 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/5/b/solution

p^{\circ} = 1 bar

K_{p} = \frac{p _{Cl F_{3}} / p ^{\circ}}{( p _{ClF} / p ^{\circ} ) ( p _{F_{2}} / p ^{\circ} )} = exp (\frac{73010}{8.3145 ( 298 )}) = 6.27 \times 1 0^{12} .

Since

p_{ClF} / p_{Cl F_{3}} = 9

, this gives

p_{F_{2}} / p^{\circ} = 1/ (9 K_{p})

. Therefore

p_{F_{2}} = 1.77 \times 1 0^{- 14} bar, p_{ClF} ≃ 0.900 bar, p_{Cl F_{3}} ≃ 0.100 bar .

More exactly, subtract the tiny fluorine pressure from

1 bar

and divide the remainder in the ratio

9 : 1

. The very small fluorine activity reconciles the reactant-rich ClF ratio with the very large equilibrium constant.

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