= Solution
“Low NO” means $k_2[\mathrm{NO}]\ll k_{-1}$, or $[\mathrm{NO}]\ll6.5\times10^{-6}\,\mathrm M$. Then
$$
\frac{d[\mathrm{NO_2}]}{dt}\simeq\frac{2k_1k_2}{k_{-1}}[\mathrm{NO}]^2[\mathrm{O_2}].
$$
\b[The effective orders are two in NO and one in oxygen, giving overall order three.] For the production-rate convention,
$$
\boxed{k_{\rm eff,prod}=\frac{2(189)(1.0\times10^9)}{6.5\times10^3}=5.82\times10^7\,\mathrm{M^{-2}\,s^{-1}}.}
$$
If the conventional reaction rate $v=\tfrac12d[\mathrm{NO_2}]/dt$ is used, its coefficient is $2.91\times10^7\,\mathrm{M^{-2}\,s^{-1}}$. These are the same kinetics under explicitly different normalizations. At high NO the denominator changes, so the effective NO order approaches one rather than remaining two.
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