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

The dominant OH-to-HO

_{2}

pseudo-first-order rate is

k_{5} [O_{3}] = 90 s^{- 1}

, corresponding to $0.0111 s$ . Recycling HO

_{2}

through atomic oxygen and NO has rate

k_{6} [O] + k_{8} [NO] = 0.0218 s^{- 1}

, corresponding to $45.9 s$ for an individual HO

_{2}

radical. A fixed-total two-state exchange relaxes at the sum of forward and reverse rates, giving about

0.0111 s

for the partition perturbation. Its fast partition is therefore

\frac{[ H O _{2} ]}{[ OH ]} ≃ \frac{k _{5} [ O _{3} ]}{k _{6} [ O ] + k _{8} [ NO ]} = 4.13 \times 1 0^{3} .

(1)

This supports a steady-state approximation in chemical kinetics for rapidly recycled intermediates if the parent concentrations vary slowly. It does not by itself establish stationarity of the entire HOx family. The later source–termination estimate gives a family turnover time of about 16 hours, far longer than partition relaxation; a stationary absolute abundance additionally assumes sufficiently persistent radiation and background conditions. This is the distinction between fast family partition and slow family balance.

ii

 0  0

Solution

 0  0

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}

. Therefore

L = 2 v_{6} + v_{4} ≃ 2 k_{6} [H O_{2}] [O] .

(1)

The 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