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Lecture 2sf

# Lecture 2sf - Vapor Pressure of Solutions Assume the...

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Vapor Pressure of Solutions Assume the solution consists of solvent A and solute B, with mole fractions x A and x B . Then the vapor pressure of each component proportional to the mole fraction and the vapor pressure of the pure component. The relationship is known as Raoult’s Law. P A = x A P o A P B = x B P o B where P A = vapor pressure of component A in the solution P o A = vapor pressure of pure A x A = the mole fraction of A with corresponding definitions for component B. The vapor pressure of the entire solution is the sum of P A + P B . A solution is prepared by mixing 1.00 mol of benzene and 1.00 mol of toluene at 25 o C Pure benzene: vapor pressure = 95.1 Torr Pure toluene: vapor pressure = 28.4 Torr What is the partial pressure of each component? What is the total vapor pressure of the solution? What is the mole fraction of each component in the vapor? Let A = benzene P o A = 95.1 Torr x A = 0.50 B = toluene P o B = 28.4 Torr x B = 0.50 P A = x A P o A = 0.50 (95.1) = 47.6 torr P B = x B P o B = 0.50 (28.4) = 14.2 torr Total vapor pressure = P A + P B = 47.6 + 12.2 = 61.8 torr

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Composition of vapor x vapor A = total A P P = 8 . 61 6 . 47 = 0.770 x vapor B = total B P P = 8 . 61 2 . 12 = 0.230 Note that the vapor has a different composition than the liquid. The benzene (component A) started out as mole fraction 1/2 in the liquid, but more than half (0.770) in the vapor. Benzene has a higher vapor pressure than toluene. In general, the vapor will be more enriched in the more volatile component (the component with the higher vapor pressure). If we went through a succession of evaporations and condensations, each time the vapor would be more enriched in the more volatile component. After a number of such successive distillations of a benzene/toluene mixture, we would get almost pure benzene.
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