The reaction mechanism for the formation of $\ce{NO2}$ is:

\begin{align} \ce{NO + NO &<=>[$k_1$][$k_1'$] N2O2} & &\text{ (slow)} \\[0.2cm] \ce{N2O2 + O2 &->[$k_2$] NO2 +NO2} & &\text{ (fast)} \end{align}

What is the equation for rate of change of the intermediate $\ce{N2O2}$? My solution is as follows: The rate of formation of intermediate $\ce{N2O2}$ is given by:

\begin{align} \frac{\mathrm{d}[\ce{N2O2}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{NO}]^2 - \color{red}{2}k_1'[\ce{N2O2}] - k_2[\ce{N2O2}][\ce{O2}] \approx 0 \end{align} But the answer in the text book says:

\begin{align} \frac{\mathrm{d}[\ce{N2O2}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{NO}]^2 - k_1'[\ce{N2O2}] - k_2[\ce{N2O2}][\ce{O2}] \approx 0 \end{align}

Can not figure out why my answer is wrong.

  • 1
    $\begingroup$ Write the rate law for the rate of reaction of creating $\ce{NO}$ from $\ce{N2O2}$ (reverse of slow reaction). The is the rate of consuming $\ce{N2O2}$ for that component reaction. $\endgroup$
    – Zhe
    May 25, 2021 at 15:16

2 Answers 2


Split the scheme into three parts and write down each step. Start by definition with

$$\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=+\frac{d[\ce{N2O2}]}{dt}$$

then the first step $\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=k_1[\ce{NO}]^2\quad$ or $\quad\displaystyle \frac{d[\ce{NO}]}{dt}=-2k_1[\ce{NO}]^2$

and so $\ce{N_2O_2}$ is produced as

$$\displaystyle \frac{d[\ce{N2O2}]}{dt} = k_1[\ce{NO}]^2$$

In the reverse equilibrium step $\ce{N2O2}$ is lost (by dissociatiing) so $\displaystyle \frac{d[\ce{N2O2}]}{dt} = -k_1^{'}[\ce{N2O2}]$

and in the last reaction it is also lost by reaction with oxygen therefore $\displaystyle \frac{d[\ce{N2O2}]}{dt}=-k_2[\ce{N2O2}][\ce{O2}]$.

Adding up all three rates for $\ce{N_2O_2}$ gives

$$\displaystyle \frac{d[\ce{N2O2}]}{dt}=k_1[\ce{NO}]^2-k_1^{'}[\ce{N2O2}]-k_2[\ce{N2O2}][\ce{O2}]$$

You can now find the steady-state conditions.

(In your last equation there is $2k_1$ and it is quite common to see it written this way, all it does is to change $k_1$.)

  • $\begingroup$ Thank you! My final solution is:$$\displaystyle \frac{1}{2}\frac{d[NO_2]}{dt}=-\frac{d[N_2O_2]}{dt}$$ $$\displaystyle \frac{1}{2}\frac{d[NO_2]}{dt}=k_2[NO_2]^2\quad$$ That means: $$\displaystyle +\frac{d[N_2O_2]}{dt}={2}k_2[N_2O_2][O_2]$$ So, the overall rate law: $$\displaystyle \frac{\color{red}{2}k_1k_2[O_2][NO]^2\quad}{k’_{-1} + k_2[O_2]}$$ And we can simplify to: $$\displaystyle \frac{\color{red}{2}k_1k_2[O_2][NO]^2\quad}{k’_{-1}}$$ $\endgroup$
    – Tara
    May 26, 2021 at 15:56

If you apply Goldberg & Waage's Law of Mass Action to the rate determining (slowest) step, we will get: $$\frac {d[\ce{N2O2}]}{dt}=-k_2 [\ce{N2O2}][\ce{O2}]$$ Moreover, we have: $$\frac {k_1}{k'_1}=\frac {[\ce{N2O2}]}{[\ce{NO}]^2}$$ If this is substituted to the rate equation obtained before, the rate expression involves concentration of products and reactions, rather than intermediates: $$\frac {d[\ce{N2O2}]}{dt}=-\frac {k_1 k_2 [\ce{NO}]^2[\ce{O2}]}{k'_1}$$ This should be the final expression. Your expression is correct too, as $2k_1[\ce{NO}]^2-2k'_1[\ce{N2O2}]=0$, hence my expression and yours are equivalent.

I would love to hear feedback.

  • $\begingroup$ Thank you! You are right for given conditions, pre-equilibrium approximation with fast first step. Problem in textbook call for steady-state approach. Sorry for misleading! $\endgroup$
    – Tara
    May 25, 2021 at 14:23

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