Chemical Thermodynamics 2013/2014 11 th Lecture: Thermodynamics of Simple Mixtures Valentim M B...

ChemicalChemical ThermodynamicsThermodynamics2013/20142013/2014

11th Lecture: Thermodynamics of Simple MixturesValentim M B Nunes, UD de Engenharia

2

IntroductioIntroductionn

In this lecture we will begin the study of simple non-reactive mixtures. To do that we will introduce the concept of partial molar properties.

At this stage we deal mainly with binary mixtures, that is:

121 xx

We shall also consider mainly non-electrolyte solutions, where the solute is not present as ions.

3

Partial Molar VolumePartial Molar Volume

To understand the concept of partial molar quantities, the easiest property to visualize is the “volume”. When we mix two liquids A and B, we have three possibilities:•Volume contraction•Volume dilatation•No total volume change.

In the last case if we mix, for instance, 20 cm3 of liquid A and 80 cm3 of liquid B, we will obtain 100 cm3 of solution. This mean that A…B interactions are equal to A…A or A…B molecular interactions!

In most cases that is not going to happen! For instance the molar volume of water is 18 cm3.mol-1. But if we add 1 mol of water to a huge amount of ethanol the volume will only increase by 14 cm3. The quantity 14 cm3 is the partial molar partial molar volume volume of water.

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Partial Molar VolumePartial Molar Volume

The partial molar volumes change with composition because molecular environment also change. For water/ethanol system we have:

The partial molar volume of a substance is defined as:

ij nnTpii n

VV

,,

Once we know the partial molar volume of two components of a binary mixture we can calculate the total volume:

BBAA VnVnV

5

Measuring Partial Molar VolumeMeasuring Partial Molar Volume

One method is measuring the dependence of volume on composition and determine the slope dV/dn:

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Method of Intercepts Method of Intercepts

Dividing both members of last equation (slide 4) by the total number of moles, n, we obtain:

BBAAm VxVxV

Differentiating:

B

mABAB

B

m

BBAAm

dx

dVVVVV

dx

dV

dxVdxVdV

then

Substituting:

B

mBAAAAm

B

mABAAm

dx

dVxxVVxV

dx

dVVxVxV

1

See next part of lecture!

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Method of Method of Intercepts Intercepts Finally we obtain:

B

mBAm dx

dVxVV

8

Partial Molar Gibbs Partial Molar Gibbs FunctionFunction

Another partial molar property, already introduced, is the partial molar Gibbs function or the chemical potential.chemical potential.

ij nnTpii n

G

,,

Then, the total Gibbs function of a mixture is:

BBAA nnG

BBAABBAA dndndndndG

But we have seen before that (recall the fundamental equations for open systems):

BBAA dndndG

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Gibbs – Duhem EquationGibbs – Duhem Equation

Since G is a state function the previous equations must be equal, and then we obtain:

0 BBAA dndn Generalizing:

0 ii

idn

This is the Gibbs – Duhem EquationGibbs – Duhem Equation. It tell us that partial molar quantities cannot change independently: in a binary mixture if one increases the other should decrease! It can also be applied to the partial molar volume.

10

Thermodynamics of Mixing Perfect Thermodynamics of Mixing Perfect Gases Gases We already now that systems evolves spontaneously to lower Gibbs energy. If we put together two gases they will spontaneously mix, then G should decrease.

Let us consider two perfect gases in two containers at temperature T and pressure p. The initial total Gibbs energy is:

pRTnpRTnG

nnG

BBAAinitial

BBAAinitial

lnln 00

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Gibbs energy of MixingGibbs energy of Mixing

After mixing we have:

BBBAAAfinal pRTnpRTnG lnln 00

The difference is the Gibbs energy of mixing and therefore:

p

pRTn

p

pRTnG B

BA

Amix lnln

Using Dalton´s Law, yi=pi/p, we finally obtain:

BBAAmix yyyynRTG lnln

Since molar fractions are always lower than 1, the Gibbs energy of mixing is always negative!

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Other Thermodynamic mixing functionsOther Thermodynamic mixing functions

Since ST

G

p

we easily obtain:

BBAAmix yyyynRS lnln

Now since ΔH = ΔG + TΔS:

0 mixH

And since for perfect gases ΔGmix is independent of pressure, then:

0 mixV

Purely entropic

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SummarySummary

mixG

mixS

mixH

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Chemical potential of liquidsChemical potential of liquids

In order now to study the equilibrium properties of liquid solutions we need to calculate the chemical potential of a liquid. To do this, we will use the fact that the chemical potential of a substance present as a dilute vapor must be equal to the chemical potential of the liquid, at equilibrium.

Important note: We will denote quantities relating to pure substances by the superscript “*” so the chemical potential of a pure liquid A it will be written as μ*A(l).

Remember also that it is usual, in order to characterize a given solution, to distingue between the solvent (usually the substance in bigger quantity or in the same physical state of solution) and solute (in lower quantity or in a different state of aggregation)

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Ideal Ideal SolutionsSolutionsFor a pure liquid in equilibrium with the vapor we can write:

*0* ln)( AAA pRTl

If another substance is present (solution!) then the chemical potential is:

AAA pRTl ln)( 0

Combining the two equations leads to:

** ln)()(

A

AAA p

pRTll ?

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Ideal Ideal SolutionsSolutionsFrench chemist François Raoult found that the ratio pA/pA

* is equal to mole fraction of A in the liquid, that is:

*AAA pxp

This equation is known as the Raoult´s LawRaoult´s Law.

It follows that for the chemical potential of liquid:

AAA xRTll ln)()( *

This important equation is the definition of ideal solutionsideal solutions. Alternatively we may also state that an ideal solution is the on in which all the components obey to the Raoult´s Law.

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Raoult`s LawRaoult`s Law

BA ppp

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ExamplesExamples

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Ideal Dilute SolutionsIdeal Dilute Solutions

In ideal solutions the solute, as well as the solvent, obeys Raoult´s Law. But in some cases, in non-ideal solutions, the partial pressure of solute is proportional to the mole fraction but the constant of proportionality is not the pure component vapor pressure but a constant, known as the Henry´s constant, KB:

BBB Kxp

This is known as the Henry`s Law. Mixtures obeying to the Henry`s Law are called ideal dilute solutions.

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Example Example

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Ideal Mixture of Liquids Ideal Mixture of Liquids

When two liquids are separated we have:

)()( ** lnlnG BBAAinitial

When they are mixed we have:

BBBAAAfinal xRTlnxRTlnG ln)(ln)( **

As a consequence:

BBAAmix xxxxnRTG lnln

This equation is the same as that for two perfect gases! All the other conclusions are equally valid. The “driving” force for mixing is the increase of entropy.

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Excess Excess FunctionsFunctionsReal solutions have different molecular interactions between A…A, A…B and B…B particles. As a consequence ΔH≠0 and ΔV≠0. For instance if ΔH is positive (endothermic) and ΔS negative (“clustering”) ΔG is positive and the liquids are immiscible. Alternatively the liquids can be only partially miscible.

Thermodynamic properties of real solutions can be expressed in terms of excess functionsexcess functions (GE, SE, etc.). For example for entropy:

BBAAmixE xxxxnRSS lnlnreal

In this context the concept of regular solutions is very important. In this solutions HE≠0 but the SE=0.

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Examples Examples

These figure shows experimental excess functions at 25 ºC. HE values for benzene/ cyclohexane show that the mixing is endothermic (since ΔH=0 for an ideal solution).

VE for tetrachloroethene/cyclopentane shows that there is a contraction at lower mole fractions of tetrachloroethene but expansion at higher mole fractions (because ΔV=0 for an ideal solution)

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Solutions containing non-volatile solutes Solutions containing non-volatile solutes

Let us now consider a solution in which the solute is non-volatile, so it does not contribute to the vapor pressure, and it does not dissolve in solid sovent. Recalling that the chemical potential of liquid A in solution is:

AAA xRTll ln)()( *

We can conclude that the chemical potential in the liquid solvent is lower that the chemical potential of pure liquid (remember that xA<1). These give origin to a set of properties of this solutions that are called colligative propertiescolligative properties. For instance the vapor pressure loweringvapor pressure lowering:

** )1( ABAAA pxpxp

or

**AAAA pxpp

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Colligative Colligative Properties Properties All colligative properties have in common that they only depend on the amount of solute but not on the particular substance. The other colligative properties are: Boiling point Boiling point elevationelevation, Freezing point depressionFreezing point depression and Osmotic pressureOsmotic pressure.

Graphically we can observe the first two properties:

Effect of the solute

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The elevation of boiling pointThe elevation of boiling point

The heterogeneous equilibrium that maters when considering boiling is the solvent vapor and the solvent in the solution.

The equilibrium condition is:

AAA xRTlg ln*)()( **

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The elevation of boiling pointThe elevation of boiling point

This rearranges to:

R

S

RT

H

RT

G

RT

lgx vapvapvapAAB

)()(1ln

**

When xB=0 , the boiling point is that of pure liquid, Tb, and:

R

S

RT

H

b

vap

1ln

The difference between the two equations is

b

vapB TTR

Hx

11)1ln(

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The elevation of boiling pointThe elevation of boiling point

Assuming now dilute solutions, that is, xB<<1, then ln(1-xB) ≈ -xB, so:

TTR

Hx

b

vapB

11

Since T ≈ Tb we can also write: 2

11

bb

b

b T

T

TT

TT

TT

Which gives:

Bvap

b xH

RTT

2

This means that the elevation of boiling point, ΔT, depends on xB (no reference to the identity of solute, the other properties are related to the solvent!)

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The elevation of boiling pointThe elevation of boiling point

We can also calculate the elevation of boiling point in terms of the molality of the solute, mB. Since the mole fraction of solute is small nA >> nB, so:

M

kg 1 and

A

A

B

BA

BB n

n

n

nn

nx

Therefore we obtain: MmMn

n

nx B

B

A

BB

__

kg 1

Boling point elevation can now be calculated by:

B

vap

b mH

MRTT

__2

or

Bb mKT__

Kb is the molal molal ebullioscopic constantebullioscopic constant.

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The depression of freezing The depression of freezing pointpointThe equilibrium of interest now is between the pure solid solvent and the solution with dissolved solute:

At freezing point the equilibrium condition is:

AAA xRTls ln)()( **

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The depression of freezing The depression of freezing pointpoint

All the previous calculations are equal, so we can write directly:

Bf

f xH

RTT

2

or

Bf mKT___

Kf is the molal cryoscopic constantmolal cryoscopic constant.

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Constants of Common LiquidsConstants of Common Liquids

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Osmosis Osmosis

We have shown that boiling point and freezing point depends on the equilibrium between the solvent in solution and in the solid or vapor state. The last possibility is to have an equilibrium between the solvent in solution and the pure solvent. This is the basis of the phenomenon of osmosisosmosis. This consists in the passage of a pure solvent into the solution separated from it by a semipermeable membrane.

h

μA*(p,T) μA(p+π,T)

Π = ρhg is the osmotic osmotic pressurepressure

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Osmotic Osmotic pressure pressure The equilibrium condition is:

p

p AmAAAAAA xRTdpVpxRTpxpp ln)(ln)(),()( ***

From these equation we can easily obtain:

dpVxRTp

p mA

ln

For dilute solutions ln xA = ln (1-xB) ≈ -xB, so, assuming that Vm is constant:

mB VRTx Now, since xB ≈ nB/nA and nAVm = V, we finally obtain:

RTnV BThis is the van`t Hoff equation van`t Hoff equation for osmotic pressure. Note the remarkable similitude with the perfect gas equation!

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Applications Applications

Van`t Hoff equation can be rewritten as:

RTBThis is a very useful equation for the determination of molar mass of polymers and other macromolecules, assuming dilute solutions and incompressible solutions.

Other application is the reverse or inverse osmosisinverse osmosis, If we apply from the solution side a pressure bigger than the osmotic pressure we will reverse the passage of water molecules. This can be used to obtain fresh water from sea water ( π≈ 30 to 50 atm)

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Ideal solubility of Ideal solubility of solids solids Although is not strictly a colligative property the solubility of a solid can be estimated by the same techniques of previous slides. In contact with a liquid the solid will dissolve until the solution is saturated. The equilibrium condition is:

BBB xRTls ln)()( **

From this starting point and using similar deduction we will obtain:

TTR

Hx

fus

fusB

11ln

*

Being the properties related only to the solute.