Steady-state approach - how to find rate equation for intermediate? The reaction mechanism for the formation of $\ce{NO2}$ is:
\begin{align}
\ce{NO + NO &amp;&lt;=&gt;[$k_1$][$k_1'$] N2O2} &amp; &amp;\text{ (slow)} \\[0.2cm]
\ce{N2O2 + O2 &amp;-&gt;[$k_2$] NO2 +NO2} &amp; &amp;\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.
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 $k_1$.)