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Question

Study the mechanism of the following reversible reaction carefully
2A + 2B A2B2
Mechanism A + B Kf1Kb1 AB
AB + B Kf2Kb2 AB2 (slowest step)
AB2 + A Kf3Kb3 A2B2
Rate constant for reaction is:

A
k=kf1kb1×kf2
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B
k=kf1kb1×kb3kb3
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C
k=kf1kb1×kb2
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D
k=kf1kf2×kb3kb2
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Solution

The correct option is C k=kf1kb1×kf2
You have to use a steady state assumption for this mechanism
for the AB2 transition state:
ϑformation(AB2)=ϑanhillation(AB2)
kf2[AB][B]=kb2[AB2]+kf3[AB2][A]+kb3[A2B2](1)
now for the transition state AB:
kf1[A][B]+kb2[AB2]=kb1[AB2]kf2[AB][B](2)
Now, due to (1) and (2) being a linear system of equations, we solve for [AB] and [AB2]. Substituting into final stop:
k=kf1kf2kb1 and ϑ=k[A][B]

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