Iodine Clock That Ignores the Iodine
A university laboratory in Barcelona studies the acid-catalysed iodination of propanone at 298 K by following the fading of the yellow-brown iodine colour in a spectrophotometer. The stock solutions are propanone 1.00 M, HCl 1.00 M and I2 0.0200 M; the ionic strength is kept constant with KCl. Four runs give initial rates: (1) [propanone] = 0.100 M, [H⁺] = 0.100 M, [I2] = 0.0050 M, rate = 6.00 × 10⁻⁵ M s⁻¹; (2) [propanone] = 0.200 M, [H⁺] = 0.100 M, [I2] = 0.0050 M, rate = 1.20 × 10⁻⁴ M s⁻¹; (3) [propanone] = 0.100 M, [H⁺] = 0.200 M, [I2] = 0.0050 M, rate = 1.20 × 10⁻⁴ M s⁻¹; (4) [propanone] = 0.100 M, [H⁺] = 0.100 M, [I2] = 0.0100 M, rate = 6.00 × 10⁻⁵ M s⁻¹. A repeat at 308 K with the run-1 concentrations gives 1.20 × 10⁻⁴ M s⁻¹. A student proposes a one-step mechanism in which propanone and I2 collide directly, and is surprised that the colour fades at a rate that does not depend on how much iodine is present.
1. Determine the experimental rate law and the value of the rate constant at 298 K, including its units.
Read the worked answer
Compare runs while changing one concentration at a time.
- Runs 1→2: [propanone] doubles, [H⁺] and [I2] fixed; rate doubles (6.00 → 1.20 × 10⁻⁴). Order in propanone = 1.
- Runs 1→3: [H⁺] doubles; rate doubles. Order in H⁺ = 1.
- Runs 1→4: [I2] doubles; rate unchanged. Order in I2 = 0.
Rate law: rate = k[CH3COCH3][H⁺], first order overall in each of the two species, zero order in iodine (second order overall).
k = rate/([propanone][H⁺]) = 6.00 × 10⁻⁵/(0.100 × 0.100) = 6.00 × 10⁻³ M⁻¹ s⁻¹.
Check run 2: 6.00 × 10⁻³ × 0.200 × 0.100 = 1.20 × 10⁻⁴ ✓; run 3 gives the same ✓; run 4 gives 6.00 × 10⁻⁵ ✓. Units follow from the rate law: M s⁻¹/(M²) = M⁻¹ s⁻¹. Note that the iodine concentration is not part of the law, which is why the fading rate is the same even though the solution starts darker.
2. Explain why the observed zero order in I2 rules out the student's one-step collision mechanism, and outline a mechanism consistent with the rate law.
Read the worked answer
A single elementary step 'propanone + I2 → products' would have rate = k[propanone][I2], i.e. first order in iodine. Run 4 doubles [I2] with no change in rate, so iodine cannot be present in the rate-determining step. The student's mechanism is inconsistent with the data and must be rejected regardless of how plausible the collision seems.
A consistent scheme is acid-catalysed enolisation followed by fast iodination:
- CH3COCH3 + H⁺ ⇌ [CH3COHCH3]⁺ (fast pre-equilibrium, K)
- [CH3COHCH3]⁺ → CH3C(OH)=CH2 + H⁺ (slow, k2) - enol formation
- enol + I2 → iodopropanone + H⁺ + I⁻ (fast)
Step 2 is rate-determining: rate = k2[protonated ketone] = k2K[propanone][H⁺]. This matches the observed first order in each of propanone and H⁺, and predicts no dependence on I2 because iodine only enters after the slow step. The protonated ketone and the enol are intermediates, not catalysts; H⁺ is a catalyst because it is consumed in step 1 and regenerated in step 2/3. The mechanism also explains why the reaction is much faster in strong acid and why the measured rate would fall if the acid were partly neutralised.
3. Calculate the activation energy from the 298 K and 308 K data, predict k at 303 K, and comment on the common 'rate doubles every 10 K' rule.
Read the worked answer
Using the Arrhenius ratio with k(298) = 6.00 × 10⁻³ and k(308) = 1.20 × 10⁻² M⁻¹ s⁻¹: ln(k2/k1) = (Ea/R)(1/T1 − 1/T2) ln 2 = (Ea/8.314)(1/298 − 1/308) 0.693 = (Ea/8.314)(1.089 × 10⁻⁴) Ea = 0.693 × 8.314/1.089 × 10⁻⁴ = 5.29 × 10⁴ J mol⁻¹ ≈ 53 kJ mol⁻¹.
At 303 K: ln(k303/k298) = −(Ea/R)(1/303 − 1/298) = −(6364)(−5.539 × 10⁻⁵) = 0.3525, so k303 = 6.00 × 10⁻³ × e^0.3525 = 6.00 × 10⁻³ × 1.42 = 8.5 × 10⁻³ M⁻¹ s⁻¹.
Comment: doubling over 10 K here is exact only because the data were chosen that way; over a 5 K step the factor is 1.42, so the rule is a rough approximation valid near room temperature for Ea ≈ 50 kJ mol⁻¹. It fails badly at high T or for very large or very small Ea, since the true factor is exp[(Ea/R)(ΔT/T1T2)], which decreases as T rises.