Question

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

$Rate =K\left[NOBr_{2}\right] \left[NO\right]$ ...(1)
But $NOBr_{2}$ is in equilibrium.
$K_{eq}=\frac{\left[NOBr_{2}\right]}{\left[NO\right]\left[Br_{2}\right]}$
$\left[NOBr_{2}\right]=K_{eq} \left[NO\right] \left[Br_{2}\right]$ ...(2)
Putting the $\left[NOBr_{2}\right]$ in (1)
Rate =$K\cdot K_{eq} \left[NO\right] \left[Br_{2}\right] \left[NO\right]$
Hence Rate$=K\cdot K_{eq} \left[NO\right]^{2} \left[Br_{2}\right] $
Rate =$K' \left[NO\right]^{2} \left[Br_{2}\right]$
where $K' =K\cdot K_{eq\cdot}$