Collision Theory and Rates of Reaction

For a chemical reaction to occur, reactant particles must collide. But not just any collision will do -- the particles must collide with sufficient energy and in the correct orientation. This is the foundation of collision theory, and it explains why reactions happen at the rates they do.

The Requirements for a Successful Collision

Collision theory states that a reaction occurs when two conditions are met simultaneously:

The particles must collide with energy equal to or greater than the activation energy (Ea). The activation energy is the minimum energy required to break the bonds in the reactants and initiate the reaction. Collisions with less energy than Ea simply result in the particles bouncing apart unchanged.
The particles must collide with the correct orientation. Even if particles have sufficient energy, the reactive parts of the molecules must be aligned appropriately. For example, in the reaction between an OH⁻ ion and a halogenoalkane, the hydroxide must approach the carbon atom bonded to the halogen -- not some other part of the molecule.

Only collisions that satisfy both conditions are called successful collisions (or effective collisions). The rate of reaction depends on the frequency of these successful collisions per unit time.

Quantifying Successful Collisions

The rate of reaction can be expressed as:

Rate = collision frequency × fraction with E ≥ Ea × steric factor

Factor	What it represents	Affected by
Collision frequency	Number of collisions per unit time per unit volume	Concentration, pressure, temperature
Fraction with E ≥ Ea	Proportion of collisions exceeding the energy barrier	Temperature, catalyst (via lowered Ea)
Steric factor	Fraction of collisions with correct orientation	Molecular geometry, complexity

The steric factor is typically much less than 1 for complex molecules (many possible orientations, few correct) and closer to 1 for simple atoms or small molecules.

Activation Energy and Energy Profiles

The activation energy can be visualised on an enthalpy profile diagram. For an exothermic reaction, the products are at a lower energy level than the reactants, but there is an energy "hump" that the reactants must overcome first. This hump represents Ea.

For an endothermic reaction, the products are at a higher energy level, and the activation energy is measured from the reactant energy level to the top of the energy barrier.

The higher the activation energy, the fewer particles have enough energy to react at a given temperature, and the slower the reaction.

Reference Table: Typical Activation Energies

Reaction	Ea / kJ mol⁻¹	Relative rate at 298 K
H₂ + I₂ → 2HI	170	Very slow
2NOCl → 2NO + Cl₂	100	Slow
C₂H₅Br + OH⁻ → C₂H₅OH + Br⁻	90	Moderate
2NO₂ → 2NO + O₂	111	Slow
Enzyme-catalysed reactions	20--50	Very fast

Notice that reactions with Ea below about 40 kJ mol⁻¹ tend to be fast at room temperature, while those above 100 kJ mol⁻¹ are extremely slow without heating or a catalyst.

The Maxwell-Boltzmann Distribution

At any given temperature, the particles in a gas or solution have a range of kinetic energies. Some are moving slowly, some are moving very fast, and most have intermediate energies. The Maxwell-Boltzmann distribution shows the spread of these energies.

Key features of the distribution curve:

The curve starts at the origin (no particles have zero energy in practice, but the probability tends to zero).
It rises to a peak -- the most probable energy -- then tails off asymptotically to the right.
The area under the curve represents the total number of particles and remains constant at a fixed number of moles.
The area to the right of the activation energy (Ea) represents the proportion of particles with sufficient energy to react.

This last point is critical: only particles in the shaded region beyond Ea can undergo successful collisions.

The following Mermaid diagram describes the key features:

graph LR
    A["Number of<br/>particles"] -->|y-axis| B["Energy"]
    B -->|x-axis| C["Curve rises<br/>to peak"]
    C --> D["Most probable<br/>energy (Emp)"]
    D --> E["Curve tails off<br/>asymptotically"]
    E --> F["Ea marked<br/>on x-axis"]
    F --> G["Shaded area<br/>right of Ea =<br/>particles that<br/>can react"]

Effect of Temperature on the Distribution

When temperature increases:

The peak of the distribution shifts to the right and becomes lower (flatter).
The total area under the curve remains the same (the number of particles has not changed).
Crucially, the area to the right of Ea increases significantly.

Even a modest temperature rise (say 10 °C) can dramatically increase the proportion of particles with energy ≥ Ea. This is the primary reason why increasing temperature increases the rate of reaction -- there are far more particles capable of overcoming the activation energy barrier.

Worked Example: Estimating the Temperature Effect

At 300 K, the Boltzmann factor for Ea = 50 kJ mol⁻¹ is:

e^(−Ea/RT) = e^(−50000 / (8.314 × 300)) = e^(−20.05) = 1.97 × 10⁻⁹

At 310 K:

e^(−Ea/RT) = e^(−50000 / (8.314 × 310)) = e^(−19.40) = 3.72 × 10⁻⁹

Ratio = 3.72 × 10⁻⁹ / 1.97 × 10⁻⁹ = 1.89 ≈ approximately doubling

This confirms the "rate doubles for every 10 °C" rule of thumb for reactions with moderate activation energies.

Common misconception: Students often say "increasing temperature simply makes particles move faster." While this is true, the key effect is the change in the shape of the distribution -- the proportion of high-energy particles increases disproportionately. The exponential nature of the Boltzmann factor means small temperature changes produce large changes in the fraction exceeding Ea.

Factors Affecting the Rate of Reaction

Concentration (or Pressure for Gases)

Increasing the concentration of reactants in solution (or increasing the pressure of gaseous reactants) means there are more particles per unit volume. This increases the frequency of collisions, which increases the frequency of successful collisions, and therefore increases the rate.

Note: increasing concentration does not change the proportion of particles with energy ≥ Ea. It only changes how often they collide.

Temperature

As discussed above, increasing temperature increases the proportion of particles with energy ≥ Ea. It also slightly increases collision frequency (particles move faster), but the dominant effect is the increased proportion above Ea.

Surface Area

For reactions involving solids, breaking the solid into smaller pieces increases the surface area exposed to the other reactant. More surface area means more collisions can occur per unit time. This is why powdered reactants react faster than lumps.

Catalysts

A catalyst provides an alternative reaction pathway with a lower activation energy. On a Maxwell-Boltzmann diagram, this shifts the Ea marker to the left. With a lower Ea, a much larger proportion of particles now have sufficient energy to react -- so the rate increases without changing the temperature.

Importantly, a catalyst:

Is not consumed in the reaction (it is regenerated).
Does not change the position of equilibrium.
Does not change ΔH for the reaction.
Does not change the Maxwell-Boltzmann distribution itself -- it changes where Ea sits on the energy axis.

Summary of Factor Effects

Factor	Effect on collision frequency	Effect on fraction ≥ Ea	Effect on rate
Increase concentration	Increases	No change	Increases
Increase temperature	Slight increase	Large increase	Large increase
Increase surface area	Increases (at surface)	No change	Increases
Add catalyst	No change	Effectively increases (lower Ea)	Increases
Add inert gas (constant V)	No change	No change	No change

Common Mistakes to Avoid

"A catalyst gives particles more energy." Wrong -- a catalyst lowers the energy barrier; it does not change the energy distribution of particles.
"Increasing temperature increases the activation energy." Wrong -- Ea is a property of the reaction, not of the conditions. Temperature changes the fraction of particles exceeding the fixed Ea.
"At higher temperature, the peak of the Maxwell-Boltzmann curve shifts left." Wrong -- the peak shifts RIGHT (to higher energy). The curve becomes broader and flatter.
"Increasing concentration increases the proportion of particles with energy ≥ Ea." Wrong -- concentration changes collision frequency, not the energy distribution.
"The area under the Maxwell-Boltzmann curve increases at higher temperature." Wrong -- the total area (representing total particles) remains constant. The curve flattens but conserves area.

Quantitative Link: The Rate Equation Preview

While collision theory provides a qualitative understanding of reaction rates, the quantitative relationship between concentration and rate is described by the rate equation, which you will study in detail in Lesson 3. For now, understand that collision theory is the conceptual foundation upon which the mathematical treatment of kinetics is built.

Summary

The rate of a reaction depends on the frequency of successful collisions -- those with energy ≥ Ea and correct orientation. Temperature affects the energy distribution of particles (exponentially, via the Boltzmann factor); concentration affects collision frequency; surface area affects the number of collisions at the solid surface; and catalysts lower the energy barrier. The Maxwell-Boltzmann distribution is the key tool for understanding why temperature and catalysts have such a dramatic effect on rate.

A-Level Deep Dive: Collision Theory and Rates of Reaction

Spec mapping

Edexcel 9CH0 specification, Topic 9 — Kinetics I sets out collision theory, the role of activation energy $E_a$ , the Maxwell–Boltzmann distribution of molecular energies, and the qualitative effect of temperature, concentration, surface area and catalysts on the rate of reaction (refer to the official specification document for exact wording). Although Topic 9 is taught in Year 1, every Maxwell–Boltzmann argument resurfaces in Topic 16 (Kinetics II) when the Arrhenius equation is introduced, and again in Topic 11 (Equilibrium II) when reaction profiles for forward and reverse reactions are compared. Collision-theory reasoning is examined principally on Paper 1 (Section A — Topics 1–10) and Paper 3 (synoptic). The Edexcel data booklet does not list collision-theory equations — students must reproduce the Maxwell–Boltzmann sketch from memory, with the y-axis labelled "fraction of molecules with a given energy" and the x-axis labelled "energy".

Worked example with full mark scheme

Question (8 marks):

(a) State three conditions that must be satisfied for a collision between two molecules to result in reaction. (2)

(b) Explain, with reference to the Maxwell–Boltzmann distribution, why an increase in temperature from 25 °C to 35 °C approximately doubles the rate of many reactions, even though the mean molecular kinetic energy increases by less than 4%. (4)

(c) Sketch on the same axes the Maxwell–Boltzmann distribution at 25 °C and at 35 °C, marking $E_a$ and shading the region of molecules able to react. (2)

Solution with mark scheme:

(a) M1 — at least two of: (i) molecules must collide; (ii) collide with energy $\geq E_a$ ; (iii) collide with the correct geometry / orientation. A1 — all three stated explicitly. Common error: writing only "molecules must collide" and leaving $E_a$ implicit; that earns M1 but not A1.

(b) M1 — mean kinetic energy is proportional to absolute temperature: from 298 K to 308 K is only a ~3.4% increase in $\langle E_k \rangle$ , so the bulk of molecules are barely faster. M1 — but the fraction of molecules with energy above $E_a$ lies on the high-energy tail of the Maxwell–Boltzmann distribution. M1 — the high-energy tail rises disproportionately on heating because the distribution shifts right and broadens, so the fraction able to clear $E_a$ roughly doubles. A1 — rate is proportional to that fraction (collision-theory result), so rate roughly doubles even though mean energy barely changed. The argument is not "molecules collide more often" — that effect is small (~2% for a 10 K rise) and is a distractor that loses the explanatory mark.

(c) M1 — both curves correct shape (skewed right, starting at origin, peaking, decaying asymptotically to zero). A1 — the higher-T curve is shifted right with a lower peak, the area beyond $E_a$ is visibly larger for the higher-T curve, and $E_a$ is marked on the energy axis. The area under each curve must be the same (constant total number of molecules) — examiners check this implicitly by noting whether the high-T curve has been "pushed flatter" rather than just "made bigger".

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks): Hydrogen peroxide decomposes according to $2\,\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\,\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})$ . The reaction is slow at room temperature but rapid in the presence of solid $\text{MnO}_2$ .

(a) Explain in terms of collision theory and activation energy why the catalysed reaction is faster at the same temperature. (3)

(b) State and explain how, if at all, the value of $E_a$ for the uncatalysed pathway is changed by the presence of $\text{MnO}_2$ . (3)

Mark scheme decomposition by AO:

Part	Mark	AO	Earned by
(a) M1	1	AO1	The catalyst provides an alternative pathway with a lower $E_a$ .
(a) M1	1	AO2	A larger fraction of molecules has energy ≥ catalysed- $E_a$ at this T.
(a) A1	1	AO2	Therefore more successful collisions per second; rate increases.
(b) M1	1	AO1	The catalyst does not alter the uncatalysed $E_a$ .
(b) M1	1	AO2	The catalyst opens a parallel lower- $E_a$ route; the original route still exists with its original $E_a$ .
(b) A1	1	AO3	Catalyst affects rate by changing the distribution of pathways, not the energy landscape of any single pathway.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. Edexcel uses Topic 9 for predominantly AO1/AO2 work; AO3 marks appear when candidates are asked about the epistemology of catalysis — what catalysts do and do not change.

Synoptic links

Connects to:

Topic 8 — Energetics I. Activation energy $E_a$ is closely related to (but not identical to) bond enthalpies of bonds broken in the rate-determining step. Reactions with strong bonds to break (e.g. $\text{N} \equiv \text{N}$ , 945 kJ mol⁻¹) have characteristically high $E_a$ — explaining why nitrogen fixation is industrially difficult.
Topic 16 — Kinetics II / Arrhenius equation. Maxwell–Boltzmann analysis is made quantitative through $k = A\exp(-E_a/RT)$ . The "fraction of molecules with energy ≥ $E_a$ " is exactly the Boltzmann factor $\exp(-E_a/RT)$ .
Topics 10–11 — Equilibrium I and II. Catalysts increase forward and reverse rates equally because both pathways share the same transition state, so equilibrium position is unchanged but the system reaches equilibrium faster.
Topic 19 — Transition metals. Heterogeneous catalysts (Fe in Haber, V₂O₅ in Contact, Ni in hydrogenation, Pt-Rh in catalytic converters) all rely on adsorption energy lowering the activation barrier on the surface.
A-Level Physics — kinetic theory of gases. The Maxwell–Boltzmann distribution is derived in Year 13 statistical mechanics; A-Level Chemistry candidates are expected to use the result without derivation but should know it follows from "particles in thermal equilibrium adopt a Boltzmann-distributed energy".

Mark-scheme literacy

AO	Typical share on Topic 9 questions	Earned by
AO1 (knowledge)	30–40%	Stating collision-theory conditions, naming axes on a Maxwell–Boltzmann sketch, recalling that rate ∝ fraction with energy ≥ $E_a$
AO2 (application)	45–55%	Applying collision theory to predict the effect of a change in T, [c], surface area or catalyst on rate; interpreting Maxwell–Boltzmann sketches
AO3 (analysis)	10–20%	Discussing why temperature dominates over concentration; comparing catalyst vs reactant effects; reasoning about reaction profiles for multi-step mechanisms

Examiner-rewarded phrasing: "the fraction of molecules with energy ≥ $E_a$ increases", "the distribution flattens and shifts right", "the catalyst provides an alternative pathway of lower activation energy", "the rate is proportional to the number of effective collisions per second". Phrases that lose marks: "molecules collide harder" (vague), "molecules have more energy" (misses the distribution argument), "the catalyst lowers the activation energy" without clarifying that it is a different pathway (factually loose — the original $E_a$ is unchanged).

A specific Edexcel pattern to watch: when asked about the effect of temperature on rate, candidates frequently invoke an increase in collision frequency as the dominant cause. It is not. The dominant effect — by an order of magnitude — is the increase in the fraction of molecules with energy above $E_a$ . State both, but identify which is dominant.

Grade-band model answers

3-mark question

Question: Explain why powdered calcium carbonate reacts faster with hydrochloric acid than a single block of the same mass.

Grade C response (~120 words):

The powder has a larger surface area than the block. A larger surface area means more particles are exposed to acid and so more collisions can happen between the calcium carbonate and the H⁺ ions. More collisions per second leads to more successful collisions per second, and so the rate is higher.

Examiner commentary: Full marks (3/3). The candidate identifies surface area as the variable, links it to collision frequency, and connects collision frequency to rate. Concise and correct, though the answer does not distinguish between total collisions and successful collisions — at A-Level depth this is a small gap.

Grade A response (~190 words):*

Reaction occurs only at the solid–liquid interface. With a single block, the vast majority of the calcium carbonate is internal and inaccessible to acid molecules; only the outer surface participates. Powdering the same mass exposes the previously internal carbonate to the acid, raising the surface area by orders of magnitude.

By collision theory, the rate of reaction is proportional to the number of effective collisions per unit time between H⁺ ions and exposed CaCO₃ surface sites. Increasing surface area increases the number of accessible sites, and therefore the number of collisions per second between H⁺ ions and the solid; the fraction of those collisions which are effective (energy ≥ $E_a$ , correct geometry) is unchanged because temperature is unchanged. So total successful collisions per second rises in proportion to surface area, and rate rises in proportion.

Examiner commentary: Full marks (3/3). The A* response separates "more collisions" from "more effective collisions" and explicitly notes that the effective-fraction is unchanged — a subtlety that distinguishes the candidate as analytically careful.

6-mark question

Question: A 10 °C rise from 25 °C to 35 °C approximately doubles the rate of many reactions. (a) Explain this in terms of the Maxwell–Boltzmann distribution. (b) Why is the doubling only approximate, and for what kinds of reactions might it be very different from a factor of two?

Grade A response (~310 words):*

(a) The Maxwell–Boltzmann distribution gives the fraction of molecules with each energy. Only molecules with energy ≥ $E_a$ can react. From 298 K to 308 K, the mean kinetic energy rises by only ~3.4% (proportional to T), so the bulk of the distribution barely shifts. However, the high-energy tail (above $E_a$ ) rises disproportionately because the distribution shifts right and broadens. The fraction in the tail roughly doubles. Since rate ∝ effective collisions per second ∝ fraction with energy ≥ $E_a$ × collision frequency, and collision frequency rises by only ~2%, the rate is dominated by the tail-fraction increase and roughly doubles.

(b) The factor of two is approximate because the Boltzmann factor $\exp(-E_a/RT)$ is sensitive to the value of $E_a$ . For reactions with $E_a \approx 50$ kJ mol⁻¹ (typical of many solution-phase reactions), the doubling rule works well in the 25–35 °C range. For reactions with very low $E_a$ (≲ 10 kJ mol⁻¹, e.g. radical–radical recombination, almost diffusion-controlled), the tail-fraction is already most of the distribution, so heating barely changes it; the rate rises only slightly. For very high $E_a$ (≳ 200 kJ mol⁻¹, e.g. $\text{N}_2$ dissociation), the tail-fraction is tiny but rises more than twofold for a 10 K rise; rate could quadruple or more.

Examiner commentary: Full marks (6/6). The candidate names the Boltzmann factor as the source of $E_a$ -sensitivity and gives concrete $E_a$ ranges for radical recombination versus high-barrier dissociation — exactly the kind of A-Level synoptic link from kinetics to mechanism that distinguishes A* from A.

9-mark question

Question: A student claims that a catalyst speeds up a reaction by "giving the molecules extra energy". Critically evaluate this claim, with reference to (i) the energy of individual molecules, (ii) the Maxwell–Boltzmann distribution at the reaction temperature, and (iii) the activation energy of the catalysed and uncatalysed pathways.

Grade A response (~430 words):*

The student's claim is incorrect, though it captures a partial intuition.

(i) Energy of individual molecules. A catalyst does not transfer energy to reactant molecules. The molecules' kinetic energy distribution is determined by the temperature of the system; at constant T the molecular energy distribution is identical with or without catalyst. A catalyst at the same T as its surroundings cannot make individual molecules "hotter" — that would violate the second law. The catalyst's role is to provide a different pathway, not to energise molecules.

(ii) Maxwell–Boltzmann distribution. Because temperature is unchanged, the Maxwell–Boltzmann distribution is unchanged. The graph is fixed. What changes is the position of $E_a$ on that graph: the catalysed pathway has a lower $E_a$ , so a vertical line marking "energy needed to react via the catalysed route" is to the left of the line for the uncatalysed route. The shaded area to the right of the catalysed- $E_a$ — the fraction of molecules able to react via the catalysed pathway — is therefore much larger than the area beyond the uncatalysed- $E_a$ .

(iii) Activation energies of the two pathways. Crucially, the catalyst does not lower the $E_a$ of the uncatalysed pathway; that pathway is unchanged in energy. What the catalyst does is open a parallel pathway with its own (lower) $E_a$ . Both pathways operate simultaneously — the catalysed pathway is just so much faster that it dominates the observed rate. If the catalyst were removed, the uncatalysed pathway would still be there, with the same $E_a$ .

The student's intuition is partly right in that the catalysed reaction appears to behave as if molecules had more energy — because the same Maxwell–Boltzmann distribution now overlaps a much larger "able to react" region. But the energy of the molecules is unchanged; the barrier has been lowered, not the energy raised. This distinction matters because it explains why catalysts do not shift equilibrium: forward and reverse pathways share the same lowered barrier, and $\Delta H$ — determined by the energy difference between reactants and products — is unchanged.

Examiner commentary: Full marks (9/9). The candidate refutes the claim cleanly, then unpacks each subpart with the relevant Maxwell–Boltzmann or pathway argument, and closes with the synoptic point about equilibrium. The "partial intuition" framing is examiner-aware: it acknowledges the source of the misconception rather than dismissing it.

A-Level-depth misconceptions

The errors that distinguish A from A* on collision-theory questions:

"Catalysts lower $E_a$ " is loose phrasing. Catalysts open a parallel pathway with a different $E_a$ . The original pathway, with its original $E_a$ , still exists. A-Level questions reward the parallel-pathway framing.
Conflating more collisions with more effective collisions. Heating raises both, but the increase in effective-fraction dominates by an order of magnitude. Stating only "more collisions" loses the AO2 mark for explaining why a 10 K rise roughly doubles rate.
Mis-sketching the Maxwell–Boltzmann curve. The curve must start at the origin (no molecules with zero kinetic energy), peak, and decay asymptotically to zero. Drawing a Gaussian (symmetric bell) loses marks because the distribution is right-skewed. The high-energy tail is thicker than a Gaussian would predict.
Forgetting area conservation. When sketching distributions at two temperatures, the area under each curve must be equal (same total number of molecules). The high-T curve is shifted right and flattened — not just "made wider".
Treating $E_a$ as a property of the reactants alone. $E_a$ depends on the pathway, including any catalyst. The same reactants can have many $E_a$ values across different mechanisms.
Geometry / orientation as an afterthought. Even at energy ≥ $E_a$ , collisions only react if the geometry is correct. For multi-atomic molecules (e.g. esterification, SN2), the steric factor can reduce the effective rate by orders of magnitude. Not stating geometry as a separate condition costs M1 marks.
Concentration treated as identical to pressure. For gases, raising pressure raises [c] and so raises collision frequency. For solids, "concentration" is meaningless and surface area becomes the relevant variable. Conflating these on a heterogeneous-catalysis question loses application marks.

Common errors and mark-loss patterns

Endemic Paper 1 patterns:

Energy axis mis-labelled. The Maxwell–Boltzmann x-axis is energy, not speed or temperature. The y-axis is fraction of molecules with that energy (or number of molecules with that energy — both accepted), not rate. Labelling the y-axis as "rate" is an immediate M0.
Two-curve sketches with mismatched areas. Candidates draw the higher-T curve simply taller and wider, doubling the area. Examiners check area visually; the higher-T curve must be flatter with a lower peak so areas match.
Stating "rate doubles for every 10 °C" as a law. It is a rule of thumb that holds for typical $E_a$ values around 50 kJ mol⁻¹ at room temperature, not a general result. Candidates citing it as universal lose AO3 marks for failing to qualify.
Treating "effective collision" as redundant. Examiners reserve marks for candidates who explicitly write "effective collision" or "successful collision" rather than just "collision". Use the precise term.

Going further — university and academic signposting

Collision theory points toward several undergraduate trajectories:

Physical chemistry (Year 1): the simple collision theory $z = n_A n_B \sigma \sqrt{8 k_B T / (\pi \mu)}$ (where $\sigma$ is the collision cross-section and $\mu$ the reduced mass) provides a quantitative collision frequency. Combining with the Boltzmann factor $\exp(-E_a/RT)$ gives a first-principles rate constant that typically overestimates by 10–10⁶, motivating the steric factor correction.
Transition state theory (Year 2): the Eyring equation $k = (k_B T / h) \exp(-\Delta G^\ddagger / RT)$ replaces collision theory's frequency factor $A$ with a more rigorous statistical-mechanical treatment of the transition state. $\Delta G^\ddagger = \Delta H^\ddagger - T \Delta S^\ddagger$ — the entropy of activation explains why steric factors are small for tight transition states.
Molecular dynamics (Year 3): modern reaction kinetics is studied by simulating molecular trajectories explicitly; rate constants are extracted from the fraction of trajectories that cross the transition state.
Femtochemistry (Year 4 / research): the 1999 Nobel Prize work by Zewail used femtosecond lasers to image transition-state structures directly, confirming the validity of collision/transition-state pictures and revealing previously invisible intermediates.

Oxbridge interview prompt: "If a 10 K temperature rise typically doubles reaction rate, why doesn't it also double the rate of a diffusion-controlled reaction? What sets the upper limit on rate in solution?"

Required practical reference

Edexcel A-Level Chemistry Core Practical 5 (CP5) — investigation of the rates of hydrolysis of halogenoalkanes is the canonical Year 1 rates practical for collision theory. Students measure the time for a silver halide precipitate to appear when a halogenoalkane (chloro-, bromo-, iodo-) reacts with aqueous silver nitrate in ethanol; the inverse of the time is proportional to the initial rate. Repeating at two or more temperatures lets students verify that rate roughly doubles per 10 K rise — direct evidence for the Maxwell–Boltzmann argument. CP8 — finding the activation energy of a reaction (the iodine clock with peroxodisulfate) extends this by collecting rate data at multiple temperatures so that an Arrhenius plot (Lesson 5) can be constructed. Both practicals depend on understanding collision theory qualitatively before quantification. Students should be able to describe how they controlled concentration and total volume, why the ethanol/water ratio matters for CP5 (solvent effect on $E_a$ ), and how they ensured thermostatting accuracy at each temperature.

Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Chemistry (9CH0) specification, Paper 1 — Topic 9: Kinetics I. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

graph TD
    A["Reactant molecules<br/>at temperature T"] --> B["Maxwell-Boltzmann<br/>distribution of energies"]
    B --> C{"Collision occurs?"}
    C -->|"Yes"| D{"Energy ≥ E_a?"}
    C -->|"No"| Z["No reaction"]
    D -->|"Yes"| E{"Correct geometry?"}
    D -->|"No"| Z
    E -->|"Yes"| F["Effective collision<br/>→ products"]
    E -->|"No"| Z
    G["Increase T"] --> B
    H["Increase [c]"] --> C
    I["Increase surface area"] --> C
    J["Add catalyst"] --> K["New pathway<br/>lower E_a"]
    K --> D

    style F fill:#27ae60,color:#fff
    style K fill:#3498db,color:#fff