 Solution

Nitrogen dioxide is consumed in reactions 1 and 2 and regenerated in reaction $-2$, soUsing $r_2-r_{-2}\simeq r_1$ from the nitrate-radical quasi-steady-state approximation givesThe factor of two is essential: an initiating event uses one nitrogen dioxide molecule, and the ensuing net radical-trapping event uses a second one. The exact nitrogen balance before making this approximation is $d([\mathrm{N}{\mathrm{O}}_2]+[\mathrm{N}{\mathrm{O}}_3]+2[{\mathrm{N}}_2{\mathrm{O}}_5])/dt=0$.