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)

2

Answer 2

$Rate =K\left[NOBr_{2}\right] \left[NO\right]$ ...(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 $K' =K\cdot K_{eq\cdot}$