Problema 3E.29

Para as reações de vaporização de cada componente, $$\begin{aligned} \ce{ CCl4(l) &-> CCl4(g) } &&\Delta G_1 = \pu{-61 kJ//mol} - (\pu{-65 kJ//mol}) = \pu{4 kJ//mol} \\ \ce{ C6H6(l) &-> C6H6(g) } &&\Delta G_2 = \pu{129 kJ//mol} - \pu{124 kJ//mol} = \pu{5 kJ//mol} \end{aligned}$$

De $\Delta G = -RT \ln(P^\star/P^\circ)$ com $P^\circ = \pu{760 Torr}$, $$\begin{aligned} P_{\ce{CCl4}}^\star &= \pu{760 Torr} \times e^{-\Delta G_1/RT} = \pu{760 Torr} \times e^{-4000/(8{,}314 \times 298)} \approx \pu{150 Torr} \\ P_{\ce{C6H6}}^\star &= \pu{760 Torr} \times e^{-\Delta G_2/RT} = \pu{760 Torr} \times e^{-5000/(8{,}314 \times 298)} \approx \pu{100 Torr} \end{aligned}$$

Como as quantidades são iguais, $x_{\ce{CCl4}} = x_{\ce{C6H6}} = \pu{0,5}$. De $P_\text{total} = x_{\ce{CCl4}} P_{\ce{CCl4}}^\star + x_{\ce{C6H6}} P_{\ce{C6H6}}^\star$, $$P_\text{total} = \pu{0,5} \times \pu{150 Torr} + \pu{0,5} \times \pu{100 Torr} = \boxed{ \pu{125 Torr} }$$

De $y_{\ce{C6H6}} = P_{\ce{C6H6}}/P_\text{total}$, $$y_{\ce{C6H6}} = \dfrac{ \pu{0,5} \times \pu{100 Torr} }{ \pu{125 Torr} } = \boxed{ \pu{0,4} }$$