 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-id/2/d/i/solution

Put

X = [O_{3}]

and

u = T^{- 1}

. The fast oxygen–ozone partition obeys

k_{4} [O] [O_{2}] [M] = J_{5} X, [O] = \frac{J _{5} X}{k _{4} [ O _{2} ] [ M ]} .

Dividing the production–loss balance by two and substituting gives

P = A X^{2} + BX, P = J_{1} [O_{2}], A = \frac{k _{2} J _{5}}{k _{4} [ O _{2} ] [ M ]}, B = k_{3} [NO] .

Under the specified restriction, all temperature dependence lies in the three rate coefficients: hold the photodissociation rates and the concentrations of

O_{2}

M

and NO fixed. Then

d ln A / d u = - 2100 - 510 = - 2610 K

and

d ln B / d u = - 1200 K

. Defining

Y = d ln X / d u

0 = A X^{2} (- 2610 + 2 Y) + BX (- 1200 + Y) .

Therefore

\frac{d ln [ O _{3} ]}{d ( T ^{- 1} )} = \frac{2610 L _{C} + 1200 L _{N}}{2 L _{C} + L _{N}}, L_{C} = k_{2} [O] [O_{3}], L_{N} = k_{3} [NO] [O_{3}] .

The factor two in the denominator multiplying

L_{C}

comes from the additional dependence

[O] \propto X / k_{4}

; ignoring that dependence would give the wrong sensitivity. This derivative has units of kelvin.

 New to topics? Read the docs here!