Question

The reaction $2NO+B{r}_2\to 2NOBr$, is supposed to follow the following mechanism
(i) $NO+B{r}_2\stackrel{fast}{\rightleftharpoons }NOB{r}_2$ (ii) $NOB{r}_2+NO\stackrel{slow}{\to }$
Determine rate of the reaction:

• A. $r=K^{\prime }\left[NO{]}^2\right[B{r}_2]$
• B. $r=K^{\prime }\left[NO{]}^{-2}\right[B{r}_2]$
• C. $r=K^{\prime }\left[NO{]}^2\right[B{r}_2{]}^{-1}$
• D. None of these

Answer 1

The correct option is A $r=K^{\prime }\left[NO{]}^2\right[B{r}_2]$

$2NO+B{r}_2\to 2NOBr$

(i) $NO+B{r}_2\rightleftharpoons NOB{r}_2\left(Fast\right)$

(ii)$NOB{r}_2+NO\rightleftharpoons [NOBr{]}^2${Slow}

Rate is determined by the slowest step

$\Rightarrow$$Rate=k\left[NOB{r}_2\right]\left[NO\right]$

But $NOB{r}_2$ is a visible of Intermediate

$k_2=\frac{\left[NOB{r}_2\right]}{\left[NO\right]\left[B{r}_2\right]}\Rightarrow Rate=k_1\left[NO{]}^2\right[B{r}_2]$

$\Rightarrow :3rdorderR\times N$