Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / 42 / c / ii / Solution

Let $v_i$ denote the rate of reaction $i$. For odd oxygen defined as O plus O${}_3$, reactions 5 and 6 each consume one unit, while NO${}_2$ photodissociation in reaction 7 restores one. Thus the loss from reactions 4–8 is $L=v_5+v_6-v_7$. HO${}_2$ stationarity gives $v_5=v_6+v_8+v_4$, and NO${}_2$ stationarity within this restricted recycling loop gives $v_7=v_8$. ThereforeThe requested expression neglects the small termination correction $v_4$ in the odd-oxygen budget, while retaining it to determine the much slower total-radical balance. This is consistent when cycling greatly exceeds termination. The net main catalytic cycle is O plus O${}_3$ forming two O${}_2$ molecules; the NO-assisted route mainly interconverts O and O${}_3$ and cancels from this restricted net loss after NO${}_2$ recycling.