Simple Mixtures

Thermodynamic Description of Mixtures

Partial Molar QuantitiesPartial Molar VolumePartial Molar Gibbs EnergiesSignificance of Chemical PotentialGibbs-Duhem Equation

Thermodynamics of MixingGibbs Energy of MixingOther Thermodynamic Mixing Functions

Chemical Potentials of LiquidsIdeal SolutionsIdeal Dilute Solutions

Atkins 7th, Ch. 7: Sections 7.1-7.3; 8th, Ch. 5: Sections 5.1-5.3

Last updated: Dec. 11, 2006; minor correction to slide 21

pA % pB % ...' (xA % xB % ...)p ' p

Simple MixturesWe now go from the behavior of simple single substance systems tomixtures of substances. In general, these will still be non-reactingsubstances and we will be dealing mostly with the energetics of themixing process itself and to a lesser extent with the types of nonidealinteractions which can take place between different substances. We willalso start to learn how to deal with liquid solutions, which will increase thearsenal of systems that we know how to deal with.

Mainly, we will deal with non-reactive binary mixtures, which havecomponents A and B, making use of the relation:

xA + xB = 1

We have already used such mole fractions to describe the partialpressures of mixtures of gases which give rise to a total pressure

Partial Molar VolumeThe partial molar volume is the contribution that one component in amixture makes to the total volume of a sample

H2O EtOH

Add 1.0 mol H2O Add 1.0 mol H2O

Volume increasesby 18 cm3 mol-1

Volume increasesby 14 cm3 mol-1Large

volumesMolar volume of H2O:18 cm3 mol-1

Partial molar volume ofH2O in EtOH: 14 cm3 mol-1

The different increase in total volume in the H2O/EtOH example dependson the identity of the molecules that surround the H2O. The EtOHmolecules pack around the water molecules (i.e., solvate) differentlythan how water molecules pack around water molecules: the increase inthe molar volume is only 14 cm3 mol-1!

Partial molar volume of substance A in a mixture is the change in volumeper mole of A added to the large volume of the mixture

Partial Molar Volume, 2The partial molar volume of components of a mixture vary as themixture goes from pure A to pure B - that is because the molecularenvironments of each molecule change (i.e., packing, solvation, etc.)

Partial molar volumes of a water-ethanol binary mixture are shownat 25 oC across all possiblecompositions

The partial molar volume, VJ, of asubstance J is defined as

VJ 'MVMnJ p,T,n )

where n& signifies that all othersubstances present are constant

Partial Molar Volume, 3The partial molar volume is the slope of a plot of total volume as theamount of J in the sample is changed (volume vs. composition)

For a binary mixture, the composition can bechanged by addition of dnA of A and dnB of B,with the total volume changing by

dV 'MVMnA p,T,nB

dnA %MVMnB p,T,nA

dnB

' VAdnA % VBdnB

If partial molar volumes are known for the twocomponents, then at some temperature T, thetotal volume V (state function, always positive)of the mixture is

V ' nAVA % nBVB

Partial molar volumes vary with composition (different slopes atcompositions a and b) - partial molar volume at b is negative (i.e., theoverall sample volume decreases as A is added)

Composition, nA

Partial Molar Volume, 4The partial molar volume can be measured by fitting the observedvolume (as function of composition) using a curve fitting program (findthe parameters that give the best fit of a particular function to theexperimental data)

If the function is found, the slope at any point (any composition) can bedetermined by differentiation; for example, if V has the following function

V ' A % BnA % C (n 2A & 1)

with constants A, B and C (note this is not A and B), then the partialmolar volume of A at any composition is

VA 'MVMnA p,T,nB

' B % 2CnA

and the partial molar volume of component B is obtained by rearranging V ' nAVA % nBVB:

VB 'V & nAVA

nB

'A & (n 2

A % 1)CnB

Partial Molar Gibbs EnergiesThe concept of partial molar quantity can be applied to any extensivestate function. For example, for a pure substance in a mixture, thechemical potential can be defined as the partial molar Gibbs energy

µJ 'MGMnJ p,T,n )

So chemical potential of a substance is the slope of the total Gibbsenergy of a mixture wrt amount of substance of interest, J

Total Gibbs energy isG ' nA µA % nB µB

where µJ are chemical potentials for that particularcomposition of the mixture

Each individual chemical potential of thesubstances in a mixture make a contribution to thetotal Gibbs energy of the substance

Partial Molar Gibbs Energies, 2Gibbs energy depends on composition, pressure and temperature (so, Gmay change if any of these variables change - which they may!) For asystem with components A, B, ...

dG ' Vdp & SdT % µAdnA % µBdnB % ...

which is a fundamental equation of classical thermodynamicsAt constant temperature and pressure,

dwe, max ' µAdnA % µBdnB % ...

dG ' µAdnA % µBdnB % ...

In Chapter 4, we saw that under these conditions, dG = dwe,max (themaximum non-expansion work)

What does this mean? Non-expansion work can happen just fromchanging system composition at constant pressure and temperature(e.g., battery, chemical rxn in two sites called electrodes, and the workthe battery produces comes from reactants going to products)

More on Chemical PotentialChemical potential tells us even more than just about variation in G

G ' U % pV & TS

so for an infinitesimal change in U, we can writedU ' &pdV & Vdp % SdT % TdS % dG

' &pdV & Vdp % SdT % TdS % (Vdp&SdT%µAdnA%µBdnB% ...)' &pdV % TdS % µAdnA % µBdnB % ...

and at constant volume and entropy

dU ' µAdnA % µBdnB % ... µJ 'MUMnJ S,V,n )

and if that’s not enough...what about H and A??? These too also dependupon the composition of a mixture! Chemical potential is IMPORTANT!

(a) µJ 'MHMnJ S,p,n )

(b) µJ 'MAMnJ T,V,n )

Gibbs-Duhem Equation†

Since , and µJ depend on composition, for a binarymixture G may change by

dG ' µAdnA % µBdnB % nAdµA % nBdµB

G ' nA µA % nB µB

But at constant temperature and pressure, since G is a state function

jJnJ dµJ ' 0

nAdµA % nBdµB ' 0

This special case holds for multiple components, and is the Gibbs-Duhem equation

So, for our binary mixture, if one partial molar quantity increases, theother must decrease

dµB ' &nA

nB

dµA

This holds for all partial molar quantities - in binary mixtures, the partialmolar quantity of one component can be determined from measurementsof the other Example 7.1/5.1 - in class

Thermodynamics of MixingDependence of Gibbs energy on mixture composition is

G ' nA µA % nB µB

and at constant T and p, systems tend towards a lower Gibbs energyThe simplest example of mixing: What is the Gibbs free energy if wetake two pure ideal gases and mix them together? Consider gas A andgas B, both in separate containers at pressure p at temperature T.

The chemical potentials are at their “purevalues” at this point. Gibbs energy is

Gi ' nA µA % nB µB

' nA µoA % RT ln p

p o% nB µo

B % RT ln pp o

We can simplify things by letting p denote thepressure relative to po, writing

Gi ' nA AµoA % RT ln p % nB9 Aµo

B % RT ln p

A B

Gibbs Energy of Mixing*

After mixing, the partial pressures of the gases are pA and pB, where thetotal pressure is p = pA + pB. The total Gibbs energy is then

Gf ' nA9 AµoA % RT ln pA % nB9 Aµo

B % RT ln pB

The difference in Gibbs energies, Gf - Gi, is the Gibbs energy of mixing

)mixG ' nART lnpA

p% nBRT ln

pB

p

We use mole fractions, replacing nJ with xJn:

)mixG ' nRT (xA ln xA % xB ln xB )

Since the mole fractions are never greater than 1,the ln terms are negative, and )mixG < 0

This allows is to conclude that mixing processesare spontaneous, and gases mix spontaneously inall proportions

*Mixing of gases and effusion

Calculating Gibbs Energy of MixingTwo containers of equal volume are partitioned from one another, withone containing 3.0 mol H2 and the other 1.0 mol N2 at 25oC. Calculatethe Gibbs energy of mixing when the partition is removedWe assume ideal gas behaviour, with pressure of N2 being p, pressure ofH2 being 3p. Volume of each gas will double, & partial pressures fall by 2

Gi ' (3.0 mol)9 AµoH2

% RT ln 3p % (1.0 mol)9 AµoN2

% RT ln p

Gf ' (3.0 mol) µoH2

% RT ln 32p % (1.0 mol) µo

N2% RT ln 1

2p

Gibbs energy of mixing is the differencebetween the above quantities

)mixG ' (3.0 mol) RT ln

32p

3p% (1.0 mol) RT ln

12p

p' &(3.0 mol) RT ln 2 & (1.0 mol) RT ln 2' &(4.0 mol) RT ln 2 ' &6.9 kJ

Other Thermodynamic Mixing FunctionsSince (MG/MT)p,n = -S and )mixG ' nRT (xA ln xA % xB ln xB )it follows that for a mixture of perfect gases, the entropy of mixing is

)mixS ' &M)mixGMT p,nA,nB

' &nR (xA ln xA%xB ln xB)

Since ln x < 0, then )mixS > 0 for mixtures of all compositionsWe expect this increase in disorder, sincedispersal of one gas into another implies greaterdisorder in the systemExample: )mixG = -(4.0 mol)RT ln 2, so

)mixS = (4.0 mol)R ln 2 = +23 J K-1

The isothermal isobaric enthalpy of mixing of twogases is found from )G = )H - T)S, and we find

)mixH ' 0

This is expected for a system in which there are nointeractions between the molecules in the mixture

Chemical Potentials of LiquidsQuantities relating to pure substances will be denoted with asterisks, sothe chemical potential of pure liquid A is µA*(l). Since the vapourpressure of pure liquid A is pA*, the chemical potential of A in the vapouris µA

o + RT ln pA* (pA* as relative pressure pA*/po)

Chemical potentials are at equilibrium:µ(

A ' µoA % RT ln p (

A

If another substance is present (e.g., a solute inthe liquid), chemical potential of A in the liquidis µA and vapour pressure is pA

µA ' µoA % RT ln pA

Combine to eliminate the standard potential:

µA ' µ(

A % RT lnpA

p (

A

The chemical potential of A depends on its partial vapour pressure, itfollows that chemical potential of liquid A is related to its partial pressure

Raoult’s LawAfter experimenting with mixtures of similar liquids, the French chemistFrancois Raoult found that the ratio of the partial vapour pressure ofeach component to its vapour pressure as a pure liquid, pA/pA*, isapproximately equal to the mole fraction of A in the liquid mixture

pA ' xAp(

A

Some mixtures obeyRaoult’s law veryclosely, especiallywhen they arestructurally similar -these are called idealsolutions, and theyobey

Mixture of benzeneand toluene

µA ' µ(

A % RT ln xA

Ideal binary mixture

Molecular Interpretation of Raoult’s LawWe have to consider the rate at which molecules leave and entersolution to understand Raoult’s lawThe presence of a second component hinders molecules from leavingthe solution, but does not inhibit the rate at which they return

The rate at which A molecule leave the surface isproportional to the number of them at the surface,which is proportional to their mole fraction

rate of vapourization ' kxA

where k is a proportionality constant. The rate atwhich molecules return is proportional to their gasphase concentration, which is proportional theirpartial pressure

rate of condensation ' k )pA

At equilibrium, rates of vapourization and condensation are equal, sokxA ' k )pA )

pA 'kkxA

For a pure liquid, xA = 1, so p (

A 'kk&

Departure from Raoult’s Law, Dilute SolutionsSome solution mixtures behave very differently from Raoult’s law,notably with mixtures of structurally and chemically dissimilar liquids

Mixture ofCS2 andacetone:

Ideal dilutesolution, considerthe solute (B):Henry’s law

Ideal solution,consider thesolvent (A):Raoult’s law

# When the solvent is nearly pure (i.e., B is the major component), it hasa vapour pressure proportional to the mole fraction with slope pB*

# When B is the minor component (the solute) the vapour pressure hasa different constant of proportionality

pA ' xAp(

A

pB ' xBKB

Henry’s LawThe English chemist William Henry found that for real solutions at lowconcentrations, although the vapour pressure of the solute is proportionalto the mole fraction, the constant of proportionality is not the vapourpressure of the pure substance

pB ' xBKB

Here xB is the mole fraction of the solute B, and KB is an empiricallydetermined constant with pressure dimensions. KB is chosen so that aplot of vapour pressure of B against mole fraction is tangent to theexperimental curve at xB = 0

Such solutions are called ideal-dilute solutions

Solvent behaves like a slightly modified pureliquid

Solute behaves entirely differently from its purestate (since its molecules are surrounded bysolvent molecules), unless the molecules of thetwo components are very similar

The Validity of Raoult’s & Henry’s LawsConsider a acetone (A) and chloroform (C) mixture at 35oC

xC 0 0.20 0.40 0.60 0.80 1pC (Torr) 0 35 82 142 200 273pA (Torr) 347 250 175 92 37 0

K = 175 Torr for (A) and K = 165 Torr for (C)

When mole fractions of (A) or (C) are near 1,Raoult’s Law predicts the partial pressures

When mole fractions of (A) or (C) are near 0,Henry’s Law predicts the partial pressures(ideal dilute solutions)

In both cases, there are some deviations fromthis predicted behaviour

Using Henry’s Law*

Estimate the molar solubility of oxygen in water at 25oC and partialpressure of 160 Torr, its partial pressure at sea levelThe amount of O2 dissolved is very small, and xJ = pJ/K, so

x(O2) 'n(O2)

n(O2) % n(H2O).

n(O2)n(H2O)

n(O2) . x(O2)n(H2O) 'p(O2)n(H2O)

K

. (160 Torr) × (55.5 mol)3.33 × 107 Torr

' 2.69 × 10&4 mol

So the molar solubility of oxygen is 2.69 x 10-4 mol kg-1 corresponding toconcentration of 2.7 x 10-4 mol L-1

Henry’s law constants are very well known by biochemists for treatmentof gas behaviour in fats and lipids. Henry’s law is also very important inunderstanding respiration processes: especially when partial pressure ofO2 is abnormal, such as in mountaineering, scuba diving, and in theuse of gaseous anaesthetics