2-step reaction mechanism for the decomposition of NO2Br
It is more probable  like
$$\begin{align}
\ce{NO2Br &amp;-&gt; NO2 + Br} \\
\ce{NO2Br + Br &amp;-&gt; NO2 + Br2} \\
\ce{2 Br &amp;-&gt;  Br2} \\
\end{align}$$

The last reaction is a minor one in case concentration of $\ce{Br}$ is low.

The reaction rate order can be concentration dependent and need not be the integer.

In fact, it is rather mathematical parameter, related to solution of differential equations for a complex reaction system.

If the 2nd reaction is fast enough, the overall reaction rate is given by the slow rate of generation  of $\ce{Br}$, which fast reacts to form $\ce{Br2}$

If the 2nd reaction is slow enough, 
it's rate $$k_{\rm 2}\cdot [\ce{NO2Br}][\ce{Br}]$$ can be written as $$k_{\rm 2a}\cdot [\ce{NO2Br}]^2$$

The exact solution is to solve system of differential equations for the rates of the concentration changes.



$$\frac{\mathrm{d}[\ce{Br}]}{\mathrm{d}t}=k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}] - k_3 [\ce{Br}]^2$$

For the dynamic equilibrium of the steady concentration of $\ce{Br}$:

$$\begin{align}
0&amp;=-k_1.[\ce{NO2Br}] + k_2.[\ce{NO2Br}][\ce{Br}] + k_3 [\ce{Br}]^2 \\
[\ce{Br}]&amp;=[ -k_2.[\ce{NO2Br}]+\sqrt((k_2.[\ce{NO2Br}])^2+4.k_3.k_1.[\ce{NO2Br}])]/(2.k_3) \\
\frac{\mathrm{d}[\ce{NO2Br}]}{\mathrm{d}t}&amp;=-k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}]\\
\end{align}$$