Reaction Kinetics (4) Second Order Kinetics: The Iodination of propanone

In my previous blog I showed you that the order of this reaction with respect to Iodine was zero.

Iodine has no effect on the rate of this reaction. 

It takes part in the reaction but the rate determining steps or the slower steps do not involve it. 

So how can we be sure that the rate determining steps do involve both propanone and hydrogen ions?

Well we can use a clock reaction method to demonstrate this.

Here is a method that is tried and tested and works providing you use very clean apparatus and freshly made up solutions that day. 

I have tried it myself and provide results that I obtained in carrying out this experiment in the lab.

Here is the spec for the practical or as you Americans will call it the Lab.

A “Clock” method to determine the rate equation for the iodination of propanone.

This is a simple method for obtaining the order with respect to each species in the reaction mix.  In this experiment, a yellow colour disappears (fairly abruptly) when all the iodine initially present has been consumed.

For each experiment, put the suggested volume of 0.005M iodine solution (V_b) into a boiling tube (use a burette), and add the suggested volume of 2M propanone solution (V_a), 1M sulphuric acid (V_c) and water (V_d) in a conical flask, all from burettes.  Be very precise in the way you measure the solutions.  Add the iodine solution to the conical flask quickly and start a stop-clock.  Note the time t at which the last trace of yellow colour disappears; you should be able to judge this time to within about 5s.

Results table

Mix	Va /cm3	Vb /cm3	Vc /cm3	Vd /cm3	[CH3COCH3] mmol.dm-3	[I2] mmol.dm-3	[H+] mmol.dm-3	Time t /s	Rate 1/t /s-1
1.	4	2	4	20	260	0.33	260	189	0.0053
2.	8	2	4	16	520	0.33	260	87	0.011
3.	16	2	4	8	1040	0.33	260	50	0.020
4.	8	4	4	14	520	0.66	260	173	0.005
5.	8	1	4	17	520	0.16	260	49	0.02
6.	8	2	8	12	520	0.33	520	49	0.02
7.	8	2	16	4	520	0.33	1040	24	0.042

Questions

1.  Look at the results for experiments 2, 6, and 7 and decide how the rate of reaction depends on [H⁺].

2.  Look at the results for experiments 1, 2and 3 and decide how the rate of reaction depends on [CH₃COCH₃].

3.  Look at the results for experiments 2, 4 and 5 and decide how the rate depends on [I₂], does this result confirm what you learned from Part 1 of this experiment?

4.  Construct an overall rate equation for the reaction between iodine and propanone. 

5.            What does this rate equation suggest about the mechanism for the reaction?

The results for experiments 2, 6 and 7 show that the reaction rate is proportional to the concentration of H+ ions. 

That is first order with respect to [H⁺].

The results for experiments 1,2 and 3 show the reaction rate is proportional to the concentration of propanone. 

That is first order with respect to [CH₃COCH₃].

We do not see this pattern repeated in the case of iodine.

So we can write an overall rate equation as

Rate  =  k [CH₃COCH₃]¹. [H⁺]¹.[I₂]⁰

The overall order is 2: it is a second order reaction. 

We see that the rate determining step or the slowest step in the mechanism involves both species:  H+ and CH₃COCH₃

As we can see from this mechanism step 1 involves the protonation of the carbonyl group in propanone a fast step.

But it is the second step which is slow i.e. the conversion of the keto form of propanone into the enol form and release of the proton.

It is this second step which shows the use of H+ to be catalytic. 

Both H+ ions and propanone are involved in these two overall slow steps hence the reaction depends on the concentration of each species. 

Well if you try this experiment/lab let’s hope your results also demonstrate this second order mechanism.