Lecture 5: VLE, LLE, and SLE Phase Diagrams Physical Chemistry 2 for Engineers Lecture Refresher on asynchronous lecture • Raoult’s Law provides a relation between vapor and liquid phase compositions at a given T and P for ideal solution of volatile liquids. Four main calculations: Bubble T/P, Dew T/P 𝑥𝑖 𝑃𝑖𝑠𝑎𝑡 = 𝑦𝑖 𝑃 • Henry’s Law provides relation between vapor and liquid phase compositions at a given T and P for very dilute mixtures. Usually, the dilute component in the liquid phase follows the Henry’s Law. The dominant component in the liquid phase follows the Raoult’s Law (ideally). 𝑥𝑖 ℋ𝑖 = 𝑦𝑖 𝑃 Refresher on asynchronous lecture • Vapor pressures are dependent of temperature and are calculated using the Antoine equation. Antoine coefficients are different for each chemical species. Forms of the equation and units of temperature and pressure are usually provided with the data set. Review of phase diagrams • Shows the regions of pressure and temperature at which its various phases are thermodynamically stable. In fact, any two intensive variables such as temperature and magnetic field but we will focus on temperature and pressure • A single phase is represented by an area. • Lines separating the regions are called phase boundaries. This is where two phases are at equilibrium. • So far, the phase diagrams we tackled are for pure substances. What happens when we are dealing with solutions/mixtures? Total vapor pressure of binary solution • For a binary solution of volatile liquids A and B for a given temperature, 𝑦𝐴 𝑃 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 , 𝑦𝐵 𝑃 = 𝑥𝐵 𝑃𝐵𝑠𝑎𝑡 • If we add these: 𝑦𝐴 + 𝑦𝐵 𝑃 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 + 𝑥𝐵 𝑃𝐵𝑠𝑎𝑡 𝑃 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 + (1 − 𝑥𝐴 )𝑃𝐵𝑠𝑎𝑡 𝑃 = 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 • This equation says that the total vapor pressure (or simply pressure) is equal linear with respect to liquid phase compositions. Vapor phase compositions 𝑃 = 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 • Substituting this expression into Raoult’s Law: 𝑦𝐴 𝑃 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴 = 𝑠𝑎𝑡 𝑃𝐵 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 • This equation allows us to plot vapor phase composition 𝑦𝐴 with regards to the liquid phase composition 𝑥𝐴 . In turn, we can plot pressure-vapor composition plots. Pressure in terms of vapor composition 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴 = 𝑠𝑎𝑡 𝑃𝐵 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 𝑦𝐴 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 𝑦𝐴 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡 𝑦𝐴 𝑥𝐴 = 𝑠𝑎𝑡 𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴 𝑃= 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 = 𝑃𝐵𝑠𝑎𝑡 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡 𝑦𝐴 + 𝑠𝑎𝑡 𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴 𝑃𝐴𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡 𝑃 = 𝑠𝑎𝑡 𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴 Pressure-composition (Pxy) diagram • By having two expressions for pressure in terms of both vapor and liquid composition of a binary volatile solution, we can plot two curves for pressure corresponding to vapor and liquid compositions. 𝑃 = 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡 𝑃 = 𝑠𝑎𝑡 𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴 Pxy diagram • In these diagrams, the vapor curve lies at lower pressures compared to the liquid curve. Consistent with the notion that lowering the pressure will drive the equilibrium towards the side of the vapor. • Horizontal axis can be interpreted as liquid (𝑥), vapor (𝑦), or overall (𝑧) composition. For liquid and vapor compositions, the difference is in which curve these compositions will be projected into to find the corresponding pressure. • Vapor/liquid region is where the system is purely vapor/liquid. • Area in between liquid and vapor curves is where the system is at vapor-liquid equilibrium. • At the vapor/liquid curve, there infinitesimal amount of liquid/vapor. is an Pxy diagram VLE benzene + toluene A – benzene B – toluene 𝑃 = 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡 𝑃 = 𝑠𝑎𝑡 𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴 Pure toluene Pure benzene Using Pxy diagrams • Determine the pressure and liquid composition at 20 oC when the vapor is 60% benzene. P ≈ 38 mmHg 𝒙𝑨 ≈ 𝟎. 𝟐𝟖 Using Pxy diagrams • A vapor containing 40% benzene and 60% toluene are in a sealed container at a pressure of 20 mmHg. Determine the vapor and liquid compositions when the pressure is increased to 35 mmHg. 𝒙𝑨 ≈ 𝟎. 𝟐𝟑 𝒚𝑨 ≈ 𝟎. 𝟓𝟒 Using Pxy diagrams • For the previous example, if there are initially 10 moles of vapor, how many moles will turn to liquid when the pressure is increased to 35 mmHg? 𝒙𝑨 ≈ 𝟎. 𝟐𝟑 𝒚𝑨 ≈ 𝟎. 𝟓𝟒 𝒛𝑨 = 𝟎. 𝟒 Inverse Lever Arm Rule (ILAR) • Used in phase diagrams to measure the ratio of amounts between two phases in equilibrium. Applicable in VLE, LLE, SLE, and other phase diagrams. 𝑉 𝑧𝐴 − 𝑥𝐴 = 𝑀 𝑦𝐴 − 𝑥𝐴 𝐿 𝑉 𝑦𝐴 − 𝑧𝐴 =1− = 𝑀 𝑀 𝑦𝐴 − 𝑥𝐴 𝒙𝑨 ≈ 𝟎. 𝟐𝟑 𝒚𝑨 ≈ 𝟎. 𝟓𝟒 L Tie line Isopleth 𝒛𝑨 = 𝟎. 𝟒 M V Inverse Lever Arm Rule (ILAR) • For the previous example, if there is initially 10 moles of vapor, how many moles will turn to liquid when the pressure is increased to 35 mmHg? 𝒙𝑨 ≈ 𝟎. 𝟐𝟑 𝒚𝑨 ≈ 𝟎. 𝟓𝟒 𝐿 𝑦𝐴 − 𝑧𝐴 = 𝑀 𝑦𝐴 − 𝑥𝐴 𝐿 0.54 − 0.4 = = 0.452 10 0.54 − 0.23 L 𝐿 = 4.52 𝑚𝑜𝑙 4.52 moles of the original vapor will turn to liquid. 𝒛𝑨 = 𝟎. 𝟒 M V Temperature-composition (Txy) diagrams • A Txy diagram shows the composition of phases that are in equilibrium at various temperatures at a given pressure. • Unlike in Pxy diagrams, the vapor curve lies at higher temperatures compared to the liquid curve. • ILAR is also applicable in Txy diagrams. Txy diagrams • Although Txy diagrams can be generated analytically by expressing vapor pressures in terms of temperature, Txy diagrams are usually generated using experimental data and fitting temperature-composition data into a cubic polynomial model. Generating Txy diagrams • The following temperature-composition data were gathered for a hexane-heptane system. T (oC) 65 66 70 77 85 100 xhexane 0 0.20 0.40 0.60 0.80 1 yhexane 0 0.02 0.08 0.20 0.48 1 Generating Txy diagrams • Fit data into cubic polynomial model and plot: • Models: T vs x, and T vs y Txy diagrams • Temperature at xHx = 0 is the boiling point of pure heptane while temperature at xHx = 1 is the boiling point of pure hexane. • Below liquid curve (blue), the liquid solution is below its boiling point or saturation temperature and is therefore purely liquid (subcooled liquid). • Above the vapor curve (purple), the vapor mixture is above its saturation temperature and is therefore purely vapor (superheated vapor). • In between the two curves, there is coexistence of vapor and liquid. Txy diagrams • The inverse lever arm rule (ILAR) applies to Txy diagram as well to determine ratios between vapor and liquid phases. 𝑉 𝑧𝐴 − 𝑥𝐴 = 𝑀 𝑦𝐴 − 𝑥𝐴 𝐿 𝑉 𝑦𝐴 − 𝑧𝐴 =1− = 𝑀 𝑀 𝑦𝐴 − 𝑥𝐴 V M L Distillation • Simple distillation Vapor is withdrawn and condensed to separate a volatile liquid from a non-volatile solute or solid. • Fractional distillation Boiling and condensation cycles are successively to separate two volatile liquids. repeated Azeotropes • An azeotropic state is characterized by equal vapor and liquid compositions. • Azeotropes lie at a lower or higher temperature compared to the individual boiling points of the constituents. Ethanol-water azeotrope is at 78.2 oC. Ethanol boiling point is 78.4 oC while water boils at 100 oC. (low–boiling azeotrope) • When an azeotrope is reached during distillation, no more changes in the composition is expected due to equal compositions of vapor and liquid. Maximum concentration of ethanol in ethanolwater mixture that can be achieved through fractional distillation is 97.2% by volume due to the azeotrope. • Existence of azeotropes depend on deviation of the solution from ideality. the Immiscible liquids • For the distillation of two immiscible liquids A and B, the boiling commences when the sum of the individual pressures reaches the prevailing pressure. 𝑃 = 𝑃𝐴𝑠𝑎𝑡 + 𝑃𝐵𝑠𝑎𝑡 • For the system in (b), the two liquids will boil at different temperatures • For the system in (a), the two liquids will boil at the same temperature but at a lower temperature compared to their individual boiling points. Example • Determine the boiling point of a mixture of octane (1) and water (2) at 101.325 kPa. Antoine Equation: ln 𝑃𝑖𝑠𝑎𝑡 𝑘𝑃𝑎 = 𝐴 − 𝐵 𝑇 °𝐶 + 𝐶 Antoine coefficients: Water: 𝐴 = 16.3872, 𝐵 = 3885.70, 𝐶 = 230.170 Octane: 𝐴 = 13.9346, 𝐵 = 3123.13, 𝐶 = 209.635 Solution 𝑃 = 𝑃1𝑠𝑎𝑡 + 𝑃2𝑠𝑎𝑡 101.325 = exp 13.9346 − 3123.13 𝑇 + 230.170 + exp 16.3872 − 𝑻 = 𝟖𝟏. 𝟕𝟑 °𝑪 Octane boiling point at 1 atm: 125.6 oC Water boiling point at 1 atm: 100 oC 3885.70 𝑇 + 230.170 Liquid-liquid equilibrium (LLE) Liquid-liquid equilibrium (LLE) • Consider temperature-composition diagrams for systems that consist of pairs of partially miscible liquids. These liquids do not mix in all proportions at all temperatures. Same principles in VLE phase diagrams may be applied. • Suppose small amounts of liquid B is added to pure liquid A at temperature T There will come a time when successive additions will not dissolve but will form a second phase. This is called phase separation. • A dissolves slightly into the added B and the two phases co-exist at compositions 𝑎′ and 𝑎′′. • If more and more B is added, system crosses the phase boundary and becomes a single phase system again. Example • The phase diagram for the system nitrobenzene (N) and hexane (H) at 1 atm is shown in the figure. A mixture of 0.59 mol H and 0.41 mol N was prepared at 290 K. What are the compositions of the phases, and in what proportions do they occur? 𝑥𝑁 = 0.41 Solution 𝑙𝛽 𝛼 = 𝛼 + 𝛽 𝑙𝛼 + 𝑙𝛽 𝛽 𝑙𝛼 = 𝛼 + 𝛽 𝑙𝛼 + 𝑙𝛽 𝛼 𝛼 𝛼 + 𝛽 𝑙𝛽 0.83 − 0.41 = = = =7 𝛽 𝛽 𝑙𝛼 0.41 − 0.35 𝛼+𝛽 There is 7 times more hexane-rich phase than nitrobenzene-rich phase. 𝑥𝑁 = 0.35 𝑥𝑁 = 0.83 Critical solution temperature • The upper critical solution temperature 𝑇𝑢𝑐 is the maximum temperature at which phase separation can occur. • Some systems exhibit a minimum temperature 𝑇𝑙𝑐 or lower critical solution temperature where phase separation can occur. Ternary systems • For ternary liquid mixtures (3 partially miscible liquids), the conventional Cartesian coordinates cannot represent the system fully. • Thus, a triangular diagram will be used. • For ternary systems: 𝑥𝐴 + 𝑥𝐵 + 𝑥𝐶 = 1 Ternary systems Ternary phase diagrams • Miscibility of liquids are highly dependent on temperature. • A face of the triangular prism in the figure represents a binary system. A ternary system phase diagram is a horizontal section of this triangular prism at a given constant temperature. Ternary phase diagrams of partially miscible liquids • The figure is a ternary phase diagram of a liquid mixture. • A is fully miscible with both C and B, but C and B are only partially miscible. • In this diagram, the two phases formed are C-rich and B-rich phases. • The point P is called the plait point. This is where the compositions of the two phases become identical. • Tie lines are identified experimentally. Usually, several tie lines are included in ternary phase diagrams. P Example of a ternary phase diagram • Line a corresponds to a H2O-CHCl3 ratio of 3:2. Adding acetic acid will retain the ratio between water and trichloromethane but the system will move from the bottom to the acetic acid corner through line a. Acetic acid will dissolve in both phases which will also change the composition water-rich and trichloromethane-rich phases. After some amount is added, the system will be fully miscible. • Line b corresponds to a H2O-CHCl3 ratio of 2:3. Same goes for line b. Extension of ternary phase diagrams to solids. Solid-liquid equilibrium (SLE) Eutectics • Consider a binary liquid mixture with composition a 1. • The cooling of the binary liquid mixture will undergo several stages: a2 – B begins to precipitate. a3 – pure B solid is in equilibrium with a liquid of composition b3. a4 – The system freezes into a solid with pure A and pure B phases. • The point 𝑒2 is called the eutectic point. It is the lowest temperature where the liquid is present in the system. • Relative amounts of liquids and solids can be elucidated using ILAR. Cooling curves • The solid-liquid boundary is given by the points at which the rate of cooling changes. • Points of interest: a1 – liquid is being cooled. a2 – B begins to precipitate. a3 – eutectic begins solidifying. It takes some time to solidify the eutectic. Liquid begins freezing to produce two-phase solid. a4 – mixture is completely frozen. That’s it for the coverage of LE 3 • LE 3 schedule: October 18, 2021 (class hours) • Synchronous exam. Only those present in Zoom meeting will be given a grade. • LE policies still apply. • Multiple choice on BB. (30 minutes) • Problem solving will be passed through Gform in a PDF format. (2.5 hours)
