O3 + light → O(1D) + O2. (5)
This species can react with water vapor to form two hydroxyl radicals:
O(1D) + H2O → 2OH. (6)
"A second source of OH radicals," they said, "is from oxygenated organics. Formaldehyde (HCOH) provides the simplest example," using photolysis to produce hydroperoxyl radicals (HO2):
HCOH + light → H + HCO, (7)
These HO2 radicals themselves can interact to produce hydrogen peroxide (H2O2), which in turn can also undergo photolysis to produce OH radicals":
HO2 + HO2 → H2O2 + O2 (10)
H2O2 + light → 2OH. (11)
Thus the generation of hydroxyl radicals is understood, but less apparent is why they should generate excess ozone, since the basic equilibrium equation (4) is a function of the ratio of nitrogen dioxide to nitric oxide. The answer is in the propensity of the hydroperoxyl radicals to interact with and oxidize nitric oxide:
HO2 + NO → OH + NO2. (12)
The products of this reaction are important. Not only are more hydroxyl radicals produced, but newly formed nitrogen dioxide also pushes the numerator of the ozone equilibrium equation up, and along with it the concentration of ozone. Without these excess concentrations, it normally requires an ozone molecule to oxidize nitric oxide into nitrogen dioxide, but equation (12) describes a reaction that, unlike equation (3), "provides a way to oxidize NO without consuming an ozone molecule, which in turn implies that the equilibrium in (4) is shifted to higher concentration levels," according to McRae and Russell.
SOURCE: McRae and Russell (1990).