Chapter 7: Simple Mixtures

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Chapter 7: Simple Mixtures. Homework: Exercises(a only):7.4, 5,10, 11, 12, 17, 21 Problems: 1, 8. Chapter 7 - Simple Mixtures. Restrictions Binary Mixtures x A + x B = 1, where x A = fraction of A Non-Electrolyte Solutions Solute not present as ions. Partial Molar Quantities -Volume.

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## Chapter 7: Simple Mixtures

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### Chapter 7: Simple Mixtures

Homework:

Exercises(a only):7.4, 5,10, 11, 12, 17, 21

Problems: 1, 8

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 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
• 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
• At a given compositon and temperature, the total volume, V, is
• V = nAVA + nBVB
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
• 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
• 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
• 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
• 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
• 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
• 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

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

Gibbs Energy of Mixing

Of

Two Ideal Gases

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
• 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 of Mixing

Two Ideal Gases

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
• 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
• 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
• 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
• 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
• 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
• 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
• 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 usingexcess 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
• 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 = xAeb pA*
• Henry’s law with K = eb pA*

### Colligative Properties of Dilute Solutions

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
• 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

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
• 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
• 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
• 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
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
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
• 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) + ∫pp +Vm dp {3} [Vm = molar volume of the pure solvent]
• Combining {1} and {2} : µA *(p) = µA *(p) + ∫pp +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,

∫pp +V m dp = Vm∫pp +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
• 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
• 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)
• For ideal dilute solutions Henry’s Law ( pB = KBxB) applies
• Chemical potential µB = µB * + RTln(pB /pB*)
• µB = µB * + RTln(KBxB /pB*) = µB * + RTln(KB/pB*) + RTln(xB )
• KB and pB* are characteristics of the solute so you can combine them with µB *

µB† = µB * + RTln(KBxB /pB*)

• Thus µB = µB† + 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
• For dilute solutions x B ≈ n B /nA , and x B =  b/b°
• kappa,  , is a dimensionless constant
• For ideal-dilute solution, µB = µB† + RTln(xB ) so µB = µB† + RTln( b/b°) = µB†+ RTln() + RTln( b/b°)
• Dropping b° and combing 1st 2 terms, µB = µBø+ RTln( b)
• µBø = µB†+ 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