The Iodine Clock that Refused to Obey Its Own Rate Law
A student studies the persulphate–iodide reaction at 25 °C using the iodine clock: S2O8²⁻(aq) + 2I⁻(aq) → 2SO4²⁻(aq) + I2(aq) Every 50.0 mL run contains 1.00 mL of 0.0100 M Na2S2O3 and a few drops of starch, and the time to the first blue colour is recorded. Results: run 1, [S2O8²⁻] = 0.0200 M, [I⁻] = 0.0200 M, t = 45 s; run 2, [S2O8²⁻] = 0.0400 M, [I⁻] = 0.0200 M, t = 46 s; run 3, [S2O8²⁻] = 0.0200 M, [I⁻] = 0.0400 M, t = 23 s. The student concludes the reaction is first order in iodide and zero order in persulphate, and writes rate = k[I⁻]. A demonstrator recalls that the 0.0400 M persulphate stock used in run 2 was prepared a week earlier and stored in a warm cupboard; persulphate solutions decompose slowly with time. Accepted behaviour for this reaction under these conditions is rate = k[S2O8²⁻][I⁻].
1. Interpret the timing data. Show what the accepted rate law predicts for runs 2 and 3, and decide what the run 2 result actually tells you about the solution used.
Read the worked answer
In a clock run the blue colour appears when the fixed amount of thiosulphate is used up, so the time t is inversely proportional to the initial rate: rate ∝ 1/t. The thiosulphate charge is constant in every run.
- Run 3 vs run 1: [I⁻] doubles, everything else fixed. t falls from 45 s to 23 s, a factor of 1.96 ≈ 2, so the rate doubles → first order in I⁻. Confirms the student.
- Run 2 vs run 1: [S2O8²⁻] nominally doubles. If rate = k[S2O8²⁻][I⁻], the rate should double and t should halve to about 22.5 s. Observed t = 46 s, i.e. the rate is essentially unchanged.
So the nominally doubled persulphate produced no rate increase at all, which is impossible for a first-order reactant unless the true concentration did not change. The decomposition of persulphate in the warm cupboard is the only explanation available from the information given: the effective [S2O8²⁻] in the '0.0400 M' solution was about 0.0200 M. Run 2 is invalid as a test of order; the student's zero-order conclusion rests entirely on a decomposed reagent and must be discarded.
2. Using run 3, which the demonstrator accepts as valid, calculate the rate of the reaction in mol L⁻¹ s⁻¹ and hence the rate constant. Show how the thiosulphate stoichiometry is used.
Read the worked answer
Thiosulphate present: 1.00 × 10⁻³ L × 0.0100 mol L⁻¹ = 1.00 × 10⁻⁵ mol.
Stoichiometry: I2 + 2S2O3²⁻ → 2I⁻ + S4O6²⁻, so the iodine produced before the colour appears is 1.00 × 10⁻⁵/2 = 5.00 × 10⁻⁶ mol. Since S2O8²⁻ → I2 is 1:1, the persulphate consumed is also 5.00 × 10⁻⁶ mol.
Concentration change in the 50.0 mL run: 5.00 × 10⁻⁶ mol / 0.0500 L = 1.00 × 10⁻⁴ mol L⁻¹. t = 23 s, so average rate = 1.00 × 10⁻⁴ / 23 = 4.35 × 10⁻⁶ mol L⁻¹ s⁻¹.
With rate = k[S2O8²⁻][I⁻]: k = 4.35 × 10⁻⁶ / (0.0200 × 0.0400) = 4.35 × 10⁻⁶ / 8.00 × 10⁻⁴ = 5.4 × 10⁻³ L mol⁻¹ s⁻¹.
The value is an average over the first 23 s; because only about 0.5% of the persulphate is used up, the average is a good approximation to the initial rate, which is why the clock method works at all.
3. The student wants to repeat the whole series and asks how to make the results trustworthy. Identify the procedural changes that matter most, and justify why starch is added only near the end point in some protocols.
Read the worked answer
Priority changes:
- Prepare fresh persulphate solution (iodometric check or standardise it) and keep it away from heat; the failed run shows the cost of not doing so. Ideally verify each stock by titration with thiosulphate.
- Check the thiosulphate charge is identical in every run - the 1/t analysis collapses if the clock's 'budget' of thiosulphate varies.
- Vary one concentration at a time and repeat each run at least twice; a single 45 s vs 46 s comparison cannot distinguish zero order from experimental scatter of a few per cent.
- Keep ionic strength roughly constant by adding inert electrolyte when concentrations are changed, because the reaction between two anions is strongly sensitive to ionic strength.
- Thermostat the mixtures at 25 °C and start the clock at the moment of mixing, since the reaction has a significant temperature coefficient.
Starch: iodine–starch complex formation is sensitive to high iodine levels, and starch can be attacked or destabilised over long standing. Adding it just before the predicted end point sharpens the colour change and avoids a slow, smeared end point. It does not change the stoichiometry, so the calculation is unaffected.