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Chapter 13: Problem 55

Styrene, used in the manufacture of polystyrene plastics, is made by the extraction of hydrogen atoms from ethylbenzene. The product obtained contains about \(38 \%\) styrene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}_{2}\right)\) and \(62 \%\) ethylbenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{3}\right),\) by mass. The mixture is separated by fractional distillation at \(90^{\circ} \mathrm{C} .\) Determine the composition of the vapor in equilibrium with this \(38 \%-62 \%\) mixture at \(90^{\circ} \mathrm{C}\). The vapor pressure of ethylbenzene is \(182 \mathrm{mmHg}\) and that of styrene is \(134 \mathrm{mmHg}\).

Short Answer

Expert verified
The pressure of the vapor above the mixture of styrene and ethylbenzene at \(90^{\circ} \mathrm{C}\) is \(163.6 \mathrm{mmHg}\).

Step by step solution

01

Calculating Mole Fractions

First, the exercise provides that the mixture is made up of 38% styrene and 62% ethylbenzene by mass. However, Raoult's law uses mole fractions hence we must convert these percentages into mole fractions. The molecular weight of styrene (C6H5CH=CH2) is \(104.15\, g/mol\) and that of ethylbenzene (C6H5CH2CH3) is \(106.17\, g/mol\). If we have 100 g of the mixture, then we will have \(38 g/104.15 g/mol = 0.365\, mol\) of styrene and \(62 g/106.17 g/mol = 0.584\, mol\) of ethylbenzene. The total amount of moles is \(0.365 mol + 0.584 mol = 0.949 mol\). Therefore, the mole fractions (X) of styrene and ethylbenzene are \(0.365 mol/0.949 mol = 0.385\) and \(0.584 mol/0.949 mol = 0.615\) respectively.
02

Applying Raoult's Law

Next, Raoult's Law can be used to determine the partial pressures of styrene and ethylbenzene in the mixture. Raoult's law states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. This gives us \(P_{Styrene} = X_{Styrene} * P^0_{Styrene} = 0.385 * 134 \,mmHg = 51.6\, mmHg\) and \(P_{Ethylbenzene} = X_{Ethylbenzene} * P^0_{Ethylbenzene} = 0.615 * 182 \,mmHg = 112\, mmHg\).
03

Calculating Total Pressure

The total pressure of the vapor in equilibrium with the liquid mixture is the sum of the partial pressures of the components. Using the values obtained in Step 2: \(P_{Total} = P_{Styrene} + P_{Ethylbenzene} = 51.6\, mmHg + 112\, mmHg = 163.6\, mmHg\).

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