Question

Reaction between NO and $O_2$ to form $N{O}_2+O_2\to 2N{O}_2$ follows the following mechanism:

First step : $NO+NO\rightleftharpoons N_2{O}_2$ (in rapid equilibrium)

Here, $K_1$ is the rate constant for forward reaction while $K_{-1}$ is the rate constant for backward reaction.

Second step : $N_2{O}_2+O_2\stackrel{K_2}{\to }2N{O}_2$ (slow)

Determine the rate of reaction.

• A. $\frac{1}{2}\left(\frac{d\left[N{O}_2\right]}{dt}\right)=K\left[NO{]}^2\right[O_2]$
• B. $\frac{1}{2}\left(\frac{d\left[N{O}_2\right]}{dt}\right)=K\left[NO\right]\left[O_2\right]$
• C. $\frac{1}{2}\left(\frac{d\left[N{O}_2\right]}{dt}\right)=K\left[NO{]}^2\right[O_2{]}^2$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}\left(\frac{d\left[N{O}_2\right]}{dt}\right)=K\left[NO{]}^2\right[O_2]$

The rate of the reaction is given by the slow step.

$rate=k\left[N_2{O}_2\right]\left[O_2\right]$ ......(1)

The concentration of $\left[N_2{O}_2\right]$ is calculated from the equilibrium step.

$K^{\prime }=\frac{K_1}{K_{-1}}=\frac{\left[N_2{O}_2\right]}{[NO{]}^2}$

$\left[N_2{O}_2\right]=K^{\prime }[NO{]}^2$ ......(2)

Substitute (2) in (1).

$rate=K{k}^{\prime }\left[NO{]}^2\right[O_2]$

Hence, the rate expression is $\frac{1}{2}\left(\frac{d\left[N{O}_2\right]}{dt}\right)=K\left[NO{]}^2\right[O_2]$

where $K=k{K}^{\prime }$

Hence, the correct option is $\text{A}$