Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / 42

A warmer troposphere does not imply a warmer stratosphere: increased carbon dioxide can cool the middle and upper stratosphere by enhancing infrared emission. One direct kinetic effect is slower temperature-dependent destruction in the Chapman cycle, tending to increase upper-stratospheric ozone. Two indirect chemical effects can have the opposite sign. If the entry atmospheric cold trap warms, it can admit more water; reaction with an electronically excited oxygen atom then supplies more of the HOx family, enhancing catalytic ozone loss. In cold polar air, changes in water and temperature can favour polar stratospheric clouds, whose surfaces activate halogen reservoirs and can enhance ozone destruction when sunlight returns. These are conditional pathways, not a universal sign for total ozone: cold-trap temperature and circulation, halogen abundance and altitude all matter. Methane oxidation can also supply stratospheric water.^[ref]

For one specified channel, the photolysis rate coefficient isIt depends on spectral actinic flux, absorption cross section, and channel quantum yield. Solar zenith angle, season, overhead absorbers, clouds and aerosol scattering affect the photon field; branching and collisional quenching affect the chemical yield. Absorption alone does not guarantee the desired dissociation.

Curve A is oxygen photodissociation; curve B is ozone photodissociation. Molecular oxygen needs relatively short-wavelength ultraviolet photons, which are strongly screened at lower altitudes, so its rate increases steeply with height. Ozone can dissociate using longer wavelengths that penetrate much further, producing the much larger, less height-sensitive rate shown by B. The plotted quantity is a first-order coefficient with units ${\mathrm{s}}^{-1}$, rather than a bimolecular rate constant.

The dominant OH-to-HO${}_2$ pseudo-first-order rate is $k_5[{\mathrm{O}}_3]=90{\mathrm{s}}^{-1}$, corresponding to $0.0111\mathrm{s}$. Recycling HO${}_2$ through atomic oxygen and NO has rate $k_6[\mathrm{O}]+k_8[\mathrm{NO}]=0.0218{\mathrm{s}}^{-1}$, corresponding to $45.9\mathrm{s}$ for an individual HO${}_2$ radical. A fixed-total two-state exchange relaxes at the sum of forward and reverse rates, giving about $0.0111\mathrm{s}$ for the partition perturbation. Its fast partition is thereforeThis supports a steady-state approximation in chemical kinetics for rapidly recycled intermediates if the parent concentrations vary slowly. It does not by itself establish stationarity of the entire HOx family. The later source–termination estimate gives a family turnover time of about 16 hours, far longer than partition relaxation; a stationary absolute abundance additionally assumes sufficiently persistent radiation and background conditions. This is the distinction between fast family partition and slow family balance.

Let $v_i$ denote the rate of reaction $i$. For odd oxygen defined as O plus O${}_3$, reactions 5 and 6 each consume one unit, while NO${}_2$ photodissociation in reaction 7 restores one. Thus the loss from reactions 4–8 is $L=v_5+v_6-v_7$. HO${}_2$ stationarity gives $v_5=v_6+v_8+v_4$, and NO${}_2$ stationarity within this restricted recycling loop gives $v_7=v_8$. ThereforeThe requested expression neglects the small termination correction $v_4$ in the odd-oxygen budget, while retaining it to determine the much slower total-radical balance. This is consistent when cycling greatly exceeds termination. The net main catalytic cycle is O plus O${}_3$ forming two O${}_2$ molecules; the NO-assisted route mainly interconverts O and O${}_3$ and cancels from this restricted net loss after NO${}_2$ recycling.

Using an isothermal atmosphere with the stated scale height gives $p(30\mathrm{km})=10^5{e}^{-30/7}\mathrm{Pa}$ and $[\mathrm{M}]=p/(k_{B}T)=3.46\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}$ at $288\mathrm{K}$. The water atmospheric mixing ratio then gives $[{\mathrm{H}}_2\mathrm{O}]=6.92\times 10^{11}\mathrm{c}{\mathrm{m}}^{-3}$. Applying steady state to the short-lived electronically excited oxygen atom,Its HOx family source is $S=2k_3[\mathrm{O}({}^{1}D)][{\mathrm{H}}_2\mathrm{O}]\simeq 6.00\times 10^3\mathrm{c}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$. Balancing this with termination gives $2k_4[\mathrm{OH}][\mathrm{H}{\mathrm{O}}_2]=S$. Combining with the rapid partition, or solving both stationary radical equations including termination, givesConsequently the main catalytic odd-oxygen loss isThe Chapman cycle loss is $L_{\mathrm{Ch}}=2k_9[\mathrm{O}][{\mathrm{O}}_3]=5.40\times 10^5\mathrm{c}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$, so the given HO${}_x$ cycle is about 2.35 times larger. Both rates count two odd-oxygen units per completed O-plus-O${}_3$ destruction event. The omitted $v_4\simeq 3.00\times 10^3$ correction is only about $0.24\%$ of the main HO${}_x$ loss. The family abundance divided by its source is about $5.87\times 10^4\mathrm{s}$, explaining why its absolute stationary state is a stronger assumption than rapid OH/HO${}_2$ interconversion.

With $[{\mathrm{H}}_2]=5\times 10^{-6}[\mathrm{M}]$, a reaction of an electronically excited oxygen atom with hydrogen gives one OH and, after rapid H conversion, one HO${}_2$. The radical source becomesThis is about 4.75 times the original source. Holding water, oxygen, NO and radiation fixed keeps the fast radical partition approximately unchanged, while bimolecular termination makes total radical abundance scale as $\sqrt{S}$. Thusabout 2.18 times the original catalytic loss. This predicts a stronger ozone-loss tendency within the specified stationary box model, not a numerically determined percentage change in the ozone layer. The latter requires coupled odd-oxygen production, changing ozone concentrations, transport and the longer-term water budget; those have been held fixed here.