 Solution

The fast family is NO${}_3$ plus N${}_2$O${}_5$. Put $x=[\mathrm{N}{\mathrm{O}}_3]$, $y=[{\mathrm{N}}_2{\mathrm{O}}_5]$, and freeze the supplied parent concentrations while definingThe two equations are $\dot{x}=S-(a+d)x+by$ and $\dot{y}=ax-(b+c)y$. Their stationary solution isHere $a=15.67$, $b=0.0690$, $c=3.84\times 10^{-7}$ and $d=1.919$ in inverse seconds; $S=5.93\times 10^6\mathrm{molecules}\mathrm{c}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$. Reversible storage must not be counted as permanent family loss: only the $ac/(b+c)$ term represents the association route followed by hydrolysis.

Closed-family interconversion has ${\tau }_{\mathrm{eq}}=1/(a+b)=0.0635\mathrm{s}$. Including the supplied losses, the two linear relaxation times are approximately $0.0566\mathrm{s}$ and $133\mathrm{s}$, from the eigenvalues of $\left(\begin{array}{cc}a+d& -b\\ -a& b+c\end{array}\right)$. Fast exchange is therefore well separated from the slower total-family adjustment. However, NO is consumed by ozone on $1/(k_1[{\mathrm{O}}_3])\simeq 59.8\mathrm{s}$, comparable to and shorter than the family relaxation. Thus the quoted stationary concentration is justified for a maintained background or a sufficiently slowly varying quasi-stationary environment; it is not automatically reached in an isolated parcel initialized with the supplied composition. NO auto-oxidation is negligible at this concentration. Additional organic losses of NO${}_3$, if present, would need to be added to the model.