Problema 3F47

Cálculo da massa molar da mistura gasosa no equilíbrio: $$\rho=\frac{PM}{RT}$$ $$\pu{1,6 g L-1}=\frac{\pu{(1,8 atm)(M)}}{0,082\frac{\pu{atm L}}{\pu{mol K}}(873\,K)}$$ $$M=\pu{63,632 g mol-1}$$ Cálculo do número de mols inicial de $\ce{SO3}$: $$n=\frac{m}{M}$$ $$n=\frac{\pu{88 g}}{\pu{80 g mol-1}}=\pu{1,1mol}$$ Fazendo o quadrinho de equilíbrio: $$\begin{matrix}&\ce{SO3(g)}&\ce{<=>}&\ce{SO2(g)}&\ce{1/2O2(g)} \\ \text{início}&1,1&&0&0 \\ \text{reação}&-x &&+x&+ \frac{x}{2} \\ \text{final}&1,1-x&&x& \frac{x}{2}\end{matrix}$$ Cálculo de x a partir da massa molar da mistura: $$M=\frac{n_{\ce{SO3}}\cdot M_{\ce{SO3}}+n_{\ce{SO2}}\cdot M_{\ce{SO2}}+n_{\ce{O2}}\cdot M_{\ce{O2}}}{n_{\ce{SO3}}+n_{\ce{SO2}}+n_{\ce{O2}}}$$ $$63,632=\frac{(1,1-x)\cdot80+x \cdot64+ \frac{x}{2}\cdot32}{1,1-x+x+ \frac{x}{2}}$$ $$x=\pu{0,57mol}$$ Cálculo dos números de mols de cada gás: $$n_{\ce{SO3}}=1,1-x=\pu{0,53mol}$$ $$n_{\ce{SO2}}=x=\pu{0,57mol}$$ $$n_{\ce{O2}}=\frac{x}{2}=\pu{0,285mol}$$ $$n _\text{total}=1,1-x+x+ \frac{x}{2}=\pu{1,385mol}$$ Cálculo das frações molares de cada gás: $$x_{\ce{SO3}}=\frac{n_{\ce{SO3}}}{n _\text{total}}=\frac{0,53}{1,385}=0,38$$ $$x_{\ce{SO2}}=\frac{n_{\ce{SO2}}}{n _\text{total}}=\frac{0,57}{1,385}=0,41$$ $$x_{\ce{O2}}=1-x_{\ce{SO3}}-x_{\ce{SO2}}=1-0,38-0,41=0,21$$ Cálculo das pressões parciais de cada gás: $$\ce{P_{\ce{SO3}}}=x_{\ce{SO3}}\cdot P _\text{total}=0,38\cdot1,8=\pu{0,684atm}$$ $$\ce{P_{\ce{SO2}}}=x_{\ce{SO2}}\cdot P _\text{total}=0,41\cdot1,8=\pu{0,738atm}$$ $$\ce{P_{\ce{O2}}}=x_{\ce{O2}}\cdot P _\text{total}=0,21\cdot1,8=\pu{0,378atm}$$ Cálculo da constante de equilíbrio: $$K=\frac{\ce{P_{\ce{SO2}}\ce{(P_{\ce{O2}})^{\frac{1}{2}}}}}{\ce{P_{\ce{SO3}}}}$$ $$K=\frac{(0,738)\cdot(0,378)^\frac{1}{2}}{0,684}$$ $$\boxed{K=0,66}$$