chapter 6 chemical potential in mixtures - school of … notes... · chapter 6 chemical potential...

Duesberg, Chemical Thermodynamics

Chapter 7 : Slide 1

Chapter 6

Chemical Potential in

Mixtures

Partial molar quantities

Thermodynamics of mixing

The chemical potentials of liquids

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)


Chapter 7 : Slide 3

PARTIAL MOLAR QUANTITIES

Recall our use of partial pressures:

p = xA p + xB p + …

pA = xA p is the partial pressure

We can define other partial quantities…

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


Chapter 7 : Slide 4

PARTIAL MOLAR QUANTITIES: Volume

The contribution of one mole of a substance to the volume of a

mixture is called the partial molar volume of that component.

...,,, BA nnTpfV

AnTpA

An

VV

,,

...

B

B

A

A

dnn

Vdn

n

VdV

At constant T and p

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.


Chapter 7 : Slide 5

Partial molar volume

Very Large Mixture of A and B

Add nA of A to mixture

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 nAVA.

Likewise for addition of B.

constn

VV

AnTpA

A

,,

The total change in volume is nAVA + nBVB . (Composition is essentially unchanged).

... BBAA nVnVV

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:


Chapter 7 : Slide 6

Partial molar volume

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

of ethanol.

3

,OH

OH cm18

2

2

Tpn

VVThe change in

volume is 18cm3

3

OH)CHCH(,,OH

OH cm14

232

2

nTpn

VVThe change in

volume is 14cm3

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

large volume of water?

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

The partial molar volumes of water

and ethanol at 25C. Note the different

scales (water on the left, ethanol on

the right).

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 .

:

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

– VH2O = 17.1 cm3 mol-1;

– VEtOH = 57.4 cm3 mol-1

• 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?


Chapter 7 : Slide 9

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 decreases as a is

added.

10


• 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

m1 m2

Consider this single substance system:

dG = -m1 dn dG = +m2 dn

Total dG = 0 only if m1 = m2


Chapter 7 : Slide 11

The Partial Molar Gibbs Energy

Ap,T,nA

An

Gμ

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

...

B

B

A

A

dnn

Gdn

n

GdG

G = nAµA + nBµB

At constant T and p

... BBAA dndndG mm

(At equilibrium dG = 0)

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

iidnVdpSdTdG m

... BdndnVdpSdTdG BAA mm

Fundamental

Equation of Chemical

Thermodynamics.

...,,, BA nnTpfG



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 are positive.

iidndG m

In case of constant

temperature and pressure

reduces to: iidnVdpSdTdG m

Chemical Potential = Partial Molar Gibbs Energy



The Wider Significance of the Chemical Potential

i

npTi

dnn

GVdpSdTdG

j,,

i

npSi

dnn

HVdpTdSdH

j,,

i

nVTi

dnn

ApdVSdTdA

j,,

i

nVSi

dnn

UpdVTdSdU

j,,

mi

jnpTin

G

,,

jnpSin

H

,,

jnVTin

A

,,

jnVSin

U

,,

All extensive thermodynamic properties change with composition!

chem "chemical work"j j

j

dw dnm Example: E-chem cell:

Chemical Potential

iiadd dndGdw mmax,

In case of constant temperature

and pressure also:

j j

j

j j

j

j j

j

j j

j

dU TdS PdV dn

dH TdS VdP dn

dA SdT PdV dn

dG SdT VdP dn

m

m

m

m

, , , ,

, , , ,

j i j i

j i j i

i

i iS V n S P n

i iT V n T P n

U H

n n

A G

n n

m



2211 nnG mm

22112211 mmmm dndndndndG

The Gibbs-Duhem Equation

by differentiating:

At equilibrium: dG = 0 recall in case of constant temperature

and pressure => m1dn1 + m2dn2 = 0

22110 mm dndn

2

1

21or mm d

n

nd

The Gibbs-Duhem Equation. Mixtures

can not be change independentely: If

one is changed the other changes as

well…

A similar expression may be deduced

for all partial molar quantities

...2211 dndndG mm

(For binary systems)

iidndG m

iidnor m0



The Gibbs-Duhem Equation

0j j

j

SdT VdP n dm



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

m

m

p

pRTn

p

pRTnG

ln

ln

BB

AAi

The initial and final states considered in the

calculation of the Gibbs energy of mixing of

gases at different initial pressures.


After mixing each gas exerts a partial pressure pJ, where pA + pB = p. The

final G is given by

The Thermodynamics of mixing

m

m

p

pRTn

p

pRTnG

BBB

AAAf

ln

ln

ifmix GGG

NA, p, T NB, p, T

pA+ pB = p,T



mmp

pRTn

p

pRTn ln ln BBAA

mmp

pRTn

p

pRTnGG B

BBA

AAif ln ln

p

pRTn

p

pRTn B

BA

A lnln

Gmix

The Thermodynamics of mixing

NA, p, T NB, p, T

pA+ pB = p,T



mixG = nRT (xA ln xA + xB ln xB) mixG is always negative, since x<1

The Thermodynamics of Mixing

The Gibbs energy of mixing of two

perfect gases and (or of two liquids

that form an ideal solution). The

Gibbs energy of mixing is negative

for all compositions and

temperatures, so perfect gases mix

spontaneously in all proportions.

Writing xA for the mole fraction of

component A

nA = xA n therefore pA /p = xA,

(same for any component), so

n

n

n

n A

x

A

p

pRTn

p

pRTnGmix

BB

AA lnln

21

Gas mixtures What About the Enthalpy of

Mixing?

G = H – TS

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



)ln ln (

ΔΔ

BBAA

,,

mixmix

BA

xxxxnR

T

GS

nnp

The Entropy of Mixing

which is positive as xA and xB are <1.

Perfect gases mix spontaneously in all proportions


Exercise: at 25C, calculate the Gibbs energy change when the

partition is removed

(cannot used the previous one as different starting pressures)

pRTpRT

pRTnpRTnG

NH

BBBAAAi

ln1)3ln(3

lnln

22

00

00

mm

mm

BfBBAfAAf pRTnpRTnG ,

0

,

0lnln mm

mol 1mol, 3 BA nn

pppp BA 1, 3 pp f 2 pppp BfAf 21

,23

, ,

21

21

3

3

ln4ln1ln3

lnln

23

,,

RTRT

RTnRTnGGGp

p

Bp

p

Aifmix

BfAf

kJ8.6)69,0(4298/314.8 KKJGmix

pRTpRT NH 210

230 ln1)ln(3

22 mm

KT

KJR

298

/314.8

69.0ln 21

24

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).

Pure Liquid

= A

*

APMixture

= A = B

Vapor Pressure =

P

P

AAART*

ln* mm

For vapor phase:

APPartial Pressure =

P

P

AAART ln mm

For vapor phase:

At equilibrium…

)()( ** lg AA mm

At equilibrium…

)()( lg AA mm

P

P

AAART*

ln* mm

For liquid phase:

P

P

AAART ln mm

For solution:

Combine these expressions…

25

Total Vapour Pressure of an ideal solution

*

* lnA

AAA

P

PRT mm

*

AAA PxP AAA xRT ln* mm

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


For SOME pairs of liquids, RAOULT'S LAW that

pA = xA pA* is obeyed, so (xi= moluar fraction, * for pure substance)

RAOULT'S LAW

µA(l) = µA*(l) +RT ln pA/pA

*

pA = xA pA*

µA(l) = µA*(l) + RT ln xA .

This defines an IDEAL SOLUTION)

BAtotal PPP

**

BBAAtotal PxPxP

)( ***

ABBAtotal PPxPP

27

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:

*

AAA PxP

vapor pressure of pure A mole fraction of A in solution



The Measurement of Vapor Pressure of Solutions



Similar liquids can form an

ideal solution obeying

Raoult’s Law

Raoult's law

A pictorial representation of the molecular

basis of Raoult's law. The solute hinder the

escape of solvent molecules into the vapour, but

do not hinder their return

Large spheres = solvent molecules

small spheres = solute molecules.

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

31

Deviations from Raoult’s Law

Methanol, ethanol, propanol

mixed with water. Which one

is which? (All show positive

deviations from ideal behavior)

CS2 and dimethoxymethane: Positive

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

deviation from ideal (Raoult’s Law) behavior.

32

Raoult and Henry

1 as * AAAA xPxP

0 as , AAHAA xkxP

Raoult’s law

Henry’s law constant: *

, AAH Pk

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.

Henry’s behavior:

Henry’s law (Raoult’s Law)

33

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



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 the pure solute.

35

Gas solubility

KH/(kPa m3 mol1

)

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

Henry’s law constants for gases dissolved in water at 25°C

Concentration of 4 mg/L of oxygen is required to support aquatic life,

what partial pressure of oxygen is required for this?

Application-diving

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.

150 5.5 4.3 Joviality; reflexes slowed; idea

fixation.

200 7.1 5.5 Euphoria; impaired concentration;

drowsiness.

250 8.3 6.4 Mental confusion; inaccurate

observations.

300 10. 7.9 Stupefaction; loss of perceptual

faculties.

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.



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:

S E = mix S- mix S ideal.



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.



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

40

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)

mKT

mKT

bb

ff

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

µA*(g) = µA

*(l) + RT ln xA

Want to know T at which:

Presence of a solute at xB causes an increase in the boiling temp from

T * to T * + T where:

T = KxB H

RTK

vap

2*

T = K bb b is the molality of the solute (proportional to xB).

Kb is the ebullioscopic constant of the solvent.

Identical arguments lead to: T = K f b where Kf is the cryoscopic constant.

The Depression of Freezing Point



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.



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)



Solvent A with chemical

potential mA*(p)

Semipermeable

Membrane

Height

Proportional

to Osmotic

Pressure P

Solution

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] + . . . }

Osmosis

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



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:

mA = mA* + RT ln aA



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

aA = pA/pA*

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

aA = gA xA gA 1 and aA xA

Solvent activity

Illustration: The vapor pressure of 0.5 M KNO3 (aq) at 100oC is 749.7 Torr, what is the

activity of water in the solution?

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



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



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



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

a 0 0.13 0.30 0.53 0.73 1.0

g 0.65 0.75 0.87 0.91 1.0

a 0 0.21 0.50 0.86 1.21 1.65

g 1 1.05 1.25 1.43 1.51 1.65

Chloroform regarded as solvent

Chloroform regarded as solute

a = p / p*

a = p / KB

g = a / xC

g = a / xC



Activities in terms of molalities, bB:

mB = mB+ + RT ln xB

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

For an ideal-dilute solute we had written in terms of mole fractions:

with

Molality in terms of mole fraction:

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

xB = nB / (nA+nB) nB / nA

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

bB = xB / Mr(A) xB = bB Mr(A)




xB = bB Mr(A)

mB = mB+ + RT ln xB

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

mB = mB + RT ln bB

mB = mB at standard molality b = 1 mol kg-1



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

aB = gB bB (assuming unit-less)

Then:

mB = mB + RT ln aB

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.


55

Osmotic pressure

Van’t Hoff equation

MRTP

56

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.

57

Fractional Distillation-volatile liquids

Important in oil refining

http://science.howstuffworks.com/oil-refining4.htm

58

Exceptions-azeotropes Azeotrope: boiling without changing

High-boiling and Low-boiling

Favourable interactions between components

reduce vp of mixture

Trichloromethane/propanone

HCl/water (max at 80% water, 108.6°C)

Unfavourable interactions between

components increase vp of mixture

Ethanol/water (min at 4% water, 78°C)

59

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

60

Other examples

Water/triethylamine

Weak complex at low temperature disrupted

at higher T.

Nicotine/water

Weak complex at low temperature disrupted

at higher T. Thermal motion homogenizes

mixture again at higher T.

61

Liquid-solid phase diagrams

chapter 6 chemical potential in mixtures - school of … notes... · chapter 6 chemical potential...

Documents