Zero order kinetics - theory query

[Miffymycat]

If equimolar amounts of iodine and propanone are reacted in acidic conditions, the [propanone] vs time graph would be expected to show an exponential decay as expected from first order kinetics. The [iodine] vs time graph vs time must therefore also be an exponential decay in order to obey the stoichiometry. But this is a zero order reaction which is associated with a straight line decay - not a curve - how to rationalise this?

Most school experiments use a large molar excess of propanone in quenching (eg 25x) or colorimetric (100-400x) methods to illustrate the zero order behaviour of iodine. The [iodine] vs time graph indeed shows an apparent straight line indicating zero order behaviour. But this cannot be a true straight line as the reaction stoichiometry is then not obeyed. Surely the consumption of iodine must mirror that of propanone (ie exponential decay) or else the mass of iodine is not conserved at all times throughout the reaction. I presume therefore that the apparent straight line is simply a small section of an exponential curve, reflecting the very small extent of propanone reaction.

Has anyone investigated this reaction on an equimolar basis – if so did it produce an exponential decay for iodine as I suggest above?!

A related concern is the use of these methods to show zero order behaviour. Typically, a time series of iodometric titrations on a single reaction mixture is used to generate a “straight” line, which is therefore subject to my queries above. As argued above, the [iodine] vs time progress curve only appears straight because the other reactants are in excess (= 25x in our school method) and only a small extent of reaction is measured.

It seems it is more valid to use the method where different iodine concentrations are reacted with excess propanone/acid and the iodine absorbance followed, which produces a series of apparently parallel “straight” lines, again for the reasons outline above. This initial rates method is arguably better evidence to establish the kinetics, as several initial concentration runs are employed compared to a single run ie that simply more data is collected. With the levels of excess propanone/acid used in typical school procedures (100, 200, 300, 400x), this again equates to pseudo-zero order conditions in propanone/acid, and any effect on rate of changing [iodine] should be apparent (approx 4x, 3x, 2x based on these excesses). When my students conducted these experiments the lines were parallel within a 25% spread, which is reasonably good.
I suppose my concern is that teaching a [ ] vs time straight line is diagnostic for zero order is incorrect. It cannot be true for equimolar reactants and apparent linearity needs explaining when used in practice with pseudo-zero order co-reactants. For other than zero order behaviour, then graphing a single progress curve I’m sure is a more effective method.

So my conclusion is that a zero order reactant concentration cannot have a linear decay over time - an apparently linear fit must be due to an excess in other reactants. The only time a zero order reactant progress curve can be truly linear is in the case of catalysis by eg enzymes or metals, where active sites determine rate, which will be truly zero order when saturated.

Does anyone agree? Does anyone care?!


 

[Oxfordstudent]

The [iodine] vs time graph vs time must therefore also be an exponential decay in order to obey the stoichiometry.

This is only the case if the reaction is elementary: i.e. if there are no intermediate steps.

In reality the kinetics of this reaction are complex, due to a multi-step mechanism: see http://www.docbrown.info/page15/mech47.gif.

 

[Miffymycat]

Thanks for the post. However, I don't see why the presence of more steps makes any difference. In this case, the keto-enol tautomer only begins to be consumed at step 4 which is the same point at which iodine starts to be consumed. The propanone decay curve will mirror the formation of iodopropanone - NOT the enol tautomer - and so the iodine decay must mirror this ... n'est-ce pas?!

 

[Oxfordstudent]

the propanone decay curve will mirror the formation of iodopropanone - NOT the enol tautomer - and so the iodine decay must mirror this ... n'est-ce pas?!

No, the enolization is the rate determining step in these reactions.

 

[Miffymycat]

I agree, the enolisation is indeed the RDS, but it is a reactive intermediate which does not build up in concentration, and so the consumption of propanone is observable until its enol form reacts with iodine! The formation of the enol is not consuming the propanone - it's an equilibrium. H+ displaces the equilibrium to the right and in the presence of electrophilic I2 is then converted to product. So I2 and propanone must decay together at the same instantaneous rate ...!

Are there any kineticists out there who can pass judgement?!

 

[Ygggdrasil]

For any bimolecular (or other reaction involving more than two reactant species), the shape of the [substrate] vs time plot will always depend on the relative concentrations of the reactants. In kinetic studies, it is standard practice to have one reactant in excess, then to vary the amount of the other reactant in order to investigate the effect of changing that reactant's concentration on the overall kinetics of the reaction.

You are correct that the decay curve for iodine will be linear only when there is excess acetone. When [acetone] is comparable to [I₂], you will see the rate of iodine consumption decrease with time because the concentration of acetone is decreasing which decreases the overall reaction rate.

 

[Miffymycat]

Thank you very much! It's relief to know I was on the right track!

error in my earlier post ... "so the consumption of propanone is NOT observable until its enol form reacts with iodine"