 Solution

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.