Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2021 / Interdisciplinary paper IDP1 / 2 / b / Solution

A humidity assumption is necessary: warming does not by itself specify how much water is available. At fixed relative humidity, the Clausius-Clapeyron equation gives the first-order vapor pressure responseIf water abundance is approximated as proportional to $e_s$, the usual first-order estimate is a $26.7\%$ increase in hydroxyl radical production. Integrating the constant-latent-heat expression instead of linearizing gives $e_s(289)/e_s(285)=\exp [(45050/8.314)(1/285-1/289)]=1.3010$.

There is a small but explicit concentration correction when the square brackets are literal molecules per volume. Since $[{\mathrm{H}}_2\mathrm{O}]=\varphi e_s/(k_{B}T)$ and $P_{\mathrm{OH}}=2k[\mathrm{O}({}^{1}D)][{\mathrm{H}}_2\mathrm{O}]$,The corresponding finite-change concentration result is $(285/289)(1.3010)-1=0.2830$. The $25.3\%$ linear concentration estimate and the $26.7\%$ vapor pressure approximation express different stated approximations; neither follows without a humidity condition.

Extra water increases the fraction of oxygen singlet D atoms reacting to form hydroxyl radicals rather than being collisionally quenched and recycled into ozone. In clean, low-nitrogen oxides air this tends to increase ozone loss. In nitrogen oxides-rich air, increased hydroxyl radical can accelerate precursor oxidation and ozone production, while radical termination also changes. Greater hydroxyl radical production therefore does not imply a universally positive or negative ozone response.