# Chemistry 508

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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
Partial Molar Volumes
substance slope of the variation of
Water & Ethanol
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 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
Thermodynamics of Mixing
   For 2 Gases (A &B) in two containers, the Gibbs              Gibbs Energy of Mixing
Of
energy, Gi                                                      Two Ideal Gases
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 and Enthalpy of Mixing
    For Smix, recall G = H - TS
Entropy of Mixing
Therefore Smix = -Gmix / T                            Two Ideal Gases
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 Liquids
Ideal 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
p*
B                                          Solutions
Total Pressure
   Phenomenology of Raoult’s
Law: 2nd component inhibits
Pressure
p*       the rate of molecules leaving a
Partial                        A
solution, but not returning
Pressure of B
» rate of vaporization  XA
» rate of condensation  pA
Partial
Pressure of A                    » at equilibrium rates equal
 implies pA = XA p*A
Mole Fraction of A, x    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
K
B       solutions
B is solute
   Raoult’s law applies to
solvent in ideal dilute
Henry's Law
Pressure                                p*
B
solutions and solute &
solvent in ideal solutions
Raoult's Law
   Real systems can (and do )
B is solvent
deviate from both

Mole Fraction of B, x    B
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 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
    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]
 x Ae  
bx 2      b 1 x 2
a A  xAe      B             A

   If b = 0, this is Raoult’s Law
p                                                    If b < 0 (endothermic mixing),
But, aA  A      p *  pressure         pure A            
p*A
A
gives vapor pressures lower than
so                                                            ideal
   If b > 0 (exothermic mixing), gives
p A  x Ae  A p*
b 1 x 2
vapor pressures higher than ideal
              A
   If xA<<1, becomes pA = xA eb 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
xA
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]

1

0.8
Hfus/RT*=0.1

0.6                 H(fus)/RT*=0.3

0.4                             H(fus)/RT*=1
H(fus)/RT*=3

0.2
H(fus)/RT*=10

0
0       0.2             0.4              0.6             0.8      1
T/T*
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}
p +
 µ*A (x A,p +) = µA *(p) + ∫p       Vm dp {3} [Vm = molar volume of the pure
solvent]
p +
 Combining {1} and {2} : µA *(p) = µA *(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 +
∫p      V m dp = Vm∫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

   Determine molar mass of
Solution with macromole cule                    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 + ...}
Water                           g h                 g h = c/M RT{1 + Bc/M + ...}
h                            h/c = RT/(Mg) {1 + Bc/M + ...}

 h/c = RT/(Mg) + RTB/(M g) + ...}
2

   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
Membrane
Non-Ideality & Activities
Solvents

   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,g, such that

•   aA = g A xA
•   As xA-> 1, gA -> 1
   Thus µA =µA * + RTln(xA) + RTln(gA) {substiuting for a A}
Non-Ideality & Activities
Solutes (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 = g B xB

   As xA-> 0, aA -> xA and gA -> 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 = 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,g B,
»   aB = g B bB/b° where g B ->1, bB -> 0
    The chemical potential then becomes µ = µø + RTln( a)
    Table 7.3 in book summarizes the relationships

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