Question

The reaction between $NO$ and ${Cl}_2$ takes place in the following two steps:
$I$. $NO+{Cl}_2\stackrel{k_1}{\underset{\underset{fast}{k_2}}{\rightleftharpoons }}NO{Cl}_2$
$II$. $NO{Cl}_2+NO\underset{slow}{\overset{k_3}{\to }}2NOCl$
The rate law of overall reaction, $2NO+{Cl}_2⟶2NOCl$, can be given by:

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

Answer 1

Answer: (A)

Rate $=k{\left[NO\right]}^2\left[{Cl}_2\right]$

Rate=$k_3\left[NOC{l}_2\right]\left[NO\right]$

Now $k=\frac{k_1}{k_2}=\frac{\left[NOC{l}_2\right]}{\left[NO\right]\left[C{l}_2\right]}$

$\left[NOC{l}_2\right]=\frac{k_1}{k_2}\left[NO\right]\left[C{l}_2\right]$

Rate=$\frac{k_1{k}_3}{k_2}\left[NO{]}^2\right[C{l}_2]$

Let $\frac{k_1{k}_3}{k_2}=k$, then rate= $K\left[NO{]}^2\right[C{l}_2]$