What are the rate laws for each elementary step?
For elementary reactions, the rate law can be directly read from the chemical equation. In your last equation this is a bimolecular reaction, i.e.
\begin{align}
\ce{N + NO &amp;-&gt; N2O} &amp; r &amp;= k\cdot c(\ce{N}) \cdot c(\ce{NO})
\end{align}

To be more precise here, one molecule of $\ce{NO}$ reacts with on atom $\ce{N}$, and in a wider sense, that is two (bi) molecules reacting. The order of the reaction is two.



That being said, there are a couple of things you need to pay close attention to. An elementary reaction is always a reaction between the number of molecules given as reactants, with exactly one transition state leading to the products. If this number or reactants is greater than three it is highly unlikely to happen this way, since a concerted transition state is statistically very unlikely. An elementary reaction therefore also has to be balanced, so your first reaction needs to be:
$$\ce{2NO -&gt; NO2 + N}$$
In this case we could see $[\ce{ON\bond{~}O\bond{~}N}]^\ddagger$ as a transition state.

I very highly doubt that your second reaction as an elementary reaction, from my point of view, there are too many bond that would need to be broken. Instead, the following reactions might be possible and/or necessary decompositions:
\begin{align}
\ce{NO2 + H2 &amp;-&gt; NO + OHH}\\
\ce{OHH &amp;-&gt; OH + H}\\
\ce{NO2 + H &amp;-&gt; NO + OH}\\
\ce{2OH &amp;-&gt; H2O + O}\\
\ce{NO + O &amp;-&gt; NO2}\\\hline
\ce{NO2 + H2 &amp;-&gt; NO + H2O}
\end{align}
And there are more possibilities.

In all but one the elementary reactions are bimolecular - guess which one would be unimolecular.



As for the mechanism itself, I think these are reactions that occur, just not all of them, maybe not even the majority of them.