Question

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

The reaction obeys the following mechanism:

$NO+B{r}_2\stackrel{Fast}{\rightleftharpoons }NOB{r}_2$;

$NOB{r}_2+NO\stackrel{Slow}{\to }2NOBr$

The rate expression of the above reaction can be written as:

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

Answer 1

Answer: (A)

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

For slowest step:

$rate=k\left[NOB{r}_2\right]\left[NO\right]....\left(i\right)$

For equilibrium constant,

$K_c=\frac{k_f}{k_b}=\frac{\left[NOB{r}_2\right]}{\left[NO\right]\left[B{r}_2\right]}....\left(ii\right)$
$k_f\left[NO\right]\left[B{r}_2\right]=k_b\left[NOB{r}_2\right]...............\left(iii\right)$

By eqs. (i) and (iii),

$r=k\left[NO\right]\left[B{r}_2\right]\left[NO\right]$

Therefore,

$Rate=k\left[NO{]}^2\right[B{r}_2]$.

Hence, option $A$ is correct.