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:
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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
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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 + B pp AB po A p B p A
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.
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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
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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?
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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
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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
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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
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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
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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
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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
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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 ∂T T2
• For pure component A ∂()GTo o A =−HA ∂T T2
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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 ↑
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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 dXB 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
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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.
5. Write and expression for the variation of ∆ Gmix 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.
5. W rite and expression for the variation of ∆ Hmix 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
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