Question

In the following reaction :
$2NO+O_2\stackrel{k^{\prime }}{\to }2N{O}_2$
What is the predicted rate law, if the mechanism is given as :
$NO+O_2\stackrel{K_{eq}}{\rightleftharpoons }N{O}_3$ (Fast)
$N{O}_3+NO\stackrel{k_1}{\to }N{O}_2+N{O}_2$ (Slow)

• A. $r=k_1{K}_{eq}\left[NO\right]\left[O_2\right]\left[N{O}_3\right]$
• B. $r=k_1{K}_{eq}\left[NO\right]\left[O_2\right]$
• C. $r=k_1{K}_{eq}\left[NO{]}^2\right[O_2]$
• D. $r=k_1{K}_{eq}[NO{]}^2$

Answer 1

The correct option is C $r=k_1{K}_{eq}\left[NO{]}^2\right[O_2]$

The rate of the reaction is always decided by the slowest step of the reaction.
The rate will be
$r=k_1\left[N{O}_3\right]\left[NO\right]$ ....(1)
The first step of the reaction is in equilibrium
$\therefore K_{eq}=\frac{\left[N{O}_3\right]}{\left[NO\right]\left[O_2\right]}$
$\therefore \left[N{O}_3\right]=K_{eq}\left[NO\right]\left[O_2\right]$
Substituting the value of $\left[N{O}_3\right]ine{q}^n$ (1)
$r=k_1{K}_{eq}\left[NO\right]\left[O_2\right]\left[NO\right]$
$r=k_1{K}_{eq}\left[NO{]}^2\right[O_2$]