Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Interdisciplinary paper IDP1 / 2 / d

Put $X=[{\mathrm{O}}_3]$ and $u=T^{-1}$. The fast oxygen–ozone partition obeysDividing the production–loss balance by two and substituting givesUnder the specified restriction, all temperature dependence lies in the three rate coefficients: hold the photodissociation rates and the concentrations of ${\mathrm{O}}_2$, $\mathrm{M}$ and NO fixed. Then $d\ln A/du=-2100-510=-2610\mathrm{K}$ and $d\ln B/du=-1200\mathrm{K}$. Defining $Y=d\ln X/du$, differentiation yieldsThereforeThe factor two in the denominator multiplying $L_C$ comes from the additional dependence $[\mathrm{O}]\propto X/k_4$; ignoring that dependence would give the wrong sensitivity. This derivative has units of kelvin.

Using the stated loss ratio $L_N=3L_C$ in the derived logarithmic sensitivity givesThe positive sign means that cooling, which increases $T^{-1}$, increases the steady ozone concentration within this restricted middle-stratosphere model.

The change in inverse temperature is $\Delta u=1/290-1/300=1.14943\times 10^{-4}{\mathrm{K}}^{-1}$. Treating the sensitivity as approximately constant over the interval,Thus the predicted concentration increases by about 15.3%. Linearizing the exponential gives a rougher 14.3% estimate; a logarithmic change is not itself the exact fractional change.

An independent finite-change calculation checks this approximation without keeping the loss ratio artificially fixed. Normalize the initial Chapman and nitrogen-cycle loss terms to 1 and 3, so the initial production is 4. If $r=X_{290}/X_{300}$, the new steady balance is $e^{-2610\Delta u}{r}^2+3e^{-1200\Delta u}r=4$. Its positive root is $r=1.15342$, essentially the same result.