this is a repost, but I have a question.
Examine the reaction mechanism found below:
A + B2 → AB2 (slow)
AB2 + C → AB2C (fast)
AB2C + C → B2C2 + A (fast)
B2C2 → D + E (fast)
a) Write all the reaction intermediates
b) Write all the species that can be considered a catalyst.
c) provide the overall equation for this process.
d) provide the rate law based on this mechanism
"d) The rate law is based on the slow step.
rate = k(B2)" - DrBob222
I have a question about the rate law, I understand that it is based on the slow step, but I don't understand why is it rate = k [B2] instead of rate = k [A] [B2]? Also my teacher said that I needed to show my steps but I don't know how to since this mechanism is based on variable, how do I show my steps?
I did not include the catalyst in my response to you just because I didn't know enough about the reaction. The literature I have proclaims that the catalyst SHOULD be included in the rate law, certainly if it increases the rate of reaction. I think you're right so include it. However, the catalyst is NOT included in the final overall equation.
As for the step problem I'll take a guess that this is what you want but I'm not positive. It might help get you started on the right path if your teacher would give you the first step you need. Anyway, here is my guess.
" A + B2 → AB2 (slow)
AB2 + C → AB2C (fast)
AB2C + C → B2C2 + A (fast)
B2C2 → D + E (fast)"
So I take equation 1 and add it to equation 2 like this:
A + B2 → AB2 (slow)
AB2 + C → AB2C (fast)
----------------------------------------
A + B2 + C ==> AB2C (eqn 1 + eqn 2)
then add in equation 3, then 4, etc until all equations are accounted for. The final equation should be the overall. If this is not what you're teacher is looking for then I'm lost. Note that A is regenerated in another step, it cancels, and is not included in the final equation.