Vapor-Liquid-Liquid Equilibrium (VLLE)

This Demonstration shows the phase equilibrium of a binary system of two liquids, A and B, that are only partially miscible. Vapor-liquid equilibrium, liquid-liquid equilibrium, and vapor-liquid-liquid equilibrium regions are present on the phase diagram because of the partial miscibility. The amounts of each of the three phases present for the temperature and mole fraction of component B represented by the black point (in the diagram) are shown on the right in the bar graph. You can vary the location of the black point by changing the mole fraction of component B and the heat added. The heat added changes the temperature, except when on the horizontal line at about 77C, when three phases can exist in equilibrium: an liquid (), a liquid (), and a vapor (). The relative amounts of each phase are determined from a mole balance. One phase must evaporate (or condense if removing heat) before the temperature increases (or decreases). The amount of heat added is used to illustrate the system behavior and to show how one point on the diagram at about 77C can represent multiple phase conditions. The amount of heat added is not meant to be an accurate value for a real system, but is used to illustrate what phases can be present. The bar graph also shows the mole fraction of component B in each phase, and the mole fraction is set to zero when that phase is not present. In two phase regions, the lever rule is used to determine the amounts of each phase. The diagram also locates the mole fractions in equilibrium in the two phase regions using vertical dashed lines.


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The relative amounts of each phase in the two-phase regions are found using the lever rule. An example of using the lever rule in the liquid plus liquid region is given by the following equations:
relative amount of liquid ,
relative amount of liquid ,
where is the overall mole fraction of the mixture (the mole fraction of the point in the diagram).
When the system is in vapor-liquid-liquid equilibrium, the relative amounts of each phase are found from mass balances. For example, using 10.0% vaporization and an initial mole fraction of B of 0.700 in the diagram:
• determine the initial mole fraction of B
• determine the mole fraction of component B in each phase
• set a basis for the amount of total moles in the system
total moles = 1.00
• determine the percentage of vaporization
percent vaporization = 10.0%
• determine the amount of vapor in the system
= (percent of vaporization) × (total moles) = 0.100 × 1.00 = 0.100
• perform a mole balance on the whole system
total moles = vapor + liquid + liquid
• perform a mole balance on component B
× total moles = ( × ) + ( x ) + ( x V)
0.700 × 1.00 = (0.275 × ) + (0.810 × ) + (0.600 × 0.100)
0.700 = (0.275 × ) + (0.810 × ) + 0.060
• simultaneously solve the equations with unknown variables (the number of equations should equal the number of unknown variables)
0.700 = (0.275 × ) + (0.810 × ) + 0.060
• relative amounts of each phase
= 0.166 = relative amount of liquid phase
= 0.734 = relative amount of liquid phase
= 0.100 = relative amount of vapor
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