Rates of reaction - kinetics (A2)

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I would like to know how you can determine the rate determining step from the rate equation?
060608

Corrrie writes ....
 
The reactant species (molecules or ions) that appear in the rate equation are the ones that are involved in the rate-determining (slowest) step - that's why they affect the rate of the reaction in the way they do! Knowing which species these are from the rate equation, you can attempt to write possible steps in the mechanism of the reaction which are consistent with this.
 
For instance, if in the reaction between a haloalkane, RX, and hydroxide ions the rate equation was found to be
 
rate = k[RX]
 
then the rate-determining step must involve only RX.
 
The steps in the mechanism for the reaction RX + OH -> ROH + X^- could therefore be:
 
1. RX -> R⁺ + X^- (Slow - rate-determining, only RX involved)
 
2. R⁺ + OH^- -> ROH (Fast, OH^- does not appear in rate equation)
 
On the other hand if the rate equation was found to be
 
rate = k[RX][OH^-]
 
then the rate-determining step must involve both RX and OH^-, and the mechanism could be
 
RX + OH^- -> ROH + X^- (Only step, and therefore rate-determining step)
 
In some cases the picture can be more complicated. You'll notice I said 'involved in the rate-determining step' and not 'in the rate-determining step'. This may sound like splitting hairs to you, but let me explain with an example.
 
The reaction BrO^- + 5Br^- + 6H⁺ -> 3Br₂ + 3H₂O has a rate equation in which is second order in [H⁺] and first order in the other two reactant species.
 
It would be difficult for an average A-level candidate to come up with a set of steps in the mechanism that would be consistent with this! However, given a possible a set of steps, it should be easier to pick out the rate-determining one. Let's say possible steps are:
 
1. H⁺ + Br^- -> HBr (Fast)
2. H⁺ + BrO₃^- -> HBrO₃ (Fast)
3. HBr + HBrO₃ -> HBrO + HBrO₂ (Slow)
4. HBrO₂ + HBr -> 2HBrO (Fast)
5. HBrO + HBr -> H₂O + Br₂ (Fast)
 
How does 3 being the slowest, i.e. rate-determining, step fit with the rate equation? Well, if Step 1 is fast, then [HBr] depends on [H⁺] and [Br^-]. So the slow step (3) in turn depends on [H⁺] and [Br^-]. Similarly, if Step 2 is fast, then [HBrO₃] depends on [H⁺] and [BrO₃^-].
 
Doubling [H⁺] will double the amount of HBr and of HBrO₃ going into Step 3. Step 3 (the slowest step) will therefore be 4x faster as a result. So, the rate equation will show that the reaction is second order with respect to [H⁺].
 

 

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updated: 08 June 2008