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Topic: rate of reaction and equilibrium  (Read 3384 times)

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Offline khwcm

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rate of reaction and equilibrium
« on: October 01, 2010, 11:37:47 AM »
for a elementary reaction,
    K = kf/kb

however, for a non-elementary reaction, say Sn1:   
    A + B  :rarrow: C + D
such that
    A         :rarrow:  A* + D
    A* + B  :rarrow:  C
so is it possible to write a equlilibrium constant in terms of rate constant? thanks alot

Offline Juan R.

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Re: rate of reaction and equilibrium
« Reply #1 on: October 01, 2010, 01:19:19 PM »
And where is the equilibrium in your Sn1?
The first canonical scientist.

Offline khwcm

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Re: rate of reaction and equilibrium
« Reply #2 on: October 02, 2010, 12:03:28 AM »
how about euilibrium exist in both steps?

Offline DevaDevil

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Re: rate of reaction and equilibrium
« Reply #3 on: October 11, 2010, 06:31:58 AM »
so I assume you ask the following:

in:   A + B  ::equil:: C + D;  K = kf/kb

then what is k in?:
1)     A  ::equil::  A* + D
2)     A* + B  ::equil::  C

am I right in that assumption? ALL equilibria?

then, since you have 2 reactions, you will have 2 reaction equilibria there:
K1 = kf1/kb1
and
K2 = kf2/ kb2

if you want this into one equation you have to rewrite one of them:

K1 = [A*] [D] / [A]
K2 = [C] / [A*] [B ]

for Koverall goes: KSn1 = [C] [D] / [A] [B ] = K1 * K2

so, yes, if they are both equilibria you can combine them into one equilibrium constant by multiplying the individual constants

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