Question

Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide:
$2NO\left(g\right)+O_2\left(g\right)\to 2N{O}_2\left(g\right)$
If the mechanism of reaction is;
$NO+O_2\stackrel{k}{\rightleftharpoons }N{O}_3\left(fast\right)$
$N{O}_3+NO\stackrel{k_1}{⟶}N{O}_2+N{O}_2\left(slow\right)$
The rate law is:

• A. Rate=$k^{\prime }\left[NO\right]\left[O_2\right]$
• B. Rate=$k^{\prime }\left[NO\right][O_2{]}^2$
• C. Rate=$k^{\prime }\left[NO{]}^2\right[O_2]$
• D. Rate=$k^{\prime }\left[NO{]}^3\right[O_2]$

Answer 1

The correct option is C Rate=$k^{\prime }\left[NO{]}^2\right[O_2]$

Slow reaction is the rate determining step. So, for
$N{O}_3+NO\stackrel{k_1}{⟶}N{O}_2+N{O}_2\left(slow\right)$
$Rate=k_1\left[N{O}_3\right]\left[NO\right]\left(1\right)$
from $NO+O_2\stackrel{k}{\rightleftharpoons }N{O}_3\left(fast\right)$ we can write $k=\frac{\left[N{O}_3\right]}{\left[O_2\right]\left[NO\right]}\left(2\right)$
On substituting value of $\left[N{O}_3\right]$ from eq.(2) to eq.(1), we get $Rate=k^{\prime }\left[NO{]}^2\right[O_2]$