Solution Behavior • Goal: Understand the behavior of homogeneous systems containing more than one component. • Concepts Activity Partial molar properties Ideal solutions Non-ideal solutions Gibbs’-Duhem Equation Dilute solutions (Raoult’s law and Henry’s law) • Homework: WS2002 1 1 Raoult’s Law • Describes the behavior of solvents with containing a small volume fraction of solute • The vapor pressure of a component A in a solution at temperature T is equal to the product of the mole fraction of A in solution and the vapor pressure of pure A at temperature T. • Assumes: A, B evaporation rates are independent of composition A-B bond strength is the same as A-A and B-B or A-B bond strength is the mean of A-A and B-B • Example of liquid A in evacuated tank Initially, A evaporates at re(A) • At equilibrium Evaporation/condensation same rate WS2002 2 2 Raoult’s Law • Add a small amount of B (solute) to A (solvent) in tank • If the composition of the surface is the same as the bulk, then the fraction of surface sites occupied by A atoms is XA, the mole fraction • The evaporation rate will decrease accordingly along with the vapor pressure of A above the liquid. Condensation should be the same. Ideal Raoultian Solution = o pXp BBB po pp+ B AB o pA p B pA Vapor Pressure A B XB WS2002 3 3 Henry’s Law • Describes the behavior of a solute in a dilute solution • The solvent obeys Raoult’s law in the same region where the solute follows Henry’s law. • The vapor pressure of B over a dilute solution of B in A is proportional to XB. • If A-B bonds are stronger than A-A and B-B (or the mean A-A/B-B strength) then every B atom on the surface will be surrounded by A. On average, B will be more strongly bound than A and its vapor pressure will be lower than predicted on the basis of XB. • If A-B bonds are weaker than A-A and B-B bonds, then B will evaporate at a faster rate than it would if it were pure B. WS2002 4 4 Henry’s Law =γo pX BBB ′ > r ′(B) < r (B) r e (B) re (B) e e Raoult's Law po po B B Henry's Law Vapor Pressure Vapor Pressure Henry's Law Raoult's Law 0 1 0 1 XB XB WS2002 5 5 Activity in Solution • The thermodynamic activity of a component A in solution, aA, is related to the difference in behavior of component A in the solution and its behavior in its pure (reference) state. µ−µ=µ=o ∆ AA AARTln a • For gases, activity is often defined in terms of fugacity, f. = fA aA o fA • Ideally, the activity of a component of a solution is related to its concentration, given as mole fraction for solids and liquids, partial pressure for gases. ==pA fA aA o o gases pA fA = a AA X liquids/ solids WS2002 6 6 Example Problem • 1 mole of Cr is added to a large quantity of Cr-Fe melt (solution). The temperature of the solid and the melt is 1600°C and XFe = 0.8. • Asssuming Raoultian behavior, what are the heat and entropy changes in the solution? WS2002 7 7 Partial Molar Quantities • Often, properties of only one component in a solution can be measured. • The rest are calculated using partial molar quantities = QQTPnnn( , ,12 , ,...c ) ∂Q ∂Q ∂Q ∂Q dQ = dT + dP + dn + dn + ... ∂ ∂ ∂ 1 ∂ 2 T Pn, ,...P Tn , ,... n1 n2 ≠ 11TPn,,22 ,... TPn ,, k n • If T and P are held constant ∂Q ∂Q dQ = dn + dn + ... ∂ 1 ∂ 2 n1 n2 ≠ TPn,,22 ,... TPn ,, k n ∂Q Q = dn i ∂ i ni ≠ TPn,, ki n • Q can be any extensive state function, G, S, H, U, or V WS2002 8 8 Partial Molar Quantities • If we examine the Gibbs’ free energy of a binary A-B mixture, then G is our extensive state function. = GGTPnn(,,AB , ) Holding T& P con tan t ∂G ∂G dG = dn + dn ∂ A ∂ B nA nB TPn,,BATPn ,, • From the definition of partial molar free energy ∂G G = A ∂ nA TPn,,B • This can be used to derive one form of the Gibbs’-Duhem equation += ndG AA ndG BB0 WS2002 9 9 Free Energy of Solution • If the partial molar free energy is known for one component, the other properties of the solution can be calculated • In an A-B binary solution dG dG GGX=+ or G=+− G()1 X AB AA dXA dXA dG dG GGX=+ or G=+− G()1 X BA dX BB dX B B WS2002 10 10 How Does Mixing Change G? • What is the free energy change in the system due to mixing? • Examine two processes to form the same solution Soute Added Directly to Solvent + Solid A Liquid B Homogeneous Solution VA VB A dissolved in B V = VA + VB Soute Evaporated and Then Condensed into Solvent ∆ Gb A Evaporates A Condenses ∆G ∆ c Ga Solid A Liquid B Homogeneous Solution VA VB A dissolved in B V = VA + VB WS2002 11 11 Properties of Ideal Solutions • For an ideal solution, activity equal to the mole fraction = aX AA • The free energy change due to mixing is given by ∆=mix, ID + GRTXaXa()AAln BB ln ∆=mix, ID + GRTXXXX()AABBln ln ∆=∆+∆mix, ID mix,, ID mix ID GGGA B ∆=mix, ID = GA RTln XAA RT ln a WS2002 12 12 Volume Change • For an ideal solution, ∆Vmix = 0 = o VVAA o V B = o VVBB Vo =+ A VXVXVAA BB or o o =+ Molar Volume VXVXV AA BB A B XB WS2002 13 13 Heat of Mixing • Gibbs’-Helmholz Equation ∂()GT ∂()∆GT ∆ =−H =− H ∂T T22∂T T • For component A of an A-B solution ∂()GT A =−HA 2 ∂T T • For pure component A ∂()GTo o A =−HA 2 ∂T T WS2002 14 14 Non-Ideal Solutions • Same analysis, but a ≠ 1 • Activity coefficient, γ • γ = 1 is described as ideal behavior • γ > 1 positive deviation from ideal behavior • γ < 1 negative deviation from ideal behavior • As the temperature increases, deviation from ideal behavior decreases, i.e., γ → 1 as T ↑ WS2002 15 15 Graphical Methods • If the free energy of mixing is known as a function of composition, then the partial molar free energy of each component can be determined graphically at any composition • Set the composition of interest, XB • Draw a tangent and note the intercepts ∆G = G • The intercept with the axis at mole mix m fraction of B = 0 is the partial molar G°A free energy of A ( G A), likewise the G°B GA other intercept is the partial molar free energy of B. dG • The slope of the tangent is M dX B rel − o GB • The difference GG A A is GA 0 1 XB WS2002 Mole Fraction of B 16 16 Heat of Mixing • γ > 1, endothermic mixing γ decreases as T increases Heat is absorbed during the dissolution process The solute tends to cluster or phase separate • γ < 1, exothermic mixing Indicates a tendency toward A-B compound formation A-B bonds are stronger than A-A or B-B bonds There is a tendency toward ordering A atoms only want B atoms as nearest neighbors WS2002 17 17 Example Problem The partial pressure of a gas A over an A-B alloy was measured. The data are: XA 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 -6 -6 -6 -6 -6 -6 -6 -6 -6 pA 5 x10 4.4 x10 3.75x10 2.9 x10 1.8 x10 1.1 x10 0.8 x10 0.6 x10 0.4 x10 pA po A Determine 1. The pressure range where Henry's law is valid. 2. The Henry's law constant in this range. ∆ mix 3. GA as a function of XA in the composition range where Henry's law is obeyed. ∆ mix 4. Evaluate GA at XA = 0.4. ∆ mix 5. Write and expression for the variation of G with composition in the Henrian range. mix 6. Evaluate ∆ G at XA = 0.4. 18 Example Problem The partial pressure of a gas A over an A-B alloy was measured. The data are: XA 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 -6 -6 -6 -6 -6 -6 -6 -6 -6 pA 5 x10 4.4 x10 3.75x10 2.9 x10 1.8 x10 1.1 x10 0.8 x10 0.6 x10 0.4 x10 pA o pA Determine 1. T he pressure range where Henry's law is valid. 2. T he Henry's law constant in this range. mix 109. 3 3. ∆H in the composition range where Henry's law is obeyed if γ o =− − A log A 0. 2886 T ∆ mix 4. E valuate HA at XA = 0.4. ∆ mix 5. W rite and expression for the variation of H with composition in the Henrian range. mix 6. E valuate ∆ H at XA = 0.4. 19 Quasi-Chemical Model • Predicts ∆Hmix is proportional to the number of unlike bonds No limits on number of bonds of any types Assumes that bond energy is not a function of composition • A-B binary solution • P is the number of bonds of a specific type • z is the number of nearest neighbors ∆ mix • U = PAB[eAB - 1/2(eAA + eBB)] • ∆Hmix = ∆Umix + P∆Vmix ≈ ∆Umix WS2002 20 20.