= Solution
The dominant OH-to-HO$_2$ pseudo-first-order rate is $k_5[\mathrm O_3]=90\,\mathrm{s^{-1}}$, corresponding to \b[$0.0111\,\mathrm s$]. Recycling HO$_2$ through atomic <oxygen> and NO has rate $k_6[\mathrm O]+k_8[\mathrm{NO}]=0.0218\,\mathrm{s^{-1}}$, corresponding to \b[$45.9\,\mathrm s$] for an individual HO$_2$ radical. A fixed-total two-state exchange relaxes at the sum of forward and reverse rates, giving about $0.0111\,\mathrm s$ for the partition perturbation. Its fast partition is therefore
$$
\frac{[\mathrm{HO_2}]}{[\mathrm{OH}]}
\simeq\frac{k_5[\mathrm O_3]}{k_6[\mathrm O]+k_8[\mathrm{NO}]}
=4.13\times10^3.
$$
This supports a <steady-state approximation in chemical kinetics> for rapidly recycled intermediates if the parent <concentrations> vary slowly. It does not by itself establish stationarity of the entire <HOx family>. The later source–termination estimate gives a family turnover time of about 16 hours, far longer than partition relaxation; a stationary absolute abundance additionally assumes sufficiently persistent radiation and background conditions. This is the distinction between <fast family partition and slow family balance>.
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