AQA A-Level Chemistry: Kinetics and Equilibrium — Complete Revision Guide (7405)

Kinetics and equilibrium are the twin pillars of physical chemistry on AQA 7405. Taken together they answer the three questions every chemist asks of a reaction: does it happen at all, how fast does it go, and how far does it proceed before stopping. Thermodynamics tells you whether a reaction is feasible in principle. Kinetics tells you how quickly the molecules find each other and clear the activation barrier. Equilibrium tells you the final ratio of products to reactants when the forward and reverse rates have balanced. Four equations underpin the whole section: Rate = k[A]^m[B]^n for the rate law, k = A·exp(−Ea/RT) for the temperature dependence of the rate constant, Kc = [products]/[reactants] for the equilibrium constant, and ΔG° = −RT ln K linking equilibrium to free energy. Master those four and almost every kinetics or equilibrium question becomes a substitution exercise.

This topic is also one of the most synoptic on the specification. The activation energy in the Arrhenius equation is the same Ea that controls reaction profiles in energetics, and the link ΔG° = −RT ln K ties equilibrium constants directly to Gibbs free energy. The acid dissociation constant Ka in acids and buffers is just a Kc for a proton-transfer equilibrium. The cell potentials in redox and electrochemistry feed through ΔG° = −nFE° into the same equilibrium constant. Even the industrial Haber process for ammonia and the Contact process for sulfuric acid are nothing more than applied Le Chatelier, with pressure, temperature and catalyst chosen to balance equilibrium yield against kinetic rate. Time spent here pays back across the rest of the course.

This guide walks through the kinetics and equilibrium content in 7405 sub-topic by sub-topic. It covers collision theory and the factors affecting rate; rate equations and orders of reaction; the Arrhenius equation; the rate-determining step and what it tells you about mechanism; chemical equilibria and Le Chatelier's principle; the equilibrium constant Kc; the equilibrium constant Kp for gaseous systems; and the three required practicals (RP3, RP6, RP7) that examine this material in the lab. For each sub-topic you will find the core ideas, common pitfalls, a worked example and a link into the LearningBro Kinetics and Equilibrium course.

What the AQA 7405 Specification Covers

AQA A-Level Chemistry (7405) is examined through Paper 1 (Inorganic and Physical, 2h, 105 marks), Paper 2 (Organic and Physical, 2h, 105 marks) and Paper 3 (Synoptic and Practical, 2h, 90 marks). Kinetics and equilibrium are split across the AS and A2 content: §3.1.5 covers AS kinetics (collision theory, Maxwell-Boltzmann), §3.1.6 covers Kc and Le Chatelier at AS, §3.1.9 extends to A2 rate equations, orders and Arrhenius, and §3.1.10 introduces Kp for gaseous equilibria. All four sections are examinable on Paper 1, and the topic is a heavy contributor to Paper 3 through the practical-skills questions.

Sub-topic	Spec area	Typical paper weight
Collision theory and rates	§3.1.5	3-5 marks
Rate equations and orders	§3.1.9	6-10 marks
Arrhenius equation	§3.1.9	4-6 marks
Rate-determining step	§3.1.9	4-6 marks
Le Chatelier's principle	§3.1.6	3-5 marks
Kc calculations	§3.1.6	4-8 marks
Kp calculations	§3.1.10	4-8 marks
Required practicals (RP3, RP6, RP7)	Practical	4-8 marks

These weights are estimates, modelled on recent 7405 papers. What is reliable is that an extended-response rate-equation or Kp question — typically eight to ten marks — appears on essentially every Paper 1.

Collision Theory and Rates of Reaction

The starting point for all kinetics is collision theory. For a reaction to occur, particles must collide, with energy greater than or equal to the activation energy Ea, and in the correct orientation. Only a small fraction of collisions in any sample meet both conditions. The rate of reaction is therefore controlled by the frequency of successful collisions, not just total collisions.

The Maxwell-Boltzmann distribution shows how molecular kinetic energies are spread out at a given temperature. The curve starts at zero, rises to a peak (the most probable energy), then tails off. The area under the curve to the right of Ea represents the fraction of molecules with enough energy to react. Increasing temperature stretches and flattens the curve, shifting the peak to higher energy and dramatically increasing the high-energy tail — which is why a 10 K rise can roughly double a reaction rate.

The four standard factors affecting rate are concentration (more particles per unit volume, more collisions), pressure for gases (equivalent to concentration), surface area for solids (more exposed particles to collide with), and temperature (faster particles and a larger high-energy tail). A catalyst provides an alternative reaction pathway with a lower Ea, increasing the fraction of molecules able to react without being consumed itself.

Worked example. A reaction has Ea = 50 kJ mol^-1 at 298 K. Sketching the Maxwell-Boltzmann curve, the area beyond 50 kJ mol^-1 is small. Raising the temperature to 308 K stretches the tail; the area beyond 50 kJ mol^-1 roughly doubles, explaining a doubling of rate.

A common pitfall is to say that temperature increases the average kinetic energy "and therefore the rate doubles" without mentioning the Maxwell-Boltzmann distribution and the high-energy tail. Examiners want both the energy and the distribution argument. See the collision theory and rates lesson.

Rate Equations and Orders of Reaction

At A2 the rate is quantified by the rate equation: Rate = k[A]^m[B]^n, where m and n are the orders with respect to A and B, and k is the rate constant. Crucially, the orders are determined experimentally — they cannot be read off the stoichiometric equation. The overall order is m + n.

A zero-order species has no effect on rate (doubling its concentration leaves rate unchanged). A first-order species doubles rate when its concentration doubles. A second-order species quadruples rate when its concentration doubles. The units of k depend on the overall order: s^-1 for first order overall, mol^-1 dm^3 s^-1 for second order overall, mol^-2 dm^6 s^-1 for third order overall.

Orders are usually found by the initial rates method: vary one reactant while keeping others constant, measure initial rate, deduce the order. Concentration-time graphs offer an alternative: a constant half-life implies first order, a half-life that doubles each cycle implies second order, and a linear concentration-time plot implies zero order.

Worked example. For 2 A + B → products, experiments give: doubling [A] doubles rate (first order in A); doubling [B] quadruples rate (second order in B). Rate = k[A][B]^2, overall third order, k in mol^-2 dm^6 s^-1.

A common pitfall is to read orders from the equation 2 A + B (and so claim "second order in A"). Another is to forget that k has different units at different overall orders, dropping marks on the units mark. See the rate equations lesson.

The Arrhenius Equation

The temperature dependence of k is captured by the Arrhenius equation: k = A·exp(−Ea/RT), where A is the pre-exponential factor (roughly, the frequency of correctly oriented collisions) and R = 8.314 J K^-1 mol^-1. Taking natural logs gives the linear form ln k = ln A − Ea/(RT), so a plot of ln k against 1/T yields a straight line with gradient −Ea/R and intercept ln A. This is the standard route to Ea from experimental rate data.

A small change in T produces a large change in k because of the exponential. For Ea ~ 50 kJ mol^-1 around 300 K, a 10 K rise typically doubles k — the rule of thumb that "rate doubles for every 10 °C". The effect is far stronger for higher activation energies, which is why low-Ea catalysed pathways are so dramatic.

Worked example. A reaction has k = 2.0 × 10^-3 s^-1 at 300 K and k = 8.0 × 10^-3 s^-1 at 320 K. ln(8/2) = (Ea/R)(1/300 − 1/320). 1.386 = (Ea/8.314)(2.083 × 10^-4). Ea = 1.386 × 8.314 / 2.083 × 10^-4 = 5.53 × 10^4 J mol^-1 ≈ 55 kJ mol^-1.

A common pitfall is to forget to use kelvin for T (not °C). Another is to mishandle units of R — using 8.314 J K^-1 mol^-1 gives Ea in J mol^-1, which must be converted to kJ mol^-1 for the final answer. A third is to confuse the gradient of ln k vs 1/T (which is −Ea/R) with +Ea/R. See the Arrhenius equation lesson.

Rate-Determining Step and Reaction Mechanisms

Most reactions occur through a sequence of elementary steps. The rate-determining step (RDS) is the slowest step — its rate sets the rate of the overall reaction. Species in the rate equation are those involved in the RDS (or in any fast equilibrium preceding it). Species that appear in the overall equation but not in the rate equation are involved only in fast steps after the RDS.

For example, consider a two-step mechanism: Step 1 (slow) A + B → C; Step 2 (fast) C + D → products. The rate equation is Rate = k[A][B], because only A and B feature in the RDS. D appears in the overall equation but not in the rate equation. The same logic explains why nucleophilic substitution at a tertiary carbon (SN1) is first order in haloalkane only, while the same reaction at a primary carbon (SN2) is second order overall — the mechanism differs.

The position of the transition state on the reaction profile is described by Hammond's postulate: an exothermic step has an "early" transition state resembling reactants; an endothermic step has a "late" transition state resembling products.

Worked example. NO2 + CO → NO + CO2 has the experimentally determined rate law Rate = k[NO2]^2, with no [CO] dependence. The mechanism proposed is: Step 1 (slow) 2 NO2 → NO3 + NO; Step 2 (fast) NO3 + CO → NO2 + CO2. The RDS involves two NO2 molecules, consistent with the second-order dependence on NO2.

A common pitfall is to deduce the mechanism by reading orders off the overall equation. Another is to forget that intermediates (like NO3 above) appear and disappear within the mechanism but not in the overall equation. See the rate-determining step lesson.

Chemical Equilibria and Le Chatelier's Principle

A dynamic equilibrium is reached when the forward and reverse rates of a reversible reaction become equal. Macroscopic concentrations are constant, but molecular interconversion continues. Equilibrium can only be established in a closed system.

Le Chatelier's principle states that if a system at equilibrium is disturbed, the position of equilibrium shifts to oppose the change. The three standard disturbances are concentration (adding a reactant shifts equilibrium toward products; removing a product also shifts forward), pressure for gaseous systems (increasing pressure shifts toward the side with fewer moles of gas), and temperature (increasing T shifts toward the endothermic direction).

Crucially, a catalyst does not shift the position of equilibrium — it speeds up forward and reverse reactions equally, so equilibrium is reached more quickly but at the same composition. This is the single most-tested misconception in the topic.

In the Haber process N2 + 3 H2 ⇌ 2 NH3, ΔH = −92 kJ mol^-1. High pressure favours ammonia (4 moles of gas to 2), and low temperature favours ammonia (exothermic forward reaction). But low temperature also makes the rate impractically slow, so the industrial compromise is around 450 °C and 200 atm with an iron catalyst, balancing kinetics against equilibrium yield.

A common pitfall is to claim that a catalyst increases yield. Another is to apply Le Chatelier to pressure changes when both sides have equal moles of gas (no shift). See the Le Chatelier lesson.

The Equilibrium Constant Kc

For the general equilibrium a A + b B ⇌ c C + d D, the equilibrium constant is Kc = [C]^c [D]^d / ([A]^a [B]^b), with concentrations in mol dm^-3. Kc depends only on temperature; concentration, pressure and catalyst changes do not alter it. Units of Kc depend on the overall change in moles: dimensionless if Δn = 0, mol dm^-3 if Δn = 1, and so on.

Kc tells you the position of equilibrium: large Kc (>>1) means equilibrium favours products; small Kc (<<1) means it favours reactants; Kc ≈ 1 means a mixed equilibrium. The value can change dramatically with temperature, in the direction predicted by Le Chatelier: for an exothermic forward reaction, raising T decreases Kc; for an endothermic reaction, raising T increases Kc. The quantitative form is the van't Hoff equation.

ICE tables (Initial, Change, Equilibrium) are the workhorse for Kc problems. Set up initial moles, define the change x, write equilibrium moles in terms of x, convert to concentrations using the volume, and substitute into the Kc expression.

Worked example. For H2 + I2 ⇌ 2 HI, 1.0 mol H2 and 1.0 mol I2 are placed in a 1.0 dm^3 vessel at 700 K. At equilibrium, 1.56 mol HI is present. So 0.78 mol of each H2 and I2 has reacted; [H2] = [I2] = 0.22, [HI] = 1.56. Kc = (1.56)^2 / (0.22 × 0.22) = 2.43 / 0.0484 = 50.2 (no units, Δn = 0).

A common pitfall is to forget volume when going from moles to concentrations. Another is to assign wrong units (or forget them when Δn ≠ 0). See the Kc lesson.

The Equilibrium Constant Kp

For gaseous equilibria it is often more convenient to use partial pressures instead of concentrations. The partial pressure of a gas is its mole fraction multiplied by the total pressure: p(A) = x(A) × P_total, where x(A) = n(A)/n_total. Kp is then defined the same way as Kc but in pressures: for a A + b B ⇌ c C + d D, Kp = p(C)^c p(D)^d / (p(A)^a p(B)^b). Units depend on Δn for the gas-phase species; for Δn = 0, Kp is dimensionless. Solids and liquids are not included in Kp (their "partial pressures" are not defined in the same way).

For the Haber process N2 + 3 H2 ⇌ 2 NH3 at 450 °C, Kp ≈ 1.4 × 10^-5 atm^-2. Although the equilibrium constant is small (so ammonia yield at equilibrium is modest), the high pressure of 200 atm pushes the position toward ammonia, and continuous removal of NH3 prevents the reverse reaction. Kc and Kp are related through Kp = Kc(RT)^Δn for gases — but you are not required to use that relation; AQA always specifies one or the other.

Worked example. 2 SO2 + O2 ⇌ 2 SO3 at 800 K with total pressure 5.0 atm. At equilibrium, mole fractions are SO2 0.10, O2 0.05, SO3 0.85. p(SO2) = 0.50 atm, p(O2) = 0.25 atm, p(SO3) = 4.25 atm. Kp = (4.25)^2 / ((0.50)^2 × 0.25) = 18.06 / 0.0625 = 289 atm^-1.

A common pitfall is to forget the Δn convention for units of Kp. Another is to include condensed phases — pure solids and pure liquids do not contribute. A third is to confuse mole fraction with partial pressure (always multiply by total pressure). See the Kp lesson.

Required Practicals: Rates and Equilibria (RP3, RP6, RP7)

Three AQA required practicals examine this topic on Paper 3 and the practical endorsement. RP3 investigates how the rate of a reaction changes with temperature using the disappearing-cross (sodium thiosulfate + dilute HCl) experiment: a flask is placed over a printed cross, reagents are mixed, and the time taken for the cross to disappear is recorded at several temperatures. A plot of ln(1/t) against 1/T mimics the Arrhenius linearisation and gives Ea from the gradient.

RP6 measures the rate of an iodine clock reaction (typically H2O2 + KI + acid, with thiosulfate and starch) at varying concentrations, allowing the order with respect to each reactant to be determined by initial rates. The starch indicator turns blue-black at a known iodine concentration, fixing the end of the "clock" interval; 1/t is proportional to initial rate.

RP7 determines an equilibrium constant for an esterification (ethanoic acid + ethanol ⇌ ethyl ethanoate + water) by titrating the equilibrium mixture against standard NaOH. Initial moles are known, equilibrium moles of acid are obtained from titre, and Kc is calculated from the resulting equilibrium concentrations.

Practical questions reliably test understanding of independent versus controlled variables, the assumptions underlying initial-rates kinetics, the use of ln k vs 1/T to extract Ea, and the use of titration to follow concentration changes. Examiners also push on uncertainty: the major source in RP3 is often the subjective judgement of when the cross has "disappeared". See the required practicals lesson.

Common Mark-Loss Patterns

• Quoting orders from the stoichiometric equation rather than experimental data.
• Forgetting to use kelvin in the Arrhenius equation.
• Misidentifying the gradient of ln k vs 1/T as +Ea/R rather than −Ea/R.
• Claiming a catalyst shifts the position of equilibrium.
• Applying Le Chatelier to pressure changes for systems with equal gas moles on each side.
• Forgetting to drop solids and liquids from Kp expressions.
• Confusing mole fraction with partial pressure.
• Missing the units of k for higher-order rate equations.
• Quoting Kc or Kp without temperature (the value depends on T).
• Not mentioning the Maxwell-Boltzmann high-energy tail when explaining the temperature effect on rate.

How to Revise This Topic

• Drill rate-equation determination from initial-rates tables. Ten different tables in a sitting until reading orders becomes automatic.
• Practise the Arrhenius linearisation by hand: gradient = −Ea/R, intercept = ln A. Convert J to kJ at the end.
• Sketch Maxwell-Boltzmann curves for two temperatures on the same axes, marking Ea and the high-energy tails. Sketch the same with and without a catalyst.
• Build a flashcard set of the four standard equilibria: Haber, Contact, esterification, dinitrogen tetroxide dissociation. For each, know ΔH, Δn(gas), the effect of T and P on yield, and the industrial conditions.
• Drill ICE tables for Kc until they are mechanical.
• Memorise the three required practicals by aim, key reagents, what is measured, and the major uncertainty.
• Use the LearningBro practice quizzes to test under timed conditions, with AI tutor feedback on extended-response answers.

Linking to Other Topics

Kinetics and equilibrium underpin most of the rest of physical chemistry. The activation energy in Arrhenius is the same Ea that shapes reaction profiles in energetics and thermodynamics, and ΔG° = −RT ln K is the direct bridge between Gibbs free energy and equilibrium constants. The Ka, Kb and Kw of acids and buffers are all special-case Kc values for proton transfer; pH calculations are equilibrium calculations in disguise. The cell potentials in redox and electrochemistry feed through ΔG° = −nFE° into the same K. The shapes and bonding from atomic structure determine reaction orientations and so the pre-exponential factor A. Even the SN1 versus SN2 distinction in organic foundations is decided by kinetics: tertiary haloalkanes go via a unimolecular RDS and show first-order kinetics in haloalkane only; primary haloalkanes go via a bimolecular RDS and show second-order overall kinetics. Time spent here is time spent on the unifying language of every other physical chemistry topic.

Final Word

Kinetics and equilibrium is a calculation-heavy topic, but the calculations sit on a small and stable set of relations: the rate equation, the Arrhenius equation, Le Chatelier's principle, and the two equilibrium constants Kc and Kp. The combinations are limited; the techniques are formulaic; the mark schemes are predictable once you have drilled enough variants. Pair the rate-equation arithmetic with confident Maxwell-Boltzmann sketches, drill the Arrhenius linearisation until it is automatic, learn the four canonical equilibria cold, and the three required practicals will write themselves. The full LearningBro Kinetics and Equilibrium course walks through every sub-topic with worked examples, ICE tables, and AI tutor feedback on extended-response answers. Get this section fluent and a substantial fraction of the physical chemistry on every AQA Paper 1 and Paper 3 becomes routine.