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SN1 Reaction Mechanism

Study the mec...

Question

Study the mechanism of the following reversible reaction carefully
$2 A + 2 B ⇌ A_{2} B_{2}$
Mechanism $A + B_{f_{1}}$
$A B + B_{f_{2}}$ (slowest step)
$A B_{2} + A_{f_{3}}$
Rate constant for reaction is:

A

k = \frac{k_{f_{1}}}{k_{b_{1}}} \times k_{f_{2}}

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B

k = \frac{k_{f_{1}}}{k_{b_{1}}} \times \frac{k_{b_{3}}}{k_{b_{3}}}

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C

k = \frac{k_{f_{1}}}{k_{b_{1}}} \times k_{b_{2}}

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D

k = \frac{k_{f_{1}}}{k_{f_{2}}} \times \frac{k_{b_{3}}}{k_{b_{2}}}

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Solution

The correct option is C $k = \frac{k_{f_{1}}}{k_{b_{1}}} \times k_{f_{2}}$
You have to use a steady state assumption for this mechanism
for the $A B_{2}$ transition state:
$ϑ_{f o r m a t i o n} (A B_{2}) = ϑ_{a n h i l l a t i o n} (A B_{2})$
$k_{f_{2}} [A B] [B] = k_{b_{2}} [A B_{2}] + k_{f_{3}} [A B_{2}] [A] + k_{b_{3}} [A_{2} B_{2}] \to (1)$
now for the transition state AB:
$k_{f_{1}} [A] [B] + k_{b_{2}} [A B_{2}] = k_{b_{1}} [A B_{2}] - k_{f_{2}} [A B] [B] \to (2)$
Now, due to (1) and (2) being a linear system of equations, we solve for $[A B]$ and $[A B_{2}]$ . Substituting into final stop:
$k = \frac{k_{f_{1}} k_{f_{2}}}{k_{b_{1}}}$ and $ϑ = k [A] [B]$

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