Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2021 / Interdisciplinary paper IDP1 / 1 / a

Write $D=[\mathrm{C}{\mathrm{l}}_2{\mathrm{O}}_2]$ 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: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 $f=J_2/(J_2+k_3[M])=0.7$, givingThis is the sunlit chlorine monoxide dimer cycle, with net conversion $2{\mathrm{O}}_3\to 3{\mathrm{O}}_2$. 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.

Low temperature favours association into the dimer and, especially, suppresses its thermally activated dissociation back to chlorine monoxide. More of the dimer can therefore follow photodissociation instead of the chemically unproductive thermal return. The supplied branching fraction is a local condition, not a temperature-independent universal yield.

Cold polar air also permits polar stratospheric clouds and efficient heterogeneous conversion of chlorine reservoirs into photolabile forms. The stratospheric polar vortex isolates and cools that air. Darkness permits activation and storage, but rapid catalytic ozone loss requires the return of sunlight. These chemical and meteorological conditions explain the springtime maximum; low temperature alone does not supply the photodissociation photons.

The two initial reservoirs each contain one chlorine atom, so complete activation supplies a total chlorine mixing ratio of $2.2\times 10^{-9}$. Under the specified approximation that active chlorine in sunlight is predominantly chlorine monoxide,The daytime loss is consequentlyFor twelve sunlit hours per day, the daily ozone decrement isThe full-day mean concentration loss is $5.269\times 10^5\mathrm{molecules}\mathrm{c}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$. Maintaining that rate would consume the initial ozone in about $79.1$ days. This is a fixed-active-chlorine estimate: reservoir reformation, transport and chlorine partitioning change during a real depletion event, and the approximation fails when insufficient ozone remains to recycle chlorine rapidly.