Edexcel International A Level Chemistry: Kinetics (YCH11) – Revision Notes

These notes condense Topic 11: Kinetics of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 11.1 to 11.13, including Core Practicals 9a, 9b and 10. The topic sits in Unit 4: Rates, Equilibria and Further Organic Chemistry, an International A2 (IA2) unit, and builds on Unit 2, Topic 9A. The kinetics study guide has the full teaching and worked examples.

When these feel secure, try the kinetics practice questions. Tick topics off on a printable checklist, try a free diagnostic, or browse other topics from the course hub.

Definitions to learn (11.1)

Term	Meaning
Rate of reaction	How fast a reactant or product concentration changes (change ÷ time)
Rate equation	rate = k[A]ᵐ[B]ⁿ, with m and n equal to 0, 1 or 2; determined experimentally
Order with respect to A	The power of [A] in the rate equation
Overall order	m + n
Rate constant, k	The k that links rate to concentrations; constant at a fixed temperature
Half-life	Time needed for a reactant’s concentration to drop to 50% of its starting value
Rate-determining step	The slowest step of a mechanism
Activation energy	Minimum energy needed by colliding particles for reaction
Homogeneous catalyst	Same phase as the reactants
Heterogeneous catalyst	Different phase from the reactants

Units of k at a glance

Example rate equation	Overall order	k in
rate = k[A]²[B]	3	mol⁻² dm⁶ s⁻¹
rate = k[A][B]	2	mol⁻¹ dm³ s⁻¹
rate = k[A]	1	s⁻¹
rate = k	0	mol dm⁻³ s⁻¹

Small worked reminder. Rate = k[A][B]. With rate = 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹, [A] = 0.20 mol dm⁻³ and [B] = 0.050 mol dm⁻³: k = 3.0 × 10⁻⁴ ÷ (0.20 × 0.050) = 0.030 mol⁻¹ dm³ s⁻¹.

Graph shapes (11.2, 11.5)

Order	Concentration-time	Rate-concentration	Half-life
0	Straight line, negative gradient	Horizontal line	Falls as [A] falls
1	Curve, levels off	Straight line through origin	Constant
2	Steep curve with long tail	Upward curve through origin	Rises as [A] falls

Method in steps: order from initial rates (11.5 iii)

1. Pick a pair of runs in which a single concentration is altered.
2. Work out the factor change in concentration and in rate.
3. Factor in rate = (factor in concentration)ⁿ. Rate unchanged: n = 0. Same factor: n = 1. Factor squared: n = 2.
4. Repeat for each reactant. Write the rate equation, then calculate k from any one experiment, with units.

Method in steps: half-life from a graph (11.2)

1. Choose a starting concentration; read the time.
2. Read the time at half that concentration. The difference is t½.
3. Repeat from a different starting point. Same t½ both times: first order.

Collecting rate data (11.3, 11.4)

• Titration: remove samples, quench, titrate (iodine with thiosulfate).
• Colorimetry: a coloured species; absorbance tracks concentration with no sampling.
• Mass change: a heavy gas (CO₂) lost from an open flask on a balance.
• Gas volume: gas syringe; best for light gases such as hydrogen.
• Other: conductivity when the number of ions changes; pH meter when [H⁺] changes.
• Initial-rate method: separate runs, one concentration changed each time. A clock reaction gives rate ∝ 1/t.
• Continuous monitoring: one run, readings over time, to plot concentration-time or volume-time graphs.

Core Practicals 9a and 9b (11.12)

9a, iodine-propanone by titration. Propanone and acid in large excess. Samples quenched in sodium hydrogencarbonate (removes the H⁺ catalyst). Titrate iodine with sodium thiosulfate and starch. A straight-line titre-time graph shows zero order in iodine.

Rate = k[CH₃COCH₃][H⁺]. The slow step involves propanone and H⁺; iodine reacts after it. H⁺ appears in the rate equation but is regenerated, so it is a catalyst (11.6).

9b, Harcourt-Esson iodine clock. H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O, with a little thiosulfate and starch. The blue-black colour appears when the thiosulfate is used up. Keep total volume constant by adding water, so concentration is proportional to the volume of each solution.

Mechanisms (11.7, 11.8, 11.9)

Must-know distinctions

• Stoichiometric equation tells you what reacts overall. Rate equation tells you what is in (or before) the slow step.
• Intermediate: made in one step, used in a later one. Catalyst: used in one step, regenerated later.
• SN1 (tertiary): rate = k[RX]; slow ionisation to a carbocation, then fast attack by OH⁻.
• SN2 (primary): rate = k[RX][OH⁻]; one step, through a transition state with OH⁻ and the leaving halogen on opposite sides of carbon.

Checks for a proposed mechanism

1. The slow step contains the species in the rate equation, in the right numbers (counting any species used to make an intermediate in an earlier fast step).
2. Zero-order species enter only after the slow step.
3. All steps add up to the overall equation.

Activation energy (11.10, 11.13)

Item	What to remember
Arrhenius equation	k = Ae^(−Ea/RT); given if needed
Straight-line form	ln k = −(Ea/R)(1/T) + ln A
Graph	ln k (y) against 1/T (x), with T in kelvin
Gradient	−Ea/R, so Ea = −gradient × R
Only two runs	ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)
Core Practical 10	Time the same amount of reaction at several temperatures; 1/t ∝ k, so plot ln(1/t) against 1/T

R = 8.31 J mol⁻¹ K⁻¹, so Ea comes out in J mol⁻¹. Divide by 1000 for kJ mol⁻¹.

Method in steps: Ea from a graph

1. Convert every temperature to kelvin and work out 1/T.
2. Work out ln k (or ln(1/t)) for each run.
3. Plot, draw the best straight line, and read two well-separated points on that line for the gradient.
4. Ea = −gradient × 8.31, then convert to kJ mol⁻¹.

Further suggested practicals (11.13): marble chips and hydrochloric acid (mass change or gas volume); magnesium and hydrochloric acid to find Ea; the iodine-propanone reaction followed by colorimetry; cobalt(II) catalysing potassium sodium tartrate with hydrogen peroxide; urease acting on urea and thiourea.

Solid catalysts (11.11)

Reactant gases bond weakly (adsorb) at active sites on the solid. Their bonds weaken and they meet in a good orientation, so Ea falls. Products desorb and free the sites. Adsorption must be neither too weak (nothing happens) nor too strong (products stay stuck). A large surface area (powder, mesh or thin coating on a support) gives more sites.

Quick self-test

1. What are the units of k when rate = k[T]²[V]?
2. Rate = k[M][N]². [M] is tripled and [N] is doubled. By what factor does the rate change?
3. [Z] falls from 0.48 to 0.24 mol dm⁻³ in 35 s and from 0.24 to 0.12 mol dm⁻³ in the next 35 s. State the order in Z and find [Z] after 105 s.
4. A rate-concentration graph for reactant G is a horizontal line. State the order in G.
5. A clock run takes 64 s. With [S] doubled and nothing else changed, it takes 32 s. What is the order in S?
6. A graph of ln k against 1/T has gradient −9.20 × 10³ K. Calculate Ea. (R = 8.31 J mol⁻¹ K⁻¹)
7. k doubles when the temperature rises from 300 K to 310 K. Calculate Ea.
8. Choose a technique to follow zinc reacting with hydrochloric acid, and justify it.
9. 2-chloro-2-methylpropane is hydrolysed with rate = k[(CH₃)₃CCl]. Name the mechanism.
10. Why are samples in Core Practical 9a run into sodium hydrogencarbonate?
11. Mechanism: 2W → W₂ (slow), then W₂ + Z → W₂Z (fast). Write the rate equation.

Answers

1. mol⁻² dm⁶ s⁻¹.
2. 3 × 2² = 12.
3. Constant half-life of 35 s, so first order. 105 s is three half-lives: 0.48 → 0.24 → 0.12 → 0.060 mol dm⁻³.
4. Zero order.
5. Time halves, so 1/t (rate) doubles: first order.
6. Ea = 9.20 × 10³ × 8.31 = 76 500 J mol⁻¹ = 76.5 kJ mol⁻¹.
7. ln 2 = (Ea/8.31)(1/300 − 1/310), so Ea = 53 600 J mol⁻¹ ≈ 53.6 kJ mol⁻¹.
8. Gas volume in a gas syringe: hydrogen forms, but it is too light for mass loss to measure accurately.
9. SN1.
10. It neutralises the acid catalyst, which stops the reaction so the iodine left at that time can be titrated.
11. Rate = k[W]².

Where marks are usually lost

• Taking orders from the stoichiometric coefficients rather than the experimental results.
• Giving k without units, or with units for the wrong overall order.
• Using two experiments where more than one concentration changes.
• Confusing “rate halves” with “time halves” in clock reactions: rate ∝ 1/t.
• Plotting ln k against T instead of 1/T, or using T in °C.
• Mixing J and kJ in the Ea calculation.
• Calling H⁺ an intermediate in the iodination of propanone; it is a catalyst.
• Saying iodine “is not involved” because it is zero order. It reacts after the slow step.
• Describing heterogeneous catalysis without the words adsorption and desorption.

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.