= Solution
Excited $\mathrm O({}^1D)$ has two competing fates. Collisional deactivation returns ground-state <oxygen>, which reforms <ozone> rapidly; reaction with <water> instead gives net <ozone> loss and two OH radicals. The <chemical branching fraction> for the latter is
$$
f_w=\frac{k_3[\mathrm{H_2O}]}{k_2[\mathrm M]+k_3[\mathrm{H_2O}]}
=\frac{3.30\times10^7}{6.38\times10^8+3.30\times10^7}
=0.0491803.
$$
Thus only about 4.92% of these <ozone> photolyses directly lead to the specified net chemical loss; treating every <photodissociation> as permanent removal would overestimate the sink. The corresponding first-order rate is $J_1f_w=9.8361\times10^{-7}\ \mathrm{s^{-1}}$. Converting the given hydroperoxyl <mixing ratio> to a <concentration> gives $[\mathrm{HO_2}]=20\times10^{-12}[\mathrm M]=4.4\times10^8\ \mathrm{cm^{-3}}$, so $k_4[\mathrm{HO_2}]=9.6800\times10^{-7}\ \mathrm{s^{-1}}$.
For a well-mixed <marine atmospheric boundary layer> of depth $H=2\times10^5\ \mathrm{cm}$, the downward surface flux is $v_{\rm dep}[\mathrm O_3]$. Dividing by the layer <volume> per unit area gives a first-order <dry deposition> rate $v_{\rm dep}/H=7.5\times10^{-7}\ \mathrm{s^{-1}}$. The air density cancels on expressing the entire balance in <ozone> <mixing ratio> $\chi$:
$$
\frac{d\chi}{dt}=P_\chi-\left(J_1f_w+k_4[\mathrm{HO_2}]+\frac{v_{\rm dep}}H\right)\chi.
$$
At <steady state (chemistry)>, with $P_\chi=0.25/3600\ \mathrm{ppb\,s^{-1}}$,
$$
\boxed{\chi_{\mathrm O_3}=\frac{0.25/3600}{2.70161\times10^{-6}}
=25.7\ \mathrm{ppb}.}
$$
The effective removal time is about 4.28 days. This calculation assumes a uniform layer of the diagnosed depth, fixed chemical conditions and a local time-independent balance.
Back to article page
