CP 9a and 9b Iodine–Propanone Kinetics and a Clock Reaction

Two thiosulfate ions react with each iodine molecule, so the titre is a direct measure of the iodine left. A worked set of results at 20 °C is shown below.

Titre against time: the six points lie about a straight line of gradient −0.80 cm³ min⁻¹ that extrapolates to 20.00 cm³ at t = 0. The rate is constant while [I₂] falls, so the reaction is zero order with respect to iodine.

The titre falls by about 2.4 cm³ every 3 minutes, and the points stay within 0.10 cm³ of the best-fit line. A straight line means a constant rate even though [I₂] is falling: the reaction is zero order with respect to iodine.

Step 1: the line extrapolates to 20.00 cm³ at t = 0, so moles of S₂O₃²⁻ at t = 0 = 0.0100 × 20.00 ÷ 1000 = 2.00 × 10⁻⁴ mol

Step 2: moles of I₂ in 10.0 cm³ = ½ × 2.00 × 10⁻⁴ = 1.00 × 10⁻⁴ mol, so [I₂]₀ = 1.00 × 10⁻⁴ ÷ 0.0100 = 0.0100 mol dm⁻³

Step 3: reading (0, 20.00) and (15, 8.00) from the line, gradient = (8.00 − 20.00) ÷ 15 = −0.80 cm³ min⁻¹

Step 4: 20.00 cm³ of titre ≡ 0.0100 mol dm⁻³, so rate = 0.80 × 0.0100 ÷ 20.00 = 4.00 × 10⁻⁴ mol dm⁻³ min⁻¹

Answer: rate = 4.00 × 10⁻⁴ mol dm⁻³ min⁻¹ = 6.67 × 10⁻⁶ mol dm⁻³ s⁻¹ (3 s.f.). The extrapolated 0.0100 mol dm⁻³ starting concentration checks: 50.0 cm³ of 0.0200 mol dm⁻³ iodine diluted to 100 cm³.

Had the line curved, flattening as the titre fell, the rate would be falling with [I₂] and the order in iodine would be first (constant half-life) or second. A straight line is the signature of zero order.

Iodine is brown, so its concentration can be followed continuously without sampling. A colorimeter measures the absorbance of the mixture in a cuvette; the more iodine, the more light is absorbed.

Filter: a blue filter, the colour complementary to the brown-yellow of iodine, gives the largest change in absorbance.

Calibration: measure the absorbance of iodine solutions of known concentration first and plot absorbance against [I₂]. The reaction mixture’s readings are then converted to concentrations with this calibration curve.

Result: absorbance falls in a straight line with time for a zero-order reaction, exactly like the titre; a reading every 30 s gives far more points than six titrations.

Advantages: no quenching, no portions to withdraw, no end point to judge, and a data logger can record the whole run automatically.

Four runs at 20 °C, each 50.0 cm³ in total and each with 10.0 cm³ of 0.0100 mol dm⁻³ Na₂S₂O₃(aq) and 2 cm³ of starch.

The four clock runs as an illustrated table: doubling [I⁻] or [H₂O₂] halves the time, doubling [H⁺] changes nothing, so rate = k[H₂O₂][I⁻].

Run 2 against run 1: [I⁻] doubled, t halved, so the rate doubled: first order in I⁻.

Run 3 against run 1: [H₂O₂] doubled, t halved: first order in H₂O₂.

Run 4 against run 1: [H⁺] doubled, t unchanged: zero order in H⁺ under these conditions.

rate = k[H₂O₂][I⁻]

Turning a time into a rate

Step 1: moles of S₂O₃²⁻ = 0.0100 × 10.0 ÷ 1000 = 1.00 × 10⁻⁴ mol, so moles of I₂ formed before the colour change = 5.00 × 10⁻⁵ mol

Step 2: Δ[I₂] = 5.00 × 10⁻⁵ ÷ 0.0500 = 1.00 × 10⁻³ mol dm⁻³ (total volume 50.0 cm³)

Step 3: initial rate (run 1) = 1.00 × 10⁻³ ÷ 48 = 2.08 × 10⁻⁵ mol dm⁻³ s⁻¹

Step 4: [H₂O₂] = [I⁻] = 0.100 × 10.0 ÷ 50.0 = 0.0200 mol dm⁻³, so k = 2.08 × 10⁻⁵ ÷ (0.0200 × 0.0200) = 0.0520 dm³ mol⁻¹ s⁻¹

Answer: initial rate 2.08 × 10⁻⁵ mol dm⁻³ s⁻¹ and k = 0.0520 dm³ mol⁻¹ s⁻¹ at 20 °C (3 s.f.).

Why 1/t works

The clock measures the time for a fixed small change: the same 5.00 × 10⁻⁵ mol of iodine forms in every run. Rate = fixed change ÷ t, so rate ∝ 1/t.

It is an average rate over the first seconds, close to the initial rate only while the concentrations have barely changed: here about 5% of the peroxide is used. Keep the thiosulfate small so that less than about 10% of any reactant reacts.

The chemistry adds errors that no burette or stopwatch can remove. For each one, say which way it moves the result and how to reduce it.

Timing uncertainty. A human reaction time of ±1 s on run 2 (24 s) is 1 ÷ 24 × 100 = 4.2%; on run 1 (48 s) it is 2.1%. Halving the concentration of one reactant would double the times and halve these percentages.

Burette uncertainty. A titre is two readings of ±0.05 cm³, so ±0.10 cm³: on the 8.05 cm³ titre at 15 min that is 0.10 ÷ 8.05 × 100 = 1.24%, but only 0.51% on the 19.60 cm³ titre at 0.5 min.

The later portions therefore carry the larger uncertainty; a more concentrated thiosulfate would make them smaller still, so a more dilute titrant, about 0.00500 mol dm⁻³, improves the late points.