Research: Science and Education
The Influence of Distillation Conditions on the Azeotropic Composition Jaime Wisniak Department of Chemical Engineering, Ben-Gurion University of the Negev, Beer-Sheva, Israel
Processes taking advantage of favorable vapor–liquid distribution of components constitute by far the most extensive list of processes in which phase equilibrium is important. Distillation is the workhorse of the chemical and petroleum industries and a classical example of an operation whose success is determined by a favorable distribution of a component between the liquid and vapor phases. A very common situation in vapor–liquid equilibria is the presence of an azeotrope (a state in which the vapor and the liquid phases have the same composition). If the distillation is carried out at constant pressure the azeotropic point will occur at the maximum or minimum of the temperature, that is, at a temperature which is higher (or lower) that the boiling temperature of each pure component. Similarly, if the distillation is carried out at constant temperature, the azeotropic point will occur at the maximum or minimum pressure, that is, at a pressure which is higher (or lower) than the vapor pressure of the pure components at the same temperature. Water solutions of an alcohol like ethanol, 2-propanol, or butanol are examples of azeotropic systems of industrial importance. The experimental information about azeotropic systems is available in several publications and data banks (1– 7) and an interesting feature of the data is that the vast majority of azeotropic systems present a minimum boiling point. An understanding of the occurrence of azeotropes is important for two reasons. First, azeotropes can make a given separation impossible by simple distillation in a given pressure range, and second, azeotropes may be used to separate mixtures not ordinarily separable by simple distillation (7). The student should understand that during an azeotropic transformation both the temperature and the pressure remain constant. It is the purpose of this paper to analyze the thermodynamic conditions for the presence of an azeotropic point in a binary system and to determine how a change in the operating conditions (pressure or temperature) may shift the composition of the azeotrope to a higher or lower value. A simplified procedure will be developed to calculate this shift, using minimum information about the pure components. Theory Consider a binary system that presents an azeotropic point with minimum or maximum boiling temperature, as shown in Figure 1. Inspection of Figure 1a (minimum boiling temperature) shows that in the range 1A the vapors are more concentrated in the more volatile compound 1 up to the azeotropic composition A, afterwards, in the range A2, the relative volatility changes and the component with the higher boiling point is now concentrated in the vapor. In the case of an azeotrope of maximum boiling point the behavior is reversed, in the range 1A the vapors are richer in the lessvolatile component. We see immediately that in azeotropic systems the concept of relative volatility is not clear-cut: the 1486
Figure 1. Typical T-x-y graph for a system forming an azeotrope.
relative volatility will change depending on whether the concentration of a particular solution is less or greater than that of the azeotrope. On the other hand, in a system that does not form an azeotrope the vapor phase will always be richer in the component with the lower boiling temperature in the full concentration range and, theoretically, it will always be possible to separate completely both components. Inspection of Figure 1 shows that this is not the case for a system that contains an azeotrope: the maximum separation that can theoretically be achieved is one phase containing one of the pure components and another phase having the azeotropic composition. A knowledge of the existence of azeotropes is then crucial in the design of distillation columns because the separation will not exceed the azeotropic composition. Thus it becomes important to know how the azeotropic composition will shift with a change in either the temperature or the pressure of distillation. Consider a binary system that can separate into two phases in equilibrium. The state of equilibrium can be described by the four parameters T, P, x1, and y1, where x and y are the compositions of the liquid and vapor phase, respectively. According to the phase rule a system of this nature will have two degrees of freedom which will normally be chosen among the pressure P, temperature T, and the composition of one of the phases. The general behavior of phase equilibrium when one of the variables is changed will be expressed by a derivative of the structure (∂a/∂b)c , where a, b, and c are variables that correspond to P, T, x, and y. Examples of these derivatives are (∂P/∂x1)T , (∂y1/∂x1)T , and (∂P/∂T)x . There are twelve derivatives in total and the method of equilibrium displacement (8) can be used to express them in terms of measurable thermodynamic parameters. All the derivatives can be obtained from the equilibrium condition µ1, L = µ1, V
(1)
µ 2, L = µ 2, V
(2)
where µi is the chemical potential.
Journal of Chemical Education • Vol. 75 No. 11 November 1998 • JChemEd.chem.wisc.edu
Research: Science & Education
The derivatives related to our purpose are as follows (8):
y1 – x 1 ∂P ∂x 1
= T
= P
∂x 12
∂x 12
P,T
(4)
y 1∆h 1 + y 2∆h 2
= P
(3)
∂2g L
x 1∆h 1 + x 2∆h 2 ∂y 1 ∂x 1
P,T
y 1∆v 1 + y 2∆v 2
T x 1 – y1 ∂T ∂x 1
∂2g L
y 1∆h 1 + y 2∆h 2
curves present an extreme value (maximum or minimum)? We see immediately that eqs 3 and 4 yield the same result, that at the particular extreme value the concentration of both phases must be equal. In other words, an extreme value in these curves points to the presence of an azeotrope. The interesting consequence is that thermodynamics provides the necessary theoretical tools that confirm the experimental evidence. The next question that can now be answered is the nature of the extremum—is it a minimum or a maximum? To do so we remember that the nature of an extreme value is determined by the sign of the second derivative. A positive sign of this parameter will indicate a minimum value, and vice versa. Performing the necessary operations we obtain, after some algebra,
∂2g L ∂x 12
P,T
∂2g V ∂y 12
∂2g L
(5) ∂2P = ∂x 12
P,T
∂x 12 P,T
–
∂2g V ∂y 12
∂2g L ∂x 12
P,T
y 2∆v 2 + y 1∆v 1
(7)
It should be noted that for a vaporization process both — — ∆vi and ∆hi will have positive values. The three derivatives given by eqs 3 to 5 will determine the pertinent change in the equilibrium state. To determine the direction (slope) of the change (increase or decrease) in a vaporization process we have to consider that stability conditions require that both (∂2gV/∂y12)P,T and (∂2gL/∂x12)P,T be positive and, as mentioned — — before, that for a vaporization process ∆vi and ∆hi are necessarily positive. Consider, for example, eqs 3 and 4. We see that their sign is determined solely by the sign of the difference (y1 – x1). For example, if component 1 is richer in the vapor phase than in the liquid phase (y1 > x1) then the difference (y1 – x1) will be positive and the slope of the change will be positive. This means that in a distillation at constant temperature an increase in the concentration of component 1 will increase the vapor pressure of the solution. Similarly, when distilling at constant pressure, an increase in the concentration of component 1 will decrease the boiling point of the solution. What happens now if the P(x1) or the T(x1) Table 1. Criteria for Determining Nature of an Azeotrope ∂2g L ∂x 22 2
∂ gV ∂y 22
∂2P ∂x 22
Azeotrope azeotrooe
Pressure
Temperature
0
Maximum
Minimum
P,T azeotrope
Inspection of eq 8 indicates that the sign of the second derivative depends only on the sign of the difference
∂2g L
___ ___ ___ ∆hi = hi,V – hi,L
(8)
∂ gV ∂y 12
where g L and g V represent the total molar Gibbs function of — — the liquid and vapor phase, and ∆vi and ∆hi the partial volume and partial enthalpy of vaporization of component i: ___ ___ ___ ∆vi = vi,V – vi,L (6)
P,T
2
∂x 12
–
∂2g V ∂y 12
P,T
at the azeotropic point and hence we can establish the criteria indicated in Table 1 for the nature of the azeotrope. We can now understand why the vast majority of the known azeotropes present a minimum boiling point: the deviations of the liquid phase are, in general, substantially larger than those of the vapor phase. The vapor phase will deviate very little from ideal behavior and the value of the derivative [ ∂ 2g V / ∂ y 12] P,T will be small. In order that it be larger than [ ∂ 2 g L / ∂ y 12] P,T the components will have to have strong interactions, such as hydrogen bonding, in the liquid phase. An example of this type of system is solutions of chloroform plus acetone, in which molecules of acetone can form hydrogen bonds with molecules of chloroform, but neither acetone nor chloroform molecules can form hydrogen bonds with themselves. In the case of aqueous systems, for example, water plus formic acid, the electrolytic dissociation of the nonaqueous component will normally yield an azeotrope of maximum boiling temperature. These interactions will result in a substantial decrease in the vapor pressure of each component in the solution, the system will present negative deviations from ideality, and [ ∂ 2g L / ∂ y 12 ] P,T will become negative. Again, we see that thermodynamics provides a solid theoretical basis for the experimental evidence. Simplified Procedure The equations derived using the equilibrium displacement method are correct from the thermodynamic viewpoint and their numerical solution will give the sign of the slope,
JChemEd.chem.wisc.edu • Vol. 75 No. 11 November 1998 • Journal of Chemical Education
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Research: Science and Education
that is, the direction of the change as well as its intensity. Unfortunately, the numerical information required to perform the integration (data on the partial volumes and enthalpies of vaporization as a function of the composition and the temperature or pressure) is not readily available and eqs 3 to 8 have only theoretical value. We will now describe an alternative empirical procedure which will allow us to give a reasonable quantitative answer to the question. Consider the vapor–liquid equilibria (VLE) data of a binary system. If we assume that the vapor phase behaves ideally then the equilibrium condition is P y i = γ i x i P i°
(9)
where x and y are the composition of the phases in equilibrium at pressure P, and γ i and P i° are respectively the activity coefficient and the vapor pressure of pure component i. At the azeotropic point yi = xi so that (10) azeotrope
If the VLE data can be represented by an empirical model such as that of Margules, Van Laar, or Wilson, then the left side of eq 10 is independent of the temperature or a very weak function of it, and thus it will be a function of the composition alone. If the operating pressure is not too high, the right side of the equation will be a function of the temperature alone. Now, if the composition and temperature of the azeotrope point are known with reasonable accuracy, the activity coefficients at this point can be used to determine the constants of the model, as follows (in all the following equations log represents log10).
Margules Model The activity coefficients are (9): log γ1 = x 22[A + 2x1(B – A)] log γ2 =
x 12[B
+ 2x 2(A – B)]
(11) (12)
Equations 11 and 12 can be solved without difficulty for the constants A and B to yield
x2 – x1
A=
x 22
B=
x1 – x2 x 12
2
x log γ2 A = 1+ 2 log γ1 x 1 log γ1 2
B= 1+
x 1 log γ1 log γ2 x 2 log γ2
(18)
Wilson Model The activity coefficients are (9) log γ1 = ᎑log (x1 + Ax 2 ) + β x 2
(19)
log γ2 = ᎑log (x 2 + Bx 1 ) – β x 1
(20)
log γ1 +
2 log γ2 x1
(13)
log γ2 +
2 log γ2 x2
(14)
β=
A B – x1 + A x2 B x1 + x1
log γ1 =
A x1 A 1+ x2 B
log γ2 =
B x 1+ 2 B x1 A
2
2
(21)
Equations 19 and 20 constitute a set of transcendental equations that can only be solved by trial and error. If we have good experimental data for the azeotropic composition, then we can use eqs 13 and 14 (or 17 and 18, or 19 and 20) to determine the values of A and B for the pertinent model, as illustrated in the examples given later. Let us analyze the significance of eq 10. The value of the left side of the equation, γ1/γ2, can only decrease as the value of x1 increases, but the value of the ratio P2°/P1° can increase or decrease with increasing temperature, depending on the relative variation with temperature of the vapor pressure of each pure component. Equation 10 can be plotted as indicated in Figure 2 and the figure can be used to calculate the shift of the azeotropic point, as illustrated in the examples that follow. (In the examples this figure will be drawn in a more compact and useful form.) Inspection of Figure 2 shows that for systems for which the value of P2°/P1° increases with temperature, an increase in the distillation temperature will enrich the azeotrope in component 1. The opposite behavior will be observed for systems for which the ratio P2°/P1° decreases with increasing temperature. In addition, it should be clear that the graphical solution will always predict the correct direction of the azeotrope shift. The quantitative prediction will depend on how
Van Laar Model The activity coefficients are (9)
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(17)
where
o
γ1 P 2 γ2 = P o 1
Again, eqs 15 and 16 can be solved for the constants A and B to yield
(15)
(16) Figure 2. Graphical solution of eq 10.
Journal of Chemical Education • Vol. 75 No. 11 November 1998 • JChemEd.chem.wisc.edu
Research: Science & Education
accurately the azeotropic conditions are known and how well the selected model describes the system in question. Example 1 Artigas et al. (10) have measured the vapor–liquid equilibrium of the system butanol (1) + chlorocyclohexane (2) at 40.0 kPa and 101.3 kPa. At 40 kPa the system has an azeotrope that boils at 365 K and contains 73.3 mole % butanol. At 101.3 kPa the azeotrope boils at 390 K and contains 84.6 mole % butanol. In other words, an increase in the distillation pressure enriches the azeotrope in butanol. Let us use the proposed method to see how it confirms these results. According to Artigas the vapor pressure of the pure components is given by o
log P 1 / kPa = 6.54743 – o
log P 2 / kPa = 5.735540 –
1338.769 T/K – 96.108
1235.480 T/K – 84.503
At 365 K we have P1° = 37.03 kPa and P2° = 21.42 kPa, so that according to eq 10, γ1 = 1.08, γ2 = 1.87, and γ1/γ2 = 0.58 at 40 kPa. According to Artigas the data are well represented by the Margules model; thus from eqs 13 and 14 we get A = 0.5214 and B = 0.4860. All this information is used to calculate the variation of γ1/γ2 with x1 and of P2°/P1° with the temperature. The pertinent results are shown in Figure 3. It is seen that for this particular system the slope of the curve γ1/γ2 is much larger than that of the curve P2°/P1°. This means that a large temperature increase will produce only a small enrichment of the azeotrope in component 1. As noted in the figure, increasing the temperature from 365 to 390 K increases the concentration of butanol in the azeotrope from 73.3% to 80%. The required distillation pressure is
The new activity coefficients, calculated from the Margules equations, are γ1 = 1.04 and γ2 = 2.09, so that P = 102.5 kPa. We see that the proposed method predicts correctly the direction of the change and that the calculated pressure and composition of the azeotropic point compare very reasonably with the experimental results. Example 2 Kurihara et al. (11) have measured the vapor–liquid equilibrium of the system benzene (1) + cyclohexane (2) at 323.25 K and 333.15 K. At 323.5 K the system has an azeotrope that boils at 40.36 kPa and contains 51.3 mole % benzene. At 333.15 K the azeotrope boils at 57.62 kPa and contains 52.2 mole % benzene. From these data we learn that increasing the distillation temperature increases the vapor pressure of the azeotrope and the concentration of the low-boiling component. Let us see how the proposed method confirms these results. According to Kurihara the vapor pressure of the pure components is given by o
log P 1 / kPa = 6.08200 – o
log P 2 / kPa = 6.17214 –
1237.868 T/K – 49.479 1318.971 T/K – 37.218
where x1 = 0.80 and the vapor pressures are calculated at the new temperature 390 K as P1° = 98.2 kPa and P2° = 49.1 kPa.
At 323.25 K we have P1° = 37.21 kPa and P2° = 36.25 kPa, so that according to eq 10, γ1 = 1.11, γ2 = 1.11, and γ1/γ2 = 1.00. Let us assume that the data follow the Van Laar model. According to eqs 17 and 18 we get A = 0.1775 and B = 0.1989. All this information is used to calculate the variation of γ1/γ2 with x1 and of P2°/P1° with the temperature. The pertinent results are shown in Figure 4. It is seen again that the slope of the curve γ1/γ2 is much larger than that of the curve P2°/P1°. This means that a large temperature increase will produce only a small enrichment of the azeotrope in component 1. As noted in the figure, increasing the temperature from 323.25 K to 333.15 K increases the concentration of
Figure 3. Solution of example 1.
Figure 4. Solution of example 2.
P = x1γ1P1° + x 2 γ 2 P2°
JChemEd.chem.wisc.edu • Vol. 75 No. 11 November 1998 • Journal of Chemical Education
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Research: Science and Education
butanol in the azeotrope from 51.5% to 52%. The azeotrope pressure is P = x1γ1P1° + x 2 γ 2 P2° where x1 = 0.52 and the vapor pressures are calculated at the new temperature 333.15 K as P1° = 52.27 kPa and P2° = 51.90 kPa. The new activity coefficients, calculated from the Van Laar equations, are γ1 = 1.11 and γ2 = 1.12, so that P = 58.03 kPa. We see that the proposed method predicts correctly the direction of the change and that the calculated pressure and composition of the azeotropic point compare very reasonably with the experimental results. In the Classroom The displacement of azeotropic composition with the operating variables has been used successfully for several years as part of the homework problems to be solved in a course teaching thermodynamics to sophomore students in chemical engineering. Each student must look up a particular azeotrope reported in the recent literature and perform the necessary calculations to determine the variation of its composition with pressure and temperature. For a course accompanied by laboratory sessions, the experimental procedures and results reported by Gibbard and Emtage (11) may be used as excellent examples. Conclusions The necessary thermodynamic conditions for the presence of an azeotropic point have been developed. It is seen that
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the method of displacement equilibrium predicts that at a maximum (minimum) in the boiling point of a binary solution both phases must have the same composition. A simplified method is illustrated to calculate how the azeotropic point shifts with changes in the temperature or pressure at which the distillation process is performed. Literature Cited 1. Lecat, M. Tables Azeotropiques, Monograph; Lamertin: Brussels, 1949. 2. Horsley, L. H. Azeotropic Data I; Advances in Chemistry Series 6, American Chemical Society: Washington, DC, 1952. 3. Horsley, L. H. Azeotropic Data II; Advances in Chemistry Series 35; American Chemical Society: Washington, DC, 1962. 4. Horsley, L. H. Azeotropic Data III; Advances in Chemistry Series 116; American Chemical Society: Washington, DC, 1973. 5. Gmehling, J.; Menke, J.; Krafczyk, J.; Fischer, K. Azeotropic Data; VCH: Weinheim, 1994. 6. Gmehling, J.; Bols, R. J. Chem. Eng. Data 1996, 41, 202. 7. Perry, R. H.; Green, D. Perry’s Chemical Engineer’s Handbook, 6th ed.; McGraw-Hill: New York, 1984. 8. Malesinski, W. Azeotropy and Other Theoretical Problems of Vapour– Liquid Equilibrium; Interscience: New York, 1965. 9. Walas, S. M. Phase Equilibrium in Chemical Engineering; Butterworth: Boston, 1985. 9. Artigas, H.; Lafuente, C.; Cea, P.; Royo, F. R.; Urieta, J. S. J. Chem. Eng. Data 1997, 42, 132. 10. Kurihara, K.; Uchiyama, M.; Kojima, K. J. Chem. Eng. Data 1997, 42, 149. 11. Gibbard, H. F.; Emptage, M. R. J. Chem. Educ. 1975, 52, 673.
Journal of Chemical Education • Vol. 75 No. 11 November 1998 • JChemEd.chem.wisc.edu
