notes ch7

Chapter 7: Simple Mixtures : 

Chapter 7: Simple Mixtures Homework: Exercises(a only):7.4, 5,10, 11, 12, 17, 21 Problems: 1, 8

Chapter 7 - Simple Mixtures : 

Chapter 7 - Simple Mixtures Restrictions Binary Mixtures xA + xB = 1, where xA = fraction of A Non-Electrolyte Solutions Solute not present as ions

Partial Molar Quantities -Volume : 

Partial Molar Quantities -Volume Partial molar volume of a substance slope of the variation of the total volume plotted against the composition of the substance Vary with composition due to changing molecular environment VJ = (V/ nJ) p,T,n’ pressure, Temperature and amount of other component constant Partial Molar Volumes Water & Ethanol

Partial Molar Quantities & Volume : 

Partial Molar Quantities & Volume If the composition of a mixture is changed by addition of dnA and dnB dV = (V/ nA) p,T,nA dnA + (V/ nB) p,T,nB dnB dV =VAdnA + VBdnB At a given compositon and temperature, the total volume, V, is V = nAVA + nBVB

Measuring Partial Molar Volumes : 

Measuring Partial Molar Volumes Measure dependence of volume on composition Fit observed volume/composition curve Differentiate Example - Problem 7.2 For NaCl the volume of solution from 1 kg of water is: V= 1003 + 16.62b + 1.77b1.5 + 0.12b2 What are the partial molar volumes? VNaCl = (∂V/∂nNaCl) = (∂V/∂nb) = 16.62 + (1.77 x 1.5)b0.5 + (0.12 x 2) b1 At b =0.1, nNaCl = 0.1 VNaCl = 16.62 + 2.655b0.5 + 0.24b = 17.48 cm3 /mol V = 1004.7 cm3 nwater = 1000g/(18 g/mol) = 55.6 mol V = nNaClVNaCl + nwaterVwater Vwater = (V - VNaCl nNaClr )/ nwater = (1004.7 -1.75)/55.6= 18.04 cm3 /mol

Partial Molar Quantities - General : 

Partial Molar Quantities - General Any extensive state function can have a partial molar quantity Extensive property depends on the amount of a substance State function depends only on the initial and final states not on history Partial molar quantity of any function is just the slope (derivative) of the function with respect to the amount of substance at a particular composition For Gibbs energy this slope is called the chemical potential, µ

Partial Molar Free Energies : 

Partial Molar Free Energies Chemical potential, µJ, is defined as the partial molar Gibbs energy @ constant P, T and other components µJ = (G/ nJ) p,T,n’ For a system of two components: G = nAµA + n B µB G is a function of p,T and composition For an open system constant composition, dG =Vdp - SdT + µA dnA + µB dn B Fundamental Equation of Thermodynamics @ constant P and T this becomes, dG = µA dnA + µB dn B dG is the the non expansion work, dwmax FET implies changing composition can result in work, e.g. an electrochemical cell

Chemical Potential : 

Chemical Potential Gibbs energy, G, is related to the internal energy, U U = G - pV + TS (G = U + pV - TS) For an infinitesimal change in energy, dU dU = -pdV - Vdp + TdS + SdT + dG but dG =Vdp - SdT + µA dnA + µB dn B so dU = -pdV - Vdp +TdS +SdT + Vdp - SdT + µA dnA + µB dn B dU = -pdV + TdS + µA dnA + µB dn B at constant V and S, dU = µA dnA + µB dn B or µJ = (U/ nJ)S,V,n’

µ and Other Thermodynamic Properties : 

µ and Other Thermodynamic Properties Enthalpy, H (G = H - TS) dH = dG + TdS + SdT dH= (Vdp - SdT + µA dnA + µB dn B) - TdS SdT dH = VdP - TdS + µA dnA + µB dn B at const. p & T : dH = µA dnA + µB dn B or µJ = (H/ nJ)p,T,n’ Helmholz Energy, A (A = U-TS) dA = dU - TdS - SdT dA = (-pdV + TdS + µA dnA + µB dn B ) - TdS - SdT dA = -pdV - SdT + µA dnA + µB dn B at const. V & T : dA = µA dnA + µB dn B or µJ = (A/ nJ)V,T,n’

Gibbs-Duhem Equation : 

Gibbs-Duhem Equation Recall, for a system of two components: G = nAµA + n B µB If compositions change infinitesimally dG = µA dnA + µB dn B + nAdµA + n Bd µB But at constant p & T, dG = µA dnA + µB dn B so µA dnA + µB dn B = µA dnA + µB dn B + nAdµA + n Bd µB or nAdµA + n Bd µB = 0 For J components, nidµi = 0 (i=1,J) {Gibbs-Duhem Equation}

Significance of Gibbs-Duhem : 

Significance of Gibbs-Duhem Chemical potentials of multi-component systems cannot change independently Two components, G-D says, nAdµA + n Bd µB = 0 means that d µB = (nA/ n B)dµA Applies to all partial molar quantities Partial molar volume dVB = (nA/ n B)dVA Can use this to determine on partial molar volume from another You do this in Experiment 2

Example Self Test 7.2 : 

Example Self Test 7.2 VA = 6.218 + 5.146b - 7.147b2 dVA = + 5.146 - 2*7.147b = + 5.146 - 14.294b db dVA/db = + 5.146b - 14.294b If *MB is in kg/mol dVB = -nA/nB (dVA); b=nA/nB*MB or b *MB = nA/nB dVB = -nA/nB (dVA ) = nA/nB dVA = b *MB dVA dVB = -b* MB (5.146 - 14.294b) db =- MB(2.573b-4.765b2) VB =VB* + MB (4.765b2 - 2.573b) from data VB* = 18.079 cm3mol-1 and MB = 0.018 kg/mol so VB = 18.079 cm3mol-1 + 0.0858b2 - 0.0463b

Thermodynamics of Mixing : 

Thermodynamics of Mixing For 2 Gases (A &B) in two containers, the Gibbs energy, Gi Gi = nAµA + nBµB But µ = µ° + RTln(p/p°) so Gi = nA(µA° + RTln(p/p°) )+ nB(µB° + RTln(p/p°)) If p is redefined as the pressure relative to p° Gi = nA(µA° + RTln(p) )+ nB(µB° + RTln(p) ) After mixing, p = pA + pB and Gf = nA(µA° + RTln(pA) )+ nB(µB° + RTln(pB) ) So Gmix = Gf - Gi = nA (RTln(pA/p) )+ nB(RTln(pB /p) Replacing nJ by xJn and pJ/p=xJ (from Dalton’s Law) Gmix = nRT(xA ln (xA ) + xBln(xB )) This equation tells you change in Gibbs energy is negative since mole fractions are always <1

Example :Self-Test 7.3 : 

Example :Self-Test 7.3 2.0 mol H2(@2.0 atm) + 4 mol N2 (@3.0 atm) mixed at const. V. What is Gmix? Initial: pH2= 2 atm;VH2= 24.5 L; pN2= 3 atm;VN2= 32.8 L{Ideal Gas} Final: VN2= VH2= 57.3 L; therefore pN2= 1.717 atm; pH2= 0.855 atm;{Ideal Gas} Gmix = RT(nA ln (pA /p) + nBln(pB /p)) Gmix = (8.315 J/mol K)x(298 K)[2mol x ( ln(0.855/2) + 4 mol x (ln(1.717/3)] Gmix = -9.7 J What is Gmix under conditions of identical initial pressures? xH2 = 0.333; xN2 = 0.667; n = 6 mol Gmix = nRT(xA ln (xA ) + xBln(xB )) Gmix = 6mol x( 8.315J/molK)x 298.15K{0.333ln0.333 +0.667ln0.667) Gmix = -9.5 J

Entropy and Enthalpy of Mixing : 

Entropy and Enthalpy of Mixing For Smix, recall G = H - TS Therefore Smix = -Gmix / T Smix = - [ nRT(xA ln (xA ) + xBln(xB ))] / T Smix = - nR(xA ln (xA ) + xBln(xB ) It follows that Smix is always (+) since xJln(xJ ) is always (-) For Hmix H = G + TS ={nRT(xA ln (xA ) + xBln(xB )} +T{- nR(xA ln (xA ) + xBln(xB )} H ={nRT(xA ln (xA ) + xBln(xB )} - {nRT(xA ln (xA ) + xBln(xB )} H = 0 Thus driving force for mixing comes from entropy change

Chemical Potentials of LiquidsIdeal Solutions : 

Chemical Potentials of LiquidsIdeal Solutions At equilibrium chem. pot. of liquid = chem. pot. of vapor, µA(l) = µA(g,p) For pure liquid, µ*A(l) = µ°A + RT ln(p *A) [1] For A in solution, µA(l) = µ°A + RT ln(p A) [2] Subtracing [1] from [2] : µA(l) - µ*A(l) = RT ln(pA) + RT ln(p *A) µA(l) - µ*A(l) = RT{ln(pA) - ln(p *A)} = RT{ln(pA/p *A)} µA(l) = µ*A(l) + RT{ln(pA/p *A)} [3] Raoult’s Law - ratio of the partial pressure of a component of a mixture to its vapor pressure as a pure substance (pA/p*A) approximately equals the mole fraction, xA pA = xA p*A Combining Raoult’s law with [3] gives µA(l) = µ*A(l) + RT{ln(xA)}

Ideal Solutions/Raoult’s Law : 

Ideal Solutions/Raoult’s Law Mixtures which obey Raoult’s Law throughout the composition range are Ideal Solutions Phenomenology of Raoult’s Law: 2nd component inhibits the rate of molecules leaving a solution, but not returning rate of vaporization  XA rate of condensation  pA at equilibrium rates equal implies pA = XA p*A

Deviations from Raoult’s Law : 

Deviations from Raoult’s Law Raoult’s Law works well when components of a mixture are structurally similar Wide deviations possible for dissimilar mixtures Ideal-Dilute Solutions Henry’s Law (William Henry) For dilute solutions, v.p. of solute is proportional to the mole fraction (Raoult’s Law) but v.p. of the pure substance is not the constant of proportionality Empirical constant, K, has dimensions of pressure pB = xBKB (Raoult’s Law says pB = xBpB) Mixtures in which the solute obeys Henry’s Law and solvent obeys Raoult’s Law are called Ideal Dilute Solutions Differences arise because, in dilute soln, solute is in a very different molecular environment than when it is pure

Applying Henry’s Law & Raoult’s Law : 

Applying Henry’s Law & Raoult’s Law Henry’s law applies to the solute in ideal dilute solutions Raoult’s law applies to solvent in ideal dilute solutions and solute & solvent in ideal solutions Real systems can (and do ) deviate from both

Applying Henry’s Law : 

Applying Henry’s Law What is the mole fraction of dissolved hydrogen dissolved in water if the over-pressure is 100 atmospheres? Henry’s constant for hydrogen is 5.34 x 107 PH2= xH2K; xH2 = PH2 /K= 100 atm x 760 Torr/atm/ 5.34 x 107 xH2 = 1.42 x 10-3 In fact hydrogen is very soluble in water compared to other gases, while there is little difference between solubility in non-polar solvents. If the solubility depends on the attraction between solute and solvent, what does this say about H2 -water interactions?

Properties of Solutions : 

Properties of Solutions For Ideal Liquid Mixtures As for gases the ideal Gibbs energy of mixing is Gmix = nRT(xA ln (xA ) + xBRTln(xB )) Similarly, the entropy of mixing is Smix = - nR(xA ln (xA ) + xBln(xB ) and Hmix is zero Ideality in a liquid (unlike gas) means that interactions are the same between molecules regardless of whether they are solvent or solute In ideal gases, the interactions are zero

Real Solutions : 

Real Solutions In real solutions, interactions between different molecules are different May be an enthalpy change May be an additional contribution to entropy (+ or - ) due to arrangement of molecules Therefore Gibbs energy of mixing could be + Liquids would separate spontaneously (immiscible) Could be temperature dependent (partially miscible) Thermodynamic properties of real solns expressed in terms of ideal solutions using excess functions Entropy: SE = Smix - Smixideal Enthalpy: HE = Smix(because Hmixideal = 0) Assume HE = nbRTxAxB where is const. b =w/RT w is related to the energy of AB interactions relative to AA and BB interactions b > 0, mixing endothermic; b < 0, mixing exothermic solvent-solute interactions more favorable than solvent-solvent or solute-solute interactions Regular solution is one in which HE  0 but SE  0 Random distribution of molecules but different energies of interactions GE = HE Gmix = nRT{(xA ln (xA ) + xBRTln(xB )) + bRTxAxB (Ideal Portion + Excess)

Activities of Regular Solutions : 

Activities of Regular Solutions Recall the activity of a compound, a, is defined a = gx where g = activity coefficient For binary mixture, A and B, consideration of excess Gibbs energy leads to the following relationships (Margules’ eqns) ln gA = bxB2 and ln gB = bxA2 [1] As xB approaches 0, gA approaches 1 Since, aA = gAxA, from [1] If b = 0, this is Raoult’s Law If b < 0 (endothermic mixing), gives vapor pressures lower than ideal If b > 0 (exothermic mixing), gives vapor pressures higher than ideal If xA<<1, becomes pA = xA eb pA* Henry’s law with K = eb pA*

Colligative Properties of Dilute Solutions : 

Colligative Properties of Dilute Solutions

Colligative Properties : 

Colligative Properties Properties of solutions which depend upon the number rater than the kind of solute particles Arise from entropy considerations Pure liquid entropy is higher in the gas than for the liquid Presence of solute increases entropy in the liquid (disorder increases) Lowers the difference in entropy between gas and liquid hence the vapor pressure of the liquid Result is a lowering chemical potential of the solvent Types of colligative properties Boiling Point Elevation Freezing Point Depression Osmotic pressure

Colligative Properties - General : 

Colligative Properties - General Assume Solute not volatile Pure solute separates when frozen When you add solute the chemical potential, µA becomes µA = µA * + RT ln(xA) where µA * = Chemical Potential of Pure Substance x A = mole fraction of the solvent Since ln(xA) in negative µA > µA*

Boiling Point Elevation : 

Boiling Point Elevation At equilibrium µ(gas) = µ(liquid) or µA(g) = µA *(l) + RTln(xA) Rearranging,(µA(g) - µA *(l))/RT = ln(xA) = ln(1- xB) But , (µA(g) - µA *(l)) = G vaporization so ln(1- xB) = G vap. /RT Substituting for G vap. (H vap. -T S vap. ) {Ingnore T dependence of H & S) ln(1- xB) = (H vap. -T S vap.)/RT = (H vap. /RT) - S vap./R When xB = 0 (pure liquid A), ln(1) = (H vap. /RTb) - S vap./R = 0 or H vap. /RTb = S vap./R where Tb= boiling point Thus ln(1- xB) = (H vap. /RT) - H vap. /RTb = (H vap. /R)(1/T- 1/Tb) If 1>> xB, (H vap. /R)(1/T- 1/Tb)  - xB and if T  Tb and T= T  Tb Then (1/T- 1/Tb) = T/Tb2 and (H vap./R) T/Tb2 = - xB so T= - xB Tb2 /(H vap./R) or T= - xB Kb where Kb = Tb2 /(H vap./R)

Boiling Point Elevation : 

Boiling Point Elevation Kb is the ebullioscopic constant Depends on solvent not solute Largest values are for solvents with high boiling points Water (Tb = 100°C) Kb = 0.51 K/mol kg -1 Acetic Acid (Tb = 118.1°C) Kb = 2.93 K/mol kg -1 Benzene (Tb = 80.2°C) Kb = 2.53 K/mol kg-1 Phenol (Tb = 182°C) Kb = 3.04 K/mol kg -1

Freezing Point Depression : 

Freezing Point Depression Derivation the same as for boiling point elevation except At equilibrium µ(solid) = µ(solid) or µA(g) = µA *(l) + RTln(xA) Instead of the heat of vaporization, we have heat of fusion Thus, T= - xB Kf where Kf = Tf2 /(H fus./R) Kf is the cryoscopic constant Water (Tf = 0°C) Kf = 1.86 K/mol kg -1 Acetic Acid (Tf = 17°C) Kf = 3.9 K/mol kg -1 Benzene (Tf = 5.4°C) Kf = 5.12 K/mol kg-1 Phenol (Tf = 43°C) Kf = 7.27 K/mol kg -1 Again property depends on solvent not solute

Temperature Dependence of Solubility : 

Temperature Dependence of Solubility Not strictly speaking colligative property but can be estimated assuming it is Starting point the same - assume @ equilibrium µ is equal for two states First state is solid solute, µB(s) Second state is dissolved solute, µB(l) µB(l) = µB*(l) + RT ln xB At equilibrium, µB(s) = µB(l) µB(s) = µB*(l) + RT ln xB Same as expression for freezing point except that use xB instead of xA

Temperature Dependence of Solubility : 

Temperature Dependence of Solubility To calculate functional form of temperature dependence you solve for mole fraction ln xB = [µB(s) - µB*(l)]/ RT = -G fusion/RT = -[H fus-T S fus]/RT ln xB = -[H fus-T S fus]/RT = -[H fus /RT] + [S fus/R] {1} At the melting point of the solute, Tm, G fusion/RTm = 0 because G fusion = 0 So [H fus-Tm S fus]/RTm = 0 or [H fus /RTm] -[S fus/R] = 0 Substituting into {1}, ln xB = -[H fus /RT] + [S fus/R]+ [H fus /RTm] -[S fus/R] This becomes ln xB = -[H fus /RT] + [H fus /RTm] Or ln xB = -[H fus /R] [1/T - 1/Tm] Factoring Tm, ln xB = [H fus /R Tm] [1 - (Tm /T)] Or xB = exp[H fus /R Tm] [1 - (Tm /T)-1] The details of the equation are not as important as functional form Solubility is lowered as temperature is lowered from melting point Solutes with high melting points and large enthalpies of fusion have low solubility Note does not account for differences in solvent - serious omission

ln xB = -[H fus /R] [1/T - 1/Tm] or xB = exp[H fus /R Tm] [1 - (Tm /T)-1] : 

ln xB = -[H fus /R] [1/T - 1/Tm] or xB = exp[H fus /R Tm] [1 - (Tm /T)-1]

Osmotic pressure : 

Osmotic pressure J. A. Nollet (1748) - “wine spirits” in tube with animal bladder immersed in pure water Semi-permiable membrane - water passes through into the tube Tube swells , sometimes bladder bursts Increased pressure called osmotic pressure from Greek word meaning impulse W. Pfeffer (1887) -quantitative study of osmotic pressure Membranes consisted of colloidal cupric ferrocyanide Later work performed by applying external pressure to balance the osmotic pressure Osmotic pressure , , is the pressure which must be applied to solution to stop the influx of solvent

Osmotic Pressure van’t Hoff Equation : 

Osmotic Pressure van’t Hoff Equation J. H. van’t Hoff (1885) - In dilute solutions the osmotic pressure obeys the relationship, V=nBRT nB/V = [B] {molar concentration of B, so =[B] RT Derivation- at equilibrium µ solvent is the same on both sides of membrane: µA *(p) = µA (x A,p +) {1} µA (x A,p +) = µ*A (x A,p +) + RTln(x A) {2} µ*A (x A,p +) = µA *(p) + ∫p p +Vm dp {3} [Vm = molar volume of the pure solvent] Combining {1} and {2} : µA *(p) = µA *(p) + ∫p p +Vm dp + RTln(x A) For dilute solutions, ln(x A) = ln(x B) ≈ - x B Also if the pressure range of integration is small, ∫p p +V m dp = Vm∫p p +dp = Vm  So 0 = V m + RT - x B or Vm= RT x B Now nA V m = V and, if solution dilute x B ≈ - n B /nA so =[B] RT Non-ideality use a virial expansion =[B] RT{1 + B[B] + ...} where B I s the osmotic virial coef. (like pressure)

Application of Osmotic Pressure : 

Application of Osmotic Pressure Determine molar mass of macromolecules =[B] RT{1 + B[B] + ...} but [B] = c/M where c is the concentration and M the molar mass so = c/M RT{1 + Bc/M + ...} g h = c/M RT{1 + Bc/M + ...} h/c = RT/(Mg) {1 + Bc/M + ...} h/c = RT/(Mg) + RTB/(M2g) + ...} Plot of h/c vs. c has intercept of RT/(Mg) so Intercept= RT/(Mg) or M= RT/(intercept xg) Units (SI) are kg/mol typical Dalton (Da) {1Da = 1g/ mol

Non-Ideality & ActivitiesSolvents : 

Non-Ideality & ActivitiesSolvents Recall for ideal solution µA =µA * + RTln(xA) µA* is pure liquid at 1 bar when xA =1 If solution does not ideal xA can be replaced with activity aA activtiy is an effective mole fraction aA = pA/pA* {ratio of vapor pressures} Because for all solns µA =µA * + RTln(pA/ p*A) As xA-> 1, aA -> xA so define activity coefficient,, such that aA =  A xA As xA-> 1, A -> 1 Thus µA =µA * + RTln(xA) + RTln(A) {substiuting for a A}

Non-Ideality & ActivitiesSolutes (Ideal & Non-Ideal) : 

Non-Ideality & ActivitiesSolutes (Ideal & Non-Ideal) For ideal dilute solutions Henry’s Law ( pB = KBxB) applies Chemical potential µB = µ B * + RTln(pB /pB*) µB = µ B * + RTln(KBxB /pB*) = µ B * + RTln(KB/pB*) + RTln(xB ) KB and pB* are characteristics of the solute so you can combine them with µ B * µB† = µ B * + RTln(KBxB /pB*) Thus µB = µ B† + RTln(xB ) Non-ideal solutes As with solvents introduce acitvity and activity coefficient aB = pB/KB*; aB =  B xB As xA-> 0, aA -> xA and A -> 1

Activities in Molalities : 

Activities in Molalities For dilute solutions x B ≈ n B /nA , and x B =  b/b° kappa,  , is a dimensionless constant For ideal-dilute solution, µB = µ B† + RTln(xB ) so µB = µ B† + RTln( b/b°) = µ B†+ RTln() + RTln( b/b°) Dropping b° and combing 1st 2 terms, µB = µ Bø+ RTln( b) µ Bø = µ B†+ RTln() µB has the standard value (µBø ) when b=b° As b ->0, µB ->infinity so dilution stabilizes system Difficult to remove last traces of solute from a soln Deviations from ideality can be accounted for by defining an activity aB and activity, B, aB =  B bB/b° where  B ->1, bB -> 0 The chemical potential then becomes µ = µø + RTln( a) Table 7.3 in book summarizes the relationships