Question

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

• A. $r=K^{\prime }\left[NO{]}^2\right[B{r}_2]$
• B. $r=K^{\prime }\left[NO{]}^2\right[B{r}_2{]}^2$
• C. $r=K^{\prime }\left[NO\right]\left[B{r}_2\right]$
• D. $r=K^{\prime }\left[NO{]}^1\right[B{r}_2]$

Answer 1

Answer: (A)

$r=K^{\prime }\left[NO{]}^2\right[B{r}_2]$

From the slow step, $rate=k\left[NOB{r}_2\right]\left[NO\right]$ ...... (1)

From the equilibrium reaction, $equilibriumconstantK=\frac{\left[NOB{r}_2\right]}{\left[NO\right]\left[B{r}_2\right]}$

$\left[NOB{r}_2\right]=K\left[NO\right]\left[B{r}_2\right]$......(2)

Substitute equation (2) in equation (1).

$rate=kK\left[NO\right]\left[B{r}_2\right]\left[NO\right]=K^{\prime }\left[NO{]}^2\right[B{r}_2]$