Question

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

• A. $2$
• B. $1$
• C. $3/2$
• D. Zero

Answer 1

Answer: (C)

$3/2$

As rate depends upon the slow step,
$Rate=k\left[A\right]\left[B_2\right]$
For the equilibrium, $A_2\rightleftharpoons 2A,K_c=\frac{[A{]}^2}{\left[A_2\right]}$

or $\left[A\right]=\sqrt{K_c\left[A_2\right]}$

$\therefore Rate=k{K}_c^{1/2}\left[A_2{]}^{1/2}\right[B_2]=k^f[A_2{]}^{1/2}\left[B_2\right]$

Hence, $order=\frac{1}{2}+1=\frac{3}{2}$.