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Chapter 14 Part IV—Liquid-liquid equilibrium. The use of the stability criteria If we test some liquid mixtures (we think they are a single liquid phase),

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Presentation on theme: "Chapter 14 Part IV—Liquid-liquid equilibrium. The use of the stability criteria If we test some liquid mixtures (we think they are a single liquid phase),"— Presentation transcript:

1 Chapter 14 Part IV—Liquid-liquid equilibrium

2 The use of the stability criteria If we test some liquid mixtures (we think they are a single liquid phase), and we discover that they do not satisfy the stability criteria Such mixtures will split into two liquid phases of different compositions This is important for many separation processes such as solvent extraction

3 LLE equilibrium criteria  and  are two liquid phases, how do we model fugacity in a liquid phase?

4 Note that the activity coefficients are functions of compositions in the respective phases, T, and P

5 LLE at constant P or at reduced temperature (weak P-dependence)— binary system How many equations and how many unknowns?

6 LLE T-x diagram (solubility diagram)

7 Computing a T-x diagram Fix T and solve for compositions of component 1 in the two liquid phases

8 Prediction of LLE is strongly dependent on the chosen G E model Since we obtain the activity coefficients from G E, the selection of the G E model is crucial to appropriately describe LLE Some G E models cannot describe LLE (for example the Wilson equation, see example 14.7) The most advanced UNIQUAC, UNIFAC are able to describe LLE

9 Example: very low miscibility Suppose the oil-water case. The amount of water dissolved in oil is extremely small, so the “oil” phase (  ) is very dilute in component 1 (water), and the “aqueous” phase (  ) is very dilute in component 2 (hydrocarbon)

10 LLE equations for two almost immiscible liquids: these equations give us estimates of the compositions in both liquid phases based on a model for G E. Usually the activity coefficients at infinite dilution are related in a straightforward way to the model parameters Alternatively, if we have measured compositions in the liquid phases we can determine the activity coefficients at infinite dilution

11 The simplest G E model that predicts LLE For two phases and LLE:

12 Using Margules 1-parameter Solubility curves are symmetric with respect to x 1 =0.5

13 Only because the curves are symmetric: we can write: A > 2  3 roots: x 1 , x 1 , ½ A=2  3 roots =1/2 A < 2  only one root =1/2 A is a function of temperature !!!

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15 Temperature dependence of A For example: The excess enthalpy and the T-dependence of A are directly related

16 H E > 0  endothermic H E <0  exothermic When A =2 we have a consolute point (could be upper or lower or both) dA/dT > 0, is a LCST dA/dT < 0, is a UCST

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18 Depending on the values of a, b, and c this equation may have 0, 1, 2, or 3 Temperature roots For example T L = 272.9 K and T U =391.2 K

19 From the graph A vs T we know if there is an UCST or a LCST A>2 dA/dT>0 A=2 A>2 dA/dT<0 H E changes sign in the temperature interval of LLE

20 Only one root: T = 346 K; it is a UCST; H E is >0 A>2, dA/dT <0 A<2, dA/dT <0

21 Only one root, T = 339.7 K, there is a LCST A 0 A>2, dA/dT >0

22 Conclusion The model G E /RT =A x 1 x 2 cannot predict LLE for values of A <2 From the stability criteria we said that

23 When x 1 = x 2 =1/2,minimum value rhs

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25 HW #10, Due Wednesday, Nov. 7th Problems 14.4; 14.8 (part a); 14.11; 14.16; 14.18; 14.20


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