= Solution
The intermediate balance is
$$
\frac{d[\mathrm{ONOO}]}{dt}=k_1[\mathrm{NO}][\mathrm{O_2}]-k_{-1}[\mathrm{ONOO}]-k_2[\mathrm{NO}][\mathrm{ONOO}].
$$
Setting this derivative approximately to zero under the <quasi-steady-state approximation> gives
$$
\boxed{[\mathrm{ONOO}]_{\rm ss}=\frac{k_1[\mathrm{NO}][\mathrm{O_2}]}{k_{-1}+k_2[\mathrm{NO}]}.}
$$
Both dissociation and reaction with NO remove the intermediate. Its steady concentration is not zero and does not require the first reversible pair to be at equilibrium.
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