Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 5 / c / iv

Table of contents
- Solution iv

Solution

Let

r_{i}

be the event rate for step

i

, so

r_{1} = k_{1} [M] [C l_{2}], r_{2} = k_{2} [Cl] [H_{2}], r_{3} = k_{3} [H] [C l_{2}], r_{4} = k_{4} [Cl]^{2} [M] .

The free radical balances, including the factors of two for initiation and termination, are

00 ≃ r_{2} - r_{3}, ≃ 2 r_{1} - r_{2} + r_{3} - 2 r_{4} .

Thus

r_{2} = r_{3}

and

r_{1} = r_{4}

[Cl] = \frac{k _{1} [ C l _{2} ]}{k _{4}}, [H] = \frac{k _{2} [ Cl ] [ H _{2} ]}{k _{3} [ C l _{2} ]} .

Both propagation events form hydrogen chloride, so

\frac{d [ HCl ]}{d t} = 2 k_{2} \frac{k _{1}}{k _{4}} [H_{2}] [C l_{2}]^{1/2} .

The orders are one in hydrogen, one half in chlorine and three halves overall, with

k_{obs} = 2 k_{2} k_{1} / k_{4}

when the observed rate is HCl formation. If the rate is instead defined as the stoichiometrically normalized reaction rate, half that formation rate, the prefactor lacks the factor two. The bath concentration cancels because the same M-assisted initiation and recombination control the free radical pool. The result does not require

k_{2} = k_{3}

; different free radical concentrations make their steady fluxes equal. For molar concentration,

k_{obs}

has units

d m^{3/2} mo l^{- 1/2} s^{- 1}

.

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