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Notes/Chemistry/Paper 4/Reaction Kinetics
CAIEA2 Level9701§26.1–26.2

Reaction Kinetics

Rate equations, orders and the rate constant: deducing orders from data and graphs, first-order half-lives, mechanisms and the rate-determining step, temperature effects, and homogeneous versus heterogeneous catalysis — putting numbers on the kinetics story you met at AS.

250 min read 8 sub-topics
194
question parts
2021–2025 · 29 papers
10 marks
per paper
≈ 10% of the paper
1.7/3
avg difficulty
moderate
#7
most examined
of 15 topics by marks

The AS Reaction Kinetics note told kinetics as a story: particles collide, only sufficiently energetic collisions succeed, catalysts offer an easier pathway. Paper 4 keeps the story but demands the numbers underneath it. Which reactant controls the rate, and by exactly how much? That question is answered by an order — 0, 1 or 2 — measured in the laboratory, never read off a chemical equation. The sensitivities are packed into a rate equation, rate=k[A]m[B]n\text{rate} = k[\text{A}]^m[\text{B}]^n, whose constant kk converts concentrations into a predicted rate, carries units you must be able to derive, and stays fixed unless the temperature changes. From there the subject opens out: half-lives you can read off graphs and calculate with, mechanisms deduced from rate equations, and catalysts classified by phase with their step-by-step action written as equations.

The route through is: §01 rate equations, order and the rate constant, §02 deducing orders from experimental data, §03 calculating k and its units, §04 orders from graphs — concentration–time and rate–concentration, §05 first-order half-life calculations, §06 mechanisms and the rate-determining step, §07 the effect of temperature on k, and §08 homogeneous and heterogeneous catalysis.

Before you start you should be able to

  • Rate of reaction as change in concentration per unit time, collision theory, and what makes a collision effective (this subject's own AS Reaction Kinetics note, §01)

  • Activation energy and the Boltzmann distribution; why raising temperature speeds a reaction up (this subject's own AS Reaction Kinetics note, §02–§03)

  • Catalysts lower Ea by providing an alternative pathway; the terms homogeneous and heterogeneous (this subject's own AS Reaction Kinetics note, §04)

  • Confident mole-and-concentration arithmetic in mol dm⁻³, and fluency with powers of ten and logs (this subject's own Atoms, Molecules and Stoichiometry note)

  • The idea of dynamic equilibrium — needed to appreciate why the rate constant k depends on temperature alone (this subject's own AS Equilibria note)

By the end of this page you can

  • Explain and use the terms rate equation, order of reaction, overall order, rate constant, half-life, rate-determining step and intermediate

  • Understand and use rate equations of the form rate = k[A]ᵐ[B]ⁿ (m and n each 0, 1 or 2), and predict quantitatively how rate responds to concentration changes

  • Deduce orders of reaction from initial-rates data, from concentration–time graphs and by the half-life method, and calculate an initial rate from concentration data

  • Interpret concentration–time and rate–concentration graphs, recognising a constant half-life as the test for first order

  • Construct a rate equation, and calculate the rate constant k from initial rates or from t½ via k = 0.693/t½, deriving and quoting its units correctly

  • Use the half-life of a first-order reaction in calculations, understanding its independence of starting concentration

  • Suggest reaction mechanisms consistent with a given rate equation and overall equation, predict orders from a mechanism and rate-determining step, and identify intermediates and catalysts

  • Describe qualitatively how temperature affects k and hence rate, and explain the modes of action of homogeneous and heterogeneous catalysts — including adsorption/bond-weakening/desorption on iron (Haber) and Pt/Pd/Rh (converters), and the homogeneous cycles of NO₂ with SO₂ and Fe³⁺/Fe²⁺ with S₂O₈²⁻/I⁻

01

Rate equations, order and the rate constant

Syllabus requirement · §26.1.1–26.1.2(a)

“

explain and use the terms rate equation, order of reaction, overall order of reaction, rate constant, half-life, rate-determining step and intermediate · understand and use rate equations of the form rate = k[A]ᵐ[B]ⁿ (for which m and n are 0, 1 or 2)

”

Rates with numbers attached

The AS Reaction Kinetics note explained rates qualitatively: particles must collide, only collisions carrying at least the activation energy succeed, and a catalyst offers an alternative route of lower EaE_a. None of that predicts anything. Ask "if I double the concentration of nitrogen monoxide, what happens to the rate?", and stories stay silent — Paper 4 wants "the rate increases by a factor of four". The tool that delivers such numbers is the rate equation: one line of algebra that converts a set of concentrations into a rate. Everything in this note — deducing orders (§02–§04), calculating kk and exploiting it (§03, §05), reading mechanisms off rate equations (§06), and the two levers that change kk itself (§07–§08) — hangs on the grammar built in this section, so every definition here is worth learning word-perfect.

The rate equation, piece by piece

First, recall what is being predicted. The rate of reaction is the change in concentration of a reactant or product per unit time — typically in mol dm−3 s−1\text{mol dm}^{-3}\,\text{s}^{-1}, though some questions work per minute. The rate equation expresses that rate through the concentrations present in the mixture:

rate=k[A]m[B]n\text{rate} = k[\text{A}]^m[\text{B}]^n

Each symbol has a precise meaning:

  • [A][\text{A}], [B][\text{B}] — concentrations in mol dm−3\text{mol dm}^{-3}. Square brackets always mean "concentration of".
  • mm and nn — the orders with respect to A and B. At this level each is exactly 0, 1 or 2 — never negative, never fractional.
  • kk — the rate constant, fixed for this reaction at this temperature, which turns the concentration terms into an actual rate.

Two structural facts follow immediately. A species of order 0 does not appear in the rate equation at all — changing its concentration cannot change the rate. And only species that appear in the equation influence the rate: usually reactants, but not always — the H+\text{H}^+ supplied by an acid catalyst features in many rate equations, including one you will deduce in §02.

rate=k[A]m[B]n\text{rate} = k[\text{A}]^m[\text{B}]^n

The general rate equation. m and n are the orders with respect to A and B — each 0, 1 or 2, and found by experiment only.

·

Overall order = m + n (+ … for further species). k is fixed for a given reaction at a given temperature and NEVER changes when concentrations change.

Order of reaction, precisely

The definitions, in mark-scheme wording. The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the rate equation. The overall order of a reaction is the sum of those powers — for rate =k[A]m[B]n= k[\text{A}]^m[\text{B}]^n it is m+nm + n.

Each order answers one practical question: how sensitively does the rate respond to this concentration?

  • Zero order (power 0): the species is absent from the rate equation. Changing its concentration leaves the rate untouched.
  • First order (power 1): rate ∝ concentration. Doubling the concentration doubles the rate; tripling triples it.
  • Second order (power 2): rate ∝ concentration². Doubling the concentration multiplies the rate by 22=42^2 = 4; tripling multiplies it by 32=93^2 = 9.

order wrt X

its term in the rate equation

[X] doubled

[X] tripled

0 (zero)

absent entirely

no change

no change

1 (first)

k[X]k[\text{X}]

×2

×3

2 (second)

k[X]2k[\text{X}]^2

×4

×9

The only three orders in this syllabus. Whatever the change: rate factor=(concentration factor)order\text{rate factor} = (\text{concentration factor})^{\text{order}}.

The three orders, written out
rate=k\text{rate} = k

zero order throughout — k alone sets the rate

rate=k[A]\text{rate} = k[\text{A}]

first order in A

rate=k[A]2\text{rate} = k[\text{A}]^2

second order in A

rate=k[A][B]\text{rate} = k[\text{A}][\text{B}]

first order in each of A and B — overall second order

The rate constant k

Once the orders are known, kk is the number that makes the whole equation true — numerically, the rate the equation delivers when every concentration is 1 mol dm−31\text{ mol dm}^{-3}. Two statements about kk earn marks year after year:

  • Concentrations do not affect kk. Raising [A][\text{A}] raises the rate through the [A]m[\text{A}]^m term; kk sits unmoved. If a calculation ever implies that kk changed between two runs at the same temperature, an arithmetic slip has happened somewhere — go back and find it.
  • Temperature does affect kk. Warm the mixture and kk rises, so the rate rises even at unchanged concentrations (§07). A catalyst changes the reaction pathway, and a different pathway has its own kk (§08).

kk also carries the units of the rate equation, and those units depend on the overall order — which is why "state the units of kk" is a standing one-mark question. Section §03 derives them line by line.

Orders are measured, never copied

The most expensive reflex in this topic is reading orders from the balanced equation. Here is a real Paper 4 reaction:

IO3−+6H++5I−→3I2+3H2O\text{IO}_3^- + 6\text{H}^+ + 5\text{I}^- \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}

Coefficients 1, 6 and 5 — yet its true rate equation is rate =k[IO3−][H+]2[I−]2= k[\text{IO}_3^-][\text{H}^+]^2[\text{I}^-]^2, deduced from experiment and worked through in §02–§03. Contrast KcK_c, where exponents DO come straight from the equation: an equilibrium constant describes the balanced equation, but a rate equation describes the mechanism — and mechanisms are discovered by experiment. Unless a question hands you the rate equation, every order must be earned from data.

Predicting how rate changes, every time
  1. 1

    Write the rate equation. Copy it if the question gives it; otherwise it must already have been deduced from data (§02).

  2. 2

    For each concentration that changes, write its factor (new ÷ old): "doubled" is ×2, "halved" is ×½, "increased three times" is ×3.

    Factors, not differences — rate questions live in multiplication.

  3. 3

    Raise each factor to that species' order. Zero order gives factor⁰ = 1: no contribution.

    This one move is the entire content of 'order'. Second order means the FACTOR is squared — not the concentration used twice.

  4. 4

    Multiply the factors together to give the overall rate factor.

    Each term of the rate equation contributes independently, which is precisely why the factors multiply rather than add.

  5. 5

    Apply the factor to the original rate — or report "rate × …" if only the factor is asked.

A clean demonstration (invented numbers)

Experiment finds that for the reaction P+Q→products\text{P} + \text{Q} \rightarrow \text{products},

rate=k[P][Q]2\text{rate} = k[\text{P}][\text{Q}]^2

and the rate is 3.0×10−4 mol dm−3 s−13.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} when [P]=0.100[\text{P}] = 0.100 and [Q]=0.200 mol dm−3[\text{Q}] = 0.200\text{ mol dm}^{-3}.

Change 1 — double [Q], keep [P]. Q's factor is ×2, raised to its order: 22=42^2 = 4. P is unchanged, contributing 11=11^1 = 1. Overall rate factor =4×1=4= 4 \times 1 = 4, so the new rate is

4×3.0×10−4=1.2×10−3 mol dm−3 s−14 \times 3.0 \times 10^{-4} = 1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}

Change 2 — halve [P] and double [Q] together. P contributes (12)1=0.5(\tfrac{1}{2})^1 = 0.5; Q contributes (2)2=4(2)^2 = 4. Rate factor =0.5×4=2= 0.5 \times 4 = 2: the rate doubles, to 6.0×10−4 mol dm−3 s−16.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} — even though one concentration went down.

Change 3 — triple the concentration of R, a zero-order species. R appears nowhere in the rate equation, so its factor is 30=13^0 = 1: the rate does not move. That sentence — absent from the rate equation, so no effect — is exactly what "zero order" means, and it is the phrasing examiners reward.

Three predictions, no new mathematics anywhere: factor → power → multiply.

Reading orders off a rate equation

9701/44 O/N 2025 Q4(a)(i)1 mark

Nitrogen monoxide reacts with hydrogen:

reaction 3:2NO+2H2→N2+2H2O\text{reaction 3:}\qquad 2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}

The rate equation for reaction 3 is shown.

rate=k[H2][NO]2\text{rate} = k[\text{H}_2][\text{NO}]^2

State: the order of reaction with respect to [H2][\text{H}_2]; the order of reaction with respect to [NO][\text{NO}]; and the overall order of the reaction.

Show full working
  1. 1

    Order wrt H2\text{H}_2: the exponent on [H2][\text{H}_2] is not written, so it is 1 → first order.

    An unwritten power means 1, exactly as xx means x1x^1 in algebra. Reading the missing exponent as zero is the classic loss here.

  2. 2

    Order wrt NO\text{NO}: the exponent is 2 → second order.

    Read from the rate equation only — the balanced equation's coefficient in front of NO is coincidence, not evidence.

  3. 3

    Overall order =1+2=3= 1 + 2 = 3.

    Overall order is the sum of the individual orders — the definition, applied once.

Answer

Order wrt H₂: 1 · Order wrt NO: 2 · Overall order: 3

Two reads and one addition. Never let the stoichiometric coefficients (2 NO, 2 H₂) leak into your orders — they come from the rate equation alone.

Predicting a rate change — one concentration altered

9701/44 O/N 2025 Q4(a)(ii)1 mark

For reaction 3 above, predict how the initial rate changes when the concentration of NO\text{NO} is halved.

Show full working
  1. 1

    Write the factor for the changed species: [NO]new[NO]old=12\frac{[\text{NO}]_{\text{new}}}{[\text{NO}]_{\text{old}}} = \frac{1}{2}

    'Halved' becomes exactly ½ before any order touches it — turning words into a factor is always step one.

  2. 2

    Raise it to NO's order: (12)2=14\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}

    Second order squares the factor. This squaring is where the tempting 'halved → half the rate' answer dies.

  3. 3

    [H2][\text{H}_2] is unchanged, contributing 11=11^1 = 1, so the rate factor is 14×1=14\tfrac{1}{4} \times 1 = \tfrac{1}{4}.

    Only the factors multiply — nothing else in the equation moved, so nothing else contributes.

Answer

rate × ¼ — the initial rate falls to one quarter of its previous value

Factor → power → multiply, one short line each. The examiner is checking the square, so show it.

Predicting a rate change — both concentrations altered

9701/44 O/N 2025 Q4(a)(iii)1 mark

For reaction 3, predict how the initial rate changes when the concentrations of NO\text{NO} and H2\text{H}_2 are both increased three times.

Show full working
  1. 1

    Write both factors: H2→×3\text{H}_2 \rightarrow \times 3 and NO→×3\text{NO} \rightarrow \times 3.

    Both concentrations changed, so both terms contribute — list the factors before touching the powers.

  2. 2

    Apply each order: H2\text{H}_2: 31=33^1 = 3; NO\text{NO}: 32=93^2 = 9.

    Each factor meets its own order — 3¹ from the first-order term, 3² from the second-order term.

  3. 3

    Multiply: rate factor =3×9=27= 3 \times 9 = 27.

    27, not 6: adding the factors (3 + 3) instead of multiplying is precisely the error this one-mark question fishes for.

Answer

rate × 27

When several concentrations change at once, their factors always MULTIPLY — never add.

Defining order — the wording that earns the mark

9701/41 M/J 2023 Q3(a)(i)1 mark

Explain what is meant by the order of reaction.

Show full working
  1. 1

    Quote the definition: the power to which the concentration of a reactant is raised in the rate equation.

    Every phrase earns its keep: 'power' (not coefficient), 'of a reactant', 'in the rate equation'. Writing 'in the (balanced) equation' scores zero — that confusion is exactly what this mark tests.

Answer

The power to which the concentration of a reactant is raised in the rate equation.

Learn definitions as full sentences. A paraphrase that drops 'power', or swaps 'rate equation' for 'equation', forfeits the mark.

In the exam
194 parts · 279 marks · 29 papers · 2021–2025 (question bank)

Reaction kinetics is a heavyweight Paper 4 topic: across 2021–2025 the bank tags 194 leaf parts worth 279 marks across 29 papers, making it the 7th most-examined of the 15 Paper 4 topics — roughly nine marks in every sitting. Its questions are unusually predictable: read an order, compare a pair of experiments, compute kk with its units, quote a half-life, classify a catalyst. Predictable means drillable — every mark type in this note can be rehearsed until it is automatic.

Common mistakes

  • Taking orders from the coefficients of the balanced equation

    Orders come only from experiment; the balanced equation's coefficients tell you nothing about rate.

    The iodate(V) reaction above has coefficients 1, 6, 5 but orders 1, 2, 2. Examiners print equations with deceptively matching numbers precisely to spring this trap.

  • Saying that doubling the concentration of a second-order reactant doubles the rate

    Doubling a second-order reactant multiplies the rate by 2² = 4.

    The order applies to the FACTOR: rate factor=(concentration factor)order\text{rate factor} = (\text{concentration factor})^{\text{order}}. Say it, write it, apply it.

  • Reading an unwritten exponent as zero order (so [H₂] looks like 'zero order')

    No written power means power 1 — first order.

    Algebra's convention carries over unchanged: x means x¹. Zero order announces itself by complete absence from the rate equation, not by a missing superscript.

  • Claiming k changes when concentrations change

    At a fixed temperature k is constant; changing concentrations alters the [A]m[B]n[\text{A}]^m[\text{B}]^n part, never k.

    k responds only to temperature (and to a changed pathway under a catalyst, §08) — a distinction mark schemes probe directly.

Your turn

Read orders, predict changes, recognise zero order — the three moves this section owns.

  1. 1

    For the reaction W+2Z→Y\text{W} + 2\text{Z} \rightarrow \text{Y}, experiment gives the rate equation

    rate=k[W][Z]2\text{rate} = k[\text{W}][\text{Z}]^2

    (i) State the order with respect to W, the order with respect to Z, and the overall order.

    (ii) Predict the factor by which the rate changes when [W][\text{W}] is tripled and [Z][\text{Z}] is doubled simultaneously.

    Stuck? Show hint

    Factor → power → multiply — and ignore the 2 in 'W + 2Z' completely.

    Show solution
    1. 1

      (i) W: exponent 1 → first order. Z: exponent 2 → second order. Overall order =1+2=3= 1 + 2 = 3.

      The coefficient 2 in 'W + 2Z' is stoichiometry, not order — the rate equation is the only authority.

    2. 2

      (ii) W contributes 31=33^1 = 3; Z contributes 22=42^2 = 4.

      Each factor raised to its own order, separately, before anything is combined.

    3. 3

      Rate factor =3×4=12= 3 \times 4 = 12: the rate becomes twelve times larger.

      The factors multiply because they arise from different terms of the rate equation.

    Answer

    (i) first order W, second order Z, overall order 3 · (ii) rate ×12

  2. 2

    A reaction of J and L has the experimentally determined rate equation

    rate=k[J]2\text{rate} = k[\text{J}]^2

    (i) State the order with respect to L, and the overall order of the reaction.

    (ii) State the effect on the rate of doubling [L][\text{L}].

    (iii) State the effect on the rate of halving [J][\text{J}].

    Stuck? Show hint

    Ask yourself what it means that L appears nowhere in the equation.

    Show solution
    1. 1

      (i) L does not appear in the rate equation → zero order with respect to L. Overall order =2+0=2= 2 + 0 = 2.

      Absence IS the zero-order signature — and zero still counts when the overall order is totalled.

    2. 2

      (ii) Doubling [L][\text{L}] contributes a factor 20=12^0 = 1 → the rate does not change.

      Zero order means exactly this: any concentration change multiplies the rate by 1, i.e. not at all.

    3. 3

      (iii) Halving [J][\text{J}] contributes (12)2=14\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4} → the rate falls to one quarter.

      J's second order operates independently of L's absence — each term stands alone.

    Answer

    (i) zero order wrt L, overall order 2 · (ii) no change · (iii) rate ×¼

  3. 3

    Define the term overall order of reaction.

    Stuck? Show hint

    Sum … of what … raised where?

    Show solution
    1. 1

      Assemble the definition: the sum of the powers to which the concentrations are raised in the rate equation.

      Three earning pieces: 'sum of the powers', 'concentrations', 'in the rate equation'. Dropping the word 'rate' drifts towards Kc wording and risks the mark.

    Answer

    The sum of the powers to which the concentrations are raised in the rate equation.

  4. 4

    For a reaction with rate equation rate=k[G]2[H]\text{rate} = k[\text{G}]^2[\text{H}], the concentrations are changed so that [G][\text{G}] is tripled while [H][\text{H}] is halved.

    Calculate the factor by which the rate changes.

    Stuck? Show hint

    One factor grows, one shrinks — each goes through its own power first.

    Show solution
    1. 1

      G contributes 32=93^2 = 9.

      Tripled, second order: the factor is squared.

    2. 2

      H contributes (12)1=0.5\left(\tfrac{1}{2}\right)^1 = 0.5.

      A decrease is still just a factor — ½ — and its order processes it identically.

    3. 3

      Rate factor =9×0.5=4.5= 9 \times 0.5 = 4.5: the rate multiplies by 4.5.

      Fractional rate factors are perfectly respectable — the multiplication does not care whether a factor sits above or below 1.

    Answer

    rate ×4.5

Practise rate equations, orders and the rate constantReal past-paper questions · Rate equations, order, rate constant, half-life

The rest of this note

Checking your access…

Can you do all of these?

  • I can define order of reaction and overall order, and read all three from any given rate equation

  • I can deduce orders from initial-rates tables using comparison pairs, including fractional factors like ×1.5, and write the complete rate equation

  • I can calculate k from experimental data with correct units for any overall order, and rearrange the rate equation to find an unknown concentration

  • I can read orders from concentration–time and rate–concentration graphs, drawing a tangent and computing its gradient where needed

  • I can use t½ = ln2/k in both directions, find fractions remaining with (½)ⁿ, and explain pseudo-first-order behaviour under a large excess

  • I can judge whether a proposed mechanism is consistent with a rate equation and an overall equation, identify the rate-determining step, intermediates and catalysts

  • I can explain the effect of temperature on k through the Boltzmann distribution, naming the greater proportion of molecules with E ≥ Ea as the dominant cause

  • I can classify a catalyst as homogeneous or heterogeneous and justify the classification with an explicit phase comparison

  • I can describe the mode of action of a heterogeneous catalyst (adsorption, bond weakening, desorption) and apply it to iron in the Haber process and Pt/Pd/Rh in converters

  • I can write the two-step homogeneous catalytic cycles for NO₂ in SO₂ oxidation and Fe³⁺/Fe²⁺ or Co³⁺/Co²⁺ in the peroxodisulfate–iodide reaction, balanced for atoms and charge, summing to the overall equation

Now do the questions
194 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes