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

For ideal gases and standard pressure $p^{\circ }=1\mathrm{bar}$,Since $p_{\mathrm{ClF}}/p_{\mathrm{Cl}{\mathrm{F}}_3}=9$, this gives $p_{{\mathrm{F}}_2}/p^{\circ }=1/(9K_p)$. ThereforeMore exactly, subtract the tiny fluorine pressure from $1\mathrm{bar}$ 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.