Iodine-clock kinetics: orders, activation energy and a proposed mechanism
The rate of iodination of propanone, CH₃COCH₃ + I₂ → CH₃COCH₂I + HI, is followed by the iodine-clock method. A student mixes fixed volumes of aqueous propanone, dilute sulfuric acid and iodine solution in a thermostat bath at 25.0 °C and records the time for the brown colour of iodine to disappear. In every run the initial iodine concentration is 0.0100 mol dm⁻³ and the total volume is the same. Results:
- Run A: [propanone] = 0.200 mol dm⁻³, [H₂SO₄] = 0.100 mol dm⁻³, t = 121 s
- Run B: [propanone] = 0.400 mol dm⁻³, [H₂SO₄] = 0.100 mol dm⁻³, t = 61 s
- Run C: [propanone] = 0.600 mol dm⁻³, [H₂SO₄] = 0.100 mol dm⁻³, t = 40 s
- Run D: [propanone] = 0.200 mol dm⁻³, [H₂SO₄] = 0.200 mol dm⁻³, t = 60 s
- Run E: [propanone] = 0.200 mol dm⁻³, [H₂SO₄] = 0.100 mol dm⁻³, at 35.0 °C, t = 30 s
The student writes: 'When the propanone concentration doubled the time halved, so the reaction is second order in propanone.' A mechanism is proposed in which the ketone is protonated quickly, the protonated ketone slowly forms its enol, and the enol then reacts rapidly with iodine.
1. Use Runs A–D to deduce the order with respect to propanone and with respect to hydrogen ions, and comment on the student's conclusion.
Read the worked answer
Because the same amount of iodine (0.0100 mol dm⁻³) is consumed to the same end-point in each run, the initial rate is proportional to 1/t. Rates in arbitrary units:
- A: 1/121 = 8.26 × 10⁻³
- B: 1/61 = 1.64 × 10⁻²
- C: 1/40 = 2.50 × 10⁻²
- D: 1/60 = 1.67 × 10⁻²
Propanone: rate(B)/rate(A) = 1.99 for a ×2 change → 2¹, first order. rate(C)/rate(A) = 3.03 for a ×3 change → 3¹, first order. Formally, order = log(3.03)/log(3) = 1.0.
The student is wrong. Halving the time means the rate has doubled, which is first order, not second order. Second order would have required the time to fall to one quarter (≈30 s) when the concentration doubled.
Hydrogen ion: Run D doubles [H₂SO₄], so the nominal [H⁺] doubles (0.200 → 0.400 mol dm⁻³). rate(D)/rate(A) = 2.02 → first order in H⁺. Strictly, HSO₄⁻ is a weak acid so [H⁺] is not exactly double, but the ratio is close enough to 2 to support first order; a strong monobasic acid would remove the ambiguity.
Rate equation: rate = k[CH₃COCH₃][H⁺]. The data say nothing yet about iodine because [I₂] was never varied.
2. Calculate the activation energy from Runs A and E and state two assumptions on which the calculation depends.
Read the worked answer
At identical concentrations k ∝ 1/t, so
k(35 °C)/k(25 °C) = t(25 °C)/t(35 °C) = 121/30 = 4.03
ln(k₂/k₁) = −(Ea/R)(1/T₂ − 1/T₁)
ln 4.03 = 1.395
1/298.15 = 3.354 × 10⁻³ K⁻¹; 1/308.15 = 3.245 × 10⁻³ K⁻¹; difference = 1.088 × 10⁻⁴ K⁻¹
Ea = (8.314 × 1.395)/(1.088 × 10⁻⁴) = 1.07 × 10⁵ J mol⁻¹ ≈ 107 kJ mol⁻¹
Assumptions
- The mixed solutions had actually reached the thermostat temperature (a cold mixture would give a spuriously long time and an over-estimate of Ea).
- 1/t remains proportional to k at both temperatures, i.e. the same iodine concentration is consumed in both runs and the reaction has not changed mechanism over a 10 K range.
- The acid is fully dissociated to the same extent in both runs, so [H⁺] is unchanged by the temperature change.
With only two temperatures no check of linearity is possible; several temperatures and an Arrhenius plot would be needed for a trustworthy value.
3. Explain how the rate data support the proposed mechanism, and describe what would be observed instead if the reaction of the enol with iodine were the rate-determining step.
Read the worked answer
If enolisation is slow, rate = k₂[protonated ketone]. The fast pre-equilibrium gives [protonated ketone] = K[CH₃COCH₃][H⁺], so
rate = k₂K[CH₃COCH₃][H⁺]
That is first order in ketone, first order in acid and zero order in iodine - exactly what the clock experiment shows. Iodine enters only after the slow step, so changing [I₂] changes how long the colour lasts but not the rate at which iodine is consumed; a plot of 1/t against [I₂] would be a horizontal line.
If the enol + I₂ step were rate-determining, the enol would build up to a steady state and the rate would become proportional to [I₂]. Experimentally one would then see the reaction become first order in iodine, and doubling [I₂] would halve t rather than double it. Acid catalysis would also become less straightforward, because the enol concentration would be set by the fast equilibrium rather than by the slow step.
The observed acid catalysis plus the independence from [I₂] therefore points to enolisation as the slow step, and explains why only ketones with α-hydrogens are iodinated at a measurable rate under these conditions.