The iodine clock reaction

The iodine clock reaction is a reaction in which a colourless mixture suddenly turns dark blue-black after a delay. Iodine is made slowly, but a small amount of sodium thiosulfate turns it straight back into iodide. When the thiosulfate runs out, free iodine appears and reacts with starch to give the colour. The time taken is a measure of the initial rate, so A-level students use the iodine clock to find orders of reaction.

How the iodine clock reaction works

The most common version, used in the Royal Society of Chemistry demonstration, mixes two colourless solutions:

• Solution A: potassium iodide, sodium thiosulfate and starch.
• Solution B: hydrogen peroxide and an acid (ethanoic acid in the RSC version).

Two reactions then run at the same time, one slow and one fast.

Reaction 1, slow: iodine is made

H₂O₂(aq) + 2I⁻(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l)

Hydrogen peroxide oxidises iodide ions to iodine. This is the reaction whose rate you are measuring.

Reaction 2, fast: thiosulfate removes the iodine

2S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2I⁻(aq)

As soon as any iodine forms, thiosulfate ions reduce it back to iodide, forming tetrathionate ions. So no iodine builds up and the starch stays colourless.

The colour change

The thiosulfate is the limiting reagent. Once it has all been used, the iodine being made has nothing left to react with. Free iodine (strictly I₃⁻ ions) builds up and forms a blue-black complex with starch. Because the change happens throughout the whole mixture at once, it looks instant, like a clock going off.

Why the clock time measures the rate

Every run uses the same amount of thiosulfate and the same total volume. So the colour always appears when the same amount of iodine has been made. If one mixture reaches that point in half the time, its rate was twice as fast:

rate ∝ 1 ÷ t, where t is the time for the blue-black colour to appear.

Only a small fraction of the hydrogen peroxide and iodide has reacted by then, so their concentrations are still close to their starting values. That makes 1/t a good measure of the initial rate. OCR describes clock reactions as an approximation to the initial rates method for exactly this reason: the time is measured before the reaction has gone too far.

Iodine clock experiment: method

To find the order with respect to iodide, change only the iodide concentration from run to run:

1. Measure the potassium iodide solution, the sodium thiosulfate solution and the starch into a beaker. Use a different volume of potassium iodide each run, and add water so the total volume is always the same.
2. Measure the hydrogen peroxide (with its acid) into a separate beaker. Keep its volume and concentration the same every run.
3. Pour the hydrogen peroxide into the first beaker and swirl, starting the stopwatch as you pour.
4. Stop the stopwatch when the mixture turns blue-black. Record the time.
5. Repeat for each iodide volume, and repeat each run to check the times agree.
6. Work out 1/t for each run and plot it against the volume (or concentration) of iodide.

Keep the temperature the same for every run, for example by standing the solutions in a water bath, because rate depends strongly on temperature. To find the order with respect to hydrogen peroxide, repeat the whole method with iodide fixed and the hydrogen peroxide changed instead.

Finding the order of reaction

The rate equation for the hydrogen peroxide and iodide reaction has the form rate = k[H₂O₂]^m[I⁻]ⁿ, and the orders m and n can only be found by experiment. Use your 1/t values like any initial rates:

A graph of 1/t against iodide concentration makes this clear: a straight line through the origin means first order, and a curve bending upwards suggests second order. The RSC states that this reaction is first order with respect to both hydrogen peroxide and iodide. See the rate equation guide for working out k and its units.

Once the orders are known, the rate equation gives the rate for any concentrations. Practise using one in this question from the Reaction Kinetics course.

Worked examples

Example 1: order with respect to iodide

A student keeps everything the same except the volume of 0.10 mol dm⁻³ potassium iodide, topping up with water to the same total volume each time. Find the order with respect to iodide.

• Because the total volume is fixed, the iodide concentration is proportional to the KI volume.
• Doubling the volume from 5 to 10 cm³ roughly doubles 1000/t, from 8.5 to 16.4, and doubling again from 10 to 20 cm³ roughly doubles it from 16.4 to 34.5.
• A straight line through (0, 0) fits the points well, so the reaction is first order with respect to iodide.

Example 2: order with respect to hydrogen peroxide

With the iodide fixed, doubling the hydrogen peroxide concentration changes the clock time from 64 s to 32 s. What is the order with respect to hydrogen peroxide, and what is the rate equation?

• The time halves, so 1/t, and so the rate, doubles.
• Rate doubles when concentration doubles: first order with respect to H₂O₂.
• Together with Example 1: rate = k[H₂O₂][I⁻], second order overall.

Example 3: activation energy from two temperatures

With every concentration fixed, the clock time is 80 s at 20 °C and 19 s at 40 °C. Estimate the activation energy. R = 8.31 J K⁻¹ mol⁻¹.

1. With fixed concentrations, rate ∝ k, and rate ∝ 1/t, so k₂ ÷ k₁ = t₁ ÷ t₂ = 80 ÷ 19 = 4.21.
2. Convert to kelvin: T₁ = 293 K and T₂ = 313 K.
3. From ln k = −E_a/RT + ln A for both temperatures: ln(k₂ ÷ k₁) = (E_a/R)(1/T₁ − 1/T₂).
4. ln 4.21 = 1.438, and 1/293 − 1/313 = 2.181 × 10⁻⁴ K⁻¹.
5. E_a = 8.31 × 1.438 ÷ (2.181 × 10⁻⁴) = 54,800 J mol⁻¹ = 54.8 kJ mol⁻¹.

With more temperatures, plot ln(1/t) against 1/T: the gradient is −E_a/R. The Arrhenius equation guide and the Rates and Equilibrium Calculator take this further.

Other clock reactions

The same trick works with other slow reactions that make iodine. A common one uses peroxodisulfate (persulfate) ions instead of hydrogen peroxide:

S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)

Thiosulfate and starch play the same parts, and the rate equation has the form rate = k[S₂O₈²⁻]^m[I⁻]ⁿ, with m and n found from clock times in the same way.

Common mistakes

• Saying starch reacts with thiosulfate. Starch only shows when free iodine is present.
• Changing the total volume. Adding more KI without taking out water changes every concentration, not just the iodide.
• Plotting t instead of 1/t. The rate is proportional to 1/t; t on its own gives a curve that hides the order.
• Forgetting that thiosulfate must be the limiting reagent. If there is too much, the reaction goes too far and 1/t no longer measures the initial rate.
• Leaving temperature uncontrolled. A few degrees change the time a lot.
• Writing the iodine equation with I instead of I₂, or forgetting the H⁺. Balance both atoms and charges.

Practise with a short timed paper

Rate equations and the initial rates method are in the second year of A-level Chemistry for AQA (3.1.9, Required practical 7) and OCR (5.1.1). The daily A-level Chemistry: Year 2 paper is a short timed paper of 8 questions, with a new one every day.

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