Chemistry · General chemistry II · Worked example

Vapor pressure of a benzene–toluene mixture

A liquid mixture contains $1.00 mol$ benzene and $2.00 mol$ toluene at $25 ° C$ , where the pure vapor pressures are $95.1 torr$ (benzene) and $28.4 torr$ (toluene). Treating the mixture as ideal, find the total vapor pressure and the mole fraction of benzene in the vapor.

P_{total} = χ_{B} P_{B}^{\circ} + χ_{T} P_{T}^{\circ}

Liquid mole fractions

The glucose example had one volatile component. Here both evaporate, so each keeps its own Raoult’s-law term.

χ_{B} = \frac{1.00}{3.00} = 0.333, χ_{T} = \frac{2.00}{3.00} = 0.667

\begin{array}{c} χ_{B} = \frac{1.00}{3.00} = 0.333 \\ χ_{T} = \frac{2.00}{3.00} = 0.667 \end{array}

Partial pressures

Multiply each mole fraction by that component’s pure vapor pressure.

P_{B} = (0.333) (95.1) = 31.7 torr, P_{T} = (0.667) (28.4) = 18.9 torr

\begin{array}{c} P_{B} = (0.333) (95.1) = 31.7 torr \\ P_{T} = (0.667) (28.4) = 18.9 torr \end{array}

Total pressure and vapor composition

Dalton’s law adds the partial pressures. The benzene mole fraction in the vapor is its share of the total.

P_{total} = 50.6 torr, y_{B} = \frac{31.7}{50.6} = 0.626

\begin{array}{c} P_{total} = 50.6 torr \\ y_{B} = \frac{31.7}{50.6} = 0.626 \end{array}

Result

The total vapor pressure is $50.6 torr$ . The vapor is $62.6 %$ benzene although the liquid is only $33.3 %$ benzene. The vapor is enriched in the more volatile component.

Your turn

1. One like the example

Reverse the amounts, $2.00 mol$ of benzene with $1.00 mol$ of toluene at $25 ° C$ . Find the total vapor pressure and benzene’s mole fraction in the vapor.

Show the answer and explanation

$72.9 torr$ , and the vapor is $0.870$ benzene.

Now $χ (benzene) = 2 / 3$ . $P (benzene)$ $= (2 / 3) (95.1)$ $= 63.4 torr$ and $P (toluene)$ $= (1 / 3) (28.4)$ $= 9.47 torr$ , so the total is $72.87 torr$ , or $72.9 torr$ . The vapor is $63.4 \div 72.87 = 0.870$ benzene.

\begin{array}{l} P_{t o t a l} = \frac{2}{3} (95.1) + \frac{1}{3} (28.4) \\ P_{t o t a l} \approx 72.9 \\ y_{B} = \frac{\frac{2}{3} (95.1)}{P_{t o t a l}} \\ y_{B} \approx 0.870 \end{array}

2. A variation

A benzene–toluene mixture at $25 ° C$ has a total vapor pressure of $50.0 torr$ . Let $x$ be benzene’s mole fraction in the liquid. Find $x$ .

Show the answer and explanation

$x \approx 0.324$ .

The total is $x (95.1) + (1 - x) (28.4)$ $= 50.0$ . Expanding gives $66.7 x + 28.4 = 50.0$ , so $66.7 x = 21.6$ and $x = 0.324$ . That is close to the $1$ : $2$ mixture, whose total was $50.6 torr$ .

\begin{array}{l} 50.0 = 95.1 x + 28.4 (1 - x) \\ 50.0 = 66.7 x + 28.4 \\ 21.6 = 66.7 x \\ x \approx 0.324 \end{array}

3. Somewhere new

The vapor above the $1$ : $2$ mixture is condensed and allowed to evaporate again at $25 ° C$ . What is benzene’s mole fraction in the new vapor?

Show the answer and explanation

About $0.849$ , richer in benzene again.

The condensed liquid is $0.626$ benzene and $0.374$ toluene. Its new vapor is $(0.626) (95.1)$ $\div [(0.626) (95.1) + (0.374) (28.4)]$ $= 59.5 \div 70.2$ $= 0.849$ benzene. Each repeat enriches the more volatile component, which is what a fractionating column does continuously.

\begin{array}{l} y = \frac{(0.626) (95.1)}{(0.626) (95.1) + (0.374) (28.4)} \\ y \approx 0.849 \end{array}

\begin{array}{c} y = \frac{0.626 \cdot 95.1}{0.626 \cdot 95.1 + 0.374 \cdot 28.4} \\ y \approx 0.849 \end{array}

Keep exploring

Open the rows in a Math box, where the mole fractions are written with $x$ instead of $χ$ , because Math does not yet read $χ$ before a bracket. Each partial pressure, the total and the vapor fraction read ✓. Change the mole fractions to see how the vapor composition moves.

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Liquid mole fractions

Partial pressures

Total pressure and vapor composition

Result

Your turn

1. One like the example

72.9 torr72.9 torr, and the vapor is 0.8700.870 benzene.

2. A variation

x≈0.324x≈0.324.

3. Somewhere new

About 0.8490.849, richer in benzene again.

Keep exploring

Study it, then try it

$72.9 torr$ , and the vapor is $0.870$ benzene.

$x \approx 0.324$ .

About $0.849$ , richer in benzene again.