Propanone reacts with iodine in alkaline conditions:
CH3COCH3+I2+OH−→CH3COCH2I+I−+H2O \text{CH}_3\text{COCH}_3 + \text{I}_2 + \text{OH}^- \rightarrow \text{CH}_3\text{COCH}_2\text{I} + \text{I}^- + \text{H}_2\text{O} CH3COCH3+I2+OH−→CH3COCH2I+I−+H2OThe rate equation for this reaction is:
Rate=k[CH3COCH3][OH−] \text{Rate} = k [\text{CH}_3\text{COCH}_3][\text{OH}^-] Rate=k[CH3COCH3][OH−]Sketch or describe a graph to show how, at constant temperature, the concentration of iodine changes during this reaction if the other reactants are not in large excess. Explain your answer.
Table 1 shows the initial rate of this reaction for three experiments using different mixtures containing propanone, iodine, and hydroxide ions.
Table 1
| Experiment | [CH3COCH3][\text{CH}_3\text{COCH}_3][CH3COCH3] / mol dm−3\text{mol dm}^{-3}mol dm−3 | [I2][\text{I}_2][I2] / mol dm−3\text{mol dm}^{-3}mol dm−3 | [OH−][\text{OH}^-][OH−] / mol dm−3\text{mol dm}^{-3}mol dm−3 | Initial rate / mol dm−3 s−1\text{mol dm}^{-3}\text{ s}^{-1}mol dm−3 s−1 |
|---|---|---|---|---|
| 1 | 2.00 × 10-2 | 3.00 × 10-2 | 2.00 × 10-2 | 3.60 × 10-11 |
| 2 | 2.00 × 10-2 | 3.00 × 10-2 | [ i ] | 1.44 × 10-10 |
| 3 | 5.00 × 10-3 | 6.00 × 10-2 | 6.00 × 10-2 | [ ii ] |
Complete Table 1 by calculating the missing values [ i ] and [ ii ].
Use the data from Experiment 1 to calculate the rate constant k k\,k for this reaction and state its units.
A mechanism is proposed where Step 1 is the reaction of propanone with hydroxide ions to form water and an enolate intermediate, and Step 2 is the reaction of the enolate intermediate with iodine. Use evidence from the rate equation to explain why Step 1 is the rate-determining step.