= Solution
For ideal gases and standard pressure $p^\circ=1\,\mathrm{bar}$,
$$
K_p=\frac{p_{\mathrm{ClF_3}}/p^\circ}{(p_{\mathrm{ClF}}/p^\circ)(p_{\mathrm{F_2}}/p^\circ)}
=\exp\!\left(\frac{73010}{8.3145(298)}\right)=6.27\times10^{12}.
$$
Since $p_{\mathrm{ClF}}/p_{\mathrm{ClF_3}}=9$, this gives $p_{\mathrm{F_2}}/p^\circ=1/(9K_p)$. Therefore
$$
\boxed{p_{\mathrm{F_2}}=1.77\times10^{-14}\,\mathrm{bar},\quad
p_{\mathrm{ClF}}\simeq0.900\,\mathrm{bar},\quad
p_{\mathrm{ClF_3}}\simeq0.100\,\mathrm{bar}.}
$$
More 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>.
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