Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: demonat0r on February 05, 2009, 07:16:59 PM
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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?
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(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 Kf/Kr?
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i know that...
kf / kr = K eq
i don't see how K eq just disappears though.
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Ok, Keq x k2 = k because constant times constant gives you a constant.
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oh ok so basically:
k(from an elementary step) x K eq = k (overall reaction)
right?
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You're correct ;D