Edexcel International A Level Chemistry: Kinetics (YCH11)

This guide teaches Topic 11: Kinetics of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers outcomes 11.1 to 11.13, with Core Practicals 9a, 9b and 10, in the International A2 (IA2) unit Unit 4: Rates, Equilibria and Further Organic Chemistry, and it builds on Unit 2, Topic 9A: Kinetics (see the Topic 9 guide).

Useful pages: a printable topic checklist; the shorter kinetics revision notes; the kinetics practice questions; a free diagnostic to spot gaps; and the course hub, which lists every unit.

Topic 11 at a glance

Outcome	Skill
11.1	Define the key kinetics terms
11.2	Find a half-life from a graph; recognise constant half-life as first order
11.3	Choose and justify a technique for collecting rate data
11.4	Describe initial-rate (including clock) and continuous monitoring methods
11.5	Deduce orders 0, 1 or 2 from graphs and initial rates
11.6	Iodination of propanone: orders, rate-determining step, mechanism
11.7, 11.8	Link rate equations, rate-determining steps and mechanisms
11.9	Rate equations as evidence for SN1 and SN2
11.10	Find activation energy by calculation and from a graph
11.11	Solid catalysts for gas-phase industrial reactions
11.12, 11.13	Core Practicals 9a, 9b and 10

11.1 The key terms

• Rate of reaction: how quickly a reactant or product concentration changes, as change ÷ time (mol dm⁻³ s⁻¹).
• Rate equation: rate = k[A]ᵐ[B]ⁿ, where m and n are 0, 1 or 2. Only experiments can give it.
• Order with respect to a substance: its power (m or n) in the rate equation.
• Overall order: the sum of the powers, m + n.
• Rate constant, k: the number k linking rate to the concentration terms. For a given reaction it is constant at a fixed temperature.
• Half-life, t½: the time for a reactant’s concentration to halve.
• Rate-determining step: the slowest step.
• Activation energy, Ea: the minimum energy colliding particles need to react.
• Homogeneous catalyst: same phase as the reactants (aqueous H⁺ in aqueous solution). Heterogeneous catalyst: different phase (a solid with gases).

Units of k

Rearrange for k and cancel units. Each extra order adds mol⁻¹ dm³: zero order gives mol dm⁻³ s⁻¹, first order s⁻¹, second order mol⁻¹ dm³ s⁻¹, third order mol⁻² dm⁶ s⁻¹.

Worked example 1: orders and k from initial rates (11.5 iii)

Bromate(V) ions oxidise bromide ions in acid: BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O. Use these data at constant temperature.

Experiment	[BrO₃⁻] / mol dm⁻³	[Br⁻] / mol dm⁻³	[H⁺] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.015	0.050	0.10	1.8 × 10⁻⁵
2	0.030	0.050	0.10	3.6 × 10⁻⁵
3	0.015	0.150	0.10	5.4 × 10⁻⁵
4	0.015	0.050	0.20	7.2 × 10⁻⁵

1. Experiments 1 and 2: [BrO₃⁻] doubles, rate doubles. First order.
2. Experiments 1 and 3: [Br⁻] triples, rate triples. First order.
3. Experiments 1 and 4: [H⁺] doubles, rate × 4 (2²). Second order.
4. Rate = k[BrO₃⁻][Br⁻][H⁺]². Overall order 4.
5. From experiment 1: k = 1.8 × 10⁻⁵ ÷ (0.015 × 0.050 × 0.10²) = 2.4 mol⁻³ dm⁹ s⁻¹.

The 6H⁺ in the equation does not mean sixth order.

11.2 and 11.5 Reading graphs

Concentration-time graphs

Order	Shape of [A] against time	Half-life
0	Straight line sloping down	Gets shorter as [A] falls
1	Curve that levels off	Constant
2	Steeper curve at first, then a long, slow tail	Gets longer as [A] falls

To find a half-life, read the time at one concentration and at half that concentration. Repeat from a second starting point. Equal half-lives mean first order.

Rate-concentration graphs

Find rates from tangents (a Topic 9A skill), then plot rate against concentration. Order 0: horizontal line. Order 1: straight line through the origin. Order 2: a curve through the origin that bends upwards; plotting rate against [A]² straightens it.

Worked example 2: half-life of a diazonium salt

Benzenediazonium chloride decomposes in warm water, releasing nitrogen. Use these data.

t / s	0	200	400	600	800	1000	1200
[C₆H₅N₂⁺] / mol dm⁻³	0.0600	0.0424	0.0300	0.0212	0.0150	0.0106	0.0075

1. From 0.0600 to 0.0300 mol dm⁻³ takes 400 s.
2. From 0.0300 to 0.0150 mol dm⁻³ takes 800 − 400 = 400 s.
3. t½ is constant at 400 s, so the reaction is first order in the diazonium ion.

11.3 and 11.4 Getting rate data

Choosing a technique (11.3)

Justify a choice by naming the property that changes.

Technique	Use it when	Example
Titration	An acid, base or iodine can be titrated in quenched samples	Iodine-propanone, titrated with thiosulfate
Colorimetry	One species is coloured	Manganate(VII) or iodine fading
Mass change	A dense gas such as CO₂ escapes	Marble chips with acid
Volume of gas	A gas forms (gas syringe)	Hydrogen from magnesium and acid; H₂ is too light for mass loss
Other	Ion numbers or [H⁺] change	Conductivity or pH meter during halogenoalkane hydrolysis

Two kinds of experiment (11.4)

• Initial-rate method: run separate experiments, changing the starting concentration of one reagent at a time and keeping everything else constant. Measure the rate at the very start. A clock reaction approximates this: you time how long a fixed small amount of reaction takes, and the initial rate is proportional to 1/time.
• Continuous monitoring: follow one mixture over time to give a concentration-time or volume-time graph.

11.6 and 11.12 The iodination of propanone (Core Practical 9a)

CH₃COCH₃ + I₂ → CH₃COCH₂I + H⁺ + I⁻, catalysed by H⁺ ions.

Method outline. Mix propanone and acid, add iodine solution and start the clock. At intervals, pipette a sample into sodium hydrogencarbonate solution, which neutralises the acid catalyst and quenches the reaction. Titrate the remaining iodine with sodium thiosulfate, adding starch near the end point. Propanone and acid are in large excess, so only the iodine order shows up.

Worked example 3: iodine is zero order

10.0 cm³ samples were quenched and titrated with 0.0100 mol dm⁻³ sodium thiosulfate. I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻.

Time / min	0	4	8	12
Titre / cm³	18.40	16.20	14.00	11.80

1. The titre falls by 2.20 cm³ every 4 min: a straight line, so the rate is constant and the reaction is zero order in iodine.
2. Gradient = −2.20 ÷ 4 = −0.550 cm³ min⁻¹.
3. Each cm³ of thiosulfate reacts with 0.0100 ÷ 1000 ÷ 2 = 5.00 × 10⁻⁶ mol I₂.
4. Rate = (0.550 × 5.00 × 10⁻⁶) ÷ 0.0100 dm³ = 2.75 × 10⁻⁴ mol dm⁻³ min⁻¹ (4.58 × 10⁻⁶ mol dm⁻³ s⁻¹).

Changing the starting [propanone] or [H⁺] alone changes the gradient in proportion, so the reaction is first order in each:

rate = k[CH₃COCH₃][H⁺]

What this tells you about the mechanism (11.6 ii, iii)

The rate-determining step involves one propanone and one H⁺. Iodine reacts in a later, fast step. A possible mechanism:

1. CH₃COCH₃ + H⁺ → CH₃C(OH⁺)CH₃ (slow; protonation of the C=O oxygen)
2. CH₃C(OH⁺)CH₃ → CH₂=C(OH)CH₃ + H⁺ (fast; forms the enol)
3. CH₂=C(OH)CH₃ + I₂ → CH₃COCH₂I + H⁺ + I⁻ (fast)

The steps add up to the overall equation. H⁺ used in step 1 is returned in step 2: a catalyst.

11.12 Clock reaction (Core Practical 9b)

The Harcourt-Esson (iodine clock) reaction is H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O. A small fixed amount of sodium thiosulfate and some starch are added. Thiosulfate turns the first iodine back into iodide; when it runs out, the mixture suddenly turns blue-black. The same amount of reaction has happened each time, so initial rate ∝ 1/t. Vary the volume of potassium iodide solution and add water to keep the total volume constant, so [I⁻] is proportional to the volume used.

Worked example 4: order in iodide

Volume of KI / cm³	10	15	20	25
t / s	89	61	45	36
1000/t / s⁻¹	11.2	16.4	22.2	27.8

(1000/t) ÷ volume is about 1.1 each time. Rate is proportional to [I⁻]: first order in iodide.

11.7 and 11.8 Mechanisms and rate equations

• Species in the rate equation appear in (or before) the rate-determining step, in the numbers given by their orders.
• A zero-order species reacts only after that step.
• A rate-equation species missing from the overall equation is usually a catalyst.
• The steps must add up to the stoichiometric equation.

Worked example 5: from rate equation to mechanism

For the clock reaction, rate = k[H₂O₂][I⁻]. The slow step uses one H₂O₂ and one I⁻; H⁺ joins later.

1. H₂O₂ + I⁻ → IO⁻ + H₂O (slow)
2. IO⁻ + H⁺ → HIO (fast)
3. HIO + I⁻ + H⁺ → I₂ + H₂O (fast)

The steps add up to the overall equation.

Worked example 6: from mechanism to rate equation

A reaction D + 2E → DE₂ is thought to go by:

1. D + E → DE (fast)
2. DE + E → DE₂ (slow)

The slow step uses DE and E, and DE comes from one D and one E. So rate = k[D][E]².

11.9 Evidence for SN1 and SN2

Hydrolysis by hydroxide ions gives different rate equations:

• Tertiary (2-bromo-2-methylpropane): rate = k[(CH₃)₃CBr]. Only the halogenoalkane is in the slow step. This fits SN1: slow heterolytic fission of C–Br to a tertiary carbocation, then fast attack by OH⁻.
• Primary (1-bromobutane): rate = k[C₄H₉Br][OH⁻]. Both species are in one step. This fits SN2: OH⁻ attacks the δ+ carbon opposite the bromine as C–Br breaks, through one transition state.

11.10 and 11.13 Activation energy (Core Practical 10)

The Arrhenius equation is given if needed:

k = Ae^(−Ea/RT), so ln k = −(Ea/R)(1/T) + ln A

Plot ln k on the y-axis against 1/T on the x-axis. The gradient equals −Ea/R, which rearranges to Ea = −R × gradient (in J mol⁻¹).

In Core Practical 10 you time the same amount of reaction at several temperatures. Then 1/t is proportional to k, so ln(1/t) against 1/T has the same gradient.

Worked example 7: Ea from a graph

Use these values for a first-order reaction and R = 8.31 J mol⁻¹ K⁻¹.

T / K	293	303	313	323
k / s⁻¹	0.0349	0.0810	0.178	0.372
(1/T) / 10⁻³ K⁻¹	3.413	3.300	3.195	3.096
ln k	−3.36	−2.51	−1.73	−0.99

1. Gradient = (−0.99 − (−3.36)) ÷ (3.096 − 3.413) × 10⁻³ = 2.37 ÷ (−3.17 × 10⁻⁴) = −7480 K.
2. Ea = 7480 × 8.31 = 62 200 J mol⁻¹ ≈ 62 kJ mol⁻¹.

With only two temperatures, use ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).

11.11 Solid catalysts in industry

Gas-phase industrial reactions often use a solid catalyst, such as iron for ammonia. It provides a surface for the reaction:

1. Reactant gas molecules bond weakly to active sites on the solid (adsorption).
2. Bonds in adsorbed molecules weaken and the molecules are held close together, lowering the activation energy.
3. Products form and leave the surface (desorption), freeing the sites.

Adsorption must weaken bonds yet still let products leave. Only the surface works, so the catalyst is spread thinly or finely divided for a large surface area.

Further suggested practicals (11.13): marble chips with acid; magnesium and acid to find Ea; iodine-propanone by colorimetry; cobalt(II) catalysis of tartrate with hydrogen peroxide; urease with urea and thiourea.

Common errors

• Taking orders from the coefficients in the balanced equation.
• Comparing experiments where two concentrations both change.
• Claiming first order from a single half-life.
• Leaving R in J but quoting Ea in kJ, or dropping the minus sign on the gradient.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 11: Kinetics, outcomes 11.1 to 11.13.