Question

A hypothetical reaction $A_2+B_2\to 2AB$ follows the mechanism as given below:
$A_2\rightleftharpoons A+A$ (fast)
$A+B_2\to AB+B$ (slow)
A + B $\to$ AB (fast)
The order of the overall reaction is:

• A. 2
• B. 1
• C. 1.5
• D. 0

Answer 1

Answer: (C)

1.5

Since (A) is the intermediate reactive species whose concentration is determined from equilibrium step.
Slow step is:
A + $B_2\to AB+B$ (slow)
r = k[A]$\left[B_2\right]$
From equilibrium step
$A_2\rightleftharpoons A+A\left(fast\right)$
$k_{eq}=\frac{[A{]}^2}{\left[A_2\right]}$
$\therefore \left[A\right]=\left(k_{eq}\right[A_2{]}^{1/2}$
Substitute the value of [A] in equation (i),
$r=k.k_{eq}{}^{1/2}\left[A_2{]}^{1/2}\right[B_2]$
Thus, order of reaction = $\frac{1}{2}+1=1\frac{1}{2}$