4.1.9Unit 4. Rates, Equilibria and Further Organic Chemistry
Acid-Catalysed Iodination of Propanone — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 18 of 56 Unit quiz
Curriculum Equation Key pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: pH + pOH = 14
Part 1 of 6The Reaction
This reaction is a standard example of using kinetics to learn about a mechanism.
Equation
- CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq), catalysed by hydrogen ions.
Experimental rate equation
- Rate = k[CH₃COCH₃][H⁺], and the reaction is zero order with respect to iodine.
Meaning
- Iodine is not in the rate-determining step, and the catalyst H⁺ does appear in the rate equation.
Overall order
- The reaction is second order overall, being first order in propanone and first order in acid.
Part 2 of 6Following the Reaction
Following the Reaction
Sampling
- Withdraw portions of the reaction mixture at known times with a pipette.
Quenching
- Add sodium hydrogencarbonate solution to neutralise the acid, which stops the reaction.
Titration
- Titrate the iodine left with sodium thiosulfate solution, using starch near the end point.
Colorimetry
- Alternatively, the fading brown colour of iodine can be followed in a colorimeter without sampling.
| Experiment changed | Observation | Order |
|---|---|---|
| Doubling [propanone] | Rate doubles | 1 |
| Doubling [H⁺] | Rate doubles | 1 |
| Doubling [I₂] | Rate unchanged | 0 |
Part 3 of 6Titration of Iodine
1
The titration equation is I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq).
2
Thiosulfate is added until the brown colour fades to pale yellow, and then starch is added to give a blue-black colour.
3
More thiosulfate is added until the blue-black colour just disappears, which is the end point.
4
The amount of iodine in the sample is half the amount of thiosulfate used.
5
A graph of iodine concentration against time is a straight line, with a gradient equal to the negative of the rate.
Part 4 of 6Mechanism Link
Rate-determining step
- This involves propanone and an H⁺ ion, which together lead to the formation of the enol form of propanone.
Iodine step
- Iodine reacts quickly with the enol after the slow step, so the iodine concentration does not affect the rate.
Zero order
- The straight-line graph shows that the rate is constant, because iodine does not appear in the rate equation.
Evidence
- Rate equation data suggest the mechanism but cannot prove it.
Part 5 of 6Worked Example
Question
- A 10.0 cm³ sample of the mixture requires 20.0 cm³ of 0.0100 mol dm⁻³ sodium thiosulfate at t = 0 and 15.0 cm³ at t = 300 s.
- Find the average rate of loss of iodine.
- At t = 0: n(S₂O₃²⁻) = 0.0100 × 0.0200 = 2.00 × 10⁻⁴ mol, so n(I₂) = 1.00 × 10⁻⁴ mol and [I₂] = 1.00 × 10⁻⁴ ÷ 0.0100 = 0.0100 mol dm⁻³.
- At t = 300 s: n(S₂O₃²⁻) = 1.50 × 10⁻⁴ mol, n(I₂) = 7.50 × 10⁻⁵ mol and [I₂] = 7.50 × 10⁻³ mol dm⁻³.
- Rate = (0.0100 − 0.00750) ÷ 300 = 8.3 × 10⁻⁶ mol dm⁻³ s⁻¹.
Answer Average rate = 8.3 × 10⁻⁶ mol dm⁻³ s⁻¹
Part 6 of 6Exam Checklist & Tips
4.1.9 – Key Points for the Exam
- Quench To stop a reaction, for example by neutralising the catalyst.
- Enol A compound containing a C=C double bond with an –OH group on one of the carbon atoms.
- Zero order The rate is independent of the concentration of that species.
- Starch indicator A reagent that gives a blue-black colour with iodine.
- Catalyst in rate equation An acid that speeds up the reaction and appears in the rate equation but not in the overall equation.
Common Mistakes
- Writing iodine in the rate equation because it appears in the overall equation.
- Forgetting to quench the sample before titrating.
- Adding starch too early, when there is too much iodine, so that the end point is hard to see.
- Using the volume of thiosulfate as the amount of iodine without halving it.
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