 Solution

For each ideal gas, the chemical potential relative to the pure gas at standard pressure is $\Delta {\mu }_i=RT\ln (p_i/p^{\circ })$. HenceAll three shifts are negative because their partial pressures are below $1\mathrm{bar}$. As a check, ${\Delta }_r{G}^{\circ }+\Delta {\mu }_{\mathrm{Cl}{\mathrm{F}}_3}-\Delta {\mu }_{\mathrm{ClF}}-\Delta {\mu }_{{\mathrm{F}}_2}=0$ at equilibrium, within rounding.