[demonat0r]

1. A + A <~~> X     (fast)
2. X + B ~~> C + Y (slow)
3. Y + B ~~> D       (fast)
Rate = k[A]^2[ B]

the slowest step is step 2 so...rate = k(2)[X][ B]
step 1 is an equilibrium so...k(f)[A][A] = k(r)[X]
solve for x...[X] = (k(f) / k(r)) * [A]^2
substitute...Rate = k(2) * (k(f) / k(r)) [A]^2[ B]

then it becomes Rate = k[A]^2 [ B]

that is how it was presented in my book. what i don't get is how the terms cancel out after you substitute. where the the k(f) / k(r) go? by the way for k(2), the 2 is a subscript. can someone please explain how you get to the last step after substituting?
(the rate should be k times [A] squared times the concentration of B. sry for the confusion but whenever i type the concentration of B with the brackets, it just won't show up for some reason.)

also a quick question. for a reaction mechanism, is the reaction with the higest activation energy also the slowest step?

[Astrokel]

(the rate should be k times [A] squared times the concentration of B. sry for the confusion but whenever i type the concentration of B with the brackets, it just won't show up for some reason.)

Lol, use [B ] with a space.

that is how it was presented in my book. what i don't get is how the terms cancel out after you substitute. where the the k(f) / k(r) go? by the way for k(2), the 2 is a subscript.

What is essentially K_f/K_r?

[demonat0r]

i know that...
kf / kr = K eq

i don't see how K eq just disappears though.

[Astrokel]

Ok, K_eq x k₂ = k because constant times constant gives you a constant.

[demonat0r]

oh ok so basically:
k(from an elementary step) x K eq = k (overall reaction)
right?

[Astrokel]

You're correct