Chapter 6

Chemical Potential in Mixtures

Partial molar quantities Thermodynamics of mixing The chemical potentials of liquids

Chapter 7 : Slide 1

Duesberg, Chemical Thermodynamics Concentration Units

• There are three major concentration units in use in thermodynamic descriptions of solutions. • These are – molarity – molality – mole fraction • Letting J stand for one component in a solution (the solute), these are represented by • [J] = nJ/V (V typically in liters) • bJ = nJ/msolvent (msolvent typically in kg) • xJ = nJ/n (n = total number of moles of all species present in sample) PARTIAL MOLAR QUANTITIES

Recall our use of partial pressures:

p = xA p + xB p + …

pA = xA p is the partial pressure

For binary mixtures (A+B): xA + xB= 1

We can define other partial quantities…

Chapter 7 : Slide 3

Duesberg, Chemical Thermodynamics PARTIAL MOLAR QUANTITIES: Volume

In a system that contains at least two substances, the total value of any extensive property of the system is the sum of the contribution of each substance to that property.

The contribution of one mole of a substance to the volume of a mixture is called the partial molar volume of that component.

V  f p,T,nA,nB...  V    VA    At constant T and p  nA  p,T ,n A  V   V      dV   dnA   dnB ...  nA   nB  Chapter 7 : Slide 4

Duesberg, Chemical Thermodynamics Partial molar volume

Add nA of A to mixture

Very Large Mixture of A and B

When you add nA of A to a large mixture of A and B, the composition remains essentially unchanged. In this case: VA can be considered constant and the volume change of the mixture is n V . A A  V    VA     const  nA  p,T ,n A Likewise for addition of B. The total change in volume is nAVA + nBVB . (Composition is essentially unchanged).

Scoop out of the reservoir a sample containing nA of A and nB of B its volume is nAVA + nBVB . Because V is a state function: V VAnA VBChapternB  7 :... Slide 5 Duesberg, Chemical Thermodynamics Partial molar volume

Example: What is the change in volume of adding 1 mol of water to a large volume of water?

  The change in  V  3 VH O  18cm 3 2  n  volume is 18cm  H2O  p,T

A different answer is obtained if we add 1 mol of water to a large volume of ethanol.

  The change in  V  3 VH O  14cm 3 2  n  volume is 14cm H2O   p,T ,n(CH 3CH 2OH)

Chapter 7 : Slide 6

Duesberg, Chemical Thermodynamics Introduction to mixtures

 Homogeneous mixtures of a solvent (major component) and solute (minor component).  Introduce partial molar property: contribution that a substance makes to overall property.

V = nAVA + nBVB

VA is not generally a constant: It is a function of composition, because the environment and therefore their interaction of the The partial molar volumes of water molecules changes with the and ethanol at 25C. Note the different composition . scales (water on the left, ethanol on : the right). Exercise Calculate the density of a mixture of 20 g of water and 100g of ethanol. MM: EtOH: 46.07g/mol d=0.789 g/ml • First calculate the mole fractions.

– 20 g H2O = 1.11 mol; – 100 g EtOH = 2.17 mol

– xH2O = 0.34; xEtOH = 0.66 • Then interpolate from the mixing curve 3 -1 – VH2O = 17.1 cm mol ; 3 -1 – VEtOH = 57.4 cm mol • Then plug the moles and partial molar volume – (1.11 mol)(17.1 cm3/mol) + (2.17 mol)(57.4 cm3/mol) =18.1 cm3 + 125 cm3 = 143.1 cm3 • Finally, the total mass is divided by the total volume: 120 g/143 cm3 = 0.84 g/ cm3 Task: You mix 300 ml alcohol with a bottle of juice (water, 0.7 l) - Will you really end up with for your x-mas party supply (1l)? -Of course not… - How much do you have to mix to get 1 l of the same strength? PARTIAL MOLAR QUANTITIES

VA is not generally a constant: It is a function of composition, because the environment and therefore their interaction of the molecules changes with the composition : :

The partial molar volume of a substance is the slope of the variation of the total volume of the sample plotted against the composition. In general, partial molar quantities vary with the composition, as shown by the different slopes at the compositions a and b. Note that

the partial molar volume at b is negative: the overall volume of the sample decreasesChapter 7 : Slide as a9 is added. Duesberg, Chemical Thermodynamics PARTIAL MOLAR QUANTITIES • We can extend the concept of partial molar properties to state functions, such as Gibbs energy, G. Or Chemical Pot m

G = nAGA + nBGB or G = nAmA + nBmB

Consider this single substance system:

m1 m2

dG = -m1 dn dG = +m2 dn

Total dG = 0 only if m1 = m2 10 The Partial Molar Gibbs Energy

The partial molar Gibbs energy is called the “chemical potential”

G  f p,T,nA,nB... At constant T and p  G   G   G  dG   dn   dn ... μA      A   B  n   nA   nB   A  p,T,n A dG  m dn  m dn ... A A B B G = nAµA + nBµB (At equilibrium dG = 0)

The implication of G = f ( p,T, ni ) is that we now write:

dG  SdT Vdp  mAdnA  mBdnB ... Fundamental dG  SdT Vdp  midni Equation of Chemical Thermodynamics.Chapter 7 : Slide 11

Duesberg, Chemical Thermodynamics Chemical Potential = Partial Molar Gibbs Energy

In case of constant temperature and pressure

dG  SdT Vdp  midni reduces to:

dG  midni

The chemical potential of a substance is the slope of the total Gibbs energy of a mixture with respect to the amount of substance of interest. In general, the chemical potential varies with composition, as shown for the two values at a and b. In this case, both chemical potentials areChapter positive. 7 : Slide 12 Duesberg, Chemical Thermodynamics The Wider Significance of the Chemical Potential

 G     G  dG  SdT Vdp    dni   ni    T , p,n j ni  T , p,n j

 H    dH  TdS Vdp    dni ni      S, p,n j H A   m   ni i ni  S , p,n j  T ,V ,n j  A    dA  SdT  pdV    dni ni  T ,V ,n j  U     ni   U  S ,V ,n j   dU  TdS  pdV    dni ni  S,V ,n j

All extensive thermodynamic properties change with composition!Chapter 7 : Slide 13

Duesberg, Chemical Thermodynamics Chemical Potential In case of constant temperature dw  dG  m dn and pressure also: add,max  i i

dw  m dn "chemical work" Example: E-chem cell: chem  jj j

dU TdS  PdV  m dn  jj UH    j mi     nnii  S,,,, V nj i   S P n j i dH TdS  VdP   m jj dn j AG     dA  SdT  PdV  m jj dn      nnii j  T,,,, V nj i   T P n j i

dG  SdT  VdP   m jj dn j The Gibbs-Duhem Equation

dG  m1dn1  m2dn2 ...

G  m1n1  m2n2 (For binary systems)

by differentiating:

dG  m1dn1  m2dn2  n1dm1  n2dm2 At equilibrium: dG = 0 recall in case of constant temperature m m and pressure dG   midni => 1dn1 + 2dn2 = 0

 0  n dm  n dm 1 1 2 2 The Gibbs-Duhem Equation. Mixtures can not be change independentely: If or 0  n dm  i i one is changed the other changes as n well… or dm   2 dm 1 2 A similar expression may be deduced n1 for all partial molar quantitiesChapter 7 : Slide 15 Duesberg, Chemical Thermodynamics The Gibbs-Duhem Equation

SdT VdP  njj dm  0 j

Chapter 7 : Slide 16

Duesberg, Chemical Thermodynamics THE THERMODYNAMICS OF MIXING

Imagine a system of two perfect gases in amounts nA and nB at equal T and p are separated by a barrier.

The initial total Gibbs energy of the system Gi is given by

Gi = nAµA + nBµB

 p  G  n  m  RT ln  i A  A    p   p   n  m  RT ln  B  B    p  The initial and final states considered in the calculation of the Gibbs energy of mixing of gases at different initial pressures.Chapter 7 : Slide 17 Duesberg, Chemical Thermodynamics The Thermodynamics of mixing

After mixing each gas exerts a partial pressure pJ, where pA + pB = p. The final G is given by  p  G  n  m  RT ln A  f A  A    p  N , p, T A NB, p, T  p   n  m  RT ln B  B  B    p  pA+ pB = p,T

mixG  Gf Gi

Duesberg, Chemical Thermodynamics The Thermodynamics of mixing

 p   p   G  G G  n m  RT ln A   n m  RT ln B  mix f i A  A   B  B    p   p   p   p   n m  RT ln   n m  RT ln  A  A   B  B    p   p 

p p  n RT ln A  n RT ln B A p B p N , p, T A NB, p, T

Chapter 7 : Slide 19 pA+ pB = p,T

Duesberg, Chemical Thermodynamics The Thermodynamics of Mixing

p p G  n RT ln A  n RT ln B mix A p B p

Writing xA for the mole fraction of n n component A  A  A n n  x

nA = xA n therefore pA /p = xA, (same for any component), so The Gibbs energy of mixing of two perfect gases and (or of two liquids that form an ideal solution). The mixG = nRT (xA ln xA + xB ln xB) Gibbs energy of mixing is negative for all compositions and mixG is always negative, since x<1 temperatures, so perfect gases mix spontaneously inChapter all proportions. 7 : Slide 20

Duesberg, Chemical Thermodynamics Gas mixtures What About the Enthalpy of Mixing? G = H – TS Compare

Gmix = nRT {xAln xA+ xB ln xB}

Smix = − nR {xAln xA+ xB ln xB} Therefore

H mix= 0

All due to the effect of entropy

21 The Entropy of Mixing

 Δ G  Δ S   mix  mix T   p,nA ,nB

 nR(xAln xA  xBln xB )

which is positive as xA and xB are <1.

Perfect gases mix spontaneously in all proportions

Chapter 7 : Slide 22

Duesberg, Chemical Thermodynamics Exercise: at 25C, calculate the Gibbs energy change when the partition is removed (cannot used the previous one as different starting pressures) 0 0 nA  3 mol,nB 1 mol Gi  nA mA  RT ln pA  nB mB  RT ln pB  0 0  3 m H2 RT ln(3p) 1 m N2  RT ln p p  3 p, p  1 p pA  3 p, pB 1 p p f  2 p f ,A 2 f ,B 2

0 0 Gf  nA mA  RT ln p f ,A  nB mB  RT ln p f ,B 

0 3 0 1  3m H2 RT ln( 2 p) 1m N2  RT ln 2 p

p f ,A p f ,B Gmix  G f Gi  nA RT ln 3 p  nB RT ln p 

3 2 1 1  3 RT ln 3 1 RT ln 2  4ln 2

1 ln 2  0.69 R  8.314J / K Gmix  8.314J / K 298K 4(0,69)  6.8kJ T  298K Duesberg, Chemical Thermodynamics Chemical Potential of Liquids We need to know how the Gibbs energy of a liquid varies with composition in order to discuss properties of liquid mixtures (like solutions).

* P Vapor Pressure = PA Pure Liquid Mixture Partial Pressure = A For vapor phase: For vapor phase: * P *  PA  A m  m  RT ln  mA  mA  RT ln   A A P  P At equilibrium… At equilibrium… m* (g)  m* (l) A A mA(g)  mA(l) = A = A = B For liquid phase: For solution: * P *  PA  A m  m  RT ln  mA  mA  RT ln   A A P  P Combine these expressions…

24 Total Vapour Pressure of an ideal solution

* PA m A  m A  RT ln * * PA mA  mA  RT ln xA * PA  xAPA

This serves to define an ideal solution if true for all values of xA

25 RAOULT'S LAW For SOME pairs of liquids, RAOULT'S LAW that * * pA = xA pA is obeyed, so (xi= moluar fraction, for pure substance)

* * µA(l) = µA (l) +RT ln pA/pA

* pA = xA pA

* µA(l) = µA (l) + RT ln xA . This defines an IDEAL SOLUTION)

Ptotal  PA  PB * * Ptotal  xAPA  xB PB P  P*  x (P*  P*) Duesberg, Chemical Thermodynamics total A B B A Ideal Mixtures

• Two types of molecules are randomly distributed • Typically, molecules are similar in size and shape • Intermolecular forces in pure liquids & mixture are similar • Examples: benzene & toluene, hexane & heptane

In ideal solutions, the partial vapor pressure of component A is simply given by Raoult’s Law: * PA  xAPA

vapor pressure of pure A mole fraction of A in solution

27 The Measurement of Vapor Pressure of Solutions

Chapter 7 : Slide 28

Duesberg, Chemical Thermodynamics Raoult's law Similar liquids can form an Large spheres = solvent molecules ideal solution obeying small spheres = solute molecules. Raoult’s Law

A pictorial representation of the molecular basis of Raoult's law. The solute hinder the escape of solvent moleculesChapter into 7the : Slide vapour, 29 but do not hinder their return Duesberg, Chemical Thermodynamics Example

• A solution is prepared by dissolving 1.5 mol C10H8 in 1.00 kg benzene. The v.p. of pure benzene is 94.6 torr at this temperature (25oC). What is the partial v.p. of benzene in the solution? • Solution: We can use Raoult’s law, but first we need to know the mole fraction of benzene. • MM benzene = 78.1 g/mol, so 1.00 kg = 12.8 mol.

• xbenz = 12.8 mol / (12.8 mol + 1.5 mol) = 0.895

• pbenz = xbenz p*benz = (0.895)(94.6 torr) = 84.7 torr

Chapter 7 : Slide 30 Deviations from Raoult’s Law

CS2 and dimethoxymethane: Positive trichloromethane/acetone: Negative deviation from ideal (Raoult’s Law) behavior. deviation from ideal (Raoult’s Law) behavior.

Methanol, ethanol, propanol mixed with water. Which one is which? (All show positive deviations from ideal behavior)

31 Raoult and Henry

* PA  xAPA as xA 1 Raoult’s law

(Raoult’s Law) Henry’s law

PA  xAkH ,A as xA  0 k  P* Henry’s behavior: Henry’s law constant: H ,A A . The Henry’s law constant reflects the intermolecular interactions between the two components.

. Solutions following both Raoult’s and Henry’s Laws are called ideal-dilute solutions. 32 Ideal-dilute solutions  Raoult’s law generally describes well solvent vapour pressure when solution is dilute, but not the solute vapour pressure  Experimentally found (by Henry) that vp of solute is proportional to its mole fraction, but proportionality constant is not the vp of pure solute.

Henry’s Law

pB = xBKB

33 Ideal-Dilute Solutions (Henry’s Law)

For real solutions at low concentrations i.e. xB << xA the vapor pressure of a the solute is proportional to its mole fraction but the proportionality

constant is not pA* but some empirical constant KB

pB = xBKB Henry’s Law

In a dilute solution, the solvent molecules are in an environment that differs only slightly from that of the pure solvent. The solute particles, however, are in an environment totally unlike that of theChapter pure 7 : Slidesolute. 34 Duesberg, Chemical Thermodynamics Gas solubility Henry’s law constants for gases dissolved in water at 25°C

3 1 KH/(kPa m mol )

Ammonia, NH3 5.69

Carbon dioxide, CO2 2.937 Helium, He 282.7

Hydrogen, H2 121.2

Methane, CH4 67.4

Nitrogen, N2 155

Oxygen, O2 74.68 Concentration of 4 mg/L of oxygen is required to support aquatic life, what partial pressure of oxygen is required for this?

35 Application-diving Gas narcosis caused by nitrogen in normal air dissolving into nervous tissue during dives of more than 120 feet [35 m] Pain due to expanding or contracting trapped gases, potentially leading to

Barotrauma. Can occur either during ascent or descent, but are potentially most severe when gases are expanding. Decompression sickness due to evolution of inert gas bubbles.

Table 1 Increasing severity of nitrogen narcosis symptoms with depth in feet and pressures in atmospheres.

Depth P Total P N2 Symptoms

100 4.0 3.0 Reasoning measurably slowed.

Joviality; reflexes slowed; idea 150 5.5 4.3 fixation. Euphoria; impaired concentration; 200 7.1 5.5 drowsiness.

Mental confusion; inaccurate 250 8.3 6.4 observations. Stupefaction; loss of perceptual 300 10. 7.9 faculties. The experimental partial vapour pressures of a mixture of chloroform (trichloromethane) and acetone (propanone) based on the data in Example 7.3. The values of K are obtained by extrapolating the dilute solution vapour pressures as explained in the Example.

Excess Functions:

We discuss properties of real solutions in terms excess functions XE. The excess entropy for example is defined as:

Chapter 7 : Slide 37 S E =  S-  S ideal. Duesberg, Chemical Thermodynamics mix mix SUMMARY

• Partial molar quantities and the Gibbs-Duhem equation. Tells us how chemical potentials vary with composition of a mixture.

• Chemical potentials µ of liquids are accessed via µ for the vapor in equilibrium.

• Raoult's Law, Henry’s Law

• Real and ideal gases  activity

• In general µ = µs + RT ln a.

Chapter 7 : Slide 38

Duesberg, Chemical Thermodynamics Colligative Properties Colligative properties are the properties of dilute solutions that depend only on the number of solute particles present.

They include:

1. The elevation of boiling point 2. The depression of boiling point 3. The osmotic pressure

All colligative properties stem from the reduction of the solvents m by the presence of the solute.

* µA(l) = µA (l) + RT ln xA

Chapter 7 : Slide 39

Duesberg, Chemical Thermodynamics Colligative properties

 Chemical potential of a solution (but not vapour or solid) decreases by a factor (RTlnxA) in the presence of solute  Molecular interpretation is based on an enhanced molecular randomness of the solution  Get empirical relationship for FP and BP (related to enthalpies of transition)

Tf  K f m

Tb  Kbm 40 Boiling-point elevation Tb = KbbB Freezing-point depression Tf = KfbB where bB is the molality of the solute B in the solution

The Elevation of Boiling Point Want to know T at which:

* * µA (g) = µA (l) + RT ln xA

Presence of a solute at xB causes an increase in the boiling temp from T * to T * + T where: RT *2 T = KxB K  vapH

b is the molality of the solute (proportional to x ). T = K b B b Kb is the ebullioscopic constant of the solvent.

The Depression of Freezing Point

Identical arguments lead to: T = K f b where Kf is the cryoscopic constant. Chapter 7 : Slide 42

Duesberg, Chemical Thermodynamics Examples:

7.8(b) Calculate the cryoscopic and ebullioscopic constants of naphthalene.

7.10(b) The addition of 5.0 grams of a compound to 250 grams of naphthalene lowered the freezing point of the solvent by 0.780 K. Calculate the molar mass of the compound.

Chapter 7 : Slide 43

Duesberg, Chemical Thermodynamics Osmosis – for Greek word “push” Spontaneous passage of a pure solvent into a solution separated from it by a semi-permeable membrane (membrane permeable to the solvent, but not to the solute) Osmotic pressure – P – the pressure that must be applied to the solution to stop the influx of the solvent

Van’t Hoff equation: P = [B] R T where [B] = nB/V

Osmometry - determination of molar mass by measurement of osmotic pressure – macromolecules (proteins and polymers)Chapter 7 : Slide 44 Duesberg, Chemical Thermodynamics Osmosis

Height Solution Proportional to Osmotic Pressure P

Solvent A with chemical Semipermeable * potential mA (p) Membrane

Chapter 7 : Slide 45

Duesberg, Chemical Thermodynamics Osmosis

It is assumed that the van’t Hoff equation is only the first term of a virial- like expression:

P = [B] R T {1 + B[B] + . . . }

EXAMPLE: Using Osmometry to determine the molar mass of a macromolecule.

Osmotic pressures of PVC in cyclohexanone are given below. The pressures are given in terms of heights of solution (with r = 0.980 g cm-3) in balance with the osmotic pressure. Determine the molar mass of the polymer.

c / (g L-1) 1.00 2.00 4.00 7.00 9.00 h / cm 0.28 0.71 2.01 5.10 8.00

c is the mass concentration Chapter 7 : Slide 46

Duesberg, Chemical Thermodynamics Activities  How can we adjust previous equations to account for deviations from ideal behavior? 1. solvent 2. solute

Solvent activity:

General form of the chemical potential of a real OR ideal solvent:

mA = mA* + RT ln (pA/pA*)

Ideal solution – Raoult’s law is obeyed:

mA = mA* + RT ln xA i.e xA = pA/pA*

Real solution – we can write:

Chapter 7 : Slide 47 mA = mA* + RT ln aA Duesberg, Chemical Thermodynamics Solvent activity

aA is the activity of A – essentially an “effective” mole fraction

aA = pA/pA*

o Illustration: The vapor pressure of 0.5 M KNO3 (aq) at 100 C is 749.7 Torr, what is the activity of water in the solution?

As xA  1 (solute concentration is nearing zero), aA  xA,

aA = gA xA gA  1 and aA  xA For a real solution, we can write:

mA = mA* + RT ln xA + RT ln gA

Standard state is the pure liquid solvent at 1 bar Chapter 7 : Slide 48 Duesberg, Chemical Thermodynamics Solute activity  approach ideal dilute (Henry’s law) behavior as xB  0

Ideal-dilute: pB = KB xB

mB = mB* + RT ln (pB/pB*) = mB* + RT ln (KB /pB*) + RT ln xB

The second term on the rhs of the above equation is composition independent, so we may define a new reference state:

+ mB = mB* + RT ln (KB /pB*)

So that: + mB = mB + RT ln xB Chapter 7 : Slide 49 Duesberg, Chemical Thermodynamics Real solutes – permit deviations from ideal-dilute behavior

+ mB = mB + RT ln aB

Where aB = pB/KB and aB = gB xB

Note: As xB  0, aB  xB and gB  1

Chapter 7 : Slide 50

Duesberg, Chemical Thermodynamics Measuring Activity Use the following information to calculate the activity and activity coefficient of chloroform in acetone at 25oC, treating it first as a

solvent and then as a solute with KB = 165 Torr.

xC 0 0.2 0.4 0.6 0.8 1.0

pC / Torr 0 35 82 142 200 273 Chloroform regarded as solvent a = p / p* a 0 0.13 0.30 0.53 0.73 1.0

g 0.65 0.75 0.87 0.91 1.0 g = a / xC Chloroform regarded as solute

a = p / KB a 0 0.21 0.50 0.86 1.21 1.65 g g = a / xC 1 1.05 1.25 1.43 1.51 Chapter1 .765 : Slide 51

Duesberg, Chemical Thermodynamics Activities in terms of molalities, bB: For an ideal-dilute solute we had written in terms of mole fractions: + mB = mB + RT ln xB

with + mB = mB* + RT ln (KB /pB*)

Molality in terms of mole fraction:

bB = nB / MA MA = nA * Mr(A)

bB = nB / ( nA * Mr(A) )

xB = nB / (nA+nB)  nB / nA

bB = xB / Mr(A)  xB = bB Mr(A) Chapter 7 : Slide 52

Duesberg, Chemical Thermodynamics Activities in terms of molalities, bB:

xB = bB Mr(A)

+ mB = mB + RT ln xB

+ mB = mB + RT ln bB + RT ln Mr(A)

 mB = mB + RT ln bB

 -1 mB = mB at standard molality b = 1 mol kg Chapter 7 : Slide 53

Duesberg, Chemical Thermodynamics Activities in terms of molalities, bB:

As bB  0, mB  -∞

In other words, as a solution becomes increasingly diluted, the solution becomes more stabilized – It becomes difficult to remove the last little bit of solute.

To allow for deviations from ideality we introduce (in the normal way)

aB = gB bB (assuming unit-less)

Then:

 mB = mB + RT ln aB

Chapter 7 : Slide 54

Duesberg, Chemical Thermodynamics Osmotic pressure

Van’t Hoff equation

P  MRT

55 Phase diagrams of mixtures

We will focus on two-component systems (F = 4 ─ P), at constant pressure of 1 atm (F’ = 3 ─ P), depicted as temperature- composition diagrams. 56 Fractional Distillation-volatile liquids

Important in oil refining

57 Exceptions-azeotropes Azeotrope: boiling without changing High-boiling and Low-boiling

Favourable interactions between components Unfavourable interactions between reduce vp of mixture components increase vp of mixture Trichloromethane/propanone Ethanol/water (min at 4% water, 78°C) HCl/water (max at 80% water, 108.6°C)

58 Liquid-Liquid (partially miscible)

 Hexane/nitrobenzene as example  Relative abundances in 2 phases given by Lever Rule n’l’ = n’’l’’  Upper critical Temperature is limit at which phase separation occurs. In thermodynamic terms the Gibbs energy of mixing becomes negative above this temperature

59 Other examples

Nicotine/water Water/triethylamine Weak complex at low temperature disrupted Weak complex at low temperature disrupted at higher T. Thermal motion homogenizes at higher T. mixture again at higher T.

60 Liquid-solid phase diagrams

61