[Roy16]

Hi, I am currently studying on the iodination of acetone. I have discovered that acetone and H⁺ both have an order of reaction of 1 while I₂ has a zero order. This results in a raw law of: k[acetone]¹[H⁺]¹. I have attached an image for reference.

However, given a reaction mechanism (attached below), the 2nd elementary step was known to be the rate determining step (rds). Since it contains an intermediate, I substituted it with [acetone] and [H⁺] (derived from the first step). But, this leaves me with a rate law of: k[acetone][H⁺][H₂O], which is not equal to the experimentally determined rate law.

Is water meant to be in the rate law? Or is there some misconception that I'm having?

[Orcio_87]

I think that water can be removed from equation. After all: H₃O⁺ = H₂O + H⁺.

But how do you get that rate reaction for I₂ is 0

[Borek]

Typically in water solutions water has a constant concentration and as such can be included in the k coefficient.

It is also rarely listed in equilibrium formulas (like K_w=[OH^-][H⁺] - technically it should have [H₂O] in the denominator).

[Orcio_87]

@Borek In this reaction concentration of water can be very different (acetone and water are completely soluble), so it cannot be taken as constant.

In my opinion reaction's speed will depend on concentration of H⁺, acetone and I₂, but i will think about it tomorrow.

[Borek]

Yes, they are miscible, but whether concentration of the water can be treated as constant or not depends on the range of concentrations covered. It is quite typical to do such experiments testing just concentrations of substances up to 1M in water solutions, and then the dependence between water concentration and kinetics just doesn't show.

We don't know enough about the experiment to rule that out.

[Orcio_87]

In my opinion:

(1) CH₃-CO-CH₃ + H⁺  [CH₃-COH-CH₃]⁺

(2) [CH₃-COH-CH₃]⁺  + H₂O CH₃-C(OH)=CH₂ + H₃O⁺

(3) [CH₃-COH-CH₃]⁺  CH₃-CO-CH₃ CH₃-C(OH)=CH₂ + [CH₃-COH-CH₃]⁺

(reactions 2 and 3, are competive each other, but 3 is not important when concentration of acetone is low)

(4) CH₃-C(OH)=CH₂ + I₂  CH₃-CO-CH₂I + HI


So speed of overall reaction:

(5) CH₃-CO-CH₃ + I₂ CH₃-CO-CH₂I + HI

Will be:

V = k₁ [CH₃-CO-CH₃] x [H⁺] x [k₂[H₂O] + k₃[CH₃-CO-CH₃]] x k₄ [I₂]

But..

- if reaction is carried in acetone as solvent then k₂[H₂O] = 0 and overall rate will be:

V = K [CH₃-CO-CH₃]² x [H⁺] x [I₂]

(K = k₁ x k₃ x k₄)

- if reaction is carried in water as solvent, and concentration of acetone is low then:

V = K [CH₃-CO-CH₃] x [H⁺] x [I₂]

(K = k₁ x k₂[H₂O] x k₃ x k₄   but k₂[H₂O] is taken as constant)

- if reaction is carried in mixed (water acetone solution) then:

V = K [CH₃-CO-CH₃]^₂ x [H⁺] x [H₂O] x [I₂]

(K = k₁ x k₂ x k₃ x k₄)