Lecture 36. the Phase Rule
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Lecture 36. The Phase Rule P = number of phases C = number of components (chemically independent constituents) F = number of degrees of freedom xC,P = the mole fraction of component C in phase P The variables used to describe a system in equilibrium: x11, x21, x31,...,xC −1,1 phase 1 x12 , x22 , x32 ,..., xC−1,2 phase 2 x1P , x2P , x3P ,...,xC−1,P phase P T,P Total number of variables = P(C-1) + 2 Constraints on the system: m11 = m12 = m13 =…= m1,P P - 1 relations m21 = m22 = m23 =…= m2,P P - 1 relations mC,1 = mC,2 = mC,3 =…= mC,P P - 1 relations 1 Total number of constraints = C(P - 1) Degrees of freedom = variables - constraints F=P(C- 1) + 2 - C(P - 1) F=C- P+2 Single Component Systems: F = 3 - P In single phase regions, F = 2. Both T and P may vary. At the equilibrium between two phases, F = 1. Changing T requires a change in P, and vice versa. At the triple point, F = 0. Tt and Pt are unique. 2 Four phases cannot be in equilibrium (for a single component.) Two Component Systems: F = 4 - P The possible phases are the vapor, two immiscible (or partially miscible) liquid phases, and two solid phases. (Of course, they don’t have to all exist. The liquids might turn out to be miscible for all compositions.) 3 Liquid-Vapor Equilibrium Possible degrees of freedom: T, P, mole fraction of A xA = mole fraction of A in the liquid yA = mole fraction of A in the vapor zA = overall mole fraction of A (for the entire system) We can plot either T vs zA holding P constant, or P vs zA holding T constant. Let A be the more volatile substance: * * PA >PB and Tb,A <Tb,B Pressure-composition diagrams Fix the temperature at some value, T. Assume Raoult’s Law: * * P=PA xA +PB xB * * * * * P=PA xA +PB (1 - xA)=PB + (PA - PB )xA 4 Composition of the vapor P x P* y = A = A A > x A * + ()* − * A P PB PA PB x A P x P* y = B = B B < x B * + ()* − * B P PB PA PB xA Point a: One phase, F = 3 (T, P, xA) Point b: Liquid starts to vaporize, F = 2 (T,P; xA not free.) xA = zb,yA = yb” Vapor is rich in A. Point c: Liquid has lost so much A that its composition is xA = xc’. The vapor is now poorer in A, yA = yc” Ratio of moles in the two phases is given by the lever rule: n ′′ liq = cc ′ nvap cc 5 Pointd:Liquidisalmostallgone,xA = xd’,yA = yd =zA For points below d, only the vapor is present (F=3). Numerical example: benzene-toluene at 20 C Exercise 8.4b. Let B = toluene and A = benzene. * * Given: PA = 74 Torr, PB = 22 Torr, zA =0.5 Q. At what pressure does the mixture begin to boil? A. At point b, P = 0.5x22 + 0.5x74 = 48 Torr. Q. What is the composition of the vapor at this point? A. PB = 0.5x22 = 11, PA = 0.5x74 = 37, yA = 37/48 = 0.77 Q. What is the composition and vapor pressure of the liquid when the last few drops of liquid boil? A. Point d. 74x y = z = 0.5 = A A A + − 74xA 22(1 xA ) xA = 0.229, xB =0.771 P = 0.229x74 + 0.771x22= 33.9 Torr 6 Lecture 37. Construction of the Temperature- Composition Diagram Fix the total pressure at some chosen value, P. Boiling does not occur at a unique temperature, but rather over a range of temperatures. (Contrast with a pure substance.) The vapor pressure curve is determined by the liquid composition: P=PA(T) + PB(T) PA(T) is the vapor pressure of pure A at temperature T. 7 Assume Raoult’s Law: = * + * P xA PA (T) xB PB (T) = * + − * xAPA (T) (1 xA )PB (T) = * + * − * PB (T ) xA (PA (T ) PB (T )) The vapor pressure pure liquid A at any temperature T is given by the Clapeyron-Clausius equation: * ∆H PA (T) = − vap,A 1 − 1 ln * PA (T0 ) R T T0 * PA (T0 ) =1atm,T0 = normal boiling point of A. The same reasoning applies to substance B. * * Once we have PA (T ) and PB (T ) , Raoult’s Law gives us xA as a unique function of T and P. This allows us to construct the liquid curve. 8 Composition of the vapor: P x P* (T) y = A = A A A P P ∆H 1 1 * − vap − y P (T ) RT T T A = A = e b, A xA P This result is equivalent to µ * + = µ * + A,vap RT ln y A A,liq RT ln xA This last result is the starting point for the boiling point elevation formula, except that there we assume yA =1. 9 Recipe for construction of the phase diagram: 1. Calculate Tb,A and Tb,B at the pressure of the diagram. 2. Choose a temperature Tb,A <T<Tb,B * * 3. Calculate PA (T ) and PB (T ). P − P* (T ) x = B 4. A * − * PA (T ) PB (T ) = * 5. y A x APA (T)/P. 10 Cooling of a vapor mixture Point a. Pure vapor, yA =zA Point b. Vapor begins to condense at xA =zb’,yA =zA =zb Point c. Comparable amounts of the two phases are n ′ vapor = cc present. Lever rule: ′′ nliquid cc Point d. Vapor is almost all gone; xA =zd’ =zA,yA =zd” Point e. Only liquid is present. 11 Distillation Point a. Mixture starts to boil, with xA =zA,yA =zb Points b-c. Vapor is condensed to form a liquid with xA=zb =zc Point c. The liquid that was collected in the previous step is boiled to form a vapor with xA=zd Condensation of the last bit of vapor produces a liquid very rich in either A (if Tb,A <Tb,B on the left) or B (if Tb,A >Tb,B on the right). 12 Non-ideal solutions Left: Impossible phase diagram, because at Pmax,where the liquid of this composition just starts to boil, there is no corresponding point on the vapor curve. (There is no tie line.) The vapor should always lie below the liquid in a pressure-composition diagram. At an extremum they must touch. Center: Vapor pressure reaches a maximum because of repulsion between A and B. This is an azeotrope, where the liquid and vapor have the same composition. (Note an error in the drawing: The vapor curve does not have a cusp, but rather is tangent to the liquid curve.) Right: Vapor pressure reaches a minimum because of attraction between A and B. This is also an azeotrope. 13 Distillation of non-ideal solutions Left diagram: low boiling azeotrope. A solution of composition za first boils at Ta. The vapor comes off with composition zb. It is condensed and then boils at Tc The final vapor to come off has the azeotropic composition, and the remaining liquid is enriched in B (or A, depending on which side of the azeotrope the process started). Right diagram: high boiling azeotrope. As before, za first boils at Ta The final vapor to come off is enriched in B (or A), and the remaining liquid has the azeotropic composition. 14 Equilibrium between immiscible liquids Upper Critical Lower Critical Double Critical Temperature Temperature Temperature The two phase region is always described by a tie line. It is a “no man’s land.” 15 Boiling of immiscible liquids Melting of immiscible solids Mixture of non-reactive Solids A and B react to solids. form compound AB. 16 Lecture 38. Equilibrium Constants Consider the following reaction: H2 +Cl2 Ø 2HCl Does it occur spontaneously? Let’s calculate DG for one mole of reaction. ∆ 0 = G f (H 2 ) 0 ∆ 0 = G f (Cl2 ) 0 ∆ 0 = − G f (HCl) 95.30kJ / mol ∆ 0 = − Gr 190.6 It seems that the reaction occurs spontaneously. But what if we start with pure HCl. Shouldn’t some of it react to form H2 and Cl2? 17 To determine when the reaction occurs spontaneously, we must take the partial pressures into account: P ∆G (Cl ) = 0 + RT ln Cl2 f 2 P o Cl2 P ∆ = − + HCl G f (HCl) 95.30 RT ln o PHCl P 2 P P ∆ = − + HCl − H 2 − Cl2 Gr 190.6 RT ln RT ln RT ln P0 Po Po At equilibrium, DGr =0.Thatis, 2 PHCl (P o ) 2 P 2 190.6 = −∆G o = RT ln = RT ln HCl r P P H 2 Cl2 PH PCl 2 2 P o P o = −∆ 0 RT ln K P Gr ln KP = 76.93 at 298 K 33 KP =2.57x10 18 So what happens if we start with pure HCl? P P = P = 0 Suppose that initially HCl =1barand H2 Cl2 P = 1− 2z P = P = z At equilibrium, HCl and H 2 Cl2 ()1− 2z 2 = 2.57x1033 z2 1 ≈ 2.57x1033 z2 z=1.97 x 10-17 This reaction goes nearly to completion. But this won’t be the case at higher T. 19 Another example: N2O4 F 2NO2 ∆ 0 = G f 97.89 and 51.31 kJ/mol ∆ 0 = − = Gr 2x51.31 97.89 47.30 ln KP = -4.730/RT = 1.909 2 PNO 2 Po P2 = = = NO2 K P 0.148 o PN O P P 2 4 N2O4 Po What is the composition of 2 bar of this material? Let x be the mole fraction of NO2.