Reaction kinetics (A Level)

• Orders come from experiment, never from the equation. Only what reacts in (or before) the slowest step appears in the rate equation.
• Orders from experiment, never from the equation. Constant half-life means first order, and k = 0.693/t½. Only what reacts in or before the slowest step appears in the rate equation.

• 26.1.1 I can explain — explain and use the terms rate equation, order of reaction, overall order of reaction, rate constant, half-life, rate-determining step and intermediate
• 26.1.2 I can — (a) understand and use rate equations of the form rate = k[A]ᵐ[B]ⁿ (for which m and n are 0, 1 or 2) (b) deduce the order of a reaction from concentration–time graphs or from experimental data relating to the initial rates method and half-life method (c) interpret experimental data in graphical form, including concentration–time and rate–concentration graphs (d) calculate an initial rate using concentration data (e) construct a rate equation
• 26.1.3 I can — (a) show understanding that the half-life of a first-order reaction is independent of concentration (b) use the half-life of a first-order reaction in calculations
• 26.1.4 I can calculate — calculate the numerical value of a rate constant, for example by: (a) using the initial rates and the rate equation (b) using the half-life, t½, and the equation k = 0.693/t½
• 26.1.5 I can — for a multi-step reaction: (a) suggest a reaction mechanism that is consistent with the rate equation and the equation for the overall reaction (b) predict the order that would result from a given reaction mechanism and rate-determining step (c) deduce a rate equation using a given reaction mechanism and rate-determining step for a given reaction (d) identify an intermediate or catalyst from a given reaction mechanism (e) identify the rate determining step from a rate equation and a given reaction mechanism
• 26.1.6 I can describe — describe qualitatively the effect of temperature change on the rate constant and hence the rate of a reaction
• 26.2.1 I can explain — explain that catalysts can be homogeneous or heterogeneous
• 26.2.2 I can describe — describe the mode of action of a heterogeneous catalyst to include adsorption of reactants, bond weakening and desorption of products, for example: (a) iron in the Haber process (b) palladium, platinum and rhodium in the catalytic removal of oxides of nitrogen from the exhaust gases of car engines
• 26.2.3 I can describe — describe the mode of action of a homogeneous catalyst by being used in one step and reformed in a later step, for example: (a) atmospheric oxides of nitrogen in the oxidation of atmospheric sulfur dioxide (b) Fe2+ or Fe3+ in the I-/S2O82- reaction

Units and significant figures are marked. The syllabus states that failure to quote units, the inclusion of units in quantities defined as ratios, and answers given to an inappropriate number of significant figures are all liable to be penalised. Give a calculated answer to the same number of significant figures as the least precise data, or one more; keep full precision in the working and round only at the end. A fifth of the qualification is experimental: Papers 3 and 5 test AO3 only, and their questions may be set in contexts outside the syllabus content, so the practical work in this chapter is set out as procedure, recording and evaluation rather than as theory.

• “For 2A + B → C the rate equation is rate = k[A]2[B].” Correct Orders come from experiment, never from the balanced equation. The coefficients tell you how much reacts overall. They tell you nothing about which species take part in the slowest step, and that is what sets the powers.
• “Rate doubled when I doubled [B], so B is second order.” Correct If doubling [B] doubles the rate, B is first order: 21 = 2. Second order gives ×4 (22), and zero order gives no change.
• “k is always in mol dm−3 s−1.” Correct That is the unit of a rate, and of k only for zero order. Derive the unit of k from the rate equation every time: s−1 for first order, mol−1 dm3 s−1 for second, mol−2 dm6 s−1 for third.
• “The half-life is constant, so k = 0.693/t½ works for any reaction.” Correct Both statements hold for first order only. A zero-order half-life gets shorter as the reaction goes on and a second-order half-life gets longer.
• “Heating the mixture raises the concentrations, so the rate rises.” Correct Warming does not change the concentrations. It increases the rate constant k, because a greater proportion of collisions have energy ≥ Ea.
• “Fe2+ is absent from the overall equation, so it must be the catalyst.” Correct Intermediates are absent from the overall equation too. Look at where the species sits in the steps. An intermediate is formed first and used later. A catalyst is used first and re-formed later.
• “The rate equation for 2A + B → C is rate = k[A]2[B].” Repair Orders come only from experiment. The coefficients in the balanced equation say nothing about them.
• “Rate doubles when [B] doubles, so the order in B is 2.” Repair Rate ∝ [B]1. Doubling gives ×2 for first order and ×4 for second.
• “k has units mol dm−3 s−1.” Repair Only for zero order. Deduce the unit from the rate equation every time: s−1 for first order, mol−1 dm3 s−1 for second.
• “The half-life is constant for every reaction, and k = 0.693/t½ applies to any order.” Repair Both are true of first order only. The half-life lengthens for second order and shortens for zero order.
• “Initial rate = (0.800 − 0.635)/10, from the first two readings.” Repair That is an average over the first ten seconds. The initial rate is the gradient of the tangent at t = 0; the chord underestimates it.
• “Raising the temperature increases the concentrations, so the rate rises.” Repair The concentrations are unchanged. The rate constant k increases, because more collisions have E ≥ Ea.
• “OH− is not in the SN1 rate equation, so it is not involved in the reaction.” Repair It reacts in the fast step after the RDS, so its concentration does not affect the rate. It is still a reactant.
• “Fe2+ is the catalyst because it is absent from the overall equation.” Repair Intermediates are absent too. An intermediate is formed first and consumed later; a catalyst is consumed first and re-formed later. Read the position in the steps.
• “A heterogeneous catalyst works by lowering the activation energy.” (and nothing more) Repair Name the stages: adsorption at active sites, bond weakening, reaction on the surface, desorption of products.
• “Fe3+ + I− → Fe2+ + I2, then S2O82− + Fe2+ → SO42− + Fe3+.” Repair Neither step balances, and they do not add to the overall equation. Write 2Fe3+ + 2I− → 2Fe2+ + I2 and S2O82− + 2Fe2+ → 2SO42− + 2Fe3+; the iron cancels.
• “In a clock reaction the time is the rate.” Repair The initial rate is proportional to 1/t, because the same amount of iodine forms in every run.
• “The largest error is human error.” Repair Name it: judging the instant of the colour change (random), or a temperature drift between runs (systematic). Then say how to reduce it.

• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.
• Interleave with the chapters that use this one. Chapter 28 (transition elements) explains why transition elements behave as catalysts, through more than one stable oxidation state and vacant d orbitals: re-answer the iodide–peroxodisulfate steps there. Chapter 24 gives the E⦵ reasoning behind the iron catalyst. Recalling a topic inside a new context is worth more than another pass over this chapter on its own; at A Level, Paper 4 assumes the whole of the AS content, so nothing here is ever finished with.

• Cambridge International AS & A Level Chemistry 9701 has five components. Topic 26 is A Level content, so it is examined in Papers 4 and 5. A Level content: examined in Paper 4 (A Level structured, which also requires the AS content) and, as practical context, Paper 5. AS Level candidates take Papers 1, 2 and 3; A Level candidates take all five, either staged over two years (Papers 1–3 in year one, Papers 4 and 5 in year two) or together in one series. Examinations are available in the June and November series, and in March in India.
• Across both the AS Level and the A Level the assessment objectives are weighted AO1 40% (knowledge and understanding), AO2 40% (handling, applying and evaluating information) and AO3 20% (experimental skills and investigations). AS candidates are graded a–e; A Level candidates A*–E.
• There is no multiple-choice paper on A Level content. Paper 4 is structured questions, and this topic supplies the definitions of 26.1.1; deductions of order from tables and graphs; a rate equation to construct; a mechanism to test against a rate equation, naming the intermediate, the catalyst and the rate-determining step; and the mode of action of a named catalyst, described in stages.
• Initial-rates tables, concentration–time graphs (tangents and half-lives) and rate–concentration graphs. You calculate an initial rate, k from one run of a table, k = 0.693/t½, fractions remaining after whole half-lives, and the unit of k, which you deduce from the rate equation. Rate constants and half-lives are supplied by the question; the Data section holds none of them.
• The syllabus's rate experiment, timing an observation after mixing, here as an iodine clock: vary one concentration, hold the rest, use 1/t as the rate, plot it, read the order, then evaluate the timing error and the temperature control. Sampling with quenching and titration is the alternative that gives a whole concentration–time curve.
• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.