Chapter 7: Simple Mixtures

1
Chapter 7 Simple Mixtures

• Homework
• Exercises(a only)7.4, 5,10, 11, 12, 17, 21
• Problems 1, 8

2
Chapter 7 - Simple Mixtures

• Restrictions
• Binary Mixtures
• xA xB 1, where xA fraction of A
• Non-Electrolyte Solutions
• Solute not present as ions

3
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
4
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

5
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

6
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

7
Partial Molar Free Energies

• Chemical potential, µJ, is defined as the partial
  molar Gibbs energy _at_ 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
• _at_ 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

8
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

9
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

10
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 (i1,J) Gibbs-Duhem Equation

11
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

12
Example Self Test 7.2

• VA 6.218 5.146b - 7.147b2
• dVA 5.146 - 27.147b 5.146 - 14.294b db
• dVA/db 5.146b - 14.294b
• If MB is in kg/mol
• dVB -nA/nB (dVA) bnA/nBMB 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

13
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/pxJ (from Daltons
  Law)
• ?Gmix nRT(xA ln (xA ) xBln(xB ))
• This equation tells you change in Gibbs energy is
  negative since mole fractions are always lt1

14
Example Self-Test 7.3

• 2.0 mol H2(_at_2.0 atm) 4 mol N2 (_at_3.0 atm) mixed
  at const. V. What is ?Gmix?
• Initial pH2 2 atmVH2 24.5 L pN2 3 atmVN2
  32.8 LIdeal Gas
• Final VN2 VH2 57.3 L therefore pN2 1.717
  atm pH2 0.855 atmIdeal 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 298.15K0.333ln0.333
  0.667ln0.667)
• ?Gmix -9.5 J

15
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

16
Chemical Potentials of LiquidsIdeal Solutions

• At equilibrium chem. pot. of liquid chem. pot.
  of vapor, µA(l) µA(g,p)
• For pure liquid, µA(l) µA RT ln(p A)
  1
• For A in solution, µA(l) µA RT ln(p A)
  2
• Subtracing 1 from 2
• µA(l) - µA(l) RT ln(pA) RT ln(p A)
• µA(l) - µA(l) RTln(pA) - ln(p A)
  RTln(pA/p A)
• µA(l) µA(l) RTln(pA/p A) 3
• Raoults Law - ratio of the partial pressure of a
  component of a mixture to its vapor pressure as a
  pure substance (pA/pA) approximately equals the
  mole fraction, xA
• pA xA pA
• Combining Raoults law with 3 gives
• µA(l) µA(l) RTln(xA)

17
Ideal Solutions/Raoults Law

• Mixtures which obey Raoults Law throughout the
  composition range are Ideal Solutions
• Phenomenology of Raoults 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 pA

18
Deviations from Raoults Law

• Raoults Law works well when components of a
  mixture are structurally similar
• Wide deviations possible for dissimilar mixtures
• Ideal-Dilute Solutions
• Henrys Law (William Henry)
• For dilute solutions, v.p. of solute is
  proportional to the mole fraction (Raoults Law)
  but v.p. of the pure substance is not the
  constant of proportionality
• Empirical constant, K, has dimensions of pressure
• pB xBKB (Raoults Law says pB xBpB)
• Mixtures in which the solute obeys Henrys Law
  and solvent obeys Raoults 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

19
Applying Henrys Law Raoults Law

• Henrys law applies to the solute in ideal dilute
  solutions
• Raoults law applies to solvent in ideal dilute
  solutions and solute solvent in ideal solutions
• Real systems can (and do ) deviate from both

20
Applying Henrys Law

• What is the mole fraction of dissolved hydrogen
  dissolved in water if the over-pressure is 100
  atmospheres?
• Henrys 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?

21
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

22
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 gt 0, mixing endothermic b lt 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)

23
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 Raoults Law
• If b lt 0 (endothermic mixing), gives vapor
  pressures lower than ideal
• If b gt 0 (exothermic mixing), gives vapor
  pressures higher than ideal
• If xAltlt1, becomes pA xA eb pA
• Henrys law with K eb pA

24
Colligative Properties of Dilute Solutions
25
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

26
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 gt µA

27
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 1gtgt 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)

28
Boiling Point Elevation

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

29
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 0C) Kf 1.86 K/mol kg -1
• Acetic Acid (Tf 17C) Kf 3.9 K/mol kg -1
• Benzene (Tf 5.4C) Kf 5.12 K/mol kg-1
• Phenol (Tf 43C) Kf 7.27 K/mol kg -1
• Again property depends on solvent not solute

30
Temperature Dependence of Solubility

• Not strictly speaking colligative property but
  can be estimated assuming it is
• Starting point the same - assume _at_ 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

31
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

32
ln xB -?H fus /R 1/T - 1/Tm or xB
exp?H fus /R Tm 1 - (Tm /T)-1
33
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

34
Osmotic Pressure vant Hoff Equation

• J. H. vant Hoff (1885) - In dilute solutions the
  osmotic pressure obeys the relationship, ?VnBRT
• 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 ?) µA (x A,p ?) RTln(x A) 2
• µA (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 RT1 BB ... where B I s the osmotic
  virial coef. (like pressure)

35
Application of Osmotic Pressure

• Determine molar mass of macromolecules
• ?B RT1 BB ... but B c/M where c
  is the concentration and M the molar mass so
• ?? c/M RT1 Bc/M ...
• ??????g h c/M RT1 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

36
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/ pA)
• As xA-gt 1, aA -gt xA so define activity
  coefficient,?, such that
• aA ? A xA
• As xA-gt 1, ?A -gt 1
• Thus µA µA RTln(xA) RTln(?A) substiuting
  for a A

37
Non-Ideality ActivitiesSolutes (Ideal
Non-Ideal)

• For ideal dilute solutions Henrys 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-gt 0, aA -gt xA and ?A -gt 1

38
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 bb
• As b -gt0, µB -gtinfinity 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 -gt1, bB -gt 0
• The chemical potential then becomes µ µø
  RTln( a)
• Table 7.3 in book summarizes the relationships