reaction rate and elementary reactions

[Farcus]

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if you are given a reaction broken down into its elementary reactions for example

I2 -> 2I (fast,equil)
I + H2 -> H2I (fast,equil)
H2I + I -> 2HI (slow)

and you're asked to find the rate of reaction then we all know its the slowest step. However the slowest step contains a reaction intermediates so am I suppose to count the composition of the intermediate? in which case would e I+H2, however I itself is another intermediate of the first reaction so then I'll need to exchange that for I2. So would this be why the rate of reaction is r=k[I2][H2]? Because none of it is the intermediate?

So basically if you have a slow step and it contains the intermediate you don't use the intermediate to formulate the reaction rate but the steps used to get the intermediate right?

 

[Craig McDermott]

Kind of. When a slow step comes after a fast step, you have to consider the concentration of the intermediates. To tackle this kind of rate problem here is the strategy:

1. Look at the SLOW step - ignore everything that comes after.
2. Set the rate like you normally would: k[reactants]. in this case, k[H2I]
3. Now check which reactants are produced in previous (fast) steps. In this problem both are. Lets tackle [H2I] first.
4. how do we get [H2I]? well we know since its fast step, and in equilibrium, the forward rate equals the reverse rate. so k1[H2] = k-1[H2I]. divide both sides by k-1 and we have [H2I]=(k1/k-1)[H2].
5. Plug this back into the equation you got in part 2: its now (k)(k1/k-1)[H2], or putting the 's together, (k)(k1/k-1)^2[H2].
6. Now, referring back to step 3, we already got [H2I] but remember was also an intermediate produced in a previous step. So we have to get the concentration for as well.
7. k2[I2] = k-2^2. Divide both sides by k-2 and you have ^2 = (k2/k-2)[I2]
8. Plug this back into the equation from step 5. You get (k)(k1/k-1)(k2/k-2)[H2][I2]. Simplify all the little k's in front (lets just call it big K) and your final answer is:
rate=K[I2][H2]

simple plug and chug. hopefully this helps