Reaction Kinetics (3) Zero Order Kinetics: The iodination of propanone

So far in these blogs on reaction kinetics I’ve been discussing first order reaction kinetics where the rate of the reaction is directly proportional to the concentration of one of the reacting species.

The graph of rate vs concentration looks like this:

And the graph of concentration against time looks like this:

You will note the constant half life for the reaction.

This is an example of an exponential decay curve.

But there are other mathematical models that fit the behaviour of species in reactions.

For example: how can we have a rate equation in which the power of the concentration term is zero?  A zero order reaction?

The general rate equation would look like this:

Rate of reaction =  k [X]⁰

Let’s think of the implications of this mathematical model.

First, any value raised to the power zero equals one.

e.g. 10⁰  =  1 or  (246.55)⁰  =  1  etc…

Second, this means that the rate  =  k, the rate constant. 

Third, the implication must be that the species X in the above rate equation has no effect on the reaction rate because [X]⁰  =  1

So the question remains to ask whether it is possible for a reactant to have no effect on the reaction rate?

What we need to do is see if there are any specific examples of reaction where a reactant has no effect on the reaction rate.

There are of course some classic examples of this phenomena that have been studied to death to illustrate this point.

Here is the example that is often used:

Investigating the effect of the concentration of iodine on the rate of the iodination of propanone.

The acid catalysed iodination of propanone (acetone)

The reaction is:

CH₃COCH₃(aq)   +  I₂(aq)     =      CH₂ICOCH₃(aq)   +   H⁺(aq)  +    I^-(aq) 

The product is a powerful lachrymator, and even the very small vapour pressure of it above the solution will make your eyes smart, unless you are careful to keep the reaction vessel securely corked except when you are actually withdrawing samples for titration.

There are two different ways of studying the rate of reaction.

We’ll look first at the titration and sampling method:

Titration method

Order with respect to iodine:

In one 250ml flask, A, put 50cm³ of a solution of 0.02M iodine made up in 0.2M KI solution.  In another 250ml flask, B, put a mixture of 25cm³ of M H₂SO₄ and 25 cm³ M propanone (in aqueous solution).  In each of several 100ml conical flasks, put 10cm³ of 0.5M NaHCO₃.

Pour the contents of flask A into flask B while starting a stop-clock, and mix well.  After one minute withdraw 10cm³ of the mixture with a pipette, run it into one of the portions of sodium hydrogen carbonate, and titrate the iodine present with 0.01M Na₂S₂O₃ and 10 drops starch solution.  At ten minute intervals, withdraw further 10cm³ portions and titrate as before.  Complete four such determinations and enter your data in the table below.

Results table

Expt	Time (t) /s	Thiosulphate Titre /cm3
1.
2.
3.
4.
5.

Questions

1.  Plot the thiosulphate titre against time.

2.  From your graph, deduce the order of reaction with respect to iodine.

3.  What is the function of the sodium hydrogen carbonate solution?

4.  Why cannot sodium hydroxide be used instead?

5.  What do these results suggest about the mechanism of the reaction between propanone and iodine|?

Conclusion and analysis

What you should find is that the titres plotted against time give straight line like this:

 

The thiosulphate titre is a measure of the concentration of iodine in the reaction mixture at any one time t.

But the [I₂] falls at a constant rate during the reaction.

Interestingly, if we double the concentration of propanone used then the slope of this graph is doubled too.  

For the reaction to be first order with respect to iodine the graph would have to have that exponential decay type curve. 

But it is a straight line telling us that the [I₂] has no effect on the rate of this reaction.

The reaction is zero order with respect to iodine. 

The implication is that the rate is controlled by other species in the reaction and not iodine.

But if the reaction involves iodine then that implies there are at least two steps in the reaction: one not involving iodine and one that does: this then is evidence of a reaction mechanism.

Further rates studies show the reaction mechanism to be:

And it takes place in three steps the first two steps being the slower and therefore rate controlling.  Iodine does not take part in these step but in the third and  fastest step so iodine is not involved controlling the reaction rate.

The function of the sodium hydrogen carbonate solution is to quench the reaction mixture.

Once the reaction mixture is pipetted into the conical flask, the sodium hydrogen carbonate neutralises the acid in the reaction mixture and prevents steps 1 and 2 of the mechanism taking place, stopping the reaction at that point.

Why not use sodium hydroxide solution then.  Well sodium hydroxide solution reacts with the iodine in the reaction mixture. 

NaOH   +  I₂   =    NaI   +   NaOI    +  H₂O

Clearly this renders the results for the concentration of iodine invalid.