Q. The following mechanism has been proposed for the reaction of NO with $Br_{2}$ to form NOBr $NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g)$ $NOBr_{2}(g) + NO(g) \rightarrow 2NOBr(g)$ If the second step is the rate determining step, the order of the reaction with respect to NO(g) is
A) 1
B) 0
C) 3
D) 2
Explanation
(d)
$Rate = k\left\lbrack NOBr_{2} \right\rbrack\lbrack NO\rbrack$ …(i)
But $NOBr_{2}$is in equilibrium
$k_{eq} = \frac{\left\lbrack NOBr_{2} \right\rbrack}{\lbrack NO\rbrack\lbrack Br_{2}\rbrack}$
$\left\lbrack NOBr_{2} \right\rbrack = k_{eq}\lbrack NO\rbrack\lbrack Br_{2}\rbrack$ …(ii)
Putting the $\left\lbrack NOBr_{2} \right\rbrack\ in\ (i)$
$rate = k.k_{eq}\lbrack NO\rbrack\lbrack Br_{2}\rbrack\lbrack NO\rbrack$
Hence, $rate = k.k_{eq}\lbrack NO\rbrack^{2}\lbrack Br_{2}\rbrack$
$rate = k'\lbrack NO\rbrack^{2}\lbrack Br_{2}\rbrack$
where, k’.Keq
the order, of reaction with respect to NO(g) is 2
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.