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9476 / 2027
Reaction Kinetics

Lesson 3 of 5

Test a mechanism and design a measurement

Connect elementary steps with observable rates and suitable signals.

A-Level 9476 (2026-2027)

  • I can explain the central kinetic terms.Syllabus

    Syllabus 9476, 8(a). Explain the central kinetic terms.

  • I can construct and use simple rate equations.Syllabus

    Syllabus 9476, 8(b). Construct and use simple rate equations.

  • I can devise a suitable rate experiment.Syllabus

    Syllabus 9476, 8(e). Devise a suitable rate experiment.

A mechanism must match both the overall equation and the rate law

Intermediates cancel from the overall equation; a proposed rate law must use measurable reactant concentrations.

A mechanism is a series of simple steps, called elementary steps. The rate-determining step is the slowest step, and it controls the overall rate. If the slow step is a collision between A and B, the rate equation is rate = k[A][B]. A later fast step can use up another reactant. That reactant then does not appear in the rate equation.

Worked exampleTest a two-step mechanism

Proposed steps are A + B → I (slow), followed by I + B → P (fast). The observed rate is k[A][B]. Is the mechanism consistent?

Try it first. Then show the worked stepsHide the worked steps
  1. Add the steps and cancel intermediate I: A + 2B → P.
  2. The elementary slow step predicts first order in A and first order in B.
  3. This matches the observed rate law even though B has coefficient 2 in the overall equation.
Answer

It is consistent with both pieces of evidence. Agreement supports the mechanism but does not prove it is the only possible mechanism.

Worked exampleUse a fast pre-equilibrium to remove an intermediate

A proposed mechanism is 2A ⇌ I (fast equilibrium), then I + B → P (slow). Predict the rate law.

Try it first. Then show the worked stepsHide the worked steps
  1. The slow elementary step gives rate = k2[I][B].
  2. For the preceding equilibrium, K = [I]/[A]2, so [I] = K[A]2.
  3. Substitute: rate = k2K[A]2[B]; at fixed temperature, combine k2K into the observed constant k.
Answer

Rate = k[A]2[B], second order in A and first order in B. This prediction relies on the stated fast-equilibrium approximation.

An elementary step cannot have an order that does not match the particles in that step. Also, zero order does not mean a substance takes no part in the reaction. It means its concentration does not control the slow step over the range measured. Always use the mechanism you are given. Do not invent a slow step from the overall coefficients.

Check your understanding

A mechanism is: step 1, A + A -> I (slow); step 2, I + B -> P (fast). What rate equation does it predict?

  1. rate = k[A][B], from the overall equation
  2. rate = k[A]2[B], from both steps
  3. rate = k[A]2, from the slow step
Show the answer

rate = k[A]2, from the slow step

The slow step controls the rate, and it involves two A particles. B reacts only in the fast step after it, so it does not appear.

Choose a signal that follows a changing species

The method must be fast enough, selective enough and controlled enough for the question.

Match a reaction to a practical rate measurement
Observable changePossible methodKey control or limitation
Gas producedRecord gas-syringe volume against time, or loss of mass if gas escapes.Check for leaks, gas solubility and temperature/pressure changes; account for mixing delay.
Coloured species consumed or formedRecord absorbance with a colorimeter at a suitable wavelength.Calibrate absorbance against concentration; use a blank and avoid interference from other coloured species.
Acid, base or redox species changesWithdraw timed aliquots, quench if needed, then titrate.Quenching must stop the reaction without changing the amount being analysed.
A small fixed amount of product triggers a colour changeUse a clock method and measure time to the same endpoint.1/time compares initial rates only when the fixed threshold is small and consistent.

To test the effect of one concentration, change only that starting concentration. Keep the total volume, the temperature and the other concentrations the same. A common way is to replace part of a solution with water. Measure the gradient at the start of each run, because concentrations fall as the reaction goes on. If a solid reacts, keep its particle size the same.

Worked examplePlan a rate measurement for calcium carbonate and acid

How could the effect of acid concentration be investigated using the carbon dioxide produced?

Try it first. Then show the worked stepsHide the worked steps
  1. Use a gas-tight flask connected to a gas syringe; use equal masses and a controlled size range of carbonate pieces.
  2. Use acid in excess and hold its total volume and temperature constant between runs.
  3. Start timing reproducibly at mixing and record gas volume at short intervals; use the initial tangent rather than the final gas volume.
  4. Repeat to assess scatter and check the apparatus for leaks.
Answer

Initial gas-volume gradients compare initial rates. The final volume mainly reflects the limiting amount, so it cannot by itself show which reaction was faster.

This is a sound way to plan an experiment on paper. The 9476 practical exam does not ask you to collect gas with a gas syringe or by displacing water. You may still be asked to interpret gas-volume data or to describe a suitable method.

Check your understanding

Which method suits following the rate of a reaction in which purple MnO4- ions are used up?

  1. Collect the gas made in a syringe at regular times
  2. Measure the final mass of solid at the end
  3. Measure absorbance with a colorimeter at regular times
Show the answer

Measure absorbance with a colorimeter at regular times

The colour fades as MnO4- is used up, so absorbance tracks its concentration. Calibrate absorbance against known concentrations.