= Solution
For each ideal gas, the <chemical potential> relative to the pure gas at standard pressure is $\Delta\mu_i=RT\ln(p_i/p^\circ)$. Hence
$$
\boxed{\begin{aligned}
\Delta\mu_{\mathrm{F_2}}&=-78.45\,\mathrm{kJ\,mol^{-1}},\\
\Delta\mu_{\mathrm{ClF}}&=-0.2611\,\mathrm{kJ\,mol^{-1}},\\
\Delta\mu_{\mathrm{ClF_3}}&=-5.705\,\mathrm{kJ\,mol^{-1}}.
\end{aligned}}
$$
All three shifts are negative because their partial pressures are below $1\,\mathrm{bar}$. As a check, $\Delta_rG^\circ+\Delta\mu_{\mathrm{ClF_3}}-\Delta\mu_{\mathrm{ClF}}-\Delta\mu_{\mathrm{F_2}}=0$ at equilibrium, within rounding.
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