All subjects · Chemistry · WCH14 (Unit 4)

Iodine-propanone and clock reactions (Core Practicals 9a and 9b)

Core Practical 9a follows the acid-catalysed reaction of propanone with iodine by titration. Core Practical 9b uses a clock reaction and the initial rate method. Both give orders that you then link to a mechanism.

The iodination of propanone (spec 11.6)

CH3COCH3(aq) + I2(aq) -> CH3COCH2I(aq) + H+(aq) + I-(aq), catalysed by H+.

Experiments show rate = k[CH3COCH3][H+]: first order in propanone, first order in H+ and zero order in iodine. So the rate-determining step involves propanone and H+, and iodine reacts only in a later, fast step. H+ appears in the rate equation even though it is a catalyst, because it takes part in the slow step and is regenerated later.

Core Practical 9a: titration method (spec 11.12)

Use propanone and acid in large excess, so their concentrations hardly change and only the effect of [I2] is seen.

1. Mix the reactants and start the clock.

2. At regular intervals withdraw a 10.0 cm3 sample with a pipette and add it to sodium hydrogencarbonate solution. This quenches the reaction by neutralising the acid catalyst.

3. Titrate the remaining iodine with standard sodium thiosulfate, adding starch near the end point (blue-black to colourless): I2 + 2S2O32- -> 2I- + S4O62-.

A graph of titre (proportional to [I2]) against time is a straight line, so the reaction is zero order in iodine.

Core Practical 9b: a clock reaction (spec 11.12)

In the iodine clock, peroxydisulfate oxidises iodide to iodine: S2O82- + 2I- -> 2SO42- + I2. A small fixed amount of thiosulfate turns the iodine back to iodide until it is all used up; then free iodine meets starch and the mixture suddenly turns blue-black. The time t is for a fixed amount of reaction, so initial rate is proportional to 1/t.

Run one series varying [S2O82-] and another varying [I-], keeping the total volume constant by adding water. Plot 1/t against volume (proportional to concentration) of the reactant varied: a straight line through the origin shows first order. A plot of log(1/t) against log(volume) has a gradient equal to the order.

Worked example

In CP9a, a 10.0 cm3 sample needs 12.0 cm3 of 0.0100 mol dm-3 sodium thiosulfate. Find [I2] in the sample.

  1. n(S2O32-) = 0.0120 x 0.0100 = 1.20 x 10-4 mol
  2. I2 : S2O32- = 1 : 2, so n(I2) = 6.0 x 10-5 mol
  3. [I2] = 6.0 x 10-5 / 0.0100 dm3

Answer: [I2] = 6.0 x 10-3 mol dm-3

Key idea: rate = k[CH3COCH3][H+]; zero order in I2, so iodine is not in the slow step. CP9a: quench samples with NaHCO3 and titrate I2 with thiosulfate. CP9b: initial rate is proportional to 1/time.

Check yourself

  1. In Core Practical 9a, why is each sample added to sodium hydrogencarbonate solution?
    • To react with the iodine
    • To act as an indicator
    • To make the solution alkaline for the titration with starch
    • To neutralise the acid catalyst and stop the reaction
    Show answer

    To neutralise the acid catalyst and stop the reaction. Removing H+ removes the catalyst, so the iodine concentration is frozen at that moment.

  2. In Core Practical 9a, a graph of titre against time is a straight line. What does this show?
    • The reaction is first order with respect to iodine
    • The reaction is zero order with respect to iodine
    • Iodine is a catalyst
    • The reaction has stopped
    Show answer

    The reaction is zero order with respect to iodine. A straight concentration-time line means the rate is constant, independent of [I2].

  3. The rate equation is rate = k[CH3COCH3][H+]. Which species are involved in the rate-determining step?
    • Propanone and iodine
    • Iodine and hydrogen ions
    • Two propanone molecules
    • One propanone molecule and one hydrogen ion
    Show answer

    One propanone molecule and one hydrogen ion. Species that appear in the rate equation (with their orders) are in, or before, the slow step.

  4. A 10.0 cm3 sample of reaction mixture needs 8.40 cm3 of 0.0100 mol dm-3 sodium thiosulfate to react with its iodine. Find the concentration of iodine in the sample.
    Show answer

    0.00420 mol dm-3. n(S2O32-) = 8.40 x 10-5 mol; n(I2) = 4.20 x 10-5 mol; [I2] = 4.20 x 10-5 / 0.0100 = 4.20 x 10-3 mol dm-3.

  5. In an iodine clock experiment, why is the initial rate taken as proportional to 1/t?
    • The total volume changes in every run
    • Starch speeds up the reaction
    • The same small amount of iodine has formed when the colour appears in every run
    • Thiosulfate is a catalyst
    Show answer

    The same small amount of iodine has formed when the colour appears in every run. Each run times the same fixed extent of reaction, so a shorter time means a proportionally faster rate.

  6. In Core Practical 9b, why is water added so that every mixture has the same total volume?
    • So the volume of each reactant is proportional to its concentration
    • To dilute the starch
    • To cool the mixture
    • To react with the thiosulfate
    Show answer

    So the volume of each reactant is proportional to its concentration. With a constant total volume, concentration is proportional to the volume of solution used.

  7. In Core Practical 9b, a graph of log(1/t) against log(volume of KI solution) has a gradient of 1.0. What is the order with respect to iodide?
    • Zero order
    • Second order
    • First order
    • Minus one
    Show answer

    First order. For rate = k[I-]n, log rate = n log[I-] + constant, so the gradient is the order.

Practise with instant feedback

Sources

Last updated 7 November 2026. Spotted a mistake? Tell us.