Iodination of propanone: a timing error that inverts a reaction order
A student studies the iodination of propanone in aqueous hydrochloric acid at 298 K, following the disappearance of the iodine colour to a visual endpoint. Each run uses 25.0 cm³ of freshly mixed solution. The recorded times are:
- Run 1: [propanone] = 0.20 M, [H⁺] = 0.10 M, [I₂] = 0.010 M → 80 s
- Run 2: [propanone] = 0.40 M, [H⁺] = 0.10 M, [I₂] = 0.010 M → 40 s
- Run 3: [propanone] = 0.20 M, [H⁺] = 0.20 M, [I₂] = 0.010 M → 40 s
- Run 4: [propanone] = 0.20 M, [H⁺] = 0.10 M, [I₂] = 0.020 M → 160 s
The student writes: 'Run 4 took twice as long as Run 1, so doubling [I₂] doubles the time; the reaction is first order in iodine. Overall order = 1 + 1 + 1 = 3.' The teacher refuses to accept the conclusion until the rate of iodine consumption in each run has been calculated, and points out that the acid concentration is not really constant during a run.
1. Calculate the rate of iodine consumption in each run, deduce the orders and the rate constant, and state the rate equation.
Read the worked answer
Because each run goes to complete decolourisation, the average rate is rate = [I₂]initial / t.
- Run 1: 0.010/80 = 1.25 × 10⁻⁴ M s⁻¹
- Run 2: 0.010/40 = 2.50 × 10⁻⁴ M s⁻¹
- Run 3: 0.010/40 = 2.50 × 10⁻⁴ M s⁻¹
- Run 4: 0.020/160 = 1.25 × 10⁻⁴ M s⁻¹
Comparing 1 and 2: doubling [propanone] doubles the rate → first order in propanone. Comparing 1 and 3: doubling [H⁺] doubles the rate → first order in H⁺. Comparing 1 and 4: doubling [I₂] leaves the rate unchanged → zero order in I₂.
Rate = k[propanone][H⁺]. From Run 1, k = 1.25 × 10⁻⁴ / (0.20 × 0.10) = 6.25 × 10⁻³ dm³ mol⁻¹ s⁻¹.
The overall order is 2, not 3, and iodine does not appear in the rate law at all.
2. Explain why the student's reasoning failed, and why a zero order in iodine is chemically reasonable here.
Read the worked answer
The student treated time as if it were rate. Time is inversely proportional to rate, so doubling the time at doubled [I₂] means the rate was unchanged, which is the signature of zero order, not first order. The correct comparison is between calculated rates, never between raw times.
Chemically, the slow step is the acid-catalysed enolisation of propanone: CH₃COCH₃ + H⁺ ⇌ [CH₃COHCH₃]⁺ → CH₃C(OH)=CH₂ + H⁺. Iodine is not involved in this step. Once the enol forms, it is iodinated very rapidly, so as long as any iodine remains the overall rate is governed entirely by enolisation. Changing [I₂] therefore changes how long the colour lasts but not how fast iodine is consumed per second.
The teacher's warning matters too: each mole of iodine consumed releases one mole of HI, so during Run 4 the acid concentration rises from 0.10 M towards 0.12 M. The 'constant [H⁺]' assumption therefore weakens at high [I₂], and the average rate over the run slightly exceeds the true initial rate. That is an additional reason for using initial rates from small extents of reaction.
3. Predict the time to decolourisation for a run with [propanone] = 0.10 M, [H⁺] = 0.050 M and [I₂] = 0.010 M, and state one assumption in your prediction.
Read the worked answer
Using Rate = k[propanone][H⁺] with k = 6.25 × 10⁻³:
Rate = 6.25 × 10⁻³ × 0.10 × 0.050 = 3.13 × 10⁻⁵ M s⁻¹
t = [I₂]/rate = 0.010 / 3.13 × 10⁻⁵ ≈ 320 s.
The prediction also follows from scaling: relative to Run 1, [propanone] is halved and [H⁺] is halved, so the rate falls to one quarter and the time becomes four times 80 s, i.e. 320 s. The iodine concentration does not enter, because the order in iodine is zero.
Assumption: the rate law and k remain valid at these lower concentrations, i.e. the acid is still in large excess relative to propanone and no side reaction with iodine becomes significant. A practical caveat is that 320 s is long enough for evaporation and for the HI produced to alter [H⁺] appreciably; a buffered acid solution would make the prediction more reliable.