chapter11
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chapter11

Course Number: CHE 327, Fall 2010

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Chapter 11 Liquid-Liquid Equilibrium When two liquids are mixed, depending on the temperature and/or composition, either they are completely miscible in each other and form a single phase or, they are partially miscible (or, totally immiscible) in each other and form two separate phases. In the case of completely miscible, i.e., single phase systems, the equations developed in Chapter 6 are applicable. What we are...

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11 Chapter Liquid-Liquid Equilibrium When two liquids are mixed, depending on the temperature and/or composition, either they are completely miscible in each other and form a single phase or, they are partially miscible (or, totally immiscible) in each other and form two separate phases. In the case of completely miscible, i.e., single phase systems, the equations developed in Chapter 6 are applicable. What we are interested in this chapter is the case when a liquid is partially miscible (or totally immiscible) in another liquid. First, the reason for making liquids partially miscible in each other will be investigated. Then, the procedure to determine the compositions in two separate liquid phases that are in equilibrium with each other will be described. Such equilibrium data are needed in various chemical and physical processes such as liquid-liquid extraction and tertiary oil recovery. 11.1 STABILITY OF MULTICOMPONENT LIQUID MIXTURES For thermodynamically allowable changes at constant temperature and pressure, the criterion (dG)T,P ≤ 0 (11.1-1) indicates that all irreversible changes taking place at constant temperature and pressure must decrease the total Gibbs energy of the system. When the system reaches equilibrium, no further changes can occur and Gibbs energy reaches its minimum value. Let us consider mixing of two liquids, 1 and 2, at constant temperature and pressure. The Gibbs energy of this binary mixture is e e Gmix = n1 G1 + n2 G2 + ∆Gmix (11.1-2) If the Gibbs energy of the mixture is lower than the summation of the Gibbs energies of pure components, then components 1 and 2 will be miscible (totally or partially) in each other. This statement is mathematically expressed as e e Gmix < n1 G1 + n2 G2 ∆Gmix < 0 which is the criterion for miscibility (partial or complete). The Gibbs energy of mixing is given by e e e ∆Gmix = ∆Hmix − T ∆Smix | {z } | {z } A B (11.1-3) or, in other words, (11.1-4) (11.1-5) Since mixing increases the degree of disorder within the system, the entropy change on mixing, e ∆Smix , is always positive. Thus, the term B in Eq. (11.1-5) is always positive. The heat of e mixing, ∆Hmix , is related to the energetic effects. In order to make a mixture of 1 and 2 from 343 pure components, it is necessary to break 1-1 (E1 ) and 2-2 (E2 ) bonds and form 1-2 (E12 ) bonds. e Depending on the magnitude of the interactions between like and unlike molecules, ∆Hmix may take positive or negative values. Therefore, it is necessary to consider the following two cases. e Case (i): Exothermic mixing (∆Hmix < 0) e When ∆Hmix < 0, interactions between unlike molecules are greater than those of like molecules, i.e., E1 + E2 (11.1-6) E12 > 2 e In this case, ∆Gmix is negative for all compositions of components 1 and 2. Therefore, liquids are completely miscible in each other and form a single phase. This situation is illustrated in Figure 11.1. 0 ~ ΔH mix ~ − T ΔS mix ~ ΔGmix 0 0 ~ − T ΔS mix ~ ΔH mix ~ ΔG mix 0 x1 1 x1 1 e Case (ii): Endothermic mixing (∆Hmix > 0) e When ∆Hmix > 0, interactions between like molecules are stronger than those of unlike molecules, i.e., E1 + E2 (11.1-7) E12 < 2 In this case, examination of Eq. (11.1-5) indicates that the terms A and B compete with each other and the magnitude of the temperature decides the dominant term and, hence, the sign e of ∆Gmix . e e a) At low temperatures, ∆Hmix À T ∆Smix and, as shown in Figure 11.2, this indicates that emix > 0. Therefore, liquids 1 and 2 prefer to remain as two separate phases throughout the ∆G whole composition range, i.e., liquids 1 and 2 are immiscible in each other. e e e e Figure 11.1 Effects of ∆Smix and ∆Hmix on ∆Gmix when ∆Hmix < 0. ~ Δ H mix ~ Δ Gmix 0 ~ − T ΔS mix 0 x1 1 e e e e Figure 11.2 Effects of ∆Smix and ∆Hmix on ∆Gmix when ∆Hmix > 0 with e e mix À T ∆Smix . ∆H 344 e b) When temperature is high enough so that ∆Gmix < 0, we may come across with two e e different situations. When T ∆Smix À ∆Hmix , liquid phases mix with each other throughout the whole composition range as shown in Figure 11.3. ~ ΔH mix 0 ~ Δ Gmix ~ − T Δ S mix 0 1 x1 e e An interesting case may happen when T ∆Smix > ∆Hmix . Under these circumstances, e variation in ∆Gmix with respect to composition may result in more than one minimum as shown in Figure 11.4. Examination of this case requires the following mathematical preliminaries. ~ Δ Gmix 0 e e e e Figure 11.3 Effects of ∆Smix and ∆Hmix on ∆Gmix when ∆Hmix > 0 with e e T ∆Smix À ∆Hmix . 0 x1 1 A function f (x) is said to be a concave function in x if and only if the points on a chord connecting two points on a function lie beneath the function. On the other hand, a function f (x) is said to be a convex function in x if and only if the points on a chord connecting two points on a function lie above the function. Concave and convex functions are shown in Figure 11.5. f(x) Concave Function f(x) Convex Function e e Figure 11.4 Variation in ∆Gmix with composition when ∆Hmix > 0 with e emix > ∆Hmix . T ∆S x x Figure 11.5 Concave and convex functions. 345 Now, consider a function f which is dependent on a single variable x. Various possibilities exist for the variation of f (x) with x as shown in Figure 11.6. Note that the functions on the left (a and c) are convex, while the functions on the right (b and d) are concave. f (x ) df >0 dx d2 f >0 dx 2 f (x ) df >0 dx d2 f <0 dx 2 x (a) (b) x f (x ) df <0 dx d2f >0 dx 2 f (x ) df <0 dx d2f <0 dx 2 x (c) (d) x Figure 11.6 Variations of the first and second derivatives of various functions. The first derivative of f (x), df (x)/dx, measures the rate of change of the function f (x). The second derivative of f (x), d2 f (x)/dx2 , measures the rate of change of the first derivative, df (x)/dx. The signs of the first and second derivatives of f (x) imply the following ½ df (x) > 0 Value of f (x) increases with increasing x (11.1-8) < 0 Value of f (x) decreases with increasing x dx ½ d2 f (x) > 0 Slope of f (x) vs x tends to increase (11.1-9) < 0 Slope of f (x) vs x tends to decrease dx2 Note that the second derivative can be used to determine whether a function is concave or convex. The second derivatives of concave and convex functions are negative and positive, respectively. Let us consider a concave function of a single variable, f (x), and draw a chord joining any two points on the function as shown in Figure 11.7. f ( x) f [λ x A + ( 1 − λ )xB ] f ( xB ) φ[λ xA + (1 − λ )x B ] f ( xA ) xA xC xB x Figure 11.7 A concave function. 346 The chord intersects the function at xA and xB . Let xc be any value between xA and xB in such a way that (11.1-10) xc = λxA + (1 − λ)xB where 0 ≤ λ ≤ 1. The equation of the chord is given by φ(x) = f (xB ) − f (xA ) (x − xA ) + f (xA ) xB − xA (11.1-11) When x = λxA + (1 − λ)xB , from Eq. (11.1-11) we have h i φ λ xA + (1 − λ) xB = λ f (xA ) + (1 − λ) f (xB ) Substitution of Eq. (11.1-13) into Eq. (11.1-12) gives ¤ £ λ f (xA ) + (1 − λ) f (xB ) < f λ xA + (1 − λ) xB From Figure 11.7 it is apparent that i h i h φ λxA + (1 − λ)xB < f λxA + (1 − λ)xB (11.1-12) (11.1-13) (11.1-14) Now, let us tackle the case shown in Figure 11.4 by drawing a common tangent line to the minima1 as shown in Figure 11.8. The so-called common tangent rule states that when the composition of the mixture is between xα and xβ , instead of having a homogeneous mixture, we 1 1 have two separate phases, α and β , that are in equilibrium with each other. The compositions of these coexisting equilibrium phases lie at the points of co-tangency, i.e., xα and xβ . The 1 1 coexisting compositions are called binodal points. The reason for the phase separation can be expressed as follows. Points of co-tangency (not minima) ~ Δ Gmix 0 Inflection points ~ d 2 Δ Gmix =0 dx12 Common tangent to the minima α x1 β x1 0 x1 1 Figure 11.8 Common tangent rule. Let x∗ be the mole fraction of component 1 between xα and xβ , i.e., 1 1 1 x∗ = λ xα + (1 − λ) xβ 1 1 1 1 (11.1-15) Keep in mind that drawing a common tangent to the minima does not imply joining the minimum points by a straight line. 347 where 0 ≤ λ ≤ 1. Note that the solution of Eq. (11.1-15) for λ gives λ= From the lever rule xβ − x∗ 1 1 − xα 1 xβ − x∗ 1 1 β x1 − xα 1 = nα nα + nβ (11.1-16) β x1 (11.1-17) which states that the combined Gibbs energies of the two separate liquid phases is lower than the Gibbs energy of a homogeneous mixture. Since the system tries to minimize its Gibbs energy, over this composition range we have two separate phases of compositions xα and xβ , 1 1 and the common tangent acts as a "tie line". e When xα < x1 < xβ , there are two inflection points at which d2 ∆Gmix /dx2 = 0. The 1 1 1 e inflection points are called spinodal points. Between the two inflection points, ∆Gmix is a 2 ∆G emix /dx2 < 0. concave function with d 1 e Figure 11.8 shows the variation of ∆Gmix as a function of composition at a fixed temperature, say T1 . By using these data, it is possible to locate binodal and spinodal compositions on e a temperature-composition (or solubility) diagram as shown in Figure 11.9. If ∆Gmix versus x1 data are known at various temperatures, then it is possible to draw binodal and spinodal curves. ~ ΔG mix 0 T = T1 Comparison of Eqs. (11.1-16) and (11.1-17) indicates that the term λ represents the mole fraction of the α-phase. According to Eq. (11.1-14) h i e e e λ ∆Gmix (xα ) + (1 − λ) ∆Gmix (xβ ) < ∆Gmix λ xα + (1 − λ) xβ (11.1-18) 1 1 1 1 0 T x1 1 Binodal curve T1 Spinodal c urve 0 x1 1 Figure 11.9 Liquid-liquid solubility diagram. 348 The binodal and spinodal curves coincide at the critical solution ( or consolute) temperature, i.e., the temperature at which two partially miscible liquids become fully miscible. In other words, when the temperature is above the critical solution temperature, the mixture is completely miscible and forms a homogeneous phase. At temperatures less than the critical solution temperature, the mixture is partially miscible and the binodal curve is the boundary between the two-phase and one-phase regions. Within the two-phase region, the region between the spinodal and binodal curves is metastable. The time it takes for the phase separation is not definite. On the other hand, within the spinodal curve, the mixture is unstable and phase separation takes place immediately. Since the end points of the spinodal curve represent inflection points, then the condition of unstability of liquid mixtures can be stated as e d2 ∆Gmix dx1 2 <0 Condition of unstability (11.1-19) where e e e ∆GIM ∆Gmix Gex mix = + RT RT RT N X e Gex xi ln xi + = RT i=1 (11.1-20) For a binary system, substitution of Eq. (11.1-20) into Eq. (11.1-19) gives e 1 d2 (Gex /RT ) + <0 2 x1 x2 dx 1 (11.1-21) One should keep in mind that the criterion given by Eq. (11.1-4) is the necessary but not e the sufficient condition for complete miscibility of components. When ∆Gmix (or ∆Gmix ) is a convex function of composition with only one minimum, components are miscible in each other over the entire composition range and form a homogeneous phase. When ∆Gmix (or e ∆Gmix ) is a convex function of composition with more than one minimum, phase separation, i.e., formation of two (or more) phases, takes place. Example 11.1 In a binary liquid mixture of 1 and 2 at constant temperature and pressure, the molar excess Gibbs energy is expressed as h i e Gex = x1 x2 1.8 + 0.3 (x1 − x2 ) − 0.85 (x1 − x2 )2 RT Therefore, the concentrations at the spinodal points are obtained from the solution of the following equation: e 1 d2 (Gex /RT ) + =0 (11.1-22) 2 x1 x2 dx 1 Determine the region of unstability. Solution e Substitution of Gex /RT into Eq. (11.1-22) and differentiation give 349 408 x4 − 852 x3 + 511 x2 − 67 x1 − 10 = 0 1 1 1 (1) R ° The solution of Eq. (1) by MATHCAD gives x1 = − 8.54 × 10−2 , x1 = 0.378, x1 = 0.682 and x1 = 1.114. Therefore, the limits of unstability are given as 0.378 < x1 < 0.682 In other words, these are the mole fractions at the spinodal points. Comment: Equation (11.1-20) gives h i e ∆Gmix = x1 ln x1 + x2 ln x2 + x1 x2 1.8 + 0.3 (x1 − x2 ) − 0.85 (x1 − x2 )2 RT 0 e The variation of ∆Gmix /RT versus x1 is given in the figure below. − 0.043 − 0.1 G( x) − 0.2 − 0.3 − 0.35 0 0.01 0.2 0.4 x 0.6 0.8 0.99 1 The common tangent rule gives the binodal compositions as x1 = 0.26 and x1 = 0.80. The simplest function to express molar excess Gibbs energy as a function of composition is the one-constant Margules model, Eq. (8.4-1), e Gex = A x1 x2 RT (11.1-23) Substitution of Eq. (11.1-23) into Eq. (11.1-20) gives Note that e ∆Gmix = x1 ln x1 + x2 ln x2 + A x1 x2 RT d dx1 à e ∆Gmix RT ! = ln µ x1 x2 ¶ + A(x2 − x1 ) (11.1-24) (11.1-25) e indicating that d(∆Gmix /RT )/dx1 = 0 at x1 = x2 = 0. On the other hand, the use of Eq. (11.1-23) in Eq. (11.1-21) indicates that phase splitting takes place when µ ¶ 1 1 (11.1-26) A> 2 x1 x2 The function 1/x1 x2 becomes + ∞ for pure components, i.e., either x1 = 1 or x2 = 1, and falls to a minimum value of 4 at x1 = 0.5. Therefore, a binary liquid mixture becomes unstable and 350 e exists as two separate phases when A > 2. A plot of Eq. (11.1-24) in the form of ∆Gmix /RT versus x1 is presented in Figure 11.10 for various values of A, i.e., 0, 1, 2, 3, 4. 0.4 0.307 0.2 G ( x1, 0) G ( x1, 1) 0 G ( x1, 2) − 0.2 G ( x1, 3) G ( x1, 4) − 0.4 − 0.6 − 0.693 − 0.8 0 1×10 −3 0.2 0.4 x1 0.6 0.8 0.999 1 e Figure 11.10 A plot of Eq. (11.1-24) in the form of ∆Gmix /RT versus x1 with A being a parameter. The parameter of the two-suffix Margules equation, A, is dependent on temperature. If A decreases with increasing temperature as shown in Figure 11.11-a, then the temperature corresponding to A = 2 is the upper critical solution temperature (UCST). At temperatures greater than UCST, a binary mixture becomes homogeneous. When A increases with increasing temperature as shown in Figure 11.11-b, then the temperature corresponding to A = 2 is the lower critical solution temperature (LCST). In this case, a binary mixture becomes homogeneous at temperatures less than LCST. If the variation of A with respect to temperature exhibits a maximum, then a binary mixture has both UCST and LCST as shown in Figure 11.11-c. Finally, if the variation of A with respect to temperature exhibits a minimum, then the corresponding solubility diagram is shown in Figure 11.11-d. 11.2 LIQUID-LIQUID PHASE EQUILIBRIUM CALCULATIONS Consider a multicomponent mixture of k species distributed in two liquid phases, α and β . The condition of equilibrium states that the fugacities of each species in α- and β -phases must be equal to each other, i.e., b b fiα (T, P, xα ) = fiβ (T, P, xβ ) i i i = 1, 2, ..., k (11.2-1) In terms of activity coefficients, Eq. (11.2-1) takes the form γ α (T, P, xα ) xα fi (T, P ) = γ β (T, P, xβ ) xβ fi (T, P ) i i i i i i or, γ α (T, P, xα ) xα = γ β (T, P, xβ ) xβ i i i i i i i = 1, 2, ..., k (11.2-3) (11.2-2) In each phase, the mole fractions are related to each other by the following equations k X i=1 xα i = 1.0 and k X i=1 xβ = 1.0 i (11.2-4) Simultaneous solution of Eqs. (11.2-3) and (11.2-4) gives the coexistence curve for the twophase system. 351 T A TWO PHASES → 2 ONE PHASE T * (UCST) ONE PHASE TW O PHAS ES T* (UCST) A T (a) x1 T TWO 2 PHASES → TWO PHASES ONE PHASE * T (LCST) T (LCST) * ONE PHASE x1 T (b) T UCST A ONE PHASE TWO 2 PHASES → TW O PHA SES ONE PHASE LCST LCST A UCST T ONE PHASE x1 (c) T TWO PHASES TWO PHASES 2 ONE PHASE → LCST ONE PHASE UCST TWO PHASES UCST LCST T (d) x1 Figure 11.11 Four different types of solubility diagrams. For a binary system, Eqs. (11.2-3) and (11.2-4) become ⎛⎞ à α! xβ γ1 1 = ln ⎝ α ⎠ ln β γ1 x1 ln à γα 2 β γ2 (11.2-5) ! In practice, one encounters two types of liquid-liquid equilibrium calculations: 352 = ln ⎝ ⎛ α 1 − x1 1 − xβ 1 ⎞ ⎠ (11.2-6) • When compositions are known, Eqs. (11.2-5) and (11.2-6) can be solved for the parameters of an activity coefficient model. • When parameters of an activity coefficient model are known, Eqs. (11.2-5) and (11.2-6) can be solved for the compositions in α- and β -phases. Example 11.2 Diethyl ether (1) and water (2) form two partially miscible liquid phases. At 308 K and atmospheric pressure, Villamanan et al. (1984) reported the following compositions of the two phases and xβ = 0.9500 xα = 0.01172 1 1 If the system is represented by the three-suffix Margules equation, estimate the parameters A and B . Solution Substitution of Eqs. (8.4-6) and (8.4-7) into Eqs. (11.2-5) and (11.2-6) gives ⎛⎞ ∙ ´2 ¸ ³ i h xβ β β2 β α2 α2 α ⎝ 1 ⎠ (1) A (1 − x1 ) − 1 − x1 + B (1 − x1 ) (4 x1 − 1) − (1 − x1 ) (4 x1 − 1) = ln α x1 ⎞ ⎛ ∙ h ³ ´2 ¸ i 1 − xβ 1⎠ + B (xα )2 (4 xα − 3) − (xβ )2 (4 xβ − 3) = ln ⎝ A (xα )2 − xβ 1 1 1 1 1 1 α 1 − x1 ⎡ (1 − xα )2 − (1 − xβ )2 1 1 (xα )2 1 − (xβ )2 1 (1 − xα )2 (4 xα − 1) − (1 − xβ )2 (4 xβ − 1) 1 1 1 1 (xα )2 (4 xα 1 1 − 3) − (xβ )2 (4 xβ 1 1 − 3) ⎡ ⎤−1 ⎦ ⎞⎤ (2) The use of matrix algebra gives the parameters A and B as µ A B ¶ =⎣ or, A= where X (Λ − Θ) − 6 Λ (xα + xβ − 1) 1 1 2 (xα − xβ )3 1 1 and B= ⎢ ln ⎝ ⎠ ⎥ ⎥ ⎢ α x1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ×⎢ ⎛ ⎞ ⎥ (3) ⎢ ⎥ 1 − xβ ⎥ ⎢ 1 ⎠⎦ ⎣ ln ⎝ α 1 − x1 (xα + xβ ) (Θ − Λ) + 2 Λ 1 1 2 (xα − xβ )3 1 1 ⎞ ⎠ ⎛ 1 − xβ 1 ⎞ ⎠ (4) ⎛ xβ 1 X = (xα + xβ )(4xα + 4xβ − 3) − 4xα xβ 1 1 11 1 1 Substitution of the numerical values into Eqs. (4) and (5) gives A = 3.854 and 353 Θ = ln ⎝ ⎛ xβ 1 xα 1 Λ = ln ⎝ 1 − xα 1 (5) B = − 0.683 Comment: Since phase separation takes place, like interactions are stronger than the unlike interactions. Example 11.3 Estimate the compositions of the coexisting liquid phases in an n-pentane (1) - sulfolane (2) mixture at 374.11 K. The system is represented by the NRTL model and Ko et al. (2007) reported the following parameters τ 12 = 2.329 Solution From Eq. (8.4-27) h i G12 = exp(− ατ 12 ) = exp − (0.3)(2.329) = 0.497 G21 Therefore, activity coefficients defined by Eqs. (8.4-29) and (8.4-30) become " # ¶2 µ 1.158 0.727 ln γ 1 = (1 − x1 )2 1.061 + 0.727 − 0.273 x1 (1 − 0.503 x1 )2 ln γ 2 = x2 1 " 0.497 2.329 1 − 0.503 x1 µ ¶2 0.771 + (0.727 + 0.273 x1 )2 # i = exp(− ατ 21 ) = exp − (0.3)(1.061) = 0.727 h τ 21 = 1.061 α = 0.3 (1) (2) Substitution of Eqs. (1) and (2) into Eqs. (11.2-5) and (11.2-6), respectively, results in two highly nonlinear equations given by ⎡ ⎤ à !2 0.727 1.158 ⎦− + (1 − xα )2 ⎣1.061 1 α (1 − 0.503 xα )2 0.727 − 0.273 x1 1 ⎡ ⎤ ⎛⎞ à !2 xβ 0.727 1.158 β 2⎣ ⎦ = ln ⎝ 1 ⎠ (3) + (1 − x1 ) 1.061 β β α 0.727 − 0.273 x1 (1 − 0.503 x1 )2 x1 (xα )2 ⎣2.329 1 ⎡ à ⎦ α (0.727 + 0.273 x1 )2 ⎡ ⎤ ⎛ ⎞ à !2 1 − xβ 0.497 0.771 1⎠ ⎦ = ln ⎝ − (xβ )2 ⎣2.329 (4) + 1 β β2 α 1 − 0.503 x1 (0.727 + 0.273 x1 ) 1 − x1 1 α − 0.503 x1 0.497 !2 + 0.771 ⎤ Simultaneous solutions of Eqs. (3) and (4) require a numerical technique2 . Using MATHR ° CAD and xβ = 0.9205 xα = 0.2341 1 1 2 Newton’s method for solving nonlinear systems of equations is explained in Problem 11.6. 354 It is much easier to determine compositions graphically by drawing a common tangent to the e minima of ∆Gmix /RT versus x1 curve. The molar excess Gibbs energy for the NRTL model is given by Eq. (8.4-26). Thus, the molar Gibbs energy change on mixing is given by e ∆Gmix = x1 x2 RT µ 0.771 1.158 + x1 + 0.727 x2 x2 + 0.497 x1 ¶ + x1 ln x1 + x2 ln x2 (8) e The values of ∆Gmix /RT as a function of x1 are given in the table below: x1 0.00 0.05 0.10 0.15 0.20 0.25 0.30 e ∆Gmix /RT 0 − 0.093 − 0.123 − 0.135 − 0.137 − 0.132 − 0.124 x1 0.35 0.40 0.45 0.50 0.55 0.60 0.65 e ∆Gmix /RT − 0.114 − 0.104 − 0.093 − 0.083 − 0.074 − 0.067 − 0.062 x1 0.70 0.75 0.80 0.85 0.90 0.95 1.00 e ∆Gmix /RT − 0.059 − 0.059 − 0.060 − 0.062 − 0.063 − 0.056 0 e The figure shown below shows the variation in ∆Gmix /RT with composition. The equilibrium compositions in the two-phase region can be determined by drawing a line tangent to the minima with the result xα = 0.23 and xβ = 0.92. 1 1 − 0.026 0 − 0.05 G x.1 () − 0.1 − 0.137 − 0.15 0 0.01 0.2 0.4 x.1 0.6 0.8 0.99 1 11.3 DISTRIBUTION OF A SOLUTE BETWEEN TWO IMMISCIBLE LIQUIDS Consider two immiscible solvents forming two separate phases, α and β . Let component 1 be the solute which is distributed between the two phases. Under equilibrium conditions bβ bα f1 = f1 (11.3-1) or, xα γ α = xβ γ β 11 11 (11.3-2) 355 The distribution of a solute between the α- and β -phases is quantified by the partition coefficient αβ or distribution coefficient, K1 , defined as αβ K1 = xα 1 xβ 1 (11.3-3) The use of Eq. (11.3-2) in Eq. (11.3-3) results in αβ K1 = γβ 1 γα 1 (11.3-4) On the other hand, the material balance for the solute is written as n1 = nα xα + nβ xβ 1 ´ ³1 αβ = nα K1 + nβ xβ 1 (11.3-5) Example 11.4 An organic acid is to be extracted from a 20 mol % acid and 80% hexane mixture by liquid-liquid extraction using water at 25 ◦ C. It is required to estimate the moles of water per mole of this mixture for removing 90% of the acid from the hexane. Assume that hexane and water are completely immiscible and that the phases leaving the extractor are in equilibrium. The following data are given at 25 ◦ C For organic acid (1) - hexane (2) mixture: For organic acid (1) - water (3) mixture: Solution e Gex /RT = 0.012 x1 x3 e Gex /RT = 0.2 x1 x2 Let α and β be the hexane and water phases, respectively. Choosing 100 mol of acid-hexane mixture as a basis, final amounts of acid and hexane in the α-phase are nα = 80 mol 2 nα = (20)(0.1) = 2 mol 1 Thus, 2 = 0.02439 82 From the given data, the activity coefficients of acid in the α- and β -phases are i i h h γ α = exp 0.2(1 − xα )2 and γ β = exp 0.012(1 − xβ )2 1 1 1 1 xα = 1 The condition of equilibrium is expressed as xα γ α = xβ γ β 11 11 or, xα exp 1 h 0.2(1 − xα )2 1 i = xβ 1 exp 0.012(1 h − xβ )2 1 i Substitution of the numerical values gives i i h h β β2 2 (0.02439) exp 0.2(1 − 0.02439) = x1 exp 0.012(1 − x1 ) 356 The solution gives xβ = 0.02917. Therefore, the number of moles of the β -phase is 1 0.02917 = The number of moles of water is nβ = 617 − 18 = 599 mol 3 The desired molar ratio is 599 = 5.99 mol of water per mol of acid-hexane mixture 100 20 − 2 nβ ⇒ nβ = 617 mol 11.3.1 Octanol-Water Partition Coefficient Octanol [CH3 (CH2 )7 OH] and water are partially immiscible and the distribution of an organic compound i between these two phases is known as the octanol-water partition coefficient, Kiow , i.e., co i (11.3-6) Kiow = w ci Since Kiow values may range from 10−4 to 108 (encompassing 12 orders of magnitude), it is usually reported as log Kiow . Octanol-water partition coefficients of various substances are given in Table 11.1. Table 11.1 Octanol-water partition coefficients of various substances3 . Substance Methanol Chloroform Benzene 1,1,2,2-Tetrachloroethane 1,1,1-Trichloroethane Naphthalene Hexachlorobenzene Chemical Formula CH3 OH CHCl3 C6 H6 C2 H2 Cl4 CH3 CCl3 C10 H8 C6 Cl6 log K ow − 0.77 1.97 2.13 2.39 2.49 3.29 6.18 Cells are mainly made of lipids and they are generally modeled as a lipid bilayer model, with a long hydrophobic (water disliking) chain and a polar hydrophilic (water liking) end. The reason for choosing n-octanol is the fact that it exhibits both hydrophobic and hydrophilic character, and its carbon/oxygen ratio is similar to lipids. In other words, n-octanol mimics the structure and properties of cells and organisms. Since octanol-water partition coefficient quantifies how a substance distributes itself between lipid and water, it is extensively used to describe lipophilic (lipid liking) and hydrophilic properties of a particular substance. Compiled from US National Library of Medicine, Hazardous Substances Data Bank (HSDB), http://toxnet.nlm.nih.gov. See also A Databank of Evaluated Octanol-Water Partition Coefficients (LOGKOW), http://logkow.cisti.nrc.ca/logkow/index.jsp. 3 357 11.4 STEAM DISTILLATION If we place two immiscible liquids in a tank, the one with a lower density will be on the top layer and solely contributes the pressure in the vapor phase. The liquid with a higher density has no contribution to the pressure in the vapor phase since it is placed in the lower layer. Therefore, when we talk about the vapor-liquid-liquid equilibrium (VLLE) calculations, we are implicitly assuming that the liquid phases are continuously agitated in such a way that there will be droplets of both liquids on the surface which is in contact with the vapor phase. A pure liquid starts to boil when its vapor pressure equals the surrounding pressure. For example, using the values given in Appendix C, the vapor pressures of toluene and water are expressed as 3096.52 vap ln Ptoluene = 9.3935 − T − 53.67 3816.44 vap ln Pwater = 11.6834 − T − 46.13 where P vap is in bar and T is in K. The normal boiling point temperatures of pure toluene and pure water are 3096.52 sat = 383.78 K Ttoluene = 53.67 + 9.3935 − ln 1.01325 3816.44 sat Twater = 46.13 + = 373.15 K 11.6834 − ln 1.01325 In the case of immiscible liquids, each component contribute to the vapor pressure of the mixture. The total pressure is simply the sum of the vapor pressures of each component, i.e., P= k X i=1 Pivap Immiscible mixture (11.4-1) The solution of above equation gives T = 357.48 K. Thus, under atmospheric pressure, a toluene-water mixture boils at 357.48 K, lower than the boiling points of pure toluene (383.78 K) and water (373.15 K). High boiling point liquids at atmospheric pressure, i.e., essential oils4 , waxes and complex fats, may decompose at high temperatures and cannot be purified by distillation. Since oils are usually insoluble in water, then the mixture of oil and steam boils at a temperature well below the boiling point of pure oil. Therefore, to decrease the boiling point, steam is directly injected into the distillation column. When the resulting vapor is condensed, the two immiscible liquid phases are separated easily. Such a process is called steam distillation. Since toluene and water are essentially immiscible as liquids, Eq. (11.4-1) is expressed as ¶ µ ¶ µ 3816.44 3096.52 + exp 11.6834 − = 1.013 exp 9.3935 − T − 53.67 T − 46.13 REFERENCES Abedinzadegan, M. and A. Meisen, 1996, Fluid Phase Equilibria 123, 259-270. Abraham, M.H., G.S. Whiting, R. Fuchs and E.J. Chambers, 1990, J. Chem. Soc. Perkin Trans. 2, 291-300. Baudot, A. and M. Marin, 1996, J. Membrane Science 120, 207-220. 4 Essential oils are the concentrated extracts of plants and herbs. 358 Furuya, T., T. Ishikawa, T. Funazukuri, Y. Takebayashi, S. Yoda, K. Otake and T. Saito, 2007, Fluid Phase Equilibria 257, 147-150. Ko, M., J. Im, J.Y. Sung and H. Kim, 2007, J. Chem. Eng. Data 52, 1464-1467. Lipinsky, C.A., F. Lombardo, B.W. Dominy and P.J. Feeney, 1996, Advanced Drug Delivery Reviews 23, 3-25. May, W.E., S.P. Wasik, M.M. Miller, Y. B. Schult, C.J., B.J. Neely, R.L. Robinson, K.A.M. Gasem and B.A. Todd, 2001, Fluid Phase Equilibria 179, 117-129. Van Ness, H.C. and M.M. Abbott, 1982, Classical Thermodynamics of Nonelectrolyte Solutions, McGraw-Hill, New York. Villamanan, M.A., A.J. Allawl and H.C. Van Ness, 1984, J. Chem. Eng. Data 29, 431-435. PROBLEMS Problems related to Section 11.1 11.1 Show that an ideal mixture always form a homogeneous phase over the entire composition range. 11.2 Start with Eq. (8.4-18) and show that e d2 (Gex /RT ) dx1 2 = Λ2 12 x1 (x1 + Λ12 x2 )2 + Λ2 21 x2 (x2 + Λ21 x1 )2 (1) for the Wilson model. Since Eq. (1) is always positive (Why?), conclude that the Wilson model cannot be used to predict phase separation. 11.3 For a binary liquid mixture of water (1) and diacetyl (2), Baudot and Marin (1996) reported the following activity coefficients at infinite dilution γ ∞ = 3.3 1 γ ∞ = 13 2 303 < T < 323 Estimate the spinodal compositions if the system is represented by the three-suffix Margules model. (Answer: x1 = 0.531 and x1 = 0.799) 11.4 When compositions, i.e., xα and xβ , are known, the parameters of an activity coefficient 1 1 model can be estimated by solving Eqs. (11.2-5) and (11.2-6). a) If the system is represented by the two-suffix Margules model, show that h i β α) ln (1 − xβ )/(1 − xα ) 1 1 ln(x1 /x1 = A= β β (1 − xα )2 − (1 − x1 )2 (xα − xβ )(xα + x1 ) 1 1 1 1 b) If the system is represented by the van Laar model, show that A=−£ (X − Y )(X Θ + Λ)2 (Y Θ + Λ)2 ¤£ ¤2 Θ (X + Y ) + 2 Λ Y Λ + X (2 Y Θ + Λ) 359 (2) (1) where X= xα 1 B=£ (X − Y )(X Θ + Λ)2 (Y Θ + Λ)2 ¤2 £ ¤ Θ (X + Y ) + 2 Λ Y Λ + X (2 Y Θ + Λ) xβ 1 1 − x1 β (3) ⎛ ⎞ ⎠ 1 − xα 1 Y= c) Abedinzadegan and Meisen (1996) studied liquid-liquid equilibrium of diethanolamine (1) and octadecane (2) mixtures and reported the following solubility values at 492 K and atmospheric pressure and xβ = 0.9924 xα = 0.0959 1 1 If the system is represented by the van Laar model, estimate the parameters A and B . (Register to View AnswerA = 2.607 B = 4.933) 11.5 A mixture of acetonitrile (1) and n-hexadecane (2) forms two partially miscible liquid phases. Liquid-liquid equilibrium data on such system are needed for the design of the oxidative desulfurization process. Furuya et al. (2007) obtained the following NRTL parameters at 333 K: τ 12 = 4.852 τ 21 = 0.2818 α = 0.2 Θ = ln ⎝ ⎛ xβ 1 x1 α ⎞ ⎠ Λ = ln ⎝ 1 − xβ 1 1 − x1 α (4) a) If the composition of acetonitrile in the acetonitrile-rich phase is 0.994, estimate its composition in the n-hexadecane-rich phase. b) Are the like interactions stronger or weaker than the unlike interactions? (Register to View Answer0.168) 11.6 Consider the following two nonlinear algebraic equations of the form f1 (x, y ) = 0 f2 (x, y ) = 0 In matrix form, Newton’s iteration equation reads ⎡ ∂f ∂f1 ⎤ 1 ∙¸ ⎢ ∂x ∂y ⎥ ∙ ∆ ¸ f ⎢ ⎥ 1 =− 1 ⎢ ⎥ f2 ⎣ ∂f ∂f2 ⎦ ∆2 2 ∂x ∂y (1) (2) (3) If the partial derivatives are difficult to evaluate analytically, numerical approximations are employed. In this respect, the forward difference approximations are f1 (x + ∆x, y ) − f1 (x, y ) ∂f1 = ∂x ∆x ∂f1 f1 (x, y + ∆y) − f1 (x, y) = ∂y ∆y f2 (x + ∆x, y ) − f2 (x, y ) ∂f2 f2 (x, y + ∆y) − f2 (x, y) ∂f2 = = ∂x ∆x ∂y ∆y Furthermore, taking ∆x = x/100 and ∆y = y/100, Eq. (3) becomes £ £ ¤ ¤⎤ ⎡ 100 f1 (1.01x, y ) − f1 (x, y ) 100 f1 (x, 1.01y ) − f1 (x, y ) ⎥∙ ¸ ⎢ ∙¸ ⎥ ⎢ x y f1 ⎥ ∆1 ⎢ ⎢ £ ¤ ¤ ⎥ ∆2 = − f2 £ ⎥ ⎢ 100 f2 (x, 1.01y ) − f2 (x, y ) ⎦ | {z } ⎣ 100 f2 (1.01x, y ) − f2 (x, y ) | {z } | x {z J (4) y Y F 360 } The iteration scheme is given as follows: • Choose initial estimates x(0) , y(0) and the error tolerance , • Initialize the iteration counter k = 0, • Loop: Increase the iteration counter by one, i.e., set k = k + 1 Use x(k−1) , y(k−1) to calculate the entries in J and F, Solve Eq. (4), i.e., J Y = − F, to obtain ∆1 and ∆2 , Update x and y from x(k) = x(k−1) + ∆1 and y (k) = y(k−1) + ∆2 , Choose maximum difference, ∆max = max {|∆1 | , |∆2 |} ª © If ∆max < then stop, the estimation is x(k) , y(k) , otherwise repeat the loop. In Example 11.3, note that Eqs. (3) and (4) can be rearranged in the form f1 (x, y) = (1 − x)2 " µ # ¶2 1.158 0.727 + 1.061 − 0.727 − 0.273 x (1 − 0.503)2 " # µ ¶2 ³y´ 0.727 1.158 = 0 (5) + − ln (1 − y)2 1.061 0.727 − 0.273 y x (1 − 0.503 y )2 µ f2 (x, y) = x2 " # ¶2 0.771 0.497 + 2.329 1 − 0.503 x (0.727 + 0.273 x)2 " # ¶2 ¶ µ µ 0.771 0.497 1−y 2 = 0 (6) − y 2.329 + − ln 1 − 0.503 y 1−x (0.727 + 0.273 y )2 xα = x 1 and 10−5 xβ = y 1 (7) where = 0.2, = 0.9, and = 1 × to begin iterations for the solution of this a) Choose system. b) For the first iteration, i.e., k = 1, show that Eq. (4) takes the form ∙ ¸ ∙ ¸∙ ¸ − 4.53664 × 10−2 1.834758 − 0.311023 ∆1 =− (8) − 5.76263 × 10−2 − 0.461567 2.95297 ∆2 c) Solve Eq. (8) to obtain ∆1 = 2.87979 × 10−2 x(1) = 0.228798 e) Show that when k = 5 x(5) = 0.234122 y(5) = 0.920495 and ∆2 = 2.40160 × 10−2 y(1) = 0.924016 x(0) y(0) d) Since ∆max = 0.03 > , continue iteration with Keep in mind that good initial estimates are extremely important in numerical solutions. Problems related to Section 11.3 11.7 Since the solubility of water in n-hexadecane is 0.0059 mole fraction and that of nhexadecane in water is 0.0072 mole fraction at 298.15 K, the n-hexadecane-water system is 361 regarded as a system containing the two pure solvents. Consider a solute (1) that is partitioned between n-hexadecane (α-phase) and water (β -phase). The condition of equilibrium states that xα γ α = xβ γ β 11 11 (1) αβ a) The hexadecane-water partition coefficient, K1 , is defined as the molar concentration of the solute in the n-hexadecane phase to the molar concentration of the solute in the water phase, i.e., cα αβ (2) K1 = 1 cβ 1 Combine Eqs. (1) and (2) to obtain αβ γ β = 16.3 K1 γ α 1 1 (3) If the solute is infinitely dilute in this system, note that Eq. (3) takes the form αβ γ β ∞ = 16.3 K1 γ α∞ 1 1 (4) If the infinite-dilution activity coefficient of solute in n-hexadecane as well as the hexadecanewater partition coefficient are known, then Eq. (4) can be used to estimate the infinite-dilution activity coefficient of solute in water. b) The infinite-dilution activity coefficient of toluene in n-hexadecane is given as a function of temperature as follows (Schult et al., 2001): Temperature ( K) γ∞ 323.2 0.941 343.2 0.870 353.2 0.846 373.2 0.808 The hexadecane-water partition coefficient of toluene is (Abraham et al., 1990) K = 575.4 Express the infinite-dilution activity coefficient in the form ln γ ∞ = A + B T and estimate the infinite-dilution activity coefficient of toluene in water at 298 K. (Answer: 9660) DATA: Component n-Hexadecane Water ρ ( kg/ m3 ) 997.05 770.20 Molecular Weight 226.44 18.02 11.8 Consider two liquid phases, α and β , in which the α-phase is almost pure 1 and the β -phase is almost pure 2. 362 a) Using the condition of equilibrium show that γ∞ = 1 1 β x1 and γ∞ = 2 1 x2 α (1) Therefore, determination of solubility enables one to estimate infinite-dilution activity coefficient. This method, known as inverse solubility, is suitable for organics that are sparingly soluble in water. b) The solubility of benzene (2) in water (1) at 298.15 K is reported by May et al. (1983) as x2 = 0.4129 × 10−3 Estimate the infinite-dilution activity coefficient of benzene in water. (Answer: 2422) 363 364

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on AIR JAM: The U.S. Federal Aviati billion Administration spent $2.6 icair-traff trying to upgrade its to cancel the control system, only dlocked skies project in 1994. Gri ay. are still with us todWe waste billions of dollars each year on entirely prev
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