Lecture 20: Two-Component Phase Equilibria

5.60 Spring 2008 Lecture #20 page 1 Two-Component Phase Equilibria Goal: Understand & predict effects of mixing of substances on vapor pressure, boiling point, freezing point, etc. Binary liquid-gas mixtures (non-reacting): Total # of variables: 4 (T, p, xA, yA) Constraints due to coexistence: 2 µA(A) = µA(g) µB(A) = µB(g) # independent variables F = 4 – 2 = 2 Only 2! e.g. knowing (T,p) uniquely determines the compositions in the liquid & gas phases Generalization: Gibbs phase rule gives # independent variables needed to describe a multi-component system where different phases coexist in equilibrium F=C–P+2 F ≡ # degrees of freedom (independent variables) C ≡ # components P ≡ # phases How do we get this? Suppose a system has C components and P phases. What are all the variables? First, T and p. 5.60 Spring 2008 Lecture #20 page 2 Then in each phase “α”, each component is specified by its mole fraction, with the constraint that C xi α ∑ i ( ) = 1. =1 So the composition of each phase is specified by (C – 1) variables. With P phases, we have P(C – 1) variables. Including T and p, the total # variables is P(C – 1) + 2. Now add constraints due to phase equilibria: Chemical potential of each component is the same in all the phases. e.g. for component “i”, µi(1) = µi(2) = K µi(P ) ⇒ P – 1 constraints For C components, it’s C(P – 1) constraints altogether So total # independent variables is F = P(C – 1) + 2 - C(P – 1) F=C–P+2 Gibbs phase rule For 1-component system: F = 3 – P P = 1 ⇒ F = 2 Can vary freely in (T,p) plane P = 2 ⇒ F = 1 Can vary along coexistence curve T(p) P = 3 ⇒ F = 0 No free variables at triple point (Tt,pt) P = 4 ⇒ Impossible! Can’t have 4 phases in equilibrium Raoult’s Law and Ideal Solutions “A” is a volatile solvent (e.g. water) “B” is a nonvolatile solute (e.g. antifreeze) pA* ≡ vapor pressure of pure A at temperature T 5.60 Spring 2008 Lecture #20 page 3 Raoult’s Law assumes a linear dependence Solvent and solute do not interact, like in mixture of ideal gases pA = xApA* = (1-xB)pA* Application: Vapor pressure lowering (our first “colligative” property) pA* - pA = pA* - xApA* = (1 - xA)pA* = xBpA* > 0 So pA < pA* Vapor pressure is lowered in the mixture Now let’s consider both A and B volatile pA = xApA * and pB = x B p B * p = pA + pB = xApA * + xBpB* (xA + xB = 1) “Ideal” solutions ≡ both components obey Raoult’s Law The diagram above shows the composition of the liquid phase It does not provide information about the gas phase composition 5.60 Spring 2008 Lecture #20 page 4 The gas phase is described by yA or yB. If T and xA are given, then yA and yB are fixed (by Gibbs phase rule). That is, if T and the composition of the liquid phase are known, then the composition of the gas phase is determined. So how do we get yA? p A = yA p (Dalton’s Law) pA = xApA * and pB = xBpB* = (1 –xA)pB* pA pA xA pA∗ yA = = = p pA + pB xA pA∗ + (1 − xA ) pB∗ yA = xA pA∗ pB∗ + ( pA∗ − pB∗ ) xA Inverting this expression ⇒ xA = yA pB∗ pA∗ + ( pB∗ − pA∗ ) yA pA xA pA∗ = Combining these two results ⇒ p = yA yA ∗ ∗ pA pB or p = ∗ pA + ( pB∗ − pA∗ ) yA This is summarized in the following diagram: (Raoult’s Law) 5.60 Spring 2008 Lecture #20 page 5 Combining both diagrams into one plot: This allows us to see the compositions of both liquid and gas phases If we know the composition of one phase at a given T, we can determined the composition of the other phase from the diagram