Problem: In a system for separating gases a tank containing a mixture of hydrogen and carbon dioxide is connected to a much larger tank where the pressure is kept very low. The two tanks are separated by a porous membrane through which the molecules must effuse. If the initial partial pressures of each gas is 5.00 atm, what will be the mole fraction of hydrogen in the tank after the partial pressure of carbon dioxide has declined to 4.50 atm?

FREE Expert Solution

Recall that Graham's Law of Effusion allows us to compare the rate of effusion of two gases. 

$$\begin{gathered} \boxed{\frac{{\mathbf{rate}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{rate}}_{{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{=}\sqrt{\frac{{\mathbf{MM}}_{{\mathbf{CO}}_{\mathbf{2}}}}{{\mathbf{MM}}_{{\mathbf{H}}_{\mathbf{2}}}}}} \\ \frac{{\mathbf{rate}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{rate}}_{{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{=}\sqrt{\frac{\mathbf{44}\mathbf{.}\mathbf{0}\cancel{\mathbf{ }\mathbf{g}\mathbf{/}\mathbf{mol}}}{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{g}\mathbf{/}\mathbf{mol}}}} \\ \frac{{\mathbf{rate}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{rate}}_{{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{69} \\ {\mathbf{rate}}_{{\mathbf{H}}_{\mathbf{2}}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{69}\mathbf{\left(}{\mathbf{rate}}_{{\mathbf{CO}}_{\mathbf{2}}}\mathbf{\right)} \end{gathered}$$

$$\begin{gathered} \mathbf{rate}\mathbf{=}\frac{\mathbf{change}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{concentration}}{\mathbf{change}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{time}}\mathbf{=}\frac{{\displaystyle \mathbf{∆}\frac{\mathbf{moles}\mathbf{ }\mathbf{\left(}\mathbf{n}\mathbf{\right)}}{\mathbf{volume}\mathbf{ }\mathbf{\left(}\mathbf{V}\mathbf{\right)}}}}{\mathbf{∆}\mathbf{t}}\mathbf{ }\mathbf{ }\mathbf{ } \\ \frac{{\mathbf{rate}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{rate}}_{{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{=}\frac{{\mathbf{\left[}\frac{{\displaystyle ∆n}}{\cancel{\mathbf{V}\mathbf{·}\mathbf{∆}\mathbf{t}}}\mathbf{\right]}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{\left[}\frac{{\displaystyle ∆n}}{\cancel{\mathbf{V}\mathbf{·}\mathbf{∆}\mathbf{t}}}\mathbf{\right]}}_{{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{=}\frac{{\left[\frac{{\displaystyle ∆P}}{\cancel{\mathrm{RT}}}\right]}_{H_2}}{{\left[{\displaystyle \frac{∆P}{\cancel{\mathrm{RT}}}}\right]}_{{\mathrm{CO}}_2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{r}\mathit{e}\mathit{c}\mathit{a}\mathit{l}\mathit{l}\mathbf{ }\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{:}\mathbf{ }\boxed{\mathbf{PV}\mathbf{=}\mathbf{nRT}}\mathbf{ }\mathbf{\to }\frac{\mathbf{P}}{\mathbf{RT}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{V}} \\ \frac{{\mathbf{rate}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{rate}}_{{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{=}\frac{\mathbf{∆}{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}}}{\mathbf{∆}{\mathbf{P}}_{{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{69} \\ \boxed{\mathbf{∆}{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{69}\mathbf{∆}{\mathbf{P}}_{{\mathbf{CO}}_{\mathbf{2}}}} \end{gathered}$$

$\boxed{{\mathbf{P}}_{\mathbf{total}}\mathbf{=}{\mathbf{P}}_{{\mathbf{CO}}_{\mathbf{2}}\mathbf{,}\mathbf{f}}\mathbf{+}{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{,}\mathbf{f}}}$  $\boxed{{\mathbf{\chi }}_{{\mathbf{H}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{n}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{n}}_{\mathbf{tot}}}\mathbf{=}\frac{{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{,}\mathbf{f}}}{{\mathbf{P}}_{\mathbf{tot}}}}$

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