Question

The following mechanism has been proposed for the reaction of NO with $B{r}_2$to form NOBr :
$NO\left(g\right)+B{r}_2\left(g\right)\text{rightleftharpoons}NOB{r}_2\left(g\right)$
$NOBr\left(g\right)+NO\left(g\right)\to 2NOBr\left(g\right)$
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

Answer 1

Answer: (D)

$Rate=K\left[NOB{r}_2\right]\left[NO\right]$ ...(1)
But $NOB{r}_2$ is in equilibrium.
$K_{eq}=\frac{\left[NOB{r}_2\right]}{\left[NO\right]\left[B{r}_2\right]}$
$\left[NOB{r}_2\right]=K_{eq}\left[NO\right]\left[B{r}_2\right]$ ...(2)
Putting the $\left[NOB{r}_2\right]$ in (1)
Rate =$K\cdot K_{eq}\left[NO\right]\left[B{r}_2\right]\left[NO\right]$
Hence Rate$=K\cdot K_{eq}{\left[NO\right]}^2\left[B{r}_2\right]$
Rate =$K^{\prime }{\left[NO\right]}^2\left[B{r}_2\right]$
where $K^{\prime }=K\cdot K_{eq\cdot }$