JEE MAIN - Chemistry (2006 - No. 36)
The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr:
NO(g) + Br2 (g) $$\leftrightharpoons$$ NOBr2 (g)
NOBr2 (g) + NO (g) $$\to$$ 2NOBr (g)
If the second step is the rate determining step, the order of the reaction with respect to NO(g) is
NO(g) + Br2 (g) $$\leftrightharpoons$$ NOBr2 (g)
NOBr2 (g) + NO (g) $$\to$$ 2NOBr (g)
If the second step is the rate determining step, the order of the reaction with respect to NO(g) is
1
0
3
2
Explanation
$$\left( i \right)\,\,\,\,\,\,\,\,\,NO\left( g \right) + B{r_2}\left( g \right)\,\,\,\rightleftharpoons\,NOB{r_2}\left( g \right)$$
$$\left( {ii} \right)\,\,\,\,\,\,\,\,\,NOB{r_2}\left( g \right) + NO\left( g \right)\,\buildrel \, \over \longrightarrow 2NOBr\left( g \right)$$
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 $$1$$st step $${K_C} = {{\left[ {NOB{r_2}} \right]} \over {\left[ {NO} \right]\left[ {B{r_2}} \right]}}$$
$$\therefore$$ $$\,\,\,\,\,\left[ {NOB{r_2}} \right] = {K_C}\left[ {NO} \right]\left[ {B{r_2}} \right]$$
$$\therefore$$ $$\,\,\,\,$$ Rate law equation $$ = k.{K_C}{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$
hence order of reaction is $$2$$ w.r.t. $$NO.$$
$$\left( {ii} \right)\,\,\,\,\,\,\,\,\,NOB{r_2}\left( g \right) + NO\left( g \right)\,\buildrel \, \over \longrightarrow 2NOBr\left( g \right)$$
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 $$1$$st step $${K_C} = {{\left[ {NOB{r_2}} \right]} \over {\left[ {NO} \right]\left[ {B{r_2}} \right]}}$$
$$\therefore$$ $$\,\,\,\,\,\left[ {NOB{r_2}} \right] = {K_C}\left[ {NO} \right]\left[ {B{r_2}} \right]$$
$$\therefore$$ $$\,\,\,\,$$ Rate law equation $$ = k.{K_C}{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$
hence order of reaction is $$2$$ w.r.t. $$NO.$$
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