= Solution
Write $D=[\mathrm{Cl_2O_2}]$ and $c=[\mathrm{ClO}]$. In the <steady-state approximation> for the dimer, <chlorine peroxide> <chlorine peroxide> formation is balanced by its two competing removal channels:
$$
\frac{dD}{dt}=k_1[M]c^2-\{J_2+k_3[M]\}D\simeq0,
\qquad D=\frac{k_1[M]c^2}{J_2+k_3[M]}.
$$
The association event rate is $k_1[M]c^2$; the associated disappearance rate of <chlorine monoxide> is twice that value. Each successful <photodissociation> event produces two <chlorine> atoms after the subsequent fast fragmentation, and their return to <chlorine monoxide> consumes two <ozone> <molecules>. <Thermal dissociation> instead returns the dimer to two <chlorine monoxide> <radicals> without destroying <ozone>. Thus the <chemical branching fraction> for an ozone-destroying event is $f=J_2/(J_2+k_3[M])=0.7$, giving
$$
\boxed{-\frac{d[\mathrm O_3]}{dt}=2J_2D
=2f k_1[M][\mathrm{ClO}]^2=1.4k_1[M][\mathrm{ClO}]^2.}
$$
This is the sunlit <chlorine monoxide dimer cycle>, with net conversion $2\mathrm O_3\to3\mathrm O_2$. It assumes that the dimer relaxes rapidly, downstream fragmentation and <chlorine>–<ozone> reactions are sufficiently fast, and competing <chlorine> sinks can be neglected. Counting the formation event twice and then multiplying by two again would overestimate loss by a factor of two.
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