= Solution
\b[The remaining fraction is $\boxed{0.120\approx12.0\%}$.] Using the tabulated $300\,\mathrm K$ value rather than replacing it by the fitted value, the constant ozone <concentration> makes nitrogen dioxide loss a <pseudo-first-order reaction>:
$$
\frac{d[\mathrm{NO_2}]}{dt}=-k_{\rm obs}[\mathrm{NO_2}],\qquad
k_{\rm obs}=2k_1[\mathrm{O_3}]
=2(23\times10^6)(1.6\times10^{-12})=7.36\times10^{-5}\,\mathrm{s^{-1}}.
$$
Integrating with initial concentration $[\mathrm{NO_2}]_0$ and $t=8(3600)=28800\,\mathrm s$ gives
$$
\frac{[\mathrm{NO_2}](t)}{[\mathrm{NO_2}]_0}
=e^{-k_{\rm obs}t}=e^{-2.11968}=0.12007.
$$
This uses the nitrate-radical <quasi-steady-state approximation> over the interval, constant <temperature> and ozone <concentration>, and no extra nitrogen dioxide sources or sinks. The stated $\mathrm{cm^3}$ and $\mathrm{mol\,cm^{-3}}$ units cancel directly; no factor of $1000$ is needed.
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