10.40 Thermodynamics: Problem Set 10, #5 (Solutions)

Problem Set 10, #5 10.40 Thermodynamics Problem Set 10 Fall 2003 Problem 15.29 Text Solution: To determine if the Margules equation is applicable for the limit of miscibility of a binary solution, we examine the behavior at the consolute point. In particular, we look at the values of A and B at the consolute point. At the consolute point, the spinodal and binodal curves meet. The limit of stability that corresponds to the spinodal curve, Eq. (15-153), applies to the consulate point. ⎛ ∂ 2 ∆Gmix ⎞ ⎜ ⎟ =0 2 ⎝ ∂x1 ⎠T .P (15-153) ⎛ ∂ 2 ∆Gmix ⎞ When ⎜ ⎟ ≥ 0 , the solution is stable, which means that only a single phase will be 2 ⎝ ∂x1 ⎠T .P present. An additional criterion of stability is applicable at the consolute point since it behaves as a critical point. From Section 7.3, at the critical point, or the consulate point in this problem, 2) y((mm−−1)( m −1)( m −1) = 0 with m = n + 2 (7-39) Following a similar derivation to Section 15.7, the second criteria for the consolute point is: ⎛ ∂ 3 ∆Gmix ⎞ ⎜ ⎟ =0 3 ⎝ ∂x1 ⎠T .P (1) EX In the problem we are given and equation of state for ∆Gmix . It can be related to ∆Gmix by applying Eq. (9-168). EX ID ∆Gmix = ∆Gmix + ∆Gmix ID = RT ( x1 ln x1 + x2 ln x2 ) where ∆Gmix (2) (9-101) Combining equations for ideal and excess properties yields, 10.40 Fall 2003 Problem Set 10 Solutions Page 1 of 4 Problem Set 10, #5 ∆Gmix 2 = Ax12 (1 − x1 ) + Bx1 (1 − x1 ) + x1 ln x1 + (1 − x1 ) ln (1 − x1 ) RT (3) Differentiation of Eq. (3) results in, ⎡ ⎛ ∆ 2Gmix ⎞ 1 1 ⎤ ⎥=0 ⎜ ⎟ = RT ⎢ A ( 2 − 6 x1 ) + B ( −4 + 6 x1 ) + + 2 x1 (1 − x1 ) ⎥⎦ ⎢⎣ ⎝ ∂x1 ⎠T , P (4) ⎡ ⎛ ∆3Gmix ⎞ 1 1 ⎤ RT A B = − 6 + 6 − + ⎢ ⎥=0 ⎜ ⎟ 3 x12 (1 − x1 )2 ⎥ ⎢⎣ ⎝ ∂x1 ⎠T , P ⎦ (5) Solving for A and B from Eqs. (4) and (5) yields their values at the consolute point as a function of solution composition. Ac = Bc = −9 x12 + 10 x1 − 2 6 x12 (1 − x1 ) 2 (6) −9 x12 + 8 x1 − 1 6 x12 (1 − x1 ) 2 The values of Ac and Bc are applicable over the entire composition range except for where x1 equals zero or one (where there will be a single pure phase of either component 1 or 2). By ⎛ ∂ 2 ∆Gmix ⎞ inspection of Eq. (4), when A and B are less than their critical values, ⎜ ⎟ > 0 , so the 2 ⎝ ∂x1 ⎠T .P solutions will be miscible and form a single phase. When A and B are greater than their critical values, the solutions will not be miscible and will form two phases. The values of A and B are constrained by Eq. (6) and are plotted below. As x1 goes to 0 or 1, the values of A and B go to negative infinity. Since the Margules equation gives conditions for solutions ranging from a single solution through the consolute point and to liquid-liquid equilibrium, it can be used for liquid-liquid equilibrium problems. 10.40 Fall 2003 Problem Set 10 Solutions Page 2 of 4 Problem Set 10, #5 Experimental determination of A and B Values of A and B can be found for a binary mixture based on values of x1 or x2 in the two liquid phases. At equilibrium between liquid 1 (α) and liquid 2 (β), the fugacity of each component in the α and β phases are equal. fˆiα = γ iα xiα fiα (T , P) = γ iβ xiβ fi β (T , P) = fˆi β (7) Since f iα and f i β is the fugacity of xi=1 at T and P of interest, they are equivalent. Eq. (7) reduces to: γ 1α x1α = γ 1β x1β γ 2α x2α = γ 2β x2β (8) (9) and can be rewritten in natural log form as: ln γ 1α − ln γ 1β = ln x1β − ln x1α α β β (10) α ln γ 2 − ln γ 2 = ln x2 − ln x2 (11) The Margules equation can be related to the activity coefficient by a combination of Eqs. (9-180) and (9-53). 10.40 Fall 2003 Problem Set 10 Solutions Page 3 of 4 Problem Set 10, #5 EX RT ln γ i = ∆G i EX ⎛ ∂∆Gmix EX = ∆Gmix − xj ⎜ ⎜ ∂x j ⎝ ⎞ for j ≠ i ⎟⎟ ⎠T , P , x[ j ,i ] (12) Plugging the Margules equation into (12) and simplifying yields: ( ) ( − x ) + 2B ( x ) − x ) = ( A + 2 ( B − A) x ) x ln γ 1 = 2 A x22 − x23 + B 2 x23 − x22 = ( B + 2 ( A − B ) x1 ) x22 (13) ln γ 2 = A 2 x13 (14) ( 2 1 2 1 3 1 2 1 2 Substituting Eqs. (13) and (14) into (10) and (11) yields two equations with two unknowns (A and B) given experimental pairs of values for xiα and xiβ . These equations can be written as: ( B + 2 ( A − B ) x )(1 − x ) − ( B + 2 ( A − B ) x )(1 − x ) = ln x − ln x ( A + 2 ( B − A) (1 − x )) ( x ) − ( A + 2 ( B − A) (1 − x )) ( x ) = ln (1 − x ) − ln (1 − x ) α α 2 β β 2 β α 1 1 1 1 1 1 α α 2 β β 2 β α 1 1 1 1 1 1 Thus, A and B can be determined experimentally using Eqs. (15) and (16). 10.40 Fall 2003 Problem Set 10 Solutions Page 4 of 4 (15) (16)