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8873 / 2027
Reaction Kinetics

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Reaction Kinetics

Infer a rate law from evidence and explain how concentration, temperature and catalysts change reaction speed.

A-Level 8873, revised syllabus (2026-2027)

01

Rate measures change per time

Name the quantity being followed before writing an equation.

Reaction rate tells you how quickly a reactant is used up or a product forms. A rate based on concentration is usually in mol dm-3 s-1. A reactant concentration falls, so its graph slopes down. Report the rate it disappears at as a positive number. If the equation has coefficients other than one, use them to link the rates for different substances.

For 2H2O2 → 2H2O + O2, two moles of hydrogen peroxide are used for every mole of oxygen made. So peroxide disappears twice as fast as oxygen forms. If peroxide disappears at 6.0 × 10-4 mol dm-3 s-1, oxygen forms at 3.0 × 10-4 mol dm-3 s-1 in the same volume.

Read a rate equation
TermMeaning
rate = k[A]m[B]nA relationship between the stated rate and the concentrations at fixed conditions.
Order with respect to A: mThe exponent showing how changing [A] affects rate while other conditions remain fixed.
Overall order: m + nThe sum of the concentration exponents.
Rate constant, kThe proportionality constant for that equation at a specified temperature and catalytic conditions.
Half-lifeTime for the amount or concentration being followed to fall to half its current value.
Activation energyThe minimum energy barrier for the reacting pathway.
CatalysisAcceleration through an alternative pathway; the catalyst is regenerated overall.

Worked exampleAn initial rate from a tangent

A tangent at t = 0 on a reactant concentration-time graph passes through (0 s, 0.160 mol dm-3) and (20.0 s, 0.0860 mol dm-3). Find the initial disappearance rate.

Try it first. Then show the worked stepsHide the worked steps
  1. Tangent slope = (0.0860 - 0.160)/(20.0 - 0) = -0.00370 mol dm-3 s-1.
  2. Take the positive magnitude for disappearance.
  3. These points are on the tangent. A chord joining distant measured points would give an average rate, not the initial rate.
Answer

Initial disappearance rate = 3.70 × 10-3 mol dm-3 s-1.

Check your understanding

For 2H2O2 -> 2H2O + O2, oxygen forms at 2.0 x 10-4 mol dm-3 s-1. How fast is H2O2 used up?

  1. 2.0 x 10-4 mol dm-3 s-1
  2. 1.0 x 10-4 mol dm-3 s-1
  3. 4.0 x 10-4 mol dm-3 s-1
Show the answer

4.0 x 10-4 mol dm-3 s-1

Two moles of H2O2 are used for each mole of O2 formed, so H2O2 disappears twice as fast as O2 appears.

02

Change one concentration at a time

The rate factor reveals the exponent.

What doubling one concentration reveals
Rate factor when [A] doublesOrder in AReason
1020 = 1
2121 = 2
4222 = 4

Compare experiments in which only one concentration changes. Keep the other concentrations, the temperature and any catalyst the same. In H1, the orders m and n are 0, 1 or 2 in simple cases. Read the orders from the evidence given. Do not copy them from the coefficients of an overall equation.

Original initial-rate dataset at constant temperature
Run[A] / mol dm-3[B] / mol dm-3Rate / mol dm-3 s-1
10.02000.1002.00 × 10-5
20.04000.1004.00 × 10-5
30.04000.2001.60 × 10-4

Worked exampleDerive, calculate and predict

Use the dataset to obtain a rate equation and predict the rate at [A] = 0.0300 and [B] = 0.200 mol dm-3.

Try it first. Then show the worked stepsHide the worked steps
  1. Runs 1 and 2: [A] doubles at fixed [B]; rate doubles, so m = 1.
  2. Runs 2 and 3: [B] doubles at fixed [A]; rate quadruples, so n = 2.
  3. rate = k[A][B]2. From run 1, k = (2.00 × 10-5)/(0.0200 × 0.1002) = 0.100.
  4. Units of k = (mol dm-3 s-1)/(mol dm-3)3 = dm6 mol-2 s-1.
  5. New rate = 0.100 × 0.0300 × 0.2002 = 1.20 × 10-4 mol dm-3 s-1.
Answer

rate = k[A][B]2; k = 0.100 dm6 mol-2 s-1; predicted rate = 1.20 × 10-4 mol dm-3 s-1.

Try it step by step

Work it out in steps: Order in A, then Order in B, then Rate constant k.

  1. Problem 1. Run 1: [A] = 0.020, [B] = 0.10 mol dm-3, rate = 0.000040 mol dm-3 s-1. Run 2: [A] = 0.040, [B] = 0.10, rate = 0.000160. Run 3: [A] = 0.040, [B] = 0.20, rate = 0.000160. Find each order and k to 3 significant figures.
  2. Problem 2. Run 1: [A] = 0.10, [B] = 0.10 mol dm-3, rate = 0.00050 mol dm-3 s-1. Run 2: [A] = 0.30, [B] = 0.10, rate = 0.0015. Run 3: [A] = 0.30, [B] = 0.20, rate = 0.0030. Find each order and k to 3 significant figures.
  3. Problem 3. Run 1: [A] = 0.050, [B] = 0.050 mol dm-3, rate = 0.00020 mol dm-3 s-1. Run 2: [A] = 0.10, [B] = 0.050, rate = 0.00040. Run 3: [A] = 0.10, [B] = 0.10, rate = 0.0016. Find the order in A, the order in B, and k to 3 significant figures.
Show the answer

Problem 1.

Order in A: Runs 1 and 2: [A] doubles and the rate rises four times, so the order in A is 2.

Order in B: Runs 2 and 3: [B] doubles and the rate does not change, so the order in B is 0.

Rate constant k: rate = k[A]2, so k = 0.000040 / 0.0202 = 0.100 dm3 mol-1 s-1.

Problem 2.

Order in A: Runs 1 and 2: [A] triples and the rate triples, so the order in A is 1.

Order in B: Runs 2 and 3: [B] doubles and the rate doubles, so the order in B is 1.

Rate constant k: rate = k[A][B], so k = 0.00050 / (0.10 x 0.10) = 0.0500 dm3 mol-1 s-1.

Problem 3.

Order in A: Runs 1 and 2: [A] doubles and the rate doubles, so the order in A is 1.

Order in B: Runs 2 and 3: [B] doubles and the rate rises four times, so the order in B is 2.

Rate constant k: rate = k[A][B]2, so k = 0.00020 / (0.050 x 0.0502) = 1.60 dm6 mol-2 s-1.

Check your understanding

A reaction is zero order in A. Does that mean no A is used up?

  1. Yes. A zero order means A has no part in the reaction, so it acts like a catalyst.
  2. No. A is still used up; the rate just does not depend on [A] under these conditions.
  3. No. It means A reacts so fast that it is used up before any other reactant starts.
Show the answer

No. A is still used up; the rate just does not depend on [A] under these conditions.

Order describes how rate depends on concentration in the rate equation. A zero-order reactant still reacts. Its concentration falls at a constant rate.

03

Recognise zero and first order from the whole curve

Constant slope differs from constant halving time.

Two concentration-time patterns

Both illustrative curves start at 0.160 mol dm-3 at constant temperature. Zero-order concentration falls linearly to zero at 120 seconds. The first-order curve halves every 30 seconds and progressively flattens.

These are different illustrative models with the same starting concentration, not two simultaneous descriptions of one experiment. Integrated rate-law formulae are not required.

A zero-order concentration-time graph is a straight line while reactant remains, because the same amount is used up every second. Each halving then takes less time than the one before. A first-order graph curves, because the rate falls as the concentration falls. Its half-life stays the same, even if you start measuring from a later point.

Read the first-order curve
Time / s0306090
[A] / mol dm-30.1600.08000.04000.0200

Worked exampleHalf-life is independent of the chosen starting concentration

Show how the table supports first order.

Try it first. Then show the worked stepsHide the worked steps
  1. 0.160 to 0.0800 takes 30 s.
  2. 0.0800 to 0.0400 also takes 30 s, despite the lower starting concentration.
  3. 0.0400 to 0.0200 again takes 30 s.
Answer

The constant 30 s halving interval supports first order. A falling curve alone is insufficient evidence: many rate laws produce falling curves.

Try it step by step

Work it out in steps: Number of half-lives, then Concentration remaining, then Percentage reacted.

  1. Problem 1. A first-order reactant has a half-life of 30.0 s and starts at 0.320 mol dm-3. Find its concentration after 90.0 s and the percentage reacted, to 3 significant figures.
  2. Problem 2. A first-order reactant has a half-life of 5.00 min and starts at 0.600 mol dm-3. Find the percentage reacted after 10.0 min, to 3 significant figures.
  3. Problem 3. A first-order reactant has a half-life of 40.0 s and starts at 0.480 mol dm-3. Find the number of half-lives in 160 s, then the concentration and percentage reacted to 3 significant figures.
Show the answer

Problem 1.

Number of half-lives: 90.0 / 30.0 = 3 half-lives.

Concentration remaining: 0.320 -> 0.160 -> 0.0800 -> 0.0400 mol dm-3.

Percentage reacted: 1/8 remains, so 7/8 has reacted: 87.5%.

Problem 2.

Number of half-lives: 10.0 / 5.00 = 2 half-lives.

Concentration remaining: 0.600 -> 0.300 -> 0.150 mol dm-3.

Percentage reacted: 1/4 remains, so 75.0% has reacted.

Problem 3.

Number of half-lives: 160 / 40.0 = 4 half-lives.

Concentration remaining: 0.480 -> 0.240 -> 0.120 -> 0.0600 -> 0.0300 mol dm-3.

Percentage reacted: 1/16 remains, so 93.75% has reacted, 93.8% to 3 significant figures.

With real data, draw a smooth curve through the points and read several halving times. Keep the temperature steady. If the mixture warms during the run, k changes and the shape of the curve is distorted. Treat an odd point with a reason. Do not delete it just to make the curve look ideal.

04

Separate more collisions from a larger successful fraction

Concentration and temperature affect different parts of the explanation.

Make a guess. You are not marked.

A reaction mixture is warmed by 10 K. What is the main reason the rate rises?

  1. More collisions have energy at or above Ea
  2. The activation energy becomes smaller when warmed
  3. The particles collide much more often than before
Show the answer

More collisions have energy at or above Ea

The energy distribution shifts to higher energies, so a much larger fraction of collisions can react. Collision frequency rises only a little.

In the collision model, particles must collide with enough energy and in a suitable orientation. A higher concentration puts more particles in each unit of volume, so reactant particles collide more often. At the same temperature, the energy distribution and k do not change. The rate changes as the rate equation predicts for that concentration.

Temperature changes the energy distribution

For the same total number of particles, the hotter distribution is broader and has a lower peak. Its area beyond the fixed activation energy is larger. The curves share the same total area over the full energy range.

Energy is in relative units in this qualitative model. The vertical quantity is number per energy interval; areas, not curve heights alone, represent particle populations.

A Boltzmann distribution shows that particles have a range of energies. The area to the right of Ea is the fraction of particles with enough energy to react. A higher temperature spreads the curve out and makes this fraction much larger. Particles also move faster, so they collide a little more often. More effective collisions each second mean a larger k and a faster reaction at the same concentrations.

Watch out for this

A student says: "Heating speeds up a reaction because it lowers the activation energy." What is wrong with this?

Show the answer

Heating does not lower the activation energy. The energy needed stays the same; what changes is how many particles have that energy. The rule that "the rate doubles every ten degrees" is only a rough guide, not a law.

Check your understanding

Why is "particles collide more often" an incomplete answer to why heating speeds up a reaction?

  1. It misses the larger fraction of collisions with energy at least Ea, which matters most.
  2. It misses that heating raises the concentration of the reactants in the flask.
  3. It misses that heating lowers the activation energy, which matters most here.
Show the answer

It misses the larger fraction of collisions with energy at least Ea, which matters most.

Heating raises collision frequency a little but greatly increases the fraction of particles with energy at or above Ea. Both lead to more effective collisions per second and a larger rate constant.

05

A catalyst changes the route, not the starting energy distribution

Connect a lowered barrier with a surface process.

A lower threshold at the same temperature

There is one distribution because temperature is unchanged. The catalytic threshold lies to the left of the original threshold, so more particles have sufficient energy. The catalyst does not supply a hotter population.

The shaded region illustrates the fraction above the lower threshold. The same-temperature distribution is unchanged.

A catalyst gives a different pathway with a lower activation energy, so the rate constant increases. The catalyst takes part in the steps but is made again by the end. It does not change the overall enthalpy change, the equilibrium constant or the final equilibrium composition. It only helps the system reach equilibrium faster.

Heterogeneous catalysis at a solid surface
  1. Adsorb reactants

    Gaseous reactants attach to active surface sites. Adsorption is attachment to the surface, not absorption into the bulk.

  2. React by an easier route

    Surface interactions weaken relevant bonds and bring reactants into suitable positions. The pathway has lower activation energy.

  3. Desorb products

    Products leave, freeing sites for another catalytic cycle.

A car catalytic converter uses a solid catalyst, while the exhaust gases react. The catalyst is in a different phase from the reactants, so this is heterogeneous catalysis. A typical reaction that removes nitrogen oxides is 2NO(g) + 2CO(g) → N2(g) + 2CO2(g). The gases adsorb on the surface, react and then desorb. NO is reduced to nitrogen while CO is oxidised. CO2 is less toxic than CO, but it is still a greenhouse gas, so do not write that all the products are harmless.

Check your understanding

Why can a contaminant that binds strongly to a solid catalyst reduce its activity?

  1. It occupies active sites, leaving fewer sites for reactants to adsorb and react.
  2. It raises the activation energy of the uncatalysed reaction in the gas phase.
  3. It reacts with the products, which moves the equilibrium back to the reactants.
Show the answer

It occupies active sites, leaving fewer sites for reactants to adsorb and react.

Heterogeneous catalysis happens on the surface. A strongly bound substance blocks active sites, so fewer reactant molecules can adsorb. This is called catalyst poisoning.

06

An enzyme has a specific active site and suitable operating conditions

Use the lock-and-key model without confusing cooling with denaturation.

Make a guess. You are not marked.

An enzyme works well at 37 degrees C. It is boiled, then cooled back to 37 degrees C. What happens to its activity?

  1. It returns to normal once the enzyme has cooled down
  2. It is higher than before, because heat activated it
  3. It stays very low, as the active site has lost its shape
Show the answer

It stays very low, as the active site has lost its shape

Boiling denatures the enzyme: the protein unfolds and the active site loses its shape. Cooling does not restore that shape.

In this course, enzymes are proteins that act as biological catalysts. Each folded protein has an active site with a shape and chemical groups that suit particular substrates. In the lock-and-key model, a matching substrate binds to the active site. The enzyme catalyses one particular reaction, then releases the products and can be used again.

A selective catalytic cycle
  1. Matching substrate

    Its shape and interactions suit the active site.

  2. Enzyme-substrate complex

    Binding enables a lower-activation-energy pathway.

  3. Products released

    The enzyme is available for another suitable substrate.

At low temperatures, particles have less kinetic energy and there are fewer effective collisions, so the enzyme works slowly. Raising the temperature first increases the rate. At high enough temperatures, the interactions that hold the protein in shape break. The active site changes shape, and activity falls. This is denaturation. Cooling a folded enzyme only slows it down, but overheating can damage it for good.

A change in pH can change the charged groups in the protein and at its active site. So each enzyme works only over a limited pH range, and extreme pH can denature it. Specificity covers both which substrate binds and which type of reaction is catalysed. An enzyme does not catalyse every reaction that its substrate could undergo.

Check your understanding

An enzyme reaction slows in a refrigerator and speeds up again when gently warmed. Is denaturation the best explanation?

  1. Yes. Cold changes the enzyme's shape, and gentle warming then refolds it back.
  2. No. Cooling lowers particle energies; the active site stays intact, so the rate recovers.
  3. No. Cooling raises the activation energy of the reaction, so fewer collisions succeed.
Show the answer

No. Cooling lowers particle energies; the active site stays intact, so the rate recovers.

Fewer particles have enough energy at low temperature, and collisions are less frequent. Recovery on warming shows the enzyme's shape was not destroyed.

Exam-style question

Paper 2 Section B style

This question draws on Chemical Energetics: Thermochemistry, and Reaction Kinetics.

Hydrogen peroxide decomposes slowly. Manganese(IV) oxide, a black solid, and the enzyme catalase both speed up the reaction.

2H2O2(aq) -> 2H2O(l) + O2(g)

Experiment 1. A student added a little manganese(IV) oxide to 50.0 cm3 of 1.00 mol dm-3 H2O2 in a polystyrene cup. The temperature rose by 21.5 °C. Take the specific heat capacity of the solution as 4.18 J g-1 K-1 and its density as 1.00 g cm-3.

Experiment 2. In a second run, the concentration of H2O2 was measured over time at a constant temperature.

Table 1. Concentration of H2O2 in Experiment 2
time / min0102030
[H2O2] / mol dm-31.000.500.250.125

Bond energies / kJ mol-1: O-H 460; O-O 150; O=O 496.

  1. (a) Explain, in terms of bond breaking and bond making, why a reaction is exothermic. [1]
  2. (b) Use Experiment 1 to calculate the enthalpy change, in kJ per mole of H2O2. [3]
  3. (c) The accepted value is about -98 kJ mol-1. Suggest two reasons why the value from (b) is less exothermic. [2]
  4. (d) Use the bond energies to calculate the enthalpy change for 2H2O2 -> 2H2O + O2, with every species treated as a gas. [3]
  5. (e) Suggest why an enthalpy change from bond energies may differ from a measured value. [1]
  6. (f) Use Table 1 to show that the reaction is first order with respect to H2O2, and write the rate equation. [3]
  7. (g) The initial rate in Experiment 2 is 0.069 mol dm-3 min-1. Calculate the rate constant, with its units. [2]
  8. (h) Explain, with reference to the Boltzmann distribution, how manganese(IV) oxide increases the rate. [3]
  9. (i) The rate of the catalase-catalysed reaction falls sharply above about 40 °C. Explain why. [2]
 [20]

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Quick revision

Revisit the essentials, then return to an explanation when you need it.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Rate of reactionThe change in concentration of a reactant or product per unit time.
Rate equationAn experimentally determined equation, rate = k[A]m[B]n, linking the rate to the concentrations of the reactants.
Order of reaction with respect to a reactantThe power to which the concentration of that reactant is raised in the experimentally determined rate equation.
Rate constant, kThe proportionality constant in the rate equation. It is constant at a given temperature (and for a given catalyst).
Half-life, t1/2The time taken for the concentration of a reactant to fall to half of its initial value.
Activation energy, EaThe minimum energy that colliding particles must have for a reaction to occur.
CatalystA substance that increases the rate of a reaction by providing an alternative pathway with a lower activation energy, and is chemically unchanged at the end of the reaction.
Rate explanations
ChangeEssential explanation
ConcentrationMore particles per volume; collision frequency changes; use the measured order.
TemperatureLarger fraction above Ea plus more frequent collisions; k increases.
CatalystLower-Ea pathway; distribution unchanged at the same temperature.
First-order half-lifeEqual times for successive halvings, independent of initial concentration.
Enzyme extremesActive-site shape/interactions change; extremes may denature the protein.

Scope

Learning outcomes

7. Reaction Kinetics (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map

  1. 7(a) Use kinetic vocabulary.

    • Rate, rate equation, order, rate constant, half-life, activation energy and catalysis.

    Rate measures change per timeChange one concentration at a time

  2. 7(b) Construct and use simple rate equations.

    • m,n = 0,1,2 for the stated H1 single-step cases.
    • (i) Initial-rate order deduction.
    • (ii) Zero/first-order concentration-time graphs.
    • (iii) Initial rate from concentration data; no integrated laws.

    Rate measures change per timeChange one concentration at a timeRecognise zero and first order from the whole curve

  3. 7(c) Explain constant first-order half-life.

    • Independence from starting concentration.

    Recognise zero and first order from the whole curve

  4. 7(d) Explain concentration effects.

    • Collision frequency at fixed temperature.

    Separate more collisions from a larger successful fraction

  5. 7(e) Explain activation energy using a distribution.

    • Boltzmann distribution and fraction above threshold.

    Separate more collisions from a larger successful fraction

  6. 7(f) Explain temperature effects on k and rate.

    • Both Boltzmann distribution and collision frequency.

    Separate more collisions from a larger successful fraction

  7. 7(g) Explain the catalytic effect.

    • (i) Alternative lower-Ea pathway and increased k.
    • (ii) Same-temperature Boltzmann interpretation.

    A catalyst changes the route, not the starting energy distribution

  8. 7(h) Explain heterogeneous catalytic action.

    • Adsorption/reaction/desorption; removal of nitrogen oxides from car exhaust.

    A catalyst changes the route, not the starting energy distribution

  9. 7(i) Explain enzyme properties.

    • Protein catalysts; substrate and reaction specificity; lock-and-key; temperature and pH sensitivity; link to 9(m), no named protein levels.

    An enzyme has a specific active site and suitable operating conditions