Cambridge International AS & A Level Chemistry 9701 · A Level topic 26

Reaction kinetics

What this chapter covers26.1–26.2

At AS you explained qualitatively why reactions go faster when they are more concentrated, hotter or catalysed. Those explanations cannot answer a practical question that matters in industry, in medicine and in the atmosphere: by how much will the rate change? A drug that decomposes in the body, a pollutant destroyed in the air, a reactor whose output must be predicted — each needs a quantitative relationship between rate and concentration, measured by experiment.

That relationship is the rate equation. It contains a rate constant, k, and an order for each reactant, and neither can be read from the balanced equation. This chapter shows how orders are found from experimental data, how the rate constant is calculated, how the special behaviour of first-order reactions — a constant half-life — is used, and how a rate equation provides evidence about the sequence of steps by which a reaction really happens. It ends with the way catalysts provide those alternative sequences, both on a solid surface and in solution.

What topic 26 asks you to do

26.1 Simple rate equations, orders of reaction and rate constants — explain and use the terms rate equation, order of reaction, overall order, rate constant, half-life, rate-determining step and intermediate; use rate equations of the form rate = k[A]^m[B]ⁿ with m and n equal to 0, 1 or 2; deduce orders from concentration–time graphs, initial rates and half-lives; interpret concentration–time and rate–concentration graphs; calculate initial rates; construct rate equations; use the constant half-life of a first-order reaction and k = 0.693/t_½; calculate rate constants; link a multi-step mechanism to its rate equation; describe qualitatively the effect of temperature on the rate constant.

26.2 Homogeneous and heterogeneous catalysts — explain the difference between them; describe the mode of action of a heterogeneous catalyst (adsorption, bond weakening, desorption) using iron in the Haber process and Pd, Pt and Rh in catalytic converters; describe the mode of action of a homogeneous catalyst (used in one step, re-formed in a later step) using oxides of nitrogen with atmospheric SO₂ and Fe²⁺/Fe³⁺ in the I⁻/S₂O₈²⁻ reaction.

What you are assumed to know already

Rate of reaction as the change in concentration of a reactant or product per unit time (topic 8.1).
Collision theory: only collisions with energy at least equal to the activation energy, E_a, and a suitable orientation are effective (topic 8.1).
The Boltzmann distribution and why a small rise in temperature greatly increases the number of effective collisions (topic 8.2).
A catalyst provides an alternative route with a lower activation energy and is chemically unchanged at the end (topic 8.3).

Measuring the rate of a reaction26.1.2(c), 26.1.2(d)

The rate of reaction is the change in concentration of a reactant or product per unit time. For a reactant A,

rate = −Δ[A] / Δt units: mol dm⁻³ s⁻¹ (or mol dm⁻³ min⁻¹ if time is in minutes)

The minus sign makes the rate positive, because [A] falls. For a product P the rate is +Δ[P]/Δt. When the stoichiometry is not 1 : 1 the numbers differ — in 2N₂O₅ → 4NO₂ + O₂, NO₂ appears twice as fast as N₂O₅ disappears — so a rate is always quoted for a named species.

Following a reaction

No experiment measures concentration directly at the particle level. Instead a property that changes as the reaction proceeds is measured at a series of times, and converted into concentration. The choice depends on what the reaction produces or uses up:

**Table 26.1** Methods of following the progress of a reaction.
property measured	suitable when	example
volume of gas (gas syringe or inverted burette)	a gas is produced	2H₂O₂ → 2H₂O + O₂
loss in mass (open flask on a balance)	a gas of reasonable M_r escapes	CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
colour intensity (colorimeter)	one species is coloured	iodine formed or used up; MnO₄⁻ decolorised
titration of samples withdrawn and quenched	an acid, alkali or oxidant changes in amount	ester hydrolysis followed by titrating the acid formed
electrical conductivity	the number or type of ions changes	(CH₃)₃CBr + H₂O → (CH₃)₃COH + H⁺ + Br⁻
pH	[H⁺] changes and is not swamped by a buffer	hydrolysis producing an acid

To follow a rate, a sequence of readings is needed; a single reading — "the time taken to collect 100 cm³ of gas" — gives only an average rate over that interval. When samples are withdrawn for titration the reaction in each sample must be stopped at a known moment (quenched), for example by cooling rapidly or by diluting, or by adding a reagent that removes the catalyst or one reactant.

AnimationMeasuring rates of reaction

Three common methods — colorimetry, mass loss and gas production — each with the property that is measured and how it relates to concentration.

Rate at an instant, and the initial rate

A concentration–time graph for a reactant is normally a curve that becomes less steep: as reactant is used up, collisions between reactant particles become less frequent and the rate falls. Because the rate changes continuously, the rate at a particular time is the gradient of the tangent to the curve at that time. The rate at t = 0 — the initial rate — is especially useful, because at that instant every concentration is known exactly (it is the concentration that was mixed) and no product is present to react back or interfere.

Figure 26.1 A concentration–time curve for a reactant A (illustrative data). The rate at any instant is minus the gradient of the tangent at that time. The initial rate, from the tangent at t = 0, is 5.75 × 10⁻⁴ mol dm⁻³ s⁻¹; at 100 s the tangent is shallower and the rate has fallen to 1.82 × 10⁻⁴ mol dm⁻³ s⁻¹.

Worked example 26.1 · Rate from a tangent

Given	Figure 26.1. The tangent at t = 0 runs from (0 s, 0.0500 mol dm⁻³) to the time axis at 87 s.
Find	the initial rate.
Relationship	rate = −gradient of the tangent = −Δ[A]/Δt, read from the tangent, not the curve.
Substitution	gradient = (0 − 0.0500) / (87 − 0) = −5.75 × 10⁻⁴ mol dm⁻³ s⁻¹
Answer	initial rate = 5.75 × 10⁻⁴ mol dm⁻³ s⁻¹
Check	Use two points far apart on the tangent to reduce reading error. A chord from t = 0 to t = 100 s gives only the average rate over that interval, (0.0500 − 0.0158)/100 = 3.4 × 10⁻⁴, which is smaller because the rate has been falling throughout.

Rate "at" a time, not "over" a time

When a question asks for the rate at 200 s, a tangent must be drawn at 200 s and its gradient calculated, with the working shown on the graph. Calculating the change in concentration between 0 and 200 s and dividing by 200 gives an average rate over the first 200 s — a common error that examiners single out. Use a ruler and draw the tangent long enough to read two widely spaced points.

The rate equation, orders and the rate constant26.1.1, 26.1.2(a)

Experiments show that, for a large number of reactions, the rate is proportional to the concentration of each reactant raised to some power. For a reaction between A and B the rate equation takes the form

rate = k[A]^m[B]ⁿ

m is the order of reaction with respect to A, and n the order with respect to B. In this syllabus each order is 0, 1 or 2.
m + n is the overall order of reaction.
k is the rate constant: the constant of proportionality linking the rate to the concentration terms. Its value is fixed at a given temperature and does not depend on the concentrations.

Definitions

A rate equation is an equation, determined by experiment, that relates the rate of a reaction to the concentrations of the reactants, each raised to a power: rate = k[A]^m[B]ⁿ.

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 reaction is the sum of the powers of the concentration terms in the rate equation.

The rate constant, k, is the constant of proportionality in the rate equation; it changes with temperature but not with concentration.

What an order means at the particle level

The order describes how strongly the rate responds when that one concentration is changed, with everything else kept the same:

**Table 26.2** The effect on the rate of changing the concentration of one reactant.
order with respect to A	term in rate equation	[A] doubled	[A] tripled	[A] halved
0	[A]⁰ = 1	rate unchanged	rate unchanged	rate unchanged
1	[A]¹	rate × 2	rate × 3	rate × ½
2	[A]²	rate × 4	rate × 9	rate × ¼

A zero order does not mean that A takes no part in the reaction — it must, because it appears in the equation. It means that, over the range studied, the rate does not depend on how much A is present. The explanation, developed in section 12, is that A reacts only in a fast step that comes after the slow step which controls the overall rate, or that some other factor (such as the number of sites on a catalyst surface) limits the rate.

Orders are not the coefficients in the equation

The orders are found by experiment. They cannot be deduced from the stoichiometric equation, and they are often different from its coefficients. Three reactions from examination questions show this:

2NO(g) + O₂(g) → 2NO₂(g): rate = k[NO]²[O₂] — here the orders happen to match the coefficients.
IO₃⁻ + 6H⁺ + 5I⁻ → 3I₂ + 3H₂O: rate = k[IO₃⁻][H⁺]²[I⁻]² — the coefficients 6 and 5 bear no relation to the orders.
2I⁻ + H₂O₂ + 2H⁺ → I₂ + 2H₂O: first order in H₂O₂ and I⁻ but zero order in H⁺.

A rate equation containing the product (rate = k[NO₂] for the first reaction) or containing an order equal to a coefficient "because it is in the equation" earns no credit.

AnimationOrders of reaction

Watch what happens to the rate of A → 2B when [A] changes, for zero-, first- and second-order reactions.

AnimationOrders of reaction: summary

Complete the table: the factor by which the rate changes when the concentration is doubled or tripled, for each order.

Units of the rate constant26.1.2(a), 26.1.4

The units of k are not fixed: they depend on the overall order, because k must convert the concentration terms into a rate in mol dm⁻³ s⁻¹. Rearranging the rate equation shows how to find them:

k = rate / ([A]^m[B]ⁿ) so units of k = (mol dm⁻³ s⁻¹) / (mol dm⁻³)^m+n

**Table 26.3** Units of the rate constant for each overall order (time in seconds).
overall order	example rate equation	units of k
0	rate = k	mol dm⁻³ s⁻¹
1	rate = k[A]	s⁻¹
2	rate = k[A]² or k[A][B]	mol⁻¹ dm³ s⁻¹
3	rate = k[A]²[B]	mol⁻² dm⁶ s⁻¹
n	—	mol¹⁻ⁿ dm³⁽ⁿ⁻¹⁾ s⁻¹

The pattern is quick to generate: for each extra order the power of mol falls by one and the power of dm rises by three. If the rates are in mol dm⁻³ min⁻¹, every "s⁻¹" becomes "min⁻¹"; the numerical value of k then differs by a factor of 60 from its value per second, so the time unit in k must match the time unit in which the rate was used.

Worked example 26.2 · Units for a fifth-order rate equation

Given	rate = k[IO₃⁻][H⁺]²[I⁻]²; rate measured in mol dm⁻³ min⁻¹.
Find	the units of k.
Relationship	overall order = 1 + 2 + 2 = 5; units = (mol dm⁻³ min⁻¹)/(mol dm⁻³)⁵
Working	= mol¹⁻⁵ dm⁻³⁺¹⁵ min⁻¹
Answer	mol⁻⁴ dm¹² min⁻¹
Check	Substitute back: mol⁻⁴ dm¹² min⁻¹ × (mol dm⁻³)⁵ = mol dm⁻³ min⁻¹ ✓

AnimationRate equations: true or false?

Decide whether each statement about rate equations, orders and the rate constant is true or false. Each wrong answer points to a common misconception.

Deducing orders by the initial-rates method26.1.2(b), 26.1.2(e), 26.1.4(a)

In the initial-rates method a reaction is run several times. Between one experiment and another only one initial concentration is changed where possible, and the initial rate is measured each time (from the tangent at t = 0, or by timing how long it takes to form a small, fixed amount of product — the basis of "clock" reactions). Comparing a pair of experiments in which only [A] changed isolates the effect of A:

rate₂ / rate₁ = ([A]₂ / [A]₁)^m → find m from the ratio: 2 → 2^m; 3 → 3^m

Because this syllabus uses only orders 0, 1 and 2, the ratio of rates is always 1, the concentration ratio, or its square, and no logarithms are needed.

Worked example 26.3 · Orders and k from initial rates

Illustrative data for 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) at constant temperature:

experiment	[NO] / mol dm⁻³	[H₂] / mol dm⁻³	initial rate / mol dm⁻³ s⁻¹
1	6.0 × 10⁻³	1.0 × 10⁻³	1.80 × 10⁻⁴
2	6.0 × 10⁻³	2.0 × 10⁻³	3.60 × 10⁻⁴
3	1.2 × 10⁻²	1.0 × 10⁻³	7.20 × 10⁻⁴

H₂	Experiments 1 → 2: [NO] constant, [H₂] × 2, rate × 2. First order in H₂.
NO	Experiments 1 → 3: [H₂] constant, [NO] × 2, rate × 4 = 2². Second order in NO.
Rate equation	rate = k[NO]²[H₂]; overall order 3.
k	k = rate/([NO]²[H₂]) = 1.80 × 10⁻⁴ / ((6.0 × 10⁻³)² × 1.0 × 10⁻³) = 5.0 × 10³ mol⁻² dm⁶ s⁻¹
Check	Experiments 2 and 3 give the same k (3.60 × 10⁻⁴/(3.6 × 10⁻⁵ × 2.0 × 10⁻³) = 5.0 × 10³). The stoichiometric coefficient of H₂ is 2 but its order is 1.

When two concentrations change at once

Sometimes no pair of experiments isolates a reactant. Deduce first the order that can be isolated, then allow for its effect before interpreting the rest of the change.

Worked example 26.4 · Separating two simultaneous changes

Illustrative data for A + 2B → products:

experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0 × 10⁻⁴
2	0.20	0.10	4.0 × 10⁻⁴
3	0.30	0.20	2.4 × 10⁻³

A	1 → 2: [A] × 2, [B] constant, rate × 2. First order in A.
B	1 → 3: rate × 12. The change in [A] (× 3) accounts for a factor of 3 because the reaction is first order in A. The remaining factor, 12 ÷ 3 = 4, comes from [B] × 2, so 2ⁿ = 4 and n = 2.
Rate equation	rate = k[A][B]²; k = 2.0 × 10⁻⁴/(0.10 × 0.10²) = 0.20 mol⁻² dm⁶ s⁻¹
Check	Experiment 3: 0.20 × 0.30 × 0.20² = 2.4 × 10⁻³ ✓

Once a rate equation and k are known, the equation is used forwards to calculate an initial rate for any starting mixture, or backwards to find a missing concentration: a concentration that is squared in the rate equation requires a square root at the end.

AnimationExample rate calculations

Step through five linked questions for A + B → C: the order for each reactant, the rate equation, the value of k and its units.

AnimationRate calculations

Six questions on a table of initial rates. Choose each answer, then reveal the working step by step.

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Graphs of rate and of concentration26.1.2(b), 26.1.2(c)

Two kinds of graph are used to find orders, and it is essential to know which is which before interpreting either.

Rate against concentration

If several initial rates are plotted against the initial concentration of one reactant (the others kept constant), the shape of the graph shows the order directly (Figure 26.2). For a zero-order reactant the line is horizontal; for a first-order reactant it is a straight line through the origin; for a second-order reactant it is a curve through the origin that gets steeper. For a first-order reactant the gradient of the straight line is k multiplied by the constant concentration terms for the other reactants.

Figure 26.2 Rate–concentration graphs for a reactant that is zero, first or second order. Only the first-order graph is a straight line through the origin.

Concentration against time

A single run followed over time gives a concentration–time graph. Its shape also depends on the order (Figure 26.3):

zero order: a straight line with a negative gradient — the rate, which is the gradient, stays constant until the reactant runs out;
first order: a curve whose gradient falls in proportion to the concentration, with a constant half-life;
second order: a curve that falls steeply at first and then flattens more markedly, with half-lives that get longer.

Figure 26.3 Concentration–time curves for zero-, first- and second-order reactions that start with the same concentration and the same initial rate. The half-life of the first-order reaction is the same (69 s) whether it is measured from 0.100 to 0.050, from 0.050 to 0.025 or from 0.025 to 0.0125 mol dm⁻³.

How to think about it · telling first order from second order on a curve

Both first- and second-order curves are "curves that level off", so the shape alone is not enough. Measure two successive half-lives on the graph. If they are equal the reaction is first order; if the second is longer than the first it is second order. Alternatively draw tangents at several points, calculate the rates and plot them against concentration: a straight line through the origin means first order.

Which graph is which?

A straight line on a concentration–time graph means zero order. A straight line through the origin on a rate–concentration graph means first order. Reading one kind of graph as though it were the other reverses the conclusion. Always check the axis labels first.

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Using the rate equation in calculations26.1.4(a)

Three calculations recur, and each starts from the rate equation written out in full:

k from one experiment: rearrange to k = rate/(concentration terms), substitute, and work out the units from the overall order.
rate for new concentrations: substitute k and the new concentrations. A reactant of order zero is left out, whatever its concentration.
a missing concentration: rearrange for the unknown term, and take a square root if it is squared.

Worked example 26.5 · Effect of pH on a rate

Given	A reaction has rate = k[X][H⁺]². It is run twice at the same temperature with the same [X], once at pH 3.0 and once at pH 2.0.
Find	rate at pH 2.0 ÷ rate at pH 3.0.
Relationship	[H⁺] = 10^−pH; rate ∝ [H⁺]² because k and [X] are unchanged.
Substitution	[H⁺] ratio = 10^−2.0/10^−3.0 = 10
Answer	rate ratio = 10² = 100
Check	A fall of one pH unit multiplies [H⁺] by 10; the second-order dependence squares that factor. Stopping at 10 is the error examiners report most often in questions of this kind.

Past-paper practice · Set 26A · Rate equations, orders and initial rates

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 26A.1[5]

Answer and marking guidance

(a) order with respect to NO = 2; with respect to O₂ = 1; overall = 3 — all three for ✔. (b)(i) k = rate/([NO]²[O₂]) = 1.51 × 10⁻⁴ ÷ (0.00300² × 0.00200) = 8.39 × 10³ ✔ (at least 2 s.f.); units mol⁻² dm⁶ s⁻¹ ✔. (ii) [NO]² = 6.05 × 10⁻⁵ ÷ (8390 × 0.00500) = 1.44 × 10⁻⁶, so [NO] = 1.20 × 10⁻³ mol dm⁻³ ✔ — remember the square root, because NO is second order. (c) the slowest step in the mechanism ✔. Examiner insight: orders were given correctly by most candidates; the majority calculated k with correct units; (b)(ii) was done successfully by most; the rate-determining step was well known.

Question 26A.2[7]

Parts (i)–(vi) of part (a). Part (b), on a mechanism for the same topic, is question 26C.4.

Answer and marking guidance

(i) the power to which the concentration of a reactant is raised in the rate equation ✔ — "in the rate equation" is essential. (ii) IO₃⁻ 1; H⁺ 2; I⁻ 2; overall 5 ✔ (all correct). (iii) [IO₃⁻]: a straight line through the origin; [I⁻]: a curve through the origin getting steeper (second order) ✔. (iv) k = 4.20 × 10⁻² ÷ (0.0400 × 0.0150² × 0.0250²) = 7.47 × 10⁶ ✔; units mol⁻⁴ dm¹² min⁻¹ ✔ (the rates are per minute). (v) 7.09 × 10⁻² = 7.47 × 10⁶ × 0.120 × [H⁺]² × 0.0125², so [H⁺] = 2.25 × 10⁻² mol dm⁻³ ✔. (vi) [H⁺] at pH 1.0 is 10 times that at pH 2.0; rate ∝ [H⁺]², so ratio = 10² = 100 ✔. Examiner insight: weaker answers omitted "in the rate equation" from the definition; the [I⁻] sketch must start at (0, 0) and must not become vertical or curve back; candidates who kept the rate in min⁻¹ found (iv) more straightforward — those who converted to s⁻¹ had to convert k consistently; (vi) was found very difficult, many finding the 10 : 1 ratio of [H⁺] but not squaring it.

Question 26A.3[4]

Only part (c)(i)–(iii) is reproduced here; parts (c)(iv)–(v) are question 26B.4 and part (d) is question 26D.4.

Answer and marking guidance

(i) moles of H₂ in 60 s = 6.4 ÷ 24 000 = 2.67 × 10⁻⁴ mol; rate = 2.67 × 10⁻⁴ ÷ 60 = 4.44 × 10⁻⁶ mol dm⁻³ s⁻¹ ✔ (the mark scheme's working divides the amount of H₂ formed by the time in seconds). (ii) experiments 1 → 2: [H₂PO₂⁻] doubles and the volume of H₂ doubles, so first order in H₂PO₂⁻ ✔; experiments 1 → 3: [H₂PO₂⁻] × 3 and [OH⁻] × ½; if second order in OH⁻ the rate should change by 3 × ¼ = ¾, and the volume does fall to ¾ (6.4 → 4.8 cm³) ✔. Substituting each experiment into the rate equation and showing that k is the same is an acceptable alternative. (iii) overall order 3: mol⁻² dm⁶ s⁻¹ ✔. Examiner insight: (i) was found challenging; the common error was 2.67 × 10⁻⁴, from omitting the division by 60. Most justified the order in H₂PO₂⁻ from experiments 1 and 2; fewer used experiment 3 correctly.

Question 26A.4[4]

Answer and marking guidance

(i) rate = k[NO][O₃] ✔. (ii) rate = 11 500 × (1.20 × 10⁻⁶)² = 1.66 × 10⁻⁸ ✔; units mol dm⁻³ s⁻¹ ✔. (iii) the half-life is not constant, because the reaction is second order overall (not first order overall) ✔. Examiner insight: a minority wrote product concentrations in the rate equation or used neither k nor 11 500; 1.7 × 10⁻⁸ and 1.66 × 10⁻⁸ were accepted but 1.6 × 10⁻⁸ was not — round correctly; (iii) was found difficult: only a reaction that is first order overall has a constant half-life.

Question 26A.5[6]

Answer and marking guidance

(i) experiments 2 → 3: [NH₃] × 2 with [ClO⁻] constant, rate × 4, so second order in NH₃ ✔; experiments 1 → 2: both concentrations × 2 and rate × 8; the change in [NH₃] accounts for × 4, leaving × 2 for [ClO⁻], so first order in ClO⁻ ✔. (ii) rate = k[NH₃]²[ClO⁻] ✔. (iii) k = 0.256 ÷ (0.200 × 0.100²) = 128 ✔; units mol⁻² dm⁶ s⁻¹ ✔. (iv) a curve (or line) showing k increasing as temperature increases ✔. Examiner insight: the best answers explained clearly how the change in [NH₃] affected the rate between experiments 1 and 2 before concluding first order in ClO⁻; some omitted k from the rate equation; some ignored the instruction to use experiment 1; the great majority showed k increasing with T — the common wrong graphs showed k constant, or the levelling-off shape of an amount-of-product–time graph.

Quick check 26.1

A reaction is second order with respect to P and zero order with respect to Q. State the effect on the rate of (i) tripling [P]; (ii) tripling [Q].
answer
(i) rate × 9; (ii) no change.
Give the units of k for rate = k[A][B]² with time in seconds.
answer
(mol dm⁻³ s⁻¹)/(mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹.
For rate = k[NO]²[O₂], k = 7.0 × 10³ mol⁻² dm⁶ s⁻¹. Calculate the initial rate when [NO] = 2.0 × 10⁻³ and [O₂] = 5.0 × 10⁻³ mol dm⁻³.
answer
7.0 × 10³ × (2.0 × 10⁻³)² × 5.0 × 10⁻³ = 1.4 × 10⁻⁴ mol dm⁻³ s⁻¹.
A concentration–time graph for reactant R is a straight line with a negative gradient. What is the order with respect to R, and what does the gradient represent?
answer
Zero order; the (constant) rate, which equals k.

Examiner's overall observation · Rate equations, orders and initial rates

Answered well: completing tables of orders from a given rate equation; writing a rate equation from stated orders; routine calculation of k from one experiment, with correct units; using the rate equation to find a missing concentration or an initial rate.

Found difficult: the definition of order of reaction when "in the rate equation" was left out; converting a volume of gas collected in a stated time into a rate in mol dm⁻³ s⁻¹ (the division by 60 for a time in seconds was often omitted); using an experiment in which two concentrations change at once; working out a rate ratio from two pH values, where the ratio of [H⁺] was found but not squared.

Recurring errors: product concentrations or [O] included in a rate equation; answers rounded too far (1.6 × 10⁻⁸ for 1.66 × 10⁻⁸); k calculated per second but units quoted per minute, or the reverse; sketch graphs of rate against concentration for a second-order reactant that became vertical or curved back.

What successful answers did: stated which experiments were compared and which concentration changed; showed the ratio of rates and the ratio of concentrations; kept the time unit of k consistent with the rate data; substituted the full rate equation before evaluating.

Half-life, and why it is constant for a first-order reaction26.1.1, 26.1.3(a)

Definition

The half-life, t_½, of a reaction is the time taken for the concentration of a reactant to fall to half of its initial value.

The definition names what is halving — the concentration (or amount) of a reactant. "The time taken for half the reaction" is not credited, because it does not say which quantity halves.

For a first-order reaction the half-life has a remarkable property: it does not depend on the starting concentration. Whatever the concentration at the start of the interval, it takes the same time to halve. The reason lies in the rate equation, rate = k[A]. At every instant the rate is proportional to the amount of A still present, so a constant fraction of the remaining A reacts in each equal interval of time. If a solution of 0.100 mol dm⁻³ takes 69 s to fall to 0.050 mol dm⁻³, the reaction is then running at half its initial rate with half as much A to use up; the two halvings cancel and the next halving also takes 69 s.

The same argument fails for other orders. In a zero-order reaction the rate stays the same while the amount to be used up shrinks, so each successive half-life is shorter. In a second-order reaction halving [A] quarters the rate, so each successive half-life is longer — Figure 26.3 shows all three.

Enrichment · where k = 0.693/t_½ comes from

For a first-order reaction the rate equation is a differential equation, −d[A]/dt = k[A]. Its solution is [A] = [A]₀e^−kt, an exponential decay. Putting [A] = ½[A]₀ gives e^−kt½ = ½, so kt_½ = ln 2 = 0.693, and t_½ = 0.693/k. [A]₀ has cancelled, which is the algebraic statement that the half-life is independent of concentration. The calculus is not required; the relationship k = 0.693/t_½ is, and the syllabus quotes 0.693.

Only first-order reactions

k = 0.693/t_½ applies only to a first-order reaction, and a "constant half-life" is evidence that a reaction is first order overall in the reactant being followed. A reaction that is first order with respect to each of two reactants, rate = k[NO][O₃], is second order overall and does not have a constant half-life when both concentrations fall together.

Calculations with the half-life26.1.3(b), 26.1.4(b)

Two relationships carry every calculation in this section:

k = 0.693 / t_½ after n half-lives, [A] = [A]₀ × (½)ⁿ

k is in s⁻¹ (or min⁻¹, h⁻¹) — the reciprocal of the time unit used for t_½. This is consistent with Table 26.3: a first-order rate constant has units of time⁻¹.
After 1, 2, 3, 4, 5 half-lives the fraction of A remaining is ½, ¼, ⅛, ¹/₁₆, ¹/₃₂. The fraction used up is therefore 50%, 75%, 87.5%, 93.75%, 96.9%.
Half-lives are successive intervals. If t_½ = 30 s, the concentration halves between 0 and 30 s, again between 30 s and 60 s, and again between 60 s and 90 s. The second half-life is 30 s, not "60 s".

Worked example 26.6 · A table of concentrations

Illustrative data for the decomposition of a compound A at constant temperature:

time / min	0	10	20	30	40
[A] / mol dm⁻³	0.160	0.113	0.080	0.057	0.040

Order	0.160 → 0.080 takes 20 min; 0.080 → 0.040 takes 20 min (from 20 to 40 min). Constant half-life, so first order in A.
Relationship	k = 0.693/t_½
Substitution	k = 0.693/20 min = 0.0347 min⁻¹
In seconds	t_½ = 1200 s, so k = 0.693/1200 = 5.8 × 10⁻⁴ s⁻¹
Check	Between 0 and 10 min the fraction remaining is 0.113/0.160 = 0.71; between 10 and 20 min it is 0.080/0.113 = 0.71. A constant fraction in equal times is the signature of first order.

Worked example 26.7 · Percentage used up

Given	In a first-order reaction 87.5% of the reactant is used up in 90 s.
Find	t_½ and k.
Relationship	87.5% used up = 12.5% remaining = ⅛ = (½)³, so 90 s is three half-lives.
Calculation	t_½ = 90/3 = 30 s; k = 0.693/30 = 0.0231 s⁻¹
Check	The commonest slip is to treat "87.5% used" as 87.5% remaining, or to divide the time by the wrong number of half-lives. Write down the fraction remaining first.

Worked example 26.8 · How long does it take? (an unfamiliar context)

Given	Without a catalyst, the decomposition of hydrogen peroxide is first order with k = 2.0 × 10⁻⁶ s⁻¹ at 298 K (a value quoted in an examination question). A solution is 0.75 mol dm⁻³.
Find	the half-life, and [H₂O₂] after 12.0 days.
Relationship	t_½ = 0.693/k
Calculation	t_½ = 0.693/2.0 × 10⁻⁶ = 3.47 × 10⁵ s = 4.0 days. 12.0 days = 3 half-lives, so [H₂O₂] = 0.75 × ⅛ = 0.094 mol dm⁻³
Reasonableness	Stock bottles of hydrogen peroxide are kept cool and dark and usually contain a stabiliser; a half-life of days at room temperature is consistent with a solution that slowly loses strength once opened.

Counting half-lives

If t_½ = 48 s, then 192 s is 192/48 = 4 half-lives and the concentration falls by (½)⁴ = ¹/₁₆. Using three half-lives, or dividing by 4 rather than by 16, are the errors reported when this calculation has been set.

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Finding orders from a single run, and pseudo-first-order conditions26.1.2(b), 26.1.2(c)

The initial-rates method needs several experiments. A single experiment followed over time can also give an order, in two ways:

Half-life method: read successive half-lives from the concentration–time graph (Figure 26.3). Constant half-life → first order; increasing half-life → second order; a straight line → zero order.
Tangent method: draw tangents at several times, calculate each rate, and plot rate against the concentration at that time. The shape (Figure 26.2) gives the order.

Both methods show the order with respect to the reactant whose concentration is changing. If a second reactant is present at the same time, its concentration changes too, and the two effects cannot be separated. The solution is to make every reactant except one present in large excess — typically at least ten times the concentration of the one being studied. The concentrations of the reactants in excess then hardly change during the run, and the rate depends only on the one that does change. A reaction studied in this way behaves as though it were first order if that reactant is first order; it is called pseudo-first-order.

Worked example 26.9 · A large excess of one reactant

Given	A reaction has rate = k[P][Q]², with k = 0.20 mol⁻² dm⁶ s⁻¹. It is run with [Q] = 1.00 mol dm⁻³, a large excess, and [P] = 0.010 mol dm⁻³.
Find	the effective rate constant, k′, and the half-life of P.
Relationship	[Q] stays ≈ 1.00, so rate = k′[P] with k′ = k[Q]².
Calculation	k′ = 0.20 × 1.00² = 0.20 s⁻¹; t_½ = 0.693/0.20 = 3.5 s
Check	Only 0.010 mol dm⁻³ of P can react, so at most 0.020 mol dm⁻³ of Q is used — 2% of the excess. The approximation that [Q] is constant is justified.

Choosing a method

**Table 26.4** The three experimental routes to an order.
method	what is measured	what gives the order	limitation
initial rates	the initial rate for several starting mixtures	ratio of rates against ratio of concentrations, one reactant at a time	several runs; each initial rate needs an accurate tangent or a clock method
half-lives	one concentration–time curve	whether successive half-lives are constant, shorter or longer	other reactants must be in large excess
rates from tangents	one concentration–time curve	shape of the rate–concentration graph	tangents are imprecise; other reactants in excess

Practical link

Colorimetry needs a calibration graph of absorbance against known concentration before readings can be converted into concentrations; the filter chosen should be the colour absorbed by the coloured species. In a "clock" reaction the time, t, for a fixed small amount of product to form is measured; because the amount is fixed, the initial rate is proportional to 1/t. Temperature must be controlled in every method, because k changes with temperature (section 15).

Past-paper practice · Set 26B · Half-life and the rate constant

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 26B.1[7]

Part (d) of the same question is question 26C.3.

Answer and marking guidance

(a) measure the volume of oxygen (or the loss in mass) at regular time intervals ✔ — a series of readings against time, not a single time. (b)(i) the time taken for the concentration of a reactant to fall to half its original value ✔. (ii) t_½ = 150 s (0.25 → 0.125 mol dm⁻³), with the construction shown on the graph ✔. (iii) no change — first-order half-life is independent of concentration ✔. (c)(i) tangent drawn at 200 s; gradient in the range 4–5 × 10⁻⁴ ✔; units mol dm⁻³ s⁻¹ ✔. (ii) k = rate ÷ [N₂O₅] = (your rate) ÷ 0.10, about 4.5 × 10⁻³ s⁻¹ ✔ — consistent with 0.693/150 = 4.6 × 10⁻³ s⁻¹. Examiner insight: many answers to (a) described a single measurement; to follow a rate, a sequence of measurements at different times is needed. Some half-life definitions did not say what was halving. In (c)(i) it was common to calculate the average rate over the first 200 s; the rate at 200 s needs a tangent.

Question 26B.2[5]

Answer and marking guidance

(a)(i) tangent drawn at t = 40 s; rate = 1.70 × 10⁻⁴ mol dm⁻³ s⁻¹ (within the accepted range) ✔. (ii) construction lines for two successive half-lives ✔; they are equal (e.g. 0 → 30 s and 30 → 60 s, each 30 s), so the half-life is constant and the reaction is first order ✔. (b) 75% consumed = 25% remaining = two half-lives, so t_½ = 320/2 = 160 s ✔; k = 0.693/160 = 4.33 × 10⁻³ s⁻¹ ✔. Examiner insight: better tangents were drawn with a ruler and pencil. Half-lives must be quoted as intervals — "the second half-life is from 30 s to 60 s, a period of 30 s" supports a constant half-life; "the first half-life is 30 s, the second 60 s" does not. "Half-time" was not credited. (b) was found very difficult: k = 0.693/t_½ was rarely used, and 213 s was a more common (wrong) half-life than 160 s.

Question 26B.3[2]

Parts (c)(i) and (c)(ii) only.

Answer and marking guidance

(i) the reaction is first order (with respect to cisplatin, and overall), so the rate is directly proportional to its concentration ✔. (ii) t_½ = 0.693/k = 0.693 ÷ 2.50 × 10⁻⁵ = 2.77 × 10⁴ s ✔. Examiner insight: (i) was usually correct; (ii) was answered well, though weaker candidates could not recall the relationship between t_½ and k.

Question 26B.4[2]

Answer and marking guidance

(iv) t_½ = 0.693 ÷ 8.25 × 10⁻⁵ = 8.4 × 10³ s ✔. (v) k₁ increases as the temperature increases ✔. Under a large excess of OH⁻, [OH⁻] is effectively constant, so the reaction behaves as first order in H₂PO₂⁻ with k₁ = k[OH⁻]². Examiner insight: most candidates answered both parts correctly.

Quick check 26.2

A first-order reaction has t_½ = 25 s. What fraction of the reactant remains after 100 s?
answer
100/25 = 4 half-lives; (½)⁴ = ¹/₁₆.
Calculate k for a first-order reaction with t_½ = 6.0 min, in s⁻¹.
answer
t_½ = 360 s; k = 0.693/360 = 1.9 × 10⁻³ s⁻¹.
The initial concentration of a first-order reactant is halved. What happens to (i) its half-life; (ii) its initial rate?
answer
(i) unchanged; (ii) halved.
Successive half-lives read from a graph are 40 s, 80 s and 160 s. What is the order?
answer
Second order — the half-life increases as the concentration falls (it doubles each time here because t_½ ∝ 1/[A]₀ for second order).

Examiner's overall observation · Half-life and the rate constant

Answered well: explaining why a first-order reaction has a constant half-life (it is first order overall; rate is proportional to concentration); showing that t_½ = 0.693/k gives a stated value; calculating t_½ from a given first-order k.

Found difficult: defining half-life without saying which quantity halves; predicting the effect of a lower initial concentration on the half-life (increase, decrease and no change were all common); finding k when a percentage of reactant had been consumed — k = 0.693/t_½ was rarely used, and a wrong half-life of 213 s was more common than 160 s; recognising that a reaction which is second order overall has no constant half-life.

Recurring errors: an average rate over an interval instead of a tangent at an instant; very poor freehand tangents; half-lives described cumulatively ("the second half-life is 60 s") rather than as equal intervals; too few half-lives counted.

What successful answers did: drew construction lines on the graph for at least two successive half-lives and stated them as intervals; used a ruler for tangents; wrote the fraction remaining before counting half-lives; recalled k = 0.693/t_½ and gave units of time⁻¹.

Multi-step reactions, the rate-determining step and intermediates26.1.1, 26.1.5(d)

A balanced equation records only where a reaction starts and where it finishes. Most reactions do not happen in the single collision that the equation seems to describe. In IO₃⁻ + 6H⁺ + 5I⁻ → 3I₂ + 3H₂O, twelve particles appear on the left; the chance of twelve particles colliding at the same instant, with enough energy and the right orientation, is effectively zero. The reaction instead proceeds through a sequence of elementary steps, each normally a collision between two particles (occasionally one particle breaking up on its own). The sequence of steps is the reaction mechanism.

The steps take place at very different rates. The overall reaction can go no faster than its slowest step, just as the output of a production line is limited by its slowest station: products of the fast steps simply wait for the slow one. The slowest step therefore controls the rate of the whole reaction.

Definitions

The rate-determining step is the slowest step in a reaction mechanism; it determines the overall rate of the reaction.

An intermediate is a species that is formed in one step of a mechanism and used up in a later step; it does not appear in the overall equation.

On a reaction pathway diagram (Figure 26.4) each step has its own activation energy, and an intermediate sits in the energy "valley" between two peaks. The rate-determining step is the step with the largest activation energy measured from the level at which that step starts.

Figure 26.4 Reaction pathway for a two-step reaction in which the first step is rate-determining. The intermediate is a real species with a finite lifetime, at a minimum between the two maxima; each maximum is a transition state, which cannot be isolated.

**Table 26.5** Intermediate, catalyst and transition state compared.
	intermediate	catalyst	transition state
first appears in the mechanism as	a product of an early step	a reactant of an early step	—
then	used up as a reactant in a later step	re-formed as a product in a later step	—
in the overall equation?	no	no (may be written over the arrow)	no
on the pathway diagram	a minimum between two peaks	changes the whole pathway	a maximum
can it be detected or isolated?	sometimes detected; has a short but real lifetime	yes: present at the start and the end	no

Worked example 26.10 · Intermediate or catalyst?

A mechanism for the decomposition of ozone in the presence of chlorine atoms:

step 1 Cl + O₃ → ClO + O₂ step 2 ClO + O → Cl + O₂

Overall	Add the steps and cancel species that appear on both sides: O₃ + O → 2O₂.
ClO	made in step 1, used in step 2 → intermediate.
Cl	used in step 1, re-formed in step 2 → catalyst.
Check	Both cancel when the steps are added, which is why neither appears in the overall equation. The order in which a species appears (product first, or reactant first) is what distinguishes them.

AnimationThe rate-determining step

A reaction that takes place in more than one step: see how the slowest step controls the rate at which products appear, whichever position it has in the sequence.

From a mechanism to a rate equation26.1.5(b), 26.1.5(c)

Because the rate-determining step controls the rate, the rate equation reflects the particles that take part in that step. This gives a simple set of rules for predicting a rate equation from a proposed mechanism:

Identify the rate-determining step (it will be stated, or marked "slow").
Write a concentration term for each particle that reacts in the rate-determining step; if two identical particles react, the term is squared.
If a particle in the rate-determining step is an intermediate made in an earlier fast step, replace it by the reactants that formed it.
Species that react only after the rate-determining step do not appear: their order is zero.
Water acting as the solvent is present in such large excess that its concentration does not change and is not included.

Worked example 26.11 · Slow first step

NO₂(g) + CO(g) → NO(g) + CO₂(g) is thought to proceed by:

step 1 (slow) NO₂ + NO₂ → NO₃ + NO step 2 (fast) NO₃ + CO → NO₂ + CO₂

Particles in the slow step	two NO₂
Rate equation	rate = k[NO₂]²; zero order with respect to CO, which reacts only after the slow step.
Check	Steps add to the overall equation: 2NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂, which cancels to NO₂ + CO → NO + CO₂ ✓. NO₃ is an intermediate.

Worked example 26.12 · Slow step after a fast step

For 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g), Worked example 26.3 found rate = k[NO]²[H₂]. One mechanism consistent with this is:

step 1 (fast) NO + NO → N₂O₂ step 2 (slow) N₂O₂ + H₂ → N₂O + H₂O step 3 (fast) N₂O + H₂ → N₂ + H₂O

Slow step	N₂O₂ + H₂; N₂O₂ is an intermediate made from 2NO in step 1.
Rate equation	replace N₂O₂ by 2NO: rate = k[NO]²[H₂], matching experiment.
Check	Sum of steps: 2NO + 2H₂ → N₂ + 2H₂O ✓. Intermediates: N₂O₂ and N₂O. The second H₂ reacts after the slow step, which is why the order in H₂ is 1 although its coefficient is 2.

A catalyst in the rate equation: the iodination of propanone

Propanone reacts with iodine in acid solution:

CH₃COCH₃(aq) + I₂(aq) —H⁺(aq)→ CH₃COCH₂I(aq) + HI(aq)

Experiment shows rate = k[CH₃COCH₃][H⁺]: first order in propanone, first order in hydrogen ions, and zero order in iodine. Two conclusions follow. Hydrogen ions take part in or before the rate-determining step although they do not appear in the overall equation — they are a catalyst, used in an early step and returned later. Iodine reacts only after the rate-determining step, so however much iodine is present the rate is unchanged. The accepted mechanism has a slow step in which protonated propanone rearranges to the enol, CH₂=C(OH)CH₃; the enol then reacts rapidly with iodine. The disappearance of the iodine colour can be used to follow this reaction, and because the reaction is zero order in I₂ the colour fades at a constant rate — a straight-line concentration–time graph for iodine.

AnimationThe rate equation and mechanisms

Three stages for the iodination of propanone: the experimental rate equation, what it says about the slow step, and a mechanism consistent with it.

Mechanisms you have already met

The same link underlies the two mechanisms of nucleophilic substitution in topic 15. For a tertiary halogenoalkane such as 2-bromo-2-methylpropane, rate = k[(CH₃)₃CBr]: the slow step is the breaking of the C–Br bond to form a carbocation, and hydroxide ions react only afterwards (S_N1 — one particle in the rate-determining step). For bromoethane, rate = k[CH₃CH₂Br][OH⁻]: both particles are involved in a single step (S_N2 — two particles). The rate equation is the experimental evidence that distinguishes the two.

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From a rate equation to a mechanism26.1.5(a), 26.1.5(e)

Running the rules backwards lets you suggest a mechanism that is consistent with a measured rate equation and the overall equation. A suggestion must satisfy every one of these conditions:

the rate-determining step contains exactly the particles in the rate equation, with the right number of each (a squared term means two of that particle, possibly delivered as an intermediate formed from them in an earlier step);
each step involves no more than two particles colliding;
each step balances for atoms and for charge;
the steps add up to the overall equation, with every intermediate cancelling;
species that are zero order appear only in steps after the rate-determining step.

Worked example 26.13 · Suggesting a two-step mechanism

Given	2NO(g) + Br₂(g) → 2NOBr(g); rate = k[NO][Br₂]; the mechanism has two steps.
Slow step	must contain one NO and one Br₂: NO + Br₂ → NOBr₂ (slow)
Fast step	must use the intermediate and the remaining NO to reach the products: NOBr₂ + NO → 2NOBr (fast)
Check	Sum: 2NO + Br₂ → 2NOBr ✓; each step bimolecular ✓; atoms balance in each step ✓; the intermediate NOBr₂ cancels ✓.

Worked example 26.14 · Identifying the rate-determining step

Given	H₂O₂ + 2H⁺ + 2I⁻ → I₂ + 2H₂O is found to be rate = k[H₂O₂][I⁻]. A proposed mechanism is: A H₂O₂ + I⁻ → IO⁻ + H₂O B IO⁻ + H⁺ → HIO C HIO + I⁻ → I₂ + OH⁻ D OH⁻ + H⁺ → H₂O
Reasoning	The rate equation contains one H₂O₂ and one I⁻ and no H⁺. Only step A has exactly those particles; H⁺ first reacts in step B, after the slow step, so its order is zero.
Answer	Step A is rate-determining. IO⁻, HIO and OH⁻ are intermediates.

Mechanism errors examiners report

a slow step that contains a species not in the rate equation, or omits one that is;
steps with three or more reacting particles, when the question says each step involves two;
"steps" that are half-equations with electrons, rather than equations between species;
steps that do not balance for atoms or charge, or introduce species that take no part in the reaction;
steps that do not add up to the overall equation.

Check each step for atoms and charge, then add the steps and cancel before moving on.

AnimationRate-determining step: questions

Three questions linking rate equations, mechanisms and the slow step. Justify each answer from the particles in the slow step.

Temperature and the rate constant26.1.6

In the rate equation, rate = k[A]^m[B]ⁿ, the concentration terms do not change when a reaction mixture is warmed — the same solution simply becomes hotter. The observed increase in rate must therefore come from an increase in the rate constant. This is why k is quoted "at a stated temperature": it is constant with respect to concentration, but not with respect to temperature.

The reason k increases is the one met at AS. Raising the temperature increases the average kinetic energy of the particles, and the Boltzmann distribution spreads to higher energies (Figure 26.5). The fraction of collisions with energy greater than or equal to the activation energy rises sharply — much more sharply than the temperature itself — so the frequency of effective collisions rises and k rises. Collisions also become slightly more frequent, but that contributes only a small part of the increase.

Figure 26.5 The Boltzmann distribution of molecular energies at two temperatures. The area beyond the activation energy (shaded, and magnified in the inset) represents the fraction of molecules able to react. In this model a 25% rise in temperature multiplies that fraction about three times, which is why the rate constant rises far more steeply than the temperature.

k increases with temperature for almost all reactions, and increasingly steeply: a graph of k against T is a curve that gets steeper as T rises.
Near room temperature, many reactions with typical activation energies roughly double their rate for a rise of 10 K. This is a rough guide, not a law.
A reaction with a larger activation energy is more sensitive to temperature, because its effective collisions come from further out in the tail of the distribution.
A catalyst also increases k, at constant temperature, by providing a route with a lower activation energy (sections 16–21).
Temperature changes k but does not change the orders, which depend on the mechanism.

Enrichment · the Arrhenius equation

The quantitative relationship is k = Ae^−E_a/RT. It is not required by this syllabus. If an examination question supplies it, it is being used as data handling: substitute the values given, keep T in kelvin and E_a in J mol⁻¹.

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Past-paper practice · Set 26C · Mechanisms, the rate-determining step and temperature

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 26C.1[3]

Answer and marking guidance

(i) the slowest step in the (overall) reaction ✔. (ii) add the four steps and cancel IO⁻, HIO and OH⁻: H₂O₂ + 2H⁺ + 2I⁻ → I₂ + 2H₂O ✔. (iii) H₂O₂ = 1; I⁻ = 1; H⁺ = 0 ✔ — H⁺ first reacts in step 2, after the rate-determining step. (No examiner report was published for this session.)

Question 26C.2[3]

This part continues from 26A.1 (the same paper and question).

Answer and marking guidance

(i) the rate-determining step must contain one S₂O₈²⁻ and one I⁻ ✔, and the steps must add to the overall equation with no species that can cancel left on either side ✔; for example step 1 (slow) S₂O₈²⁻ + I⁻ → SO₄²⁻ + SO₄I⁻; step 2 (fast) SO₄I⁻ + I⁻ → SO₄²⁻ + I₂; rate-determining step = step 1. (ii) 192/48 = 4 half-lives; [I⁻] = 0.00780 ÷ 2⁴ = 4.9 × 10⁻⁴ mol dm⁻³ ✔. Examiner insight: (i) discriminated well and many inventive mechanisms were seen, but some did not link the mechanism to the given rate equation and overall equation; the common error in (ii) was 9.75 × 10⁻⁴, from using only three half-lives.

Question 26C.3[2]

Answer and marking guidance

step 1 (slow): NO₂ + O₃ → NO₃ + O₂ ✔; step 2 (fast): NO₂ + NO₃ → N₂O₅ ✔. The rate equation rate = k[NO₂][O₃] requires exactly one NO₂ and one O₃ in the slow step; the second NO₂ reacts afterwards; NO₃ is an intermediate. Examiner insight: this discriminated well; the need for the first step to involve one O₃ and only one NO₂ was not always appreciated. Many excellent answers were seen.

Question 26C.4[3]

Answer and marking guidance

step 1: Fe³⁺ + I⁻ → FeI²⁺ ✔; step 2: FeI²⁺ + I⁻ → Fe²⁺ + I₂⁻ (or → FeI₂⁺), and this is the slowest step ✔; step 3: Fe³⁺ + I₂⁻ → Fe²⁺ + I₂ (or FeI₂⁺ + Fe³⁺ → 2Fe²⁺ + I₂) ✔. The rate-determining step (step 2) contains, directly or through the intermediate FeI²⁺, one Fe³⁺ and two I⁻, matching first order in Fe³⁺ and second order in I⁻; the second Fe³⁺ reacts after the slow step. Each step involves exactly two ions, and the three steps add to 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. Examiner insight: found difficult; few ensured two ions in each step; other errors were introducing species not involved in the reaction, writing half-equations instead of equations, and equations unbalanced for substances or for charge.

Quick check 26.3

Define intermediate.
answer
A species formed in one step of a mechanism and used up in a later step; it does not appear in the overall equation.
For 2A + B → C, the mechanism is A + B → X (slow); X + A → C (fast). Deduce the rate equation.
answer
rate = k[A][B]; the second A reacts after the slow step.
The rate equation for X + 2Y → Z is rate = k[Y]². Suggest a two-step mechanism.
answer
Y + Y → Y₂ (slow); Y₂ + X → Z (fast).
A reaction mixture is warmed from 20 °C to 30 °C. State what happens to (i) the rate constant, (ii) the order with respect to each reactant.
answer
(i) increases (for many reactions it roughly doubles); (ii) unchanged.

Examiner's overall observation · Mechanisms, the rate-determining step and temperature

Answered well: defining the rate-determining step as the slowest step; deducing orders from a given mechanism with a stated rate-determining step; many good and inventive two-step mechanisms when the rate equation was used to decide what the slow step must contain.

Found difficult: linking a suggested mechanism to both the given rate equation and the overall equation; ensuring that each step involved only two ions when told so; realising that a first step containing one O₃ and one NO₂ was required by rate = k[NO₂][O₃]; combining several equations into one overall equation.

Recurring errors: introducing species not involved in the reaction; writing half-equations instead of equations; steps unbalanced for atoms or for charge; sketches of k against temperature showing k constant at all temperatures, or with the levelling-off shape of an amount-of-product–time graph.

What successful answers did: wrote the slow step first, from the rate equation; then added fast steps that used up the intermediate and reached the products; checked atoms and charges in every step and cancelled to recover the overall equation; sketched k rising with temperature.

Homogeneous and heterogeneous catalysts26.2.1

A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy, and it is chemically unchanged at the end. At constant temperature the rate constant increases, because a larger fraction of collisions now have enough energy to react along the new route. A catalyst does not change ΔH, and for a reversible reaction it speeds up the forward and reverse reactions equally, so it does not change the position of equilibrium; it only allows equilibrium to be reached sooner.

The A Level question is how a catalyst provides a new pathway. The answer depends on whether the catalyst is in the same phase as the reactants.

Definitions

A homogeneous catalyst is in the same phase as the reactants (for example, all in aqueous solution, or all gases).

A heterogeneous catalyst is in a different phase from the reactants (usually a solid catalysing reactions of gases or of substances in solution).

**Table 26.6** Examples of the two types of catalysis. The four examples in bold are those named in the syllabus.
reaction	catalyst	phases	type
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (Haber process)	iron, Fe(s)	solid catalyst, gaseous reactants	heterogeneous
2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g) (catalytic converter)	Pt, Pd, Rh (s)	solid, gases	heterogeneous
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (Contact process)	V₂O₅(s)	solid, gases	heterogeneous
2H₂O₂(aq) → 2H₂O(l) + O₂(g)	MnO₂(s)	solid, solution	heterogeneous
SO₂(g) + ½O₂(g) → SO₃(g) in the atmosphere	NO₂(g) / NO(g)	all gases	homogeneous
S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)	Fe²⁺(aq) or Fe³⁺(aq)	all aqueous	homogeneous
CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq)	H⁺(aq)	all aqueous	homogeneous

Many of the most important catalysts are transition elements or their compounds. The reasons — more than one stable oxidation state, and vacant d orbitals that can form bonds to reactant molecules — are developed in topic 28. Both features appear in this section: iron's surface forms bonds to N₂ and H₂, and iron ions switch between +2 and +3.

How a heterogeneous catalyst works26.2.2

A heterogeneous catalyst works at its surface. The reaction takes place in a sequence of steps, each of which must be described precisely — naming which species is adsorbed, whose bonds weaken, and what leaves the surface:

Adsorption. Reactant molecules form weak bonds to atoms on the catalyst surface, at active sites. Adsorption holds the reactants close together, in an orientation favourable for reaction, and increases their effective concentration at the surface.
Bond weakening. Bonding to the surface draws electron density from bonds within the reactant molecules, so those bonds are weakened (and may break). Less energy is needed to reach the transition state: the activation energy is lower.
Reaction. New bonds form between the adsorbed species.
Desorption. The bonds between the products and the surface break, and the products leave, freeing the active sites for further reactant molecules.

Figure 26.6 The mode of action of iron in the Haber process. Nitrogen and hydrogen are adsorbed at active sites; the N≡N and H–H bonds are weakened and broken; N–H bonds form; ammonia desorbs. Every panel contains the same atoms, two nitrogen and six hydrogen.

Precise wording earns the marks

"Adsorption occurs, bonds weaken, desorption" describes nothing. The creditworthy points are: reactants adsorb onto the surface of the catalyst; bonds within the reactant molecules weaken; the reaction occurs and the products desorb. When a question includes an equation, use the formulae — "N₂O adsorbs onto the platinum; the N–O bonds in N₂O weaken; N₂ and O₂ desorb" — rather than "the gas". Note also that adsorption (onto a surface) is not absorption (into the bulk).

What makes a good heterogeneous catalyst

A large surface area. Only the surface is used, so catalysts are finely divided, porous, or spread as a thin coating on a support.
Adsorption of the right strength. Bonds to the surface must be strong enough to weaken the bonds in the reactants, but weak enough for the products to desorb. If the products stay attached, the active sites are blocked and the catalyst stops working.
Resistance to poisoning. A poison is a substance that adsorbs strongly and permanently onto active sites, so that reactants can no longer reach them. Sulfur compounds poison the iron of the Haber process, which is why the hydrogen feed is purified; lead poisons the metals of a catalytic converter.

How to think about it · zero order on a crowded surface

When a small amount of catalyst is used with a large amount of reactant, almost every active site is occupied at any instant. Supplying more reactant cannot speed things up, because there is nowhere for it to adsorb; the rate is limited by how fast the adsorbed molecules react and leave. The reaction then shows zero-order kinetics with respect to the reactant: the concentration–time graph is a straight line and the rate–time graph is horizontal until the reactant is nearly used up. The model below shows the transition from first order (few sites occupied) to zero order (surface saturated).

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Iron in the Haber process26.2.2(a)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹

Without a catalyst this reaction is extremely slow at any temperature that allows a useful equilibrium yield, because the first step would require breaking the very strong N≡N triple bond. On an iron catalyst the sequence of Figure 26.6 replaces it:

N₂ and H₂ molecules are adsorbed onto the iron surface, forming bonds to iron atoms at active sites;
the N≡N and H–H bonds are weakened and break, leaving nitrogen and hydrogen atoms bonded to the surface;
N–H bonds form in steps, NH → NH₂ → NH₃;
NH₃ is desorbed, freeing the sites.

The catalyst allows the plant to run at a moderate temperature (typically around 400–450 °C) where the rate is acceptable, rather than at the much higher temperature that would be needed without it — which would give a far lower equilibrium yield, since the forward reaction is exothermic. The iron is used in a finely divided, porous form to maximise its surface area.

Catalytic converters26.2.2(b)

A petrol engine's exhaust contains three pollutants: carbon monoxide (from incomplete combustion), oxides of nitrogen (from N₂ and O₂ combining at the high temperature in the cylinder) and unburnt hydrocarbons. In a catalytic converter the exhaust passes through a ceramic honeycomb coated with a very thin layer of platinum, palladium and rhodium (Figure 26.7). The honeycomb gives an enormous surface area for a small mass of expensive metal.

Figure 26.7 A catalytic converter. The reactions take place on the surface of the precious metals, each by adsorption, weakening of bonds within the reactants, reaction and desorption of the products.

The key reaction removes two pollutants together, carbon monoxide being oxidised as nitrogen monoxide is reduced:

2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)

Remaining CO and unburnt hydrocarbons are oxidised to CO₂ and H₂O by oxygen in the exhaust. The mode of action is the heterogeneous sequence: CO and NO are adsorbed onto the metal surface; the bonds within the NO and CO molecules are weakened; the atoms rearrange to form CO₂ and N₂; and these products are desorbed. Platinum and palladium mainly catalyse the oxidation reactions and rhodium mainly the reduction of NO.

Converter details that are often confused

The three metals are palladium, platinum and rhodium; naming only platinum is incomplete.
The converter does not remove carbon dioxide — it produces it. Its purpose is to remove CO, NO_x and unburnt hydrocarbons.
A converter works poorly when cold, and is poisoned by lead, which adsorbs permanently on the active sites.

How a homogeneous catalyst works26.2.3

A homogeneous catalyst takes part in the reaction as a reactant in one step and is re-formed as a product in a later step. The catalysed route therefore replaces one difficult step by two (or more) easier steps, each with a lower activation energy. Because the catalyst is in the same phase, it usually appears in the rate equation, like H⁺ in the iodination of propanone (section 13). Two examples are named in the syllabus.

Oxides of nitrogen and atmospheric sulfur dioxide

Sulfur dioxide released by burning fossil fuels is oxidised in the atmosphere to sulfur trioxide, which dissolves in water droplets to form sulfuric acid — a major contributor to acid rain. The direct reaction of SO₂ with oxygen is slow, but nitrogen dioxide, also present in polluted air, catalyses it:

step 1 SO₂(g) + NO₂(g) → SO₃(g) + NO(g) (NO₂ used; SO₂ oxidised)

step 2 NO(g) + ½O₂(g) → NO₂(g) (NO₂ re-formed)

overall SO₂(g) + ½O₂(g) → SO₃(g) then SO₃ + H₂O → H₂SO₄

All species are gases, so the catalysis is homogeneous. NO₂ is used in step 1 and re-formed in step 2, so it acts as a catalyst; NO is an intermediate in this sequence. (If the cycle is considered to start from NO, the roles are simply exchanged — both are "atmospheric oxides of nitrogen".) Nitrogen is oxidised in step 2 and reduced in step 1: the catalyst works by changing oxidation state, +4 → +2 → +4.

Iron ions and the reaction between iodide and peroxodisulfate

S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)

This reaction is thermodynamically very favourable but slow. The reason is the charges: both reactants are negative ions, so they repel one another, and a collision energetic enough to overcome the repulsion and react is rare. The activation energy is high.

Adding a small amount of Fe²⁺(aq) — or Fe³⁺(aq) — speeds it up greatly. The iron ions provide a route in which every step is a reaction between oppositely charged ions:

step 1 S₂O₈²⁻(aq) + 2Fe²⁺(aq) → 2SO₄²⁻(aq) + 2Fe³⁺(aq)

step 2 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq)

Fe²⁺ is oxidised in step 1 and re-formed in step 2. If Fe³⁺ is added instead, step 2 happens first and makes the Fe²⁺ for step 1; the cycle is the same (Figure 26.9). Adding the two steps gives the overall equation, with the iron cancelling.

Figure 26.8 Reaction pathways for the uncatalysed and iron-catalysed reactions between peroxodisulfate and iodide ions (schematic). The catalysed route has two steps, each with a lower activation energy than the single uncatalysed step.

Figure 26.9 The catalytic cycle. Iron switches between the +2 and +3 oxidation states and is not consumed.

Using E^⦵ to show that the catalysed steps are feasible

Standard electrode potentials (topic 24) explain why iron ions can play this role. The three half-cells involved are:

**Table 26.7** Standard electrode potentials for the peroxodisulfate–iodide system.
half-equation	E^⦵ / V
S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻	+2.01
Fe³⁺ + e⁻ ⇌ Fe²⁺	+0.77
I₂ + 2e⁻ ⇌ 2I⁻	+0.54

Step 1: S₂O₈²⁻ oxidises Fe²⁺: E^⦵_cell = +2.01 − 0.77 = +1.24 V, positive, so feasible.
Step 2: Fe³⁺ oxidises I⁻: E^⦵_cell = +0.77 − 0.54 = +0.23 V, positive, so feasible.

The general condition is that the E^⦵ of the catalyst's redox couple must lie between the E^⦵ values of the two reacting couples. A couple with E^⦵ above +2.01 V could not be oxidised by peroxodisulfate; one below +0.54 V could not oxidise iodide. E^⦵ values show whether each step is energetically feasible; they say nothing about how fast it is.

Worked example 26.15 · Could another ion catalyse the reaction?

Given	E^⦵(Cr³⁺/Cr²⁺) = −0.41 V; E^⦵(Mn³⁺/Mn²⁺) = +1.49 V.
Test	Is the E^⦵ between +0.54 V and +2.01 V?
Cr³⁺/Cr²⁺	No: Cr³⁺ cannot oxidise I⁻ (E^⦵_cell = −0.41 − 0.54 = −0.95 V), so the cycle breaks at step 2.
Mn³⁺/Mn²⁺	Yes: step 1 +2.01 − 1.49 = +0.52 V; step 2 +1.49 − 0.54 = +0.95 V. Both feasible, so Mn²⁺ could in principle act as a catalyst — provided the steps are also fast.

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Past-paper practice · Set 26D · Homogeneous and heterogeneous catalysts

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 26D.1[6]

Part (c) of the same question is question 26B.3.

Answer and marking guidance

(a)(i) a homogeneous catalyst is in the same phase (state) as the reactants AND a heterogeneous catalyst is in a different phase ✔. (ii) S₂O₈²⁻ + 2Fe²⁺ → 2Fe³⁺ + 2SO₄²⁻ ✔; 2I⁻ + 2Fe³⁺ → 2Fe²⁺ + I₂ ✔. (iii) in the uncatalysed reaction both reactants are anions, which repel each other; in the catalysed route each step is between oppositely charged ions ✔. (b)(i) rate = k[NO]²[O₂] ✔. (ii) rate = 8.60 × 10⁶ × (7.20 × 10⁻⁴)² × 1.90 × 10⁻³ = 8.47 × 10⁻³ mol dm⁻³ s⁻¹ ✔. Examiner insight: (a)(i) was mostly correct; (a)(ii) proved difficult — many could not recall the two equations; (a)(iii) was generally well answered; in (b)(i), [O] and [NO₂] were common errors.

Question 26D.2[2]

Answer and marking guidance

(i) palladium, platinum and rhodium ✔. (ii) a catalyst in a different phase (state) from the reactants ✔. Examiner insight: most knew platinum; recalling palladium and rhodium was more difficult. (ii) was usually correct.

Question 26D.3[6]

Part (a) of the same question is question 26A.4.

Answer and marking guidance

(i) a horizontal straight line (constant rate) ✔. (ii) a straight line with a negative gradient ✔ — zero order. (iii) N₂O is adsorbed onto the platinum surface ✔; the N–O bonds within the N₂O molecules weaken ✔; the reaction occurs and the products, N₂ and O₂, are desorbed ✔. (iv) all the active sites on the catalyst surface are occupied, so adding more N₂O does not increase the rate ✔. Examiner insight: (i) and (ii) were generally correct. In (iii) answers must say which species is involved at each stage: "gas" did not score, because three gases are involved; the bonds that weaken are within the N₂O molecules, not between them; the products leave the surface. (iv) was rarely credited.

Question 26D.4[2]

Answer and marking guidance

any two of: reactants adsorb onto the surface of the catalyst; bonds in the reactants weaken; the reaction occurs and the products desorb ✔ — all three ✔✔. Examiner insight: generally well answered; the common error was describing what a heterogeneous catalyst is rather than its mode of action.

Question 26D.5[4]

Answer and marking guidance

(a) relevant data: E^⦵(H₂O₂/H₂O) = +1.77 V, E^⦵(MnO₂/Mn²⁺) = +1.23 V, E^⦵(O₂/H₂O₂) = +0.68 V (so each step has a positive E_cell, 0.55 V and 0.54 V) ✔; equation 1: MnO₂ + H₂O₂ + 2H⁺ → Mn²⁺ + O₂ + 2H₂O ✔; equation 2: Mn²⁺ + H₂O₂ → MnO₂ + 2H⁺ ✔. The MnO₂/Mn²⁺ potential lies between the two H₂O₂ couples, exactly the condition used for Fe³⁺/Fe²⁺ in section 21. (b) rate = 2.0 × 10⁻⁶ × 0.75 = 1.5 × 10⁻⁶ mol dm⁻³ s⁻¹ ✔. (No examiner report was published for this session.)

Quick check 26.4

Classify each catalyst as homogeneous or heterogeneous: (i) V₂O₅ in 2SO₂ + O₂ ⇌ 2SO₃; (ii) H⁺(aq) in the hydrolysis of an ester in aqueous solution.
answer
(i) heterogeneous (solid; gaseous reactants); (ii) homogeneous (all in aqueous solution).
Write the two equations that show how NO₂ catalyses the oxidation of SO₂ in the atmosphere.
answer
SO₂ + NO₂ → SO₃ + NO; NO + ½O₂ → NO₂.
Explain why the uncatalysed reaction between S₂O₈²⁻ and I⁻ has a high activation energy.
answer
Both reactants are negatively charged ions, which repel each other.
A small piece of platinum wire catalyses the decomposition of a large amount of N₂O and the reaction is zero order. Explain.
answer
All the active sites on the platinum surface are occupied, so adding more N₂O cannot increase the rate.

Examiner's overall observation · Homogeneous and heterogeneous catalysts

Answered well: the difference between homogeneous and heterogeneous catalysts in terms of phase; explaining that the uncatalysed peroxodisulfate–iodide reaction is slow because both reactants are anions; sketching a horizontal rate–time graph for a zero-order catalysed reaction; naming platinum as a converter metal.

Found difficult: recalling the two equations for the Fe²⁺-catalysed reaction between iodide and peroxodisulfate; recalling palladium and rhodium alongside platinum; explaining zero-order kinetics in terms of all active sites on the catalyst being occupied (rarely credited).

Recurring errors: describing what a heterogeneous catalyst is when asked for its mode of action; general statements ("adsorption occurs", "bonds weaken", "desorb") that did not say which species; "gas" used when three different gases were involved; bond weakening described as between molecules rather than within the reactant molecules; the terms catalyst, reactants and products used in the wrong places.

What successful answers did: named the species at each stage — reactants adsorb onto the catalyst surface; bonds within the reactants weaken; products desorb from the surface; wrote both catalytic equations, balanced, with the iron ions regenerated.

Misconceptions and how the topic is assessed26.1–26.2

The misconceptions below recur in examiner reports on this topic. Each is set out as the incorrect idea, why it fails, the correct model and what it costs in an examination.

**Table 26.8** Recurring misconceptions.
misconception	why it is wrong	correct model	examination consequence
Orders can be read from the balanced equation.	The equation shows the overall change; the rate depends on the slow step of the mechanism.	Orders are found by experiment; they are often not the coefficients.	Wrong rate equation, and every later mark that depends on it.
Products belong in the rate equation.	The rate equation describes how the rate depends on the reactants (and any catalyst) present.	rate = k[reactants]^orders; products do not appear.	Rate-equation mark lost; [NO₂] and [O] have both been reported.
The rate "at" a time is the change up to that time divided by the time.	That is an average rate; the rate is changing throughout.	Rate at an instant = gradient of the tangent at that instant.	Tangent marks lost; value outside the accepted range.
"The second half-life is 60 s."	Half-lives are successive intervals, not cumulative times.	0 → 30 s and 30 → 60 s: both 30 s, so constant.	The evidence for first order is not credited.
Any reaction with a first-order reactant has a constant half-life.	The half-life is constant only when the reaction is first order overall in the species being followed.	rate = k[NO][O₃] is second order overall: no constant half-life when both fall.	Prediction and explanation lost.
Zero order means the reactant is not involved.	The reactant is consumed; it simply reacts after the rate-determining step (or the rate is limited by something else).	A zero-order reactant appears only in fast steps after the slow step.	Mechanisms that omit the reactant altogether.
Temperature changes the rate because it changes the concentrations.	Warming a solution does not change its concentrations.	Temperature changes k: more collisions have E ≥ E_a.	Sketch of k against T drawn flat or with the wrong shape.
A heterogeneous catalyst "absorbs" the reactants; "bonds weaken" means intermolecular forces.	Adsorption is bonding at the surface; the bonds that weaken are within the reactant molecules.	Reactants adsorb onto the surface; bonds within them weaken; products desorb.	Mode-of-action marks lost for vague or misplaced terms.
A catalyst is not involved in the reaction.	A homogeneous catalyst reacts in one step and is re-formed in another.	Two equations: catalyst used, then regenerated.	Equations for Fe²⁺/Fe³⁺ or NO₂ not written.

How the topic is assessed

**Table 26.9** Question families seen in the structured papers reviewed for this chapter.
question family	typical demand	chemistry needed
Definitions	order of reaction; half-life; rate-determining step; homogeneous and heterogeneous catalyst	"power … in the rate equation"; "concentration of a reactant … halve"; "slowest step"; "same / different phase"
Initial rates	orders from a table (sometimes two concentrations changing at once); rate equation; k with units; a missing concentration; a rate ratio from pH values	ratios; powers and roots; units from the overall order
Graphs	tangent at a time; two half-lives from a curve; sketches of rate against concentration, [X] against time, rate against time, k against T	shapes for orders 0, 1, 2; construction lines
Half-life	explain constant t_½; k = 0.693/t_½; time for a fraction to remain; concentration after n half-lives	first order only; (½)ⁿ
Mechanisms	define rate-determining step; orders from a mechanism; overall equation from steps; suggest two- or three-step mechanisms consistent with a rate equation	particles in the slow step; balanced steps; intermediates cancel
Catalysis	classify; mode of action of a heterogeneous catalyst; name the converter metals; two equations for a homogeneous catalyst; E^⦵ to justify the steps	adsorption, bond weakening, desorption; used and re-formed

Kinetics parts are frequently embedded in questions about another topic — a cisplatin hydrolysis inside a transition-element question, a catalytic converter inside an entropy question, a rate equation inside a halogenoalkane question. Recognise the kinetics demand whatever the context.

Self-test

Twelve questions across the whole unit, each with the reasoning behind the answer.

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Definitions to learn

**Table 26.10** The definitions and relationships this unit examines.
term	definition or relationship
rate of reaction	change in concentration of a reactant or product per unit time; mol dm⁻³ s⁻¹
rate equation	rate = k[A]^m[B]ⁿ, determined by experiment
order of reaction (with respect to a reactant)	the power to which the concentration of that reactant is raised in the rate equation
overall order	the sum of the powers in the rate equation, m + n
rate constant, k	the constant of proportionality in the rate equation; depends on temperature (and catalyst), not concentration
half-life, t_½	the time taken for the concentration of a reactant to fall to half its initial value; constant for a first-order reaction; k = 0.693/t_½
rate-determining step	the slowest step in a reaction mechanism
intermediate	a species formed in one step of a mechanism and used up in a later step
homogeneous catalyst	a catalyst in the same phase as the reactants
heterogeneous catalyst	a catalyst in a different phase from the reactants
adsorption / desorption	formation of weak bonds between a species and a catalyst surface / breaking of those bonds as the species leaves

Data used in this chapter

Rate constants, concentrations and rates quoted from examination questions are reproduced as printed (for example k = 2.0 × 10⁻⁶ s⁻¹ for the uncatalysed decomposition of H₂O₂ at 298 K). Data in Worked examples 26.1, 26.3, 26.4, 26.6, 26.7 and 26.9 are illustrative values chosen to show a method; they are not measurements.

The curves in Figures 26.1, 26.3 and 26.5 are calculated from the rate laws and from the Boltzmann energy distribution, not measured; Figures 26.2, 26.4 and 26.8 are schematic. The standard electrode potentials in Table 26.7 and in the catalyst model are reference values of the kind printed in the data booklet; use the values in your own data booklet in an examination.

Summary

26.1 Rate equations and orders

rate = k[A]^m[B]ⁿ; orders (0, 1 or 2) are found by experiment, not from the equation; overall order = m + n.
Order 0: rate unchanged when [A] changes; order 1: rate ∝ [A]; order 2: rate ∝ [A]².
Initial rates: compare experiments in which one concentration changes; allow for any other change before interpreting the rest.
Units of k: (mol dm⁻³)¹⁻ⁿ s⁻¹ for overall order n; match the time unit to the rate data.
Rate at an instant = gradient of the tangent; initial rate from the tangent at t = 0.
Rate–concentration graphs: horizontal (0), straight through origin (1), curve through origin (2). Concentration–time graphs: straight line (0), constant half-life (1), lengthening half-life (2).

26.1 Half-life

t_½ = time for a reactant's concentration to halve; constant only for first order; k = 0.693/t_½, units time⁻¹.
After n half-lives, fraction remaining = (½)ⁿ; half-lives are successive equal intervals.
A reactant in large excess stays at almost constant concentration: pseudo-first-order in the other.

26.1 Mechanisms and temperature

The rate-determining step is the slowest step; the rate equation contains the particles in it (intermediates traced back to the reactants that formed them).
Species reacting only after the slow step are zero order; a catalyst can appear in the rate equation.
Suggested mechanisms: two particles per step; each step balanced; steps add to the overall equation; intermediates cancel.
Intermediate: made then used. Catalyst: used then re-formed.
Increasing temperature increases k (steeply): more collisions have E ≥ E_a. Orders are unchanged.

26.2 Catalysts

Homogeneous: same phase as reactants; heterogeneous: different phase.
Heterogeneous: reactants adsorb onto the surface; bonds within them weaken; reaction; products desorb. Fe in the Haber process; Pd, Pt, Rh in catalytic converters (2CO + 2NO → 2CO₂ + N₂).
Homogeneous: used in one step, re-formed in a later step. NO₂ + SO₂ → NO + SO₃; NO + ½O₂ → NO₂. S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺; 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
The catalyst couple's E^⦵ lies between those of the two reacting couples, so both steps are feasible.

Examination checklist

Does my definition of order say "power … concentration … in the rate equation"?
Can I deduce orders from a table, including when two concentrations change together, and say which experiments I compared?
Do I write rate equations with reactants only, include k, and work out units from the overall order and the time unit used?
When I find a squared concentration, do I remember the square root?
Do I draw a ruled tangent at the stated time — not a chord from the origin?
Can I sketch rate–concentration and concentration–time graphs for orders 0, 1 and 2, starting at the origin where they should?
Do I define half-life by naming what halves, and show two successive half-lives as intervals on a graph?
Can I use k = 0.693/t_½ and (½)ⁿ, and convert "percentage used up" into a number of half-lives?
Can I predict a rate equation from a mechanism, and suggest a mechanism from a rate equation that balances, uses two particles per step and adds to the overall equation?
Can I identify intermediates and catalysts in a mechanism?
Can I describe, and sketch, how k changes with temperature, and explain it with the Boltzmann distribution?
Can I describe the mode of action of a heterogeneous catalyst, naming the species at each stage, for Fe in the Haber process and Pd/Pt/Rh in a converter?
Can I write the two equations for NO₂ with SO₂ and for Fe²⁺/Fe³⁺ with I⁻/S₂O₈²⁻, and use E^⦵ to justify them?

Knowledge organiser

idea	key facts and relationships	must-remember distinctions and common errors
Rate	−Δ[reactant]/Δt; mol dm⁻³ s⁻¹; instantaneous rate = tangent gradient	"at" a time needs a tangent; chord = average
Rate equation	rate = k[A]^m[B]ⁿ; m, n = 0, 1, 2	orders by experiment; no products; include k
Orders	× 2 conc → rate × 1, × 2, × 4 for orders 0, 1, 2	allow for a second changing concentration first
Units of k	0: mol dm⁻³ s⁻¹; 1: s⁻¹; 2: mol⁻¹ dm³ s⁻¹; 3: mol⁻² dm⁶ s⁻¹	time unit must match the data (s or min)
Graphs	rate–conc: flat / straight through 0 / curve; conc–time: straight / constant t_½ / lengthening t_½	check the axes before deciding
Half-life	first order: constant; k = 0.693/t_½; (½)ⁿ remains	intervals not cumulative times; % used ≠ % left
Large excess	[excess reagent] ≈ constant → pseudo-first-order; k′ = k[excess]^order	excess ≥ about 10 × the other
Rate-determining step	slowest step; rate equation = particles in it	intermediate in slow step → trace back to reactants
Mechanism rules	two particles per step; balanced; adds to overall equation	no half-equations; no invented species
Intermediate vs catalyst	intermediate: product first, then reactant; catalyst: reactant first, then product	neither in the overall equation
Temperature	T ↑ → k ↑ steeply (Boltzmann: more E ≥ E_a)	concentrations and orders unchanged
Heterogeneous	adsorb → bonds within reactants weaken → react → desorb; Fe (Haber); Pd, Pt, Rh (converter)	name species; adsorb ≠ absorb; poisons block sites; saturated surface → zero order
Homogeneous	used in one step, re-formed later; NO₂/NO with SO₂; Fe²⁺/Fe³⁺ with S₂O₈²⁻/I⁻	two balanced equations; E^⦵(cat) between the two couples

Reaction kinetics

What this chapter covers26.1–26.2

Measuring the rate of a reaction26.1.2(c), 26.1.2(d)

Following a reaction

Rate at an instant, and the initial rate

The rate equation, orders and the rate constant26.1.1, 26.1.2(a)

What an order means at the particle level

Units of the rate constant26.1.2(a), 26.1.4

Deducing orders by the initial-rates method26.1.2(b), 26.1.2(e), 26.1.4(a)

When two concentrations change at once

Graphs of rate and of concentration26.1.2(b), 26.1.2(c)

Rate against concentration

Concentration against time

Using the rate equation in calculations26.1.4(a)

Half-life, and why it is constant for a first-order reaction26.1.1, 26.1.3(a)

Calculations with the half-life26.1.3(b), 26.1.4(b)

Finding orders from a single run, and pseudo-first-order conditions26.1.2(b), 26.1.2(c)

Choosing a method

Multi-step reactions, the rate-determining step and intermediates26.1.1, 26.1.5(d)

From a mechanism to a rate equation26.1.5(b), 26.1.5(c)

A catalyst in the rate equation: the iodination of propanone

Mechanisms you have already met

From a rate equation to a mechanism26.1.5(a), 26.1.5(e)

Temperature and the rate constant26.1.6

Homogeneous and heterogeneous catalysts26.2.1

How a heterogeneous catalyst works26.2.2

What makes a good heterogeneous catalyst

Iron in the Haber process26.2.2(a)

Catalytic converters26.2.2(b)

How a homogeneous catalyst works26.2.3

Oxides of nitrogen and atmospheric sulfur dioxide

Iron ions and the reaction between iodide and peroxodisulfate

Using E⦵ to show that the catalysed steps are feasible

Misconceptions and how the topic is assessed26.1–26.2

How the topic is assessed

Self-test

Definitions to learn

Data used in this chapter

Summary

26.1 Rate equations and orders

26.1 Half-life

26.1 Mechanisms and temperature

26.2 Catalysts

Examination checklist

Knowledge organiser

Using E^⦵ to show that the catalysed steps are feasible