Question

The mechanism of the reaction :

$2NO+O_2⟶2N{O}_2$ is,

$NO+NO\underset{K_{-1}}{\stackrel{k_1}{\rightleftharpoons }}{N}_2{O}_2\left(fast\right)$;

$N_2{O}_2+O_2\stackrel{k_2}{\to }2N{O}_2\left(slow\right)$

The rate constant of the reaction is :

• A. $k_2$
• B. $k_2{k}_1\left(k_{-1}\right)$
• C. $k_2{k}_1$
• D. $k_2\left(\frac{k_1}{k_{-1}}\right)$

Answer 1

The correct option is D $k_2\left(\frac{k_1}{k_{-1}}\right)$

Slow step is the rate determining step (RDS) and $\left(N_2{O}_2\right)$ is the reactive intermediate.
$\therefore r=k_2\left[N_2{O}_2\right]\left[O_2\right]$
From reversible reaction, $\left[N_2{O}_2\right]$ is:
$\frac{k_1}{k_{-1}}=\frac{\left[N_2{O}_2\right]}{[NO{]}^2}$
Substitute $\left[N_2{O}_2\right]$ in equation (i),
$r=k_2\left(\frac{k_1}{k_{-1}}\right)\left[NO{]}^2\right[O_2]$
Hence, rate constant = $k_2\left(\frac{k_1}{k_{-1}}\right)$