Chapter 17, Problem 46E

Interpretation Introduction

Interpretation: A solution is made by mixing $\text{0}\text{.0300 mole}$ of ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and $\text{0}\text{.0500 mole}$ of ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ at $25°\text{C}$ . Assuming the solution to be ideal, the composition of vapor (in terms of mole fraction) at $25°\text{C}$ needs to be determined. At $25°\text{C}$ , the vapor pressure of pure ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and the pure ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ are $\text{133}\text{torr}$ and $\text{11}\text{.4 torr}$ respectively.

Concept introduction: Vapor pressure or an equilibrium vapor pressure of a substance is a pressure exerted by the vapors in thermodynamic equilibrium with its condensed phase (like liquid or solid) at the given temperature in a closed system. It can be calculated as: ${\text{P}}_{\text{soln}}\text{=}{\chi }_{\text{solvent}}\text{×}{\text{P}}_{\text{solvent}}^{\text{0}}$

Answer to Problem 46E

The mole fraction of ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ in the vapor are $\text{0}\text{.875 and 0}\text{.125}$ respectively.

Explanation of Solution

Mole fraction of each component in vapor can be calculated by taking the ratio of partial pressure of each component in vapor above the solution to the total vapor pressure of the solution.

For this, first calculate the total vapor pressure of the solution by using modified Raoult;s law;

  $\begin{array}{l}{P}_{Total}=P_{C{H}_{2}C{l}_2}+P_{C{H}_{2}B{r}_2}\\ P_{Total}={\chi }_{C{H}_{2}C{l}_2}{P}_{{}_{C{H}_{2}C{l}_2}}^0+{\chi }_{C{H}_{2}B{r}_2}{P}_{{}_{C{H}_{2}B{r}_2}}^0\mathrm{......}\left(1\right)\end{array}$

Here, ${\chi }_{C{H}_{2}C{l}_2}$ and ${\chi }_{C{H}_{2}B{r}_2}$ represents the mole fractions of ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ components.

  $P$ represents the partial pressures of the each components in vapor above the solution and $P^0$ represents the vapor pressure of the pure components.

We have a solution prepared from $\text{0300 mole}$ of ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and $\text{0}\text{.0500 mole}$ of ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ . Then, the mole fractions of ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ components can be calculated as;

Mole fraction ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ can be calculated as;

  $\begin{array}{c}{\chi }_{C{H}_{2}C{l}_2}=\frac{0.0300\text{mol}\text{of}{\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}}{0.0300\text{mol}\text{of}{\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}+0.0500\text{mol}\text{of}{\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}}\\ =0.375\end{array}$

Mole fraction ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ can be calculated as;

  $\begin{array}{c}{\chi }_{C{H}_{2}B{r}_2}=\frac{0.0500\text{mol}\text{of}{\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}}{0.0300\text{mol}\text{of}{\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}+0.0500\text{mol}\text{of}{\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}}\\ =0.625\end{array}$

Vapor pressures of the pure components are given as:

  $P_{{}_{C{H}_{2}C{l}_2}}^0=133\text{torr}$

  $P_{{}_{C{H}_{2}B{r}_2}}^0=\text{11}\text{.4}\text{torr}$

Now, substitute the values in equation (1) determine the total vapor pressure of the solution as

  $\begin{array}{c}{P}_{Total}=\left(133\text{torr}\right)\left(0.375\right)+\left(11.4\text{torr}\right)\left(0.625\right)\\ =49.9\text{torr}+7.13\text{torr}\\ =57.0\text{torr}\end{array}$

Now, the mole fraction of each component in vapor can be calculated by taking the ratio of partial pressure of the each component in vapor above the solution ( P ) to the total pressure of the solution $\left(P_{Total}\right)$ .

  $\begin{array}{c}\text{Mole}\text{fraction}\text{of}{\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{in}\text{vapor}=\frac{{\chi }_{C{H}_{2}C{l}_2}{P}_{{}_{C{H}_{2}C{l}_2}}^0}{P_{Total}}\\ =\frac{\left(133\text{torr}\right)\left(0.375\right)}{57.0\text{torr}}\\ =0.875\end{array}$

  $\begin{array}{c}\text{Mole}\text{fraction}\text{of}{\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}\text{in}\text{vapor}=\frac{{\chi }_{C{H}_{2}B{r}_2}{P}_{{}_{C{H}_{2}B{r}_2}}^0}{P_{Total}}\\ =\frac{\left(11.4\text{torr}\right)\left(0.625\right)}{57.0\text{torr}}\\ =0.125\end{array}$

Hence, the mole fraction of ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ in the vapor are 0.875 and 0.125 respectively.

Conclusion

The mole fraction of ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and ${\text{CH}}_{\text{2}}{\text{Br}}_{\text{2}}$ in the vapor are 0.875 and 0.125 respectively.