Study the mechanism of the following reversible reaction carefully 2A+2B⇌A2B2 Mechanism A+BKf1⇌Kb1AB AB+BKf2⇌Kb2AB2 (slowest step) AB2+AKf3⇌Kb3A2B2 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 Ck=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: