature rises.
- Removing the catalyst that was added
The reaction goes back to the higher-Ea pathway.
Common mistakes
Mistake 1: “Heating speeds a reaction up only because collisions are more frequent.” Collision frequency is proportional only to √T and rises very little. The rate rises sharply mainly because a much larger fraction of collisions has energy at least Ea.
Mistake 2: “At a higher temperature, the whole distribution curve rises.” The total area under the curve represents the number of particles, which doesn’t change. At a higher temperature the curve is lower and wider with its peak farther right; only the area to the right of Ea grows.
Mistake 3: “Every collision with enough energy leads to a reaction.” The orientation must also be right. In Cl + NOCl → Cl₂ + NO, a Cl atom that hits the O end can’t form a Cl–Cl bond, however much energy it has.
Quick check
Which change increases the fraction of collisions with energy at least Ea?
B, C, and D only increase the number of collisions without changing the energy distribution; heating shifts the distribution right, increasing the fraction beyond Ea.
Compared with the Maxwell–Boltzmann curve for a gas at a lower temperature, the curve at a higher temperature has:
The average kinetic energy is greater, so the peak moves right; the number of particles is unchanged, so the area is unchanged, and the curve becomes lower and wider.
For Cl + NOCl → Cl₂ + NO, which collision could form products?
The collision needs both enough energy and the right orientation: the new Cl–Cl bond has to form between the two Cl atoms.
Increasing a reactant's concentration increases the rate. Which explanation is best?
At the same temperature, the energy distribution, Ea, and k don't change; collisions are just more frequent.
Reaction 1 has Ea = 40 kJ/mol and reaction 2 has Ea = 90 kJ/mol. If both are heated from 300 K to 310 K, which rate increases by the larger factor?
A larger Ea sits farther out in the tail of the distribution, where heating increases the area by a larger factor.
Gas molecules collide an enormous number of times each second, yet most reactions aren't instantaneous. Which explanation is best?
There are many collisions, but only a small fraction has enough energy, and only some of those are correctly oriented.
A mixture of N₂ and H₂ barely reacts at room temperature. Which explanation is best?
Ammonia synthesis is exothermic (ΔH < 0), so C starts from a false premise; the molecules collide constantly, but too few collisions have enough energy.
AP-style free response
Gases A and B react in a rigid container.
(a) At constant temperature, doubling the concentration of A doubles the rate. Explain using the collision model.
(b) Describe how the Maxwell–Boltzmann distribution of the reactant molecules changes when the temperature rises from T₁ to T₂, and identify the part of the graph that represents molecules with enough energy.
(c) A 10 K temperature rise roughly doubles the rate, while the collision frequency rises by only about 2%. Explain the difference.
(d) A student says, “Any collision with energy at least Ea will form products.” Evaluate the claim using the collision model.
Model answer and scoring
(a) Doubling [A] doubles the number of A molecules per unit volume, so A–B collisions happen twice as often. The temperature is unchanged, so the fractions with enough energy and with the right orientation are unchanged; effective collisions are twice as frequent, and the rate doubles.
(b) The T₂ curve is lower and wider, with its peak farther right; the total area under the curve is the same. The area under the curve to the right of Ea represents molecules with energy at least Ea, and it is noticeably larger at T₂.
(c) The rate also depends on the fraction of collisions with enough energy. Heating shifts the distribution right, so the area to the right of Ea grows greatly (here, roughly doubling), far more than the 2% rise in collision frequency. The increase in rate comes mainly from the larger fraction of collisions with enough energy.
(d) The claim is incorrect. Besides enough energy, a collision must have the right orientation, so the atoms that will bond come into contact; a badly oriented collision just bounces off, even with enough energy.
Scoring: (a) collision frequency doubles, with the energy and orientation fractions unchanged; (b) the change in the curve, unchanged area, and the area to the right of Ea; (c) the fraction with enough energy is the main cause; (d) orientation.
Summary
- Collision model: a reaction needs particles to collide with energy at least Ea and the right orientation. Rate ∝ collision frequency × fraction with enough energy × fraction correctly oriented.
- Maxwell–Boltzmann distribution: at a higher temperature the curve is lower, wider, and shifted right, with the same area and more area to the right of Ea.
- Concentration, pressure, and surface area change only the collision frequency, not k; temperature and catalysts change the fraction with enough energy, and so change k.
- Heating speeds reactions up mainly by sharply raising the fraction of collisions with enough energy, not by increasing the number of collisions.
AP tip: In a collision-model explanation, say whether the change affects the collision frequency or the fraction of collisions with enough energy; don’t just write “more collisions.” For temperature effects, refer to the change in area to the right of Ea on the Maxwell–Boltzmann distribution.