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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]

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.

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.