Topic 11 · page 11 of 21
Iodination of propanone: data and predictions
In the acid-catalysed reaction of propanone with iodine, the rate is first order in propanone, first order in hydrogen ions and zero order in iodine. That tells you which species are in the rate-determining step.
Read first: The rate-determining step and the rate equation, Orders from initial-rate data
This is the reaction that the specification uses to bring together orders, rate equations and mechanisms. It is also the one in Core Practical 9a (page 18 and page 19).
The reaction and how to follow it 11.6(i)
Propanone reacts with iodine in aqueous solution, and the reaction is catalysed by acid.
The catalyst is hydrogen ions, from an acid such as sulfuric acid. It is in the same phase as the reactants, so it is a homogeneous catalyst. Iodine solution is yellow-brown, and propanone, iodopropanone and the acid are colourless. So the rate can be found from the loss of iodine:
- by colorimetry, as the absorbance of the iodine falls, or
- by titration of quenched samples with sodium thiosulfate (page 18).
Getting the orders 11.6(i)11.5(iii)
Three reagents can change the rate: propanone, iodine and the hydrogen ion. Each order needs its own comparison. There are two ways.
Initial-rate method. Do separate experiments in which the initial concentration of only one of the three changes. Find the initial rate of each.
Isolating iodine. To find the order with respect to iodine from a single run, use propanone and acid in large excess, so that their concentrations hardly change during the experiment. Then only the iodine concentration changes, and the shape of the iodine concentration–time graph gives its order (page 19).
Worked exampleOrders from initial rates
(Illustrative data; concentrations in mol dm⁻³.)
| Experiment | [CH₃COCH₃] | [I₂] | [H⁺] | Initial rate / 10⁻⁶ mol dm⁻³ s⁻¹ |
|---|---|---|---|---|
| 1 | 1.00 | 0.0050 | 0.50 | 5.0 |
| 2 | 2.00 | 0.0050 | 0.50 | 10.0 |
| 3 | 1.00 | 0.0100 | 0.50 | 5.0 |
| 4 | 1.00 | 0.0050 | 1.00 | 10.0 |
- 1 and 2: [propanone] ×2, rate ×2, so first order in propanone.
- 1 and 3: [I₂] ×2, rate unchanged, so zero order in iodine.
- 1 and 4: [H⁺] ×2, rate ×2, so first order in H⁺.
- rate = k[CH₃COCH₃][H⁺], second order overall. Iodine does not appear.
- k = 5.0 × 10⁻⁶ ÷ (1.00 × 0.50) = 1.0 × 10⁻⁵. Units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² = dm³ mol⁻¹ s⁻¹. So k = 1.0 × 10⁻⁵ dm³ mol⁻¹ s⁻¹.
What the rate equation predicts 11.6(ii)
Use what page 10 said about the rate-determining step. The rate equation is
- Propanone is first order, so one propanone is in the rate-determining step (or is used in a fast step before it).
- H⁺ is first order, so one H⁺ is in the rate-determining step too, even though it is a catalyst and is not in the overall equation. A catalyst that takes part in the slow step appears in the rate equation.
- Iodine is zero order, so it is not in the rate-determining step. It must react in a fast step after the slow step.
These predictions are what you use on page 12 to write a mechanism.
Check yourself
- The reaction of propanone with iodine is first order in propanone, zero order in iodine and first order in H⁺. Write the rate equation, state the overall order and give the units of k.
- Using the data below (illustrative), find the orders and calculate k, with units.
| Experiment | [CH₃COCH₃] / mol dm⁻³ | [I₂] / mol dm⁻³ | [H⁺] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|---|
| 1 | 0.80 | 0.0040 | 0.40 | 6.4 × 10⁻⁶ |
| 2 | 1.60 | 0.0040 | 0.40 | 1.28 × 10⁻⁵ |
| 3 | 0.80 | 0.0080 | 0.40 | 6.4 × 10⁻⁶ |
| 4 | 0.80 | 0.0040 | 0.80 | 1.28 × 10⁻⁵ |
- Using your value of k from question 2, calculate the initial rate when [CH₃COCH₃] = 0.40, [I₂] = 0.0020 and [H⁺] = 0.20 mol dm⁻³.
- Explain why propanone and acid are used in large excess when the order with respect to iodine is found from a single experiment.
- What do the orders tell you about the rate-determining step, and where does iodine react?
- H⁺ is not in the balanced equation, but it is in the rate equation. Explain how this is possible.
Answers cover these until you have tried the questions
- rate = k[CH₃COCH₃][H⁺]. It is second order overall (1 + 1). Units of k: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² = dm³ mol⁻¹ s⁻¹.
- 1 and 2: [propanone] ×2, rate ×2: first order. 1 and 3: [I₂] ×2, rate unchanged: zero order. 1 and 4: [H⁺] ×2, rate ×2: first order. rate = k[CH₃COCH₃][H⁺]. k = 6.4 × 10⁻⁶ ÷ (0.80 × 0.40) = 6.4 × 10⁻⁶ ÷ 0.32 = 2.0 × 10⁻⁵ dm³ mol⁻¹ s⁻¹.
- rate = 2.0 × 10⁻⁵ × 0.40 × 0.20 = 1.6 × 10⁻⁶ mol dm⁻³ s⁻¹. The iodine concentration does not matter.
- If the concentrations of propanone and H⁺ are large, only a very small fraction is used up, so they stay effectively constant. Any change in rate during the run is then due to the changing concentration of iodine only, so the iodine graph shows its order.
- The rate-determining step contains one propanone and one H⁺ (or species made from them in an earlier fast step). Iodine is zero order, so it is not in the slow step: it reacts in a fast step after it.
- H⁺ is a catalyst. It is used in the slow step and is made again in a later step, so it cancels out of the overall equation. Because it takes part in the rate-determining step, its concentration affects the rate, so it appears in the rate equation with order 1.