Kinetics of the Propanone–Iodine Reaction
A student investigates the acid-catalysed reaction between propanone and iodine in aqueous solution: CH3COCH3 + I2 → CH3COCH2I + HI. The reaction is followed by measuring the absorbance of iodine in a colorimeter. In experiment 1, the initial concentrations are [propanone] = 0.50 mol dm-3, [H+] = 0.10 mol dm-3, [I2] = 0.010 mol dm-3, and the initial rate is 2.0 × 10-5 mol dm-3 s-1. In experiment 2, [propanone] is doubled while the other concentrations are unchanged; the initial rate is 4.0 × 10-5. In experiment 3, [H+] is doubled; the initial rate is 4.0 × 10-5. In experiment 4, [I2] is doubled; the initial rate remains 2.0 × 10-5. At 308 K, with all concentrations unchanged from experiment 1, the initial rate is 1.6 × 10-4 mol dm-3 s-1. The teacher asks for the rate equation, the rate constant with units, and a mechanistic justification.
1. Deduce the rate equation and calculate the rate constant, including its units.
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Doubling [propanone] doubles the rate, so the reaction is first order with respect to propanone. Doubling [H+] doubles the rate, so it is first order with respect to H+. Doubling [I2] has no effect, so it is zero order with respect to iodine. The rate equation is rate = k[CH3COCH3][H+]. Using experiment 1: k = rate / ([CH3COCH3][H+]) = 2.0 × 10^-5 / (0.50 × 0.10) = 4.0 × 10^-4. Units: mol dm^-3 s-1 divided by (mol dm^-3)^2 gives dm3 mol-1 s-1. So k = 4.0 × 10^-4 dm3 mol-1 s-1.
2. Explain why a zero order with respect to iodine is consistent with a rate-determining enolisation step.
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The accepted mechanism involves acid-catalysed enolisation of propanone. A proton adds to the carbonyl oxygen, and the slow step is formation of the enol, accompanied by removal of an alpha-hydrogen. The enol then reacts rapidly with iodine. Because iodine is consumed in a fast step after the rate-determining step, its concentration does not appear in the rate equation. The rate depends only on the concentrations of propanone and acid catalyst. This explains the first-order dependence on both propanone and H+ and the zero-order dependence on I2.
3. Use the rates at 298 K and 308 K to calculate the activation energy. State any assumption.
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At 298 K, k1 corresponds to rate 2.0 × 10^-5. At 308 K, k2 corresponds to 1.6 × 10^-4. The concentration terms are unchanged, so k2/k1 = 1.6 × 10^-4 / 2.0 × 10^-5 = 8.0. Using ln(k2/k1) = Ea/R (1/T1 - 1/T2): ln 8.0 = 2.079. 1/298 - 1/308 = 1.089 × 10^-4 K^-1. Ea = 8.314 × 2.079 / 1.089 × 10^-4 = 1.59 × 10^5 J mol-1 = 159 kJ mol-1. Assumption: the rate constant ratio equals the initial rate ratio because all concentrations are the same, and the reaction order does not change over this temperature range.