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MKS 06160.Pdf Molecular Knowledge Systems, Inc. Page 1 of 7 Finding Binary Azeotropes Technical Note: MKS 06160 Updated: 2013-07-16 1.0 Summary This technical note explains the thermodynamic relationships used to identify binary azeotropes. A video is available from Molecular Knowledge Systems’ website that describes the step-by-step procedure for using Cranium to search for binary azeotropes. The video is available at: www.molecularknowledge.com/casestudies/azeotropes.htm 2.0 Azeotropic Behavior Figure 1 shows a qualitative depiction of a typical binary azeotrope consisting of components A and B. At low concentrations of component A the mole fraction of A in the vapor phase is greater than the mole fraction of A in the liquid phase. At high concentrations of A the mole fraction of A in the vapor phase is less than the mole fraction of A in the liquid phase. The composition at which the mole fraction of A in the vapor is equal to the mole fraction of A in the liquid is the azeotropic point. yA xA Figure 1: Typical Behavior of a Binary Azeotrope Quantitatively we can model the phase behavior of components A and B using our standard, low pressure, vapor-liquid equilibrium relationships: = (1) 푣푝 푦퐴푃 푥퐴훾퐴푃퐴 = (2) 푣푝 푦퐵푃 푥퐵훾퐵푃퐵 www.molecularknowledge.com [email protected] Molecular Knowledge Systems, Inc. Page 2 of 7 The slope of the curve in Figure 1 is given by: = 푣푝 + 푣푝 (3) 휕푦퐴 훾퐴푃퐴 푥퐴푃퐴 휕훾퐴 휕푥퐴 푃 푃 휕푥퐴 As the concentration xA approaches 0 the slope approaches: = ∞ 푣푝 (4) 휕푦퐴 훾퐴 푃퐴 � 휕푥퐴 푥퐴=0 푃 Using the relationships: = 1 (5) 푦퐴 − 푦퐵 = 1 푣푝 (6) 푥퐵훾퐵푃퐵 푦퐴 − 푃 = 푣푝 푣푝 (7) 휕푦퐴 훾퐵푃퐵 휕푥퐵 푥퐵푃퐵 휕훾퐵 − − 휕푥퐴 푃 휕푥퐴 푃 휕푥퐴 + = 0 (8) 푑푥퐴 푑푥퐵 We obtain: = 푣푝 + 푣푝 (9) 휕푦퐴 훾퐵푃퐵 푥퐵푃퐵 휕훾퐵 휕푥퐴 푃 푃 휕푥퐵 As the concentration xA approaches 1 the slope approaches: = ∞ 푣푝 (10) 휕푦퐴 훾퐵 푃퐵 � 휕푥퐴 푥퐴=1 푃 Figure 1 shows that when an azeotrope is present the slopes at each concentration limit must both be either greater than 1 or less than 1. We can thus use this observation with the slopes calculated by Equations 4 and 10 to identify binary azeotropic systems. 3.0 Temperature Values The infinite dilution activity coefficients and vapor pressures in Equations 4 and 10 are both temperature dependent properties. Equation 4 represents the slope of the equilibrium curve for a mixture containing almost pure B. Thus the properties used in Equation 4 should be evaluated at the boiling point of pure B. Similarly, Equation 10 represents the slope of the equilibrium curve for a mixture containing almost pure A. Thus the properties used in Equation 10 should be evaluated at the boiling point of pure A. www.molecularknowledge.com [email protected] Molecular Knowledge Systems, Inc. Page 3 of 7 4.0 Modeling Activity Coefficients To evaluate Equations 4 and 10 we need pure component vapor pressures and infinite dilution activity coefficients. Numerous models are available for both properties. The MOSCED model [1] is one of the easier to use infinite dilution activity coefficient models. Similar in concept to the solubility parameter model, the MOSCED model tabulates six parameters for each pure chemical: v λ τ q α β These parameters are then inserted into the following equations to generate an estimate for the infinite dilution activity coefficient. [ 3.4 Note that Equation 6 has been corrected from the original reference. For example, to calculate the infinite dilution activity coefficient of aniline in cyclohexane at 300 K we begin by compiling the pure component parameters. www.molecularknowledge.com [email protected] Molecular Knowledge Systems, Inc. Page 4 of 7 Parameter Cyclohexane (1) Aniline (2) v 108.9 91.6 λ 16.74 16.51 τ 0.00 9.41 q 1.00 0.90 α 0.00 6.51 β 0.00 6.34 Inserting these values into Equations 11 through 20 we obtain: Parameter Aniline (2) αΤ 6.388 βΤ 6.221 τΤ 9.322 POL 1.641 ξ 2.578 ψ 1.745 aa 0.953 d12 -0.01287 which gives us a final estimate of: = 56.455 (11) ∞ 훾2 5.0 Examples We use this approach to find a chemical that has an azeotrope with methanol at a pressure of 1.0 bar. We chose benzene as our first candidate. Table 1 shows the parameters for methanol and benzene taken from Reference 1. Table 1: MOSCED Model Parameters Parameter Methanol Benzene v 40.6 89.5 λ 14.43 16.71 τ 3.77 3.95 q 1.00 0.90 α 17.43 0.63 β 14.49 2.24 The boiling points of our components were taken from the NIST website [2]: Methanol: Tb = 337.8 K Benzene: Tb = 353.3 K www.molecularknowledge.com [email protected] Molecular Knowledge Systems, Inc. Page 5 of 7 We next calculate the infinite dilution activity coefficient of methanol in benzene at 353.3 K, the boiling point of benzene. Inserting the values for methanol and benzene from Table 1 into Equations 11 through 20 we obtain: Parameter Methanol (2) Benzene (1) αΤ 15.006 0.542 βΤ 12.475 1.929 τΤ 3.498 3.665 POL 1.082 ξ 1.063 ψ 1.085 aa 0.491 d12 -0.06657 which gives us an estimate of: Methanol in benzene: = 7.307 ∞ γ2 We next calculate the infinite dilution activity coefficient of benzene in methanol at 337.8 K, the boiling point of methanol. Inserting the values for benzene and methanol from Table 1 into Equations 11 through 20 we obtain: Parameter Benzene (2) Methanol (1) αΤ 0.562 15.555 βΤ 1.999 12.931 τΤ 3.731 3.561 POL 1.115 ξ 2.589 ψ 1.644 aa 0.918 d12 -0.34058 which gives us an estimate of: Benzene in methanol: = 6.314 ∞ γ2 The vapor pressures of our components are calculated using Antoine’s equation. The Antoine parameters were taken from the NIST website: Methanol: log10(Pvp) = 5.20409 – 1581.341 / (T – 33.50) Benzene: log10(Pvp) = 4.72583 – 1660.652 / (T – 1.461) www.molecularknowledge.com [email protected] Molecular Knowledge Systems, Inc. Page 6 of 7 where Pvp is in units of bar and T is in units of K. Evaluating the vapor pressure for each component at the boiling point of the other component we obtain: Methanol: Pvp(353.3) = 1.816 bar Benzene: Pvp(337.8) = 0.614 bar The slope, calculated by Equations 4 and 10, at each side of the graph is calculated to be: = 3.879 Pure methanol: ∞ vp γA PA P = 13.275 Pure benzene: ∞ vp γB PB P Since both slopes are greater than 1.0 an azeotrope is predicted. Data from Husdon and Van Winkle [3] confirm that methanol and benzene form an azeotrope at 1 bar. 1.00 0.88 0.75 0.63 0.50 0.38 0.25 Benzene, mol fraction vapor fraction mol Benzene, 0.13 0.00 0.00 0.13 0.25 0.38 0.50 0.63 0.75 0.88 1.00 Benzene, mol fraction liquid To verify our approach we can compute the slope of the curve using the first and last data points. The first point is: x: 0.027 y: 0.093 which gives us a slope of . Pure methanol: = 3.44 . 0 093 0 027 The last point is: x: 0.974 y: 0.733 which gives us a slope of www.molecularknowledge.com [email protected] Molecular Knowledge Systems, Inc. Page 7 of 7 . Pure benzene: = 10.26 . 1 0−0 733 1 0−0 974 Our calculated slopes are in good agreement with the experimental slopes. 6.0 References 1. Michael J. Lazzaroni, David Bush, Charles A. Eckert, Timothy C. Frank, Sumnesh Gupta and James D. Olson. "Revision of MOSCED Parameters and Extension to Solid Solubility Calculations." Industrial and Engineering Chemistry Research. Volume 44, number 11, page 4075-4083, 2005. 2. NIST Chemistry WebBook. http://webbook.nist.gov. Accessed November 2010. 3. John W. Hudson and Matthew Van Winkle. "Multicomponent Vapor-Liquid Equilibriums in Systems of Mixed Positive and Negative Deviations." Journal of Chemical and Engineering Data. Volume 14, number 3, page 310-318, 1969. www.molecularknowledge.com [email protected] .
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