Chemical Engineering Problem: You have a liquid-phase mixture of 0.350 mol fract

Question

Chemical Engineering Problem:

You have a liquid-phase mixture of 0.350 mol fraction ethanol and 0.650 mol fraction water at 760.0 mmHg. You'd like to calculate the bubble-point temperature of the mixture, and the composition of the vapor formed at the bubble-point temperature two different ways. For the first method, assume that the solution behaves ideally and all components follow Raoult's law.

What is the bubble-point temperature of the solution?______C

What are the mole fractions of ethanol (e) and water (w) in the vapor phase at the bubble-point temperature?

For the second method, use the Txy diagram for the ethanol-water system shown below.

http://upload.wikimedia.org/wikipedia/commons/1/1e/Vapor-Liquid_Equilibrium_Mixture_of_Ethanol_and_Water.png

According to this diagram, what is the bubble-point temperature of the 0.350 mol fraction ethanol solution? (Note the requested units.)

Explanation / Answer

water has a boiling point of 100 DegC and ethenol has a boiling point of 78.4 DegC.

Hence the bubble point temperature at atmospheric pressure will lie in the temperature range of 78.4 to 100 DegC.

Now we can find the bubble point temperature by considering an arbitrary temperature T as the bubble point in the range of 78.4 to 100 DegC.

Suppose T = 82 DegC

Now at T = 82 DegC, the vapor pressure, P0 of each component can be calculated from Antoine equation

logP0 = A - B / (C+T)

For water, logP0 = 8.07131 - 1730.63 / (233.426 + 82) = 2.5847

=> P0 = 384.3 mm Hg

For ethanol,  logP0 = 7.68117 - 1332.04 / (199.20+82) = 2.94418

=>  P0 = 879.4 mm Hg

The given liquid phase mole fractions are 0.350 ethanol and 0.650 water.

Xe = 0.350 and Xw = 0.650

Now the vapor phase mole fractions can be calculated from the following formulae.

For ehtanol the mole fraction in vapor phase, Ye = XexP0 / Pt = 0.350 x 879.4 mm Hg/ 760 mmHg = 0.405

For water the mole fraction in vapor phase, Yw = XwxP0 / Pt = 0.650 x 384.3 mm Hg / 760 mmHg = 0.329

Yt = 0.405 + 0.329 = 0.734 which is less than 1.

However at bubble point the Yt should be equals to 1. Hecnce we need to consider a higher temperature.

Let't T = 90:

For water, logP0 = 8.07131 - 1730.63 / (233.426 + 90)

=> P0 = 525.3 mm Hg

For ethanol,  logP0 = 7.68117 - 1332.04 / (199.20+90) = 2.94418

=>  P0 = 1189.1 mm Hg

The given liquid phase mole fractions are 0.350 ethanol and 0.650 water.

Xe = 0.350 and Xw = 0.650

Now the vapor phase mole fractions can be calculated from the following formulae.

For ehtanol the mole fraction in vapor phase, Ye = XexP0 / Pt = 0.350 x 1189.1 mm Hg/ 760 mmHg = 0.548

For water the mole fraction in vapor phase, Yw = XwxP0 / Pt = 0.650 x 525.3 mm Hg / 760 mmHg = 0.449

Yt = 0.548 + 0.449 = 0.997 which is nearly equals to 1.

Hence bubble point temperature is 90 DegC

mole fraction of ethanol at bubble point temperature = 0.548

mole fraction of water at bubble point temperature = 0.449

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