Edexcel International A Level Chemistry

Topic 11: Kinetics

Turns "faster" and "slower" into numbers. A rate equation says how the rate depends on concentration, the orders come from experiment, and the equation points to the slow step in the mechanism.

The one idea

The rate of a reaction depends on concentrations in a way you cannot read from the balanced equation. You find it by experiment, and write it as a rate equation. Once you have the rate equation, it tells you two things: how to predict the rate for any concentrations, and which species are in the slowest step of the mechanism. Temperature enters through the rate constant, and the Arrhenius equation turns the way k changes with temperature into the activation energy.

The topic has five jobs, and the pages follow them in order:

Before you start

This topic extends Topic 9 (collision theory, activation energy, Maxwell–Boltzmann curves, finding a rate from a tangent) and uses the curly-arrow mechanisms of Topic 10. You also need the titration calculations from Topic 1 for the core practicals.

Three constants are used throughout: R = 8.31 J mol⁻¹ K⁻¹, K = °C + 273, and ln 2 = 0.693.

How to use these notes

• Read the pages in order the first time. Later pages assume the earlier ones, especially the rate-equation pages (1, 2) and the rate-determining step (10).
• Each page opens with its one idea in a sentence. If you can explain that sentence to someone else, you know the page.
• Coloured boxes mean the same thing on every page: definitions you may be asked to quote, key ideas, common mistakes, exam technique, and beyond the spec (interesting, but you will not be examined on it).
• Each page ends with questions and fully worked answers. Cover the answers, try the questions, then mark yourself.
• Numerical data in the worked examples and the practicals are labelled "illustrative". They are invented to show a typical result, not measured values.
• Use the topic on a page the night before an exam.

Pages

1. Rate, rate equation and order

   A rate equation links the rate of a reaction to the concentrations of the reactants. The power of each concentration is its order, and the orders can only be found by experiment.

   12 min 11.1(i)11.1(ii)11.1(iii)11.1(iv)

2. The rate constant and its units

   The rate constant k turns concentrations into a rate. Its value depends on temperature and catalyst but not on concentration, and its units depend on the overall order, so you derive them each time.

   11 min 11.1(v)

3. Half-life

   The half-life is the time for a concentration to halve. Read two or more successive half-lives from a concentration-time graph, and if they are all equal the reaction is first order.

   11 min 11.1(vi)11.2

4. Activation energy and catalysts

   Activation energy is the energy barrier colliding particles must clear. A catalyst gives a route with a lower barrier. It is homogeneous if it is in the same phase as the reactants and heterogeneous if it is not.

   9 min 11.1(viii)11.1(ix)

5. Choosing a technique to follow a reaction

   To measure a rate you need something that changes as the reaction goes and that you can measure while it happens. Match the technique to the reaction, then justify it by naming the species and the property.

   14 min 11.3(i)11.3(ii)11.3(iii)11.3(iv)11.3(v)

6. Initial rates and continuous monitoring

   Two ways to collect data for finding orders. The initial-rate method uses many separate experiments and one rate from each. Continuous monitoring uses one experiment and follows the whole curve.

   13 min 11.4(i)11.4(ii)

7. Orders from initial-rate data

   Compare two experiments in which only one concentration changes. How much the initial rate changes tells you the order for that reagent. Then use any experiment to find k.

   14 min 11.5(iii)

8. Orders from concentration–time graphs

   A straight line means zero order. For a curve, measure successive half-lives. If they are constant the reaction is first order, and if each is double the last it is second order.

   13 min 11.5(i)11.2

9. Orders from rate–concentration graphs

   Plot the rate against the concentration. A horizontal line is zero order, a straight line through the origin is first order, and an upward curve is second order. Tangents on a concentration–time graph supply the rates.

   13 min 11.5(ii)11.4(ii)

10. The rate-determining step and the rate equation

    The slowest step in a mechanism controls the overall rate, so the species in the rate equation are the ones in that step. Use this to go from a mechanism to a rate equation, and back.

    14 min 11.1(vii)11.7

11. Iodination of propanone: data and predictions

    In the acid-catalysed reaction of propanone with iodine, the rate is first order in propanone, first order in hydrogen ions and zero order in iodine. That tells you which species are in the rate-determining step.

    13 min 11.6(i)11.6(ii)11.5(iii)

12. Deducing a mechanism

    Write the slow step from the rate equation, then add fast steps until the steps add up to the overall equation. Check both conditions, and remember that the rate equation can rule a mechanism out but cannot prove it.

    14 min 11.811.6(iii)

13. Primary halogenoalkanes: the SN2 evidence

    When the rate of hydrolysis of a primary halogenoalkane depends on both the halogenoalkane and hydroxide ions, both are in the rate-determining step. That points to a one-step mechanism, named SN2.

    11 min 11.9

14. Tertiary halogenoalkanes: the SN1 evidence

    When the rate of hydrolysis of a tertiary halogenoalkane does not depend on hydroxide ions, only the halogenoalkane is in the rate-determining step. That points to a two-step mechanism through a carbocation, named SN1.

    12 min 11.9

15. Activation energy from a graph

    The Arrhenius equation links k to temperature. Taking logarithms turns it into a straight line, so a graph of ln k against 1/T has a gradient of minus the activation energy divided by the gas constant.

    14 min 11.10

16. Activation energy by calculation

    Subtract the Arrhenius equation at two temperatures and the constant A cancels. The result links two rate constants, two temperatures and the activation energy, and can be rearranged for any of them.

    13 min 11.10

17. Solid catalysts in industry

    In gas-phase industrial reactions a solid catalyst provides a surface. The gases adsorb, react on the surface and desorb, by a route with a lower activation energy. A bigger surface area gives more places to react.

    11 min 11.11

18. Core Practical 9a: iodine and propanone by titration, method

    Mix propanone, acid and iodine, take samples at timed intervals, stop the reaction in each with sodium hydrogencarbonate, and titrate the iodine left with thiosulfate. The titre is proportional to the iodine concentration.

    13 min 11.1211.6(i)

19. Core Practical 9a: results and evaluation

    Titre against time is a straight line, so the rate does not depend on the iodine concentration and the reaction is zero order in iodine. Repeating with more propanone or more acid shows each is first order.

    14 min 11.1211.5(i)11.6(i)11.6(ii)

20. Core Practical 9b: the iodine clock, method

    Hydrogen peroxide oxidises iodide to iodine, and a small, fixed amount of thiosulfate removes the iodine as it forms. When the thiosulfate is used up, the starch turns blue-black. The time to that moment, t, is proportional to 1 ÷ rate.

    14 min 11.1211.4(i)

21. Core Practical 9b: results, order and activation energy

    A graph of 1 ÷ t against the volume of iodide is a straight line through the origin, so the reaction is first order in iodide. Repeating at different temperatures and plotting ln(1/t) against 1/T gives the activation energy.

    14 min 11.1211.4(i)11.5(iii)11.10

Also in this topic: Topic 11 on a page

Spec checklist

Every point you can be examined on in this topic, and where it is explained. Tick them off as you go.