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.