= Solution
The <quasi-steady-state approximation> for the nitrate radical balances its formation against its removal:
$$
\frac{d[\mathrm{NO_3}]}{dt}=r_1-r_2+r_{-2}\simeq0.
$$
Equivalently,
$$
[\mathrm{NO_3}]_{\rm ss}=\frac{k_1[\mathrm{NO_2}][\mathrm{O_3}]+k_{-2}[\mathrm{N_2O_5}]}{k_2[\mathrm{NO_2}]}.
$$
The net dinitrogen pentoxide production is $r_2-r_{-2}$. Substituting the steady-state balance gives
$$
\boxed{\frac{d[\mathrm{N_2O_5}]}{dt}\simeq k_1[\mathrm{NO_2}][\mathrm{O_3}].}
$$
The reverse rate has been retained throughout and cancels algebraically. The approximation requires sufficiently rapid radical adjustment and fails if, for example, nitrogen dioxide becomes too scarce to sustain the presumed trapping rate. It does not assume that the radical <concentration> is zero.
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