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The rate of a chemical reaction is the speed at which reactants are consumed and products are formed — quantitatively, the change in concentration of a chosen species per unit time. Chemists measure rates by following a property that varies as the reaction proceeds: volume of gas evolved, mass of the reaction vessel, the absorbance of a coloured species, the conductivity of an ionic solution, or the titre obtained from quenched samples. Underneath every measured rate sits a microscopic picture — collision theory — developed in the 1910s and 1920s through the work of W.C.M. Lewis (whose 1918 calculations first quantified collision frequencies) and Michael Polanyi (whose transition-state ideas in the 1920s would later mature into modern kinetics). Five factors govern how fast reactions go: concentration, temperature, surface area, catalysts, and (for gases) pressure. This lesson develops the qualitative collision-theory picture, the Maxwell-Boltzmann distribution of molecular energies, and the AO2/AO3 explanations the AQA examiners reward. The quantitative apparatus — rate equations, the rate constant, the Arrhenius equation, and rate-determining steps — is built in lessons 1 to 3 of this course.
Spec mapping (AQA 7405): This lesson anchors §3.1.5 (Kinetics) — specifically the qualitative collision-theory content that underpins the entire kinetics block. It cross-references lesson 1 (rate equations, where collision frequency becomes quantitative through orders of reaction), lesson 2 (the Arrhenius equation, which quantifies the exponential temperature dependence implied here), lesson 3 (rate-determining step and reaction mechanisms, where collision theory meets molecularity), §3.1.4 (Energetics — activation energy emerges as the kinetic barrier on an enthalpy-profile diagram), and §3.1.6 (Equilibrium — where the dynamic balance of forward and reverse rates defines the equilibrium position). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Definitions of rate, collision theory, activation energy, and the Maxwell-Boltzmann distribution are AO1 recall items. Sketching MB curves at two temperatures (or with/without catalyst) and explaining qualitative rate changes are AO2 application tasks that feature on every Paper 1 and Paper 2. Predicting how the shape of the MB curve and the position of the activation-energy line interact — including quantitative comparison of the shaded fraction with E ≥ Ea — is AO3, the "synthesis and evaluation" domain that separates Grade A from Grade A*.
The rate of reaction is defined as the change in concentration of a reactant or product per unit time:
rate = −Δ[reactant]/Δt = +Δ[product]/Δt
The negative sign for reactants makes the rate a positive quantity (reactant concentration decreases over time). The standard SI units are mol dm⁻³ s⁻¹ — concentration in mol per cubic decimetre, time in seconds. For very fast reactions chemists may use mol dm⁻³ ms⁻¹ or mol dm⁻³ μs⁻¹; for very slow reactions (e.g. weathering, corrosion) mol dm⁻³ year⁻¹ may be appropriate. Whatever the units, the rate has dimensions of concentration over time.
Because rate is the slope of a concentration-time graph, every rate experiment reduces to following a measurable property that varies linearly with concentration. The major techniques are:
Practical tip: For each technique, ensure that the property measured is linearly proportional to concentration of the species of interest, and that the measurement does not itself perturb the reaction. Colorimetry is non-invasive (light passes through); titrimetric sampling is invasive (an aliquot is consumed) and requires effective quenching.
A concentration-time graph is rarely linear; rate changes as the reaction proceeds. The initial rate is the gradient of the tangent drawn at t = 0. Constructing a clean tangent by eye is error-prone — examiners reward students who draw a long tangent (extending well past the curve in both directions) and use a triangle on the graph paper to read off Δy/Δx with at least three significant figures. The instantaneous rate at any other time t is similarly obtained from the tangent at that point. The mean rate between two times is the chord (straight line) gradient between the two points — useful for crude estimates but not for kinetic analysis.
Exam Tip: When asked to find an initial rate from a graph, state the method ("tangent drawn at t = 0, gradient = Δ[X]/Δt"), show the triangle clearly on the diagram, and quote units. Marks are awarded for method even if your numerical answer is slightly off the published value.
For two reactant particles A and B to react, three conditions must all be satisfied:
A collision that satisfies all three conditions is called a successful collision (or effective collision). The rate of reaction is proportional to the frequency of successful collisions per unit volume per unit time.
Key Definition: The activation energy E_a is the minimum kinetic energy that a colliding pair of particles must possess for a reaction to occur. It corresponds to the energy required to reach the transition state (the highest point on the reaction-coordinate diagram).
Even at sufficient energy, not every collision leads to reaction — the geometry must be right. For a reaction such as CH₃Cl + OH⁻ → CH₃OH + Cl⁻, the OH⁻ must approach the carbon along the C-Cl axis from the opposite side; an approach from any other direction fails. This orientation requirement (formalised as the steric factor p in undergraduate treatments) means that the rate is typically far below the maximum possible if every energetic collision reacted. At A-Level, the steric requirement is stated qualitatively — most collisions, even energetic ones, are geometrically unproductive.
Why most collisions fail. At room temperature, a typical small molecule undergoes ~10⁹ collisions per second. If every collision led to reaction, all chemistry would be over in nanoseconds. The vast majority of collisions fail either because the colliding pair lacks sufficient energy (E < E_a) or because their orientation is unproductive. The successful-collision rate is many orders of magnitude lower than the total collision frequency.
The Maxwell-Boltzmann distribution describes the spread of molecular kinetic energies in a sample of gas (or liquid) at thermal equilibrium at a given temperature. It is named for James Clerk Maxwell (who derived a velocity distribution for ideal gases in 1860) and Ludwig Boltzmann (who generalised the result in the 1870s using statistical mechanics).
The standard exam-paper sketch has:
A vertical line drawn at E = E_a divides the curve into two regions. The area to the right of the E_a line represents molecules with energy sufficient to react if they collide. This shaded fraction is the key quantity for rate explanations.
Mathematical aside. The Maxwell-Boltzmann energy distribution function for an ideal gas is N(E) ∝ √E · exp(−E/k_B T), where k_B is the Boltzmann constant. The right-hand tail therefore decays as exp(−E/k_B T) — exponentially fast in E. This exponential factor is what reappears in the Arrhenius equation as exp(−E_a/RT) and is responsible for the dramatic temperature dependence of reaction rates.
Increasing the concentration of a reactant increases the number of particles per unit volume. The frequency of A-B collisions in a bimolecular reaction is proportional to [A][B] (more strictly, to the product of the partial densities). Doubling [A] approximately doubles the collision frequency between A and B, and therefore doubles the rate (for a first-order reaction in A — see lesson 1 for the full treatment of orders).
The Maxwell-Boltzmann distribution is unchanged by concentration changes — the fraction of molecules with E ≥ E_a depends only on temperature. Concentration affects collision frequency, not the fraction of energetic collisions.
When a solid reacts with a gas or solution, only the surface particles are exposed to attack. Breaking the solid into smaller pieces (or grinding it to a powder) increases the surface-to-volume ratio, exposing more particles. The frequency of collisions per unit time at the surface increases, so the rate increases.
Example: powdered marble (CaCO₃) reacts with dilute HCl visibly faster than a single marble chip of the same mass. A chip with a mass of 1 g and side ~0.5 cm has a surface area of ~1.5 cm²; the same mass ground to 0.01 mm grains has a surface area of ~3,000 cm² — a two-thousand-fold increase.
Increasing temperature has two compounding effects on rate:
exp(−50000/(8.314 × 310)) / exp(−50000/(8.314 × 300)) = exp[(50000/8.314)(1/300 − 1/310)] ≈ exp(0.647) ≈ 1.91.
So a 10 K rise nearly doubles the fraction of molecules above the activation barrier. Combined with the 1.6% increase in collision frequency, the rate roughly doubles. This is the canonical "rule of thumb" — for many reactions with E_a near 50 kJ mol⁻¹, a 10 K temperature rise doubles the rate. For lower-E_a reactions (e.g. 25 kJ mol⁻¹) the effect is smaller; for higher-E_a reactions (e.g. 100 kJ mol⁻¹) the effect is larger.
The mechanism the AQA examiners want stated explicitly is: a temperature rise shifts the Maxwell-Boltzmann curve so that the peak moves right and lowers, the distribution broadens, and crucially the shaded area beyond E_a grows much faster than the mean kinetic energy itself. The collision-frequency increase is a minor contributor; the energy-distribution change is dominant.
Common misconception. "Particles move faster at higher temperatures, so they collide more often, so rate goes up." This earns partial credit at GCSE but is insufficient at A-Level. The dominant effect is the shift in the energy distribution — far more molecules now exceed E_a — not the modest increase in collision frequency. Marks lost on this every year.
A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy E_a', without itself being consumed overall (it is regenerated by the end of the catalytic cycle).
On a Maxwell-Boltzmann diagram:
Catalysts come in two flavours:
Key insight. A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount (it lowers the energy of the transition state, which is shared). At equilibrium, both forward and reverse rates therefore increase equally — equilibrium is reached faster, but the equilibrium position is unchanged. This is a recurring exam question (see lesson 5).
For gas-phase reactions, increasing the pressure at constant temperature is equivalent to compressing the gas into a smaller volume — which increases the concentration (mol per unit volume) of every gas species. Doubling p at constant T halves V and therefore doubles every gas concentration. Collision frequency increases; rate increases.
The Maxwell-Boltzmann distribution is unchanged by pressure (temperature is constant, so the energy distribution is the same). Pressure affects collision frequency, not the energetic fraction. This is exactly analogous to the concentration effect — pressure for gases plays the same role as concentration for solutions.
Caveat. This treatment assumes ideal-gas behaviour. At very high pressures (real-gas regime), intermolecular forces and excluded-volume effects modify the picture; this is beyond the scope of A-Level.
A student decomposes H₂O₂(aq) and measures [H₂O₂] every 30 s. The first three readings are: t = 0, [H₂O₂] = 1.00 mol dm⁻³; t = 30 s, [H₂O₂] = 0.85 mol dm⁻³; t = 60 s, [H₂O₂] = 0.72 mol dm⁻³. Estimate the initial rate.
Draw a tangent at t = 0. From the three early points, the gradient is approximately:
Δ[H₂O₂]/Δt = (1.00 − 0.85)/30 = 0.15/30 = 5.0 × 10⁻³ mol dm⁻³ s⁻¹ (initial rate, magnitude).
Strictly the gradient becomes less negative as the curve flattens; a properly-drawn tangent at t = 0 might give a slightly larger initial rate (e.g. 5.5 × 10⁻³ mol dm⁻³ s⁻¹) — the chord method underestimates initial rate.
A first-order reaction starts at [A] = 0.80 mol dm⁻³. From the graph, [A] = 0.40 mol dm⁻³ at t = 120 s and [A] = 0.20 mol dm⁻³ at t = 240 s. Show that the reaction is first order.
First half-life: 0.80 → 0.40, time taken = 120 s. Second half-life: 0.40 → 0.20, time taken = 240 − 120 = 120 s. The half-lives are equal, confirming first-order kinetics. (See lesson 1 for the formal treatment.)
In a Maxwell-Boltzmann diagram, the shaded area to the right of E_a at 300 K represents 5% of molecules. At 310 K, the shaded area represents 10%. Explain the effect on rate.
The fraction of molecules with energy ≥ E_a has doubled (5% → 10%). The collision frequency increases by a much smaller factor (~1.6%, from √(T₂/T₁)). The rate is the product of collision frequency and the energetic fraction, so the rate approximately doubles. This matches the canonical "10 K rise doubles the rate" rule.
Sketch and label a Maxwell-Boltzmann diagram showing the effect of adding a catalyst to a reaction. Explain how the catalyst increases the rate.
The diagram shows a single curve (temperature is unchanged) starting at the origin, rising to a peak, and tailing off to the right. Two vertical lines are drawn: E_a (uncatalysed, further right) and E_a' (catalysed, further left). The area to the right of E_a' is clearly larger than the area to the right of E_a. Explanation: the catalyst provides an alternative pathway with lower E_a'. The molecular energies are unchanged, but more molecules now have E ≥ E_a', so more successful collisions occur per unit time, and the rate increases.
Compare the initial rate of reaction of 5.0 g of marble chips (each ~0.5 cm side) with 5.0 g of powdered CaCO₃ (~0.01 mm grain) in excess 1.0 mol dm⁻³ HCl.
The chips and powder have the same mass and therefore the same total moles of CaCO₃ available. However, the powdered sample has a much higher surface-to-volume ratio (~3000 cm² vs ~1.5 cm² — a ~2000-fold increase). Initial rate is proportional to surface area exposed to HCl, so the powdered sample reacts approximately 2000× faster. (In practice, the powdered reaction may be so fast that diffusion of HCl through the foam of CO₂ bubbles becomes rate-limiting — a separate complication.)
A gas-phase reaction has rate r₁ at 100 kPa total pressure. The pressure is doubled to 200 kPa at constant temperature. Predict the new rate (assume the reaction is overall first order).
Doubling pressure at constant T doubles concentration of every gas species (pV = nRT, so n/V doubles when p doubles at fixed T). For a first-order reaction r = k[X], doubling [X] doubles r. So r₂ = 2r₁. Note that the MB distribution is unchanged (temperature constant), so the fraction with E ≥ E_a is unchanged — only the collision frequency is affected.
Question 1. [12 marks total]
A student studies the reaction Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) by collecting the H₂ gas evolved against time.
(a) Sketch a Maxwell-Boltzmann distribution curve for HCl(aq) at room temperature. Label the axes, the most probable energy E_mp, the activation energy E_a, and shade the region representing molecules with sufficient energy to react. [4 marks]
(b) Predict and explain the effect on the initial rate of doubling the concentration of HCl(aq) at constant temperature. Refer to collision theory in your answer. [3 marks]
(c) A second experiment is carried out at 30 °C instead of 20 °C. Explain why the initial rate is approximately twice as fast at the higher temperature, including reference to the Maxwell-Boltzmann distribution. [3 marks]
(d) A third experiment uses copper(II) chloride solution as a catalyst. Explain how the catalyst increases the rate, including a sketch (or description) of how the Maxwell-Boltzmann diagram changes. [2 marks]
(a) Maxwell-Boltzmann sketch [4 marks, AO1 + AO2]
Common errors: curve starting above zero on the y-axis (it must start at the origin); curve touching the x-axis on the right (it must approach asymptotically); E_a placed left of the peak (it should be well to the right).
(b) Doubling [HCl] [3 marks, AO2]
Reject: vague answers such as "more collisions, faster reaction" without identifying what increases (collision frequency) and why (more particles per unit volume).
(c) Temperature rise of 10 K [3 marks, AO2 + AO3]
Mark scheme rewards explicit reference to "fraction with E ≥ E_a" or "area to the right of E_a" — not just "particles move faster". The latter alone scores zero for this mark.
(d) Catalyst [2 marks, AO1 + AO2]
Accept either a sketched diagram with two E_a lines or a clear verbal description matching the above.
The three responses below cover the meaningful A-Level range: Grade C (borderline-pass floor), Grade B (solid mark-scheme coverage), and Grade A* (top-band synthesis). The commentary after each response (editorial, not a real examiner report) names the marks earned and the specific moves that differentiate from adjacent bands.
(a) [Sketch] x-axis "kinetic energy"; y-axis "number of molecules". Curve starts at the origin, rises to a peak, then falls and tails off to the right without touching the x-axis. E_mp marked at the peak. E_a marked further right on the x-axis. The area to the right of E_a is shaded.
(b) Doubling the concentration of HCl doubles the number of HCl particles in solution. More HCl particles means more collisions with the Mg surface per second. More collisions means more successful collisions, so the rate doubles.
(c) At 30 °C the particles have more kinetic energy than at 20 °C. The Maxwell-Boltzmann curve shifts so the peak is lower and further to the right. A larger area lies to the right of E_a — meaning a greater proportion of particles have energy ≥ E_a. So more successful collisions per second happen, and the rate roughly doubles.
(d) A catalyst lowers the activation energy by providing an alternative reaction pathway. On the MB diagram, the E_a line moves to the left, so a bigger area lies to the right of the new E_a value. More particles have enough energy, so the rate increases. The catalyst is not used up.
Editorial commentary (Grade C): A safe, mark-scheme-aligned response that earns most of the available marks: clear MB sketch, correct collision-frequency reasoning for concentration, and identification of the temperature effect on the energetic fraction. To progress to Grade B, the answer should explicitly state that the MB distribution is unchanged by concentration changes (not just that more particles collide), and quantify the temperature-rise rule of thumb (10 K rise ~doubles rate for typical E_a). The catalyst answer should also note that the MB distribution itself is unchanged — only the E_a line moves.
(a) [Sketch] x-axis labelled "kinetic energy / J"; y-axis "number of molecules". Curve begins at the origin, rises to a sharp peak (the most probable energy, E_mp), then descends with a long tail to the right that approaches but never touches the x-axis. E_mp labelled at the peak; E_a labelled on the x-axis well to the right of the peak. The region to the right of the E_a line is hatched.
(b) Doubling [HCl] doubles the number of HCl molecules per unit volume. The frequency of collisions between Mg surface atoms and HCl(aq) molecules doubles. The Maxwell-Boltzmann distribution is unchanged (same temperature, so the same fraction of molecules has energy ≥ E_a), but with twice as many collisions per second, the number of successful collisions per second also doubles. The initial rate therefore doubles.
(c) Increasing T by 10 K shifts the MB distribution so the peak lowers and broadens, moving to higher energy. The area under the curve is unchanged (same number of molecules). Crucially, the shaded fraction with E ≥ E_a grows significantly faster than the mean kinetic energy — for a typical E_a near 50 kJ mol⁻¹, this fraction roughly doubles. Combined with a small (~1.6%) increase in mean collision frequency from higher mean speed, the rate roughly doubles. This is the "10 K rule of thumb" for many reactions.
(d) A catalyst provides an alternative reaction pathway with a lower activation energy E_a' < E_a, without itself being consumed overall. On the MB diagram, the curve is unchanged (temperature is unchanged), but the E_a line moves to the left to E_a'. A larger fraction of molecules now have energy ≥ E_a', so more successful collisions occur per unit time. The rate increases. (Note: a catalyst speeds up forward and reverse reactions equally, so it does not shift equilibrium.)
Editorial commentary (Grade B): Solidly A-level-rigorous. Correctly identifies that the MB distribution is unchanged for both the concentration and catalyst cases, quantifies the temperature-rise rule, and adds the equilibrium-position observation for the catalyst. To progress to A*, the answer should quantify the MB tail explicitly using the exp(−E_a/RT) factor, compare the energetic fractions at the two temperatures numerically, and connect to the Arrhenius equation (lesson 2) as the formal expression of the same idea.
(a) [Sketch] x-axis labelled "kinetic energy / J molecule⁻¹"; y-axis "number of molecules per unit energy interval, N(E)". Curve starts at origin (N(0) = 0), rises as √E for small E, reaches a peak at E_mp, and falls exponentially as exp(−E/k_B T) for E ≫ k_B T (the Maxwell-Boltzmann tail). E_mp labelled at the peak (~k_B T/2 in magnitude); E_a labelled well to the right (typically several × k_B T). The tail to the right of E_a is shaded, representing the fraction with E ≥ E_a.
(b) Doubling [HCl(aq)] doubles the number of HCl molecules per unit volume. For the heterogeneous reaction Mg(s) + HCl(aq), the rate of successful collisions per unit surface area is proportional to [HCl] (first-order in HCl, since the Mg surface is in excess). The MB distribution is unchanged (T fixed, so the exp(−E_a/k_B T) tail-fraction is fixed); only the collision frequency at the Mg/HCl interface doubles. Rate doubles. Formally: r = k_app · [HCl], so doubling [HCl] doubles r.
(c) The fraction of molecules with E ≥ E_a is governed by the Maxwell-Boltzmann tail integral, which for E_a ≫ k_B T scales approximately as exp(−E_a/RT). The ratio of fractions at 303 K versus 293 K, with E_a = 50 kJ mol⁻¹: exp(−50000/(8.314 × 303)) / exp(−50000/(8.314 × 293)) = exp[(50000/8.314) · (1/293 − 1/303)] = exp(0.677) ≈ 1.97. So the tail-fraction nearly doubles for a 10 K rise. Combined with a ~1.7% increase in mean speed (√(303/293)), the total rate increase is ~2.0×. This is the quantitative basis for the "10 K doubles rate" rule and is formalised as the Arrhenius equation k = A exp(−E_a/RT).
(d) A catalyst opens an alternative reaction pathway through a lower-energy transition state with activation energy E_a' < E_a. The MB distribution is unchanged (T unchanged); the E_a line shifts left to E_a'. Quantitatively, the rate enhancement is the ratio exp(−E_a'/RT) / exp(−E_a/RT) = exp[(E_a − E_a')/RT]. For a catalyst that lowers E_a by 25 kJ mol⁻¹ at 298 K: exp(25000/(8.314 × 298)) ≈ exp(10.1) ≈ 2.4 × 10⁴ — a 24,000-fold rate enhancement. Forward and reverse barriers both fall by the same ΔE_a (the catalyst lowers the transition-state energy, which is shared); equilibrium is reached faster but is unmoved.
Editorial commentary (Grade A):* Genuinely A*: explicit MB functional form (√E · exp(−E/k_B T)), numerical calculation of the tail-ratio for the 10 K temperature rise yielding ~1.97 (matching the qualitative "doubles" claim), connection to the Arrhenius equation as the formalisation, and a quantitative catalyst rate-enhancement using exp[(E_a − E_a')/RT]. The answer demonstrates the leap from qualitative collision theory to quantitative kinetic predictions — exactly what the A* band rewards.
This content is aligned with the AQA A-Level Chemistry (7405) specification.