The following mechanism has been proposed for the reaction of $$NO$$ with $$B{r_2}$$ to form $$NOBr$$ :
$$\eqalign{
& NO\left( g \right) + B{r_2}\left( g \right) \rightleftharpoons NOB{r_2}\left( g \right) \cr
& NOB{r_2}\left( g \right) + NO\left( g \right) \to 2NOBr\left( g \right) \cr} $$
If the second step is the rate determining step, the order of the reaction with respect to $$NO\left( g \right)$$ is
A.
3
B.
2
C.
1
D.
0
Answer :
2
Solution :
$$\eqalign{
& {\text{(i)}}\,NO\left( g \right) + B{r_2}\left( g \right) \rightleftharpoons NOB{r_2}\left( g \right) \cr
& {\text{(ii)}}\,NOB{r_2}\left( g \right) + NO\left( g \right) \to 2NOBr\left( g \right) \cr} $$
Rate law equation $$ = k\left[ {NOB{r_2}} \right]\left[ {NO} \right]$$
But $$NOB{r_2}$$ is intermediate and must not appear in the rate law equation
from 1st step $${K_C} = \frac{{\left[ {NOB{r_2}} \right]}}{{\left[ {NO} \right]\left[ {B{r_2}} \right]}}$$
$$\therefore \left[ {NOB{r_2}} \right] = {K_C}\left[ {NO} \right]\left[ {B{r_2}} \right]$$
∴ Rate law equation $$ = k.\,{K_C}{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$
hence order of reaction is 2 $$w.r.t.$$ $$NO.$$
Releted MCQ Question on Physical Chemistry >> Chemical Kinetics
Releted Question 1
If uranium (mass number 238 and atomic number 92) emits an $$\alpha $$ -particle, the product has mass no. and atomic no.