Question
The reaction, $$2NO + B{r_2} \to 2NOBr,$$ obeys the following mechanism :
\[NOB{{r}_{2}}+NO\xrightarrow{\text{Slow}}2NOBr\]
The rate expression of the above reaction can be written as
A.
$$r = k{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$
B.
$$r = k\left[ {NO} \right]\left[ {B{r_2}} \right]$$
C.
$$r = k\left[ {NO} \right]{\left[ {B{r_2}} \right]^2}$$
D.
$$r = k\left[ {NOB{r_2}} \right]$$
Answer :
$$r = k{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$
Solution :
For slowest step : rate $$ = k\left[ {NOB{r_2}} \right]\left[ {NO} \right]...\left( {\text{i}} \right)$$
For equilibrium also $${K_c} = \frac{{\left[ {NOB{r_2}} \right]}}{{\left[ {NO} \right]\left[ {B{r_2}} \right]}}...\left( {{\text{ii}}} \right)$$
By eqs. (i) and (ii), $$r = k \cdot {K_c}{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$
Rate $$ = k{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$