Iodination of Propanone: Orders and Activation Energy
In a school laboratory, a student studies the acid-catalysed iodination of propanone at 25 °C by timing how long the iodine colour takes to disappear. The reaction mixture is made up so that propanone and HCl are in large excess compared with iodine. The results are: run 1, [propanone] = 0.100 M, [H+] = 0.100 M, [I2] = 0.0100 M, time = 40 s; run 2, [propanone] = 0.200 M, other concentrations as run 1, time = 20 s; run 3, [H+] = 0.200 M, others as run 1, time = 20 s; run 4, [I2] = 0.0200 M, others as run 1, time = 80 s; run 5, same concentrations as run 1 but at 35 °C, time = 17 s. The student writes the rate law as rate = k[propanone][H+][I2]. The teacher points out that the stoichiometric equation has iodine as a reactant and asks for a kinetic analysis, including the activation energy from the two temperatures.
1. Use runs 1–4 to determine the order with respect to propanone, H+, and I2. Write the correct rate law and correct the student’s claim.
Read the worked answer
Because iodine is the species whose colour is timed, a convenient rate measure is rate = [I2]0/t, provided the reaction is zero order in iodine. Check that consistency first.
Run 1: rate = 0.0100/40 = 2.50 × 10−4 M s−1. Run 2: rate = 0.0100/20 = 5.00 × 10−4 M s−1. Run 3: rate = 0.0100/20 = 5.00 × 10−4 M s−1. Run 4: rate = 0.0200/80 = 2.50 × 10−4 M s−1.
Comparing run 1 and run 2: [propanone] doubles while [H+] and [I2] are constant; rate doubles. Order with respect to propanone = 1.
Comparing run 1 and run 3: [H+] doubles; rate doubles. Order with respect to H+ = 1.
Comparing run 1 and run 4: [I2] doubles; rate is unchanged. Order with respect to I2 = 0.
The correct rate law is therefore:
rate = k[CH3COCH3][H+],
with overall order 2. The student’s inclusion of [I2] is wrong: iodine appears in the stoichiometric equation but not in the rate law. The reaction is zero order in iodine because iodine is involved after the slow, rate-determining enolisation step; changing [I2] changes how long the colour lasts but not the rate of the rate-determining step.
2. Calculate the rate constant at 25 °C, with units, and explain why the time for colour disappearance can be used as a rate measurement in this experiment.
Read the worked answer
Using the rate law rate = k[propanone][H+] and run 1:
2.50 × 10−4 M s−1 = k × (0.100 M)(0.100 M).
k = 2.50 × 10−4 / (1.00 × 10−2) = 2.50 × 10−2 M−1 s−1.
The units follow from rate/(concentration × concentration): (mol L−1 s−1)/(mol2 L−2) = L mol−1 s−1 = M−1 s−1. A check with run 2 gives the same value: 5.00 × 10−4/(0.200 × 0.100) = 2.50 × 10−2 M−1 s−1.
The time for colour disappearance is valid here because the reaction is zero order in iodine. If the rate is independent of [I2], then over the interval monitored the rate is effectively constant, and the time taken to consume a fixed initial amount of iodine is inversely proportional to the rate. That is why doubling [I2] doubles the time (run 4) without changing the calculated rate. If the reaction were first order in iodine, the rate would fall as iodine was consumed, and a simple [I2]0/t expression would not be a true initial rate; the colour-disappearance time would then need a first-order integrated-rate treatment.
3. Calculate the activation energy from the 25 °C and 35 °C runs. State one experimental reason why the 35 °C result may be less reliable.
Read the worked answer
From run 1 at 25 °C, k1 = 2.50 × 10−2 M−1 s−1.
At 35 °C, rate = 0.0100/17 = 5.88 × 10−4 M s−1. k2 = 5.88 × 10−4/(0.100 × 0.100) = 5.88 × 10−2 M−1 s−1.
Use the Arrhenius equation in the form:
ln(k2/k1) = (Ea/R)(1/T1 − 1/T2).
k2/k1 = 5.88 × 10−2 / 2.50 × 10−2 = 2.352; ln(2.352) = 0.855.
T1 = 298 K, T2 = 308 K. 1/T1 − 1/T2 = 0.003356 − 0.003247 = 1.09 × 10−4 K−1.
Ea = (0.855 × 8.314 J mol−1 K−1) / (1.09 × 10−4 K−1) = 6.5 × 10^4 J mol−1, i.e. about 65 kJ mol−1.
One reason the 35 °C result may be less reliable is that the reaction is fast: only 17 s is available, so errors in starting/stopping the timer, in mixing, and in thermal equilibration matter more. In addition, iodine is volatile and may escape from the warm solution, and the water bath may not have held the mixture exactly at 35 °C. A single temperature pair gives only an approximate Ea; several temperatures and a plot of ln k against 1/T would be better.