b18pa1 nhn vision

8/13/2019 B18PA1 NHN Vision

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B18PA1: Thermodynamics

Dr N Hendrik Nahler

email: [email protected]: G.30

2012/13


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Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

IntroductionAbout me

2012: Royal SocietyUniversity Research Fellow,Heriot-Watt University

2007 2012: RS-URF,

Durham University2005 2007: Alexander vonHumboldt Fellowship, UCSB

2003 2005: EU Marie CurieFellowship, Bristol

2002: PhD in MolecularPhysics, MPI for FluidDynamics, Gottingen

About my research

Photochemistry on water-ice surfaces at near-ambientconditions

Molecule-surface dynamics: energy- and charge-transfer

Development of spectroscopic methods

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Thermodynamics

Where does it fit in the big scheme?Chemical thermodynamics is the study of the interrelation of heatand work with chemical reactions or with physical changes.

1 How much of a product is formed?

Thermodynamics

2 How much energy will be released/used?Thermodynamics

3 How fast is it formed?Reaction Kinetics

Example

diamond

Thermodynamics Reaction Kinetics

graphite


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Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Thermodynamics and reaction kinetics

Thermodynamics all about ifIf a reaction can occur.

Applicable if a system is in stable or metastable equilibrium.

If the driving force is sufficient to enforce a favourabletransformation.

Reaction kinetics all about howHow fast a process can occur (rate of reaction).

Applicable to systems in transition from non-equilibrium toequilibrium.

How to overcome the energy barrier to go from the reactant to

the product.

Example

Graphite has a lower Gibbs free energy than diamond and istherefore the thermodynamically favourable state.

Large energy barrier to overcome. Reaction kinetics.

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Lecture schedule

Week Lecture Topics

1 1 & 2 Revision, 2nd Law of Thermodynamics3 3 & 4 2nd and 3rd Laws of Thermodynamics4 5 & 6 Free energy and equilibrium

9 7 & 8 Phase diagrams10 9 & 10 Ideal solutions11 11 & 12 Dilute solutions, equilibrium electrochemistry12 13 & 14 Equilibrium electrochemistry

Workshops in weeks 3, 4, 10, 11.

Online quizzes every week after the lecture (together count

10% towards this term, 90% exam)Kinetics lectures (weeks 5 8), quizzes and workshops: ProfKen McKendrick


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Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Reading list

Textbook for this course

2 Thermodynamics: the first law

3 Thermodynamics: applications ofthe First Law

4 Thermodynamics: the Second Law

5 Physical equilibria: pure substances

6 The properties of mixtures

7 Chemical equilibrium: the principles

8 Chemical equilibrium: equilibria insolutions

9 Chemical equilibrium:

electrochemistry

Additional readingPhysical Chemistry (Atkins & de Paula), Chemistry (Housecroft &Constable), Basic chemical thermodynamics (Smith), Principles ofthermodynamics (Kaufman)

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Revision, 2nd Law of Thermodynamics

Key concepts

Internal energy change UEnthalpy change H

Entropy change SGibbs free energy change G

Goal: describe equilibrium usingthese key concepts.

Laws of thermodynamics

1 The internal energy, U, of

an isolated system isconstant.

2 The entropy, S, of anisolated system tends toincrease.

ReadingChapter 2 Thermodynamics: the first law

Chapter 3 Thermodynamics: applications of theFirst LawChapter 4 Thermodynamics: the Second Law, pp.8393


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Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

System and Surrounding

System

The system is what we are interested in.

Examples: reaction components, engine, biological cell.

Surroundings

The surroundings is the universe apart from the system.

This is where we mostly do our measurements.

Open system

can exchange matterand energy with

surroundings.

Closed system

can only exchangeenergy with

surroundings.

Isolated system

completely insulatedfrom the

surroundings.

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The 1st Law of Thermodynamics

First law

The internal energy (U) of an isolated system is constant.

Alternative definitionIn chemical changes energy can be converted to one form or anotherbut not destroyed.

Classical mechanicsObject falling off a table

Ekin = 1

2mv2 =mgh =Epot (1)

Harmonic oscillator

Ekin=1

2mv2 =

1

2kx2 =Epot (2)


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Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Internal Energy, U

The internal energy is made up of kinetic energy, rotational energy,vibrational energy, potential energy etc.

Changes in internal energy, U

Heat transfer, q, to/from surroundingsWork,w, done on/by surroundings

U=q+w (3)

Sign convention for heat and work:q= +ve, U for surroundings systemq= ve, U for system surroundingsequivalent for work, w.

U=q+w

=q pV for an ideal gas (4)=q V = const. (5)

= 0 for an isolated system (6)

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Work: ideal gas

Expansion

When the gas expands, work is done by the systems on thesurroundings, w= veand U.If you quickly expand a gas it cools down e.g. spray can.

CompressionTo compress a gas work needs to be done by the surroundings onthe system, w= +veand U.If you quickly compress a gas it heats up e.g. bicycle pump.


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Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Work in a chemical reaction

CuCO3(s)solid

+ 2 HCl(aq)liquid

CuCl2(aq)liquid

+ H2O(l)liquid

+ CO2(g)gas

(C1)

open vessel at room temperature

ideal gas lawRemember the sign convention for work.

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Enthalpy,H

If our chemistry happens in an open system then energy put into thesystem will be partly given back to surroundings. Therefore wedefine a new quantity that avoids the complication of work ofexpansion by the system on the surroundings:

Definition of Enthalpy, H

H=U+pV (7)

Change in enthalpyAt constant pressure:

H= U+pV (8)

At constant pressure and no expansion work:

H=q (9)

Recall that the enthalpy varies with temperature, a concept which isexpressed by the heat capacity (Atkins, 2.9, pp. 56 59).


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Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

State functions and heat capacity

The internal energy Uand the enthalpy Hare state functions(in contrast to path functions like w and q).

Consequently dUand dHare exact differentials.

dU=

U

T

V

CV

dT+

U

V

T

dV (10)

CV: Heat capacity at constant volume

dH=H

T

p

Cp

dT+H

p

T

dp (11)

Cp: Heat capacity at constant pressure

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

First law problem

Calculate the change in internal energy, U, when 1 mol of zinc isdissolved in aqueous hydrochloric acid at 298 K and atmosphericpressure given that the heat that is liberated is151000J.How big is the change in enthalpy, H?


9/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Thermochemistry

Change of enthalpy in chemical reactionsexothermic process - releases heat H0

Hesss Law of constant heat summation

rH(298K) =

rH (products, 298K)

rH (reactants, 298K) (12)

Example: find fH for the reaction

2 NO(g) + O2(g) 2 NO2(g) (C 2)

given 12 N2(g) + 12 O2(g) NO(g) H1 = +90kJmol1

N2(g) + 2 O2(g) N2O4(l) H2 = 20kJmol12 NO2(g) N2O4(l) H3 = 86kJmol1

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The 2nd Law of Thermodynamics (i)

The First Law states that energy is conserved when changes occur.It is defined in terms of the state function U. However, it does nottell us about direction of change.

Examples for natural directions of change

expansion of gas into vacuum

spontaneous mixing of gases

objects of different temperature brought into contact willequalize the temperature

some reactions will occur spontaneously

a ball bouncing will stop

The Second Law is concerned with the direction of change. It isdefined in terms of a second state function, the entropy, S.The entropy, S, is a measure of the molecular disorder of asystem


10/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The 2nd Law of Thermodynamics (ii)

Second law

The entropy, S, of an isolated system tends to increase.

Alternative definition of the second law

The total entropy, Stot of a system (Ssys) and surroundings (Ssur)increases in the course of a spontaneous change.

Stot= Ssys+ Ssur >0 (13)

At equilibrium or for a reversible process, the total entropy isconstant.

Stot= 0 (14)

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Thermodynamic definition of entropy, S

For a reversible change we define:

dS= dqrev

T (15)

S=

dS= final

initial

dqrevT

(16)

Clausius inequalityThe entropy change of the system must be greater than, or equal to,the heat flow divided by temperature.

dS dQ

T (17)

This applies for a natural process (irreversible).


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Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy change: heating

From equation11we know:

Cp= dH

dT =

dq

dT (18)

dS= dqrevT

= CpT

dT (19)

S=

T2T1

CpT

dT (20)

IfCpis independent of temperature (usually only over a small range):

S=CplnT2T1

(21)

ExampleYou drink a glass of tap water (200 g, 10 C). Assuming a bodytemperature of 36 C, what is the change in entropy?(Cp,m = 75.5 JK1mol1)

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy change: expansion

The energy transferred as heat to a perfect gas when it undergoesreversible isothermal expansion is:

qrev= w= V2V1

pdV =

V2V1

nRT

V dV =nRTln

V2V1

(22)

For the change in entropy in this process follows:

S=nRlnV2V1

(23)

Examples

1 A sample of carbon dioxide that initially occupies 15.0 dm3 at250 K and 1.00 atm is compressed isothermally. Into whatvolume must the gas be compressed to reduce the entropy by10.0 J

K1? (Atkins, 4.6)

2 A monoatomic perfect gas at a temperature Ti is expandedisothermally to twice its initial volume. To what temperatureshould it be cooled to restore its entropy to its initial value?CV,m =

32 R. (Atkins, 4.12)


12/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy change: phase transition (i)

Addition of heat evaporates water.Removal of heat freezes water.Melting at melting temperature, Tf:

fusS=

fusH(Tf)

Tf (24)

Vaporization at the boiling temperature, Tb:

vapS= vapH(Tb)

Tb(25)

Troutons rule (empirical)

vapS= vapH(Tb)Tb 85 JK1mol1 (26)

Notable exceptions to the rule are water, ammonia and mercury.Hydrogen bonding and metal bonding in the latter lead to a higherorder in the liquid phase and consequently tovapS>85 JK1mol1.

Thermodynamics

N H Nahler

Lecture 1 & 2

Introduction

Revision

The Second Law

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy change: phase transition (ii)

Examples

1 Melting of ice:ice

273 K, 1 barwater (C 3)

fusH= 6000 J

mol1, Tf= 273K, fusS= 22 J

mol1

K1

2 Vaporization of water:

water373 K, 1 bar

steam (C 4)

vapH= 41 000 Jmol1, Tb= 373 K,vapS= 110Jmol1K1

3 Note that the vaporization entropy is much larger than thefusion entropy.

4 Note the difference between the vaporization entropy and thepredicted value from Troutons rule. (hydrogen bonding)

5 Estimate the enthalpy of vaporization of bromine from itsboiling temperature, 59.2 C.


13/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible changeChemicalreactions

Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

2nd and 3rd Laws of Thermodynamics

Key concepts

Entropy change S

Absolute entropyStatistics andThermodynamics

Gibbs free energy change G

Helmholtz energy change A

Goal: describe equilibrium usingthese key concepts.

Laws of thermodynamics

1 The internal energy, U, of

an isolated system isconstant.

2 The entropy, S, of anisolated system tends toincrease.

3 The entropy of a perfectlycrystalline substance atT= 0 K is S= 0.

ReadingChapter 4 Thermodynamics: the Second LawTo look further into concepts of StatisticalThermodynamics you can readChapter 22 Statistical Thermodynamics

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy change: reversible and irreversibleprocesses

Reversible change

Stot= Ssystem+ Ssurrounding= 0 (27)

Irreversible change

Stot= Ssystem+ Ssurrounding>0 (28)

To caculate Swe need to find a reversible pathway.

Example: the freezing of supercooled water at T = 10 C

H2O(l) H2O(s) (C 5)

Ice and water are not at equilibrium at 263 K. Ice is the more stablephase. Therefore the process is irreversible and

S(263K) = fusH(263K)263K


14/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Example: the freezing of supercooled water1 Construct reversible thermodynamic cycle

2 Calculate entropy changes of individual changes.

3 Calculate the entropy change of the system.

4 Calculate the entropy change of the surroundings and the totalentropy change (Stot >0).

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy change: chemical reactions

Phase changes

solids relatively ordered low entropyliquids some order medium entropygases disordered high entropy

Changes in moleculesFor a reaction with all molecules in the same phase x, the change innumber of molecules is important:

A(x) + B(x) C(x) (C 6)

Examples

NH2(g) +HCl(g) NH4Cl(s) S= veH2(g) + 12 O2(g) H2O(l) S= vePh2CHCl+(in water) Ph2CH

+(aq) + Cl (aq) S= veWould guess S= +veas number of particles increase but ions aresolvated, meaning an ordering of water molecules around the ions.


15/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy change: mixing of ideal gases (i)How can we calculate the entropy change of a mixture of idealgases?

Is there actually a change in entropy? Ideal gases do notinteract.

Mole fractions: system with J components

xJ mole fraction of component J

xJ= nJntotal

ninumber of molesi

xi=nA+nB+ +nN

ntotal= 1 (29)

Mole fractions: binary system with components A and B

xA= nA

nA+nBxB=

nB

nA+nB(30)

xA+xB= 1

xB= 1 xAThe two mole fractions are not independent from one another.The composition of a binary system is defined by one of the twomole fractions.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy change: mixing of ideal gases (ii)Both gases A and B are under the same pressure pbeforemixing.According to their mole fractions, gases A and B are describedby their partial volumes pA and pB in the mixture.Consider mixing as an expansion of an ideal gas.

mixS/nR

0

0.2

0.4

0.6

Mole fraction of A, xA

0 0.5 1

mixS=

nARln

pA

pnBRln

pB

p

(31)

= nR(xAln xA+xBln xB) (32)mixS 0 (33)

Ideal gases always mix.


16/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Mixing: statistical thermodynamics

Can we understand macroscopicbehaviour from a molecular viewpoint?

We know that there is a conceptual linkbetween molecular disorder and entropye.g. a gas is more disordered than asolid.

Macroscopic: we measure disorder byentropy.

Molecular: we measure disorder by thenumber of ways (microstates), W,molecules can organise themselves.

How can we relate S and W?

S=kBln W (34)

All we need to know to calculate S isW.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Mixing functions

Macroscopic thermodynamics

A (l) + B (l)

+

A (l) B (l)

mixS= nR(xAln xA+xBln xB)

Molecular thermodynamics

AA

B

BA

A

ABA

B

B

B

A

B

A

AA

B

B B

B

A

A B

A A

A

A

A

B

B

A

A

B

A

B

A

A What is the number of ways of placingNA molecules A into a cubic latticewith a total ofN sites?

Note that NA = Avogadros number in this context.

We express Avogrados number as NAvo..


17/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

How many arrangements are there?

Meet Jill and Jack

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Counting possibilities

Lets say we have NA Jills and N officesNumber of ways of placing . . .

...first Jill = N

...second Jill =N 1

. . . third Jill = N 2

. . .

. . . the NAth Jill = N (NA 1) =N NA+ 1

Total number of ways W =W1 W2 W3 . . .

W = (N NA+ 1) (N NA+ 2) (N 1) N1 2 3 (N NA) (N NA+ 1) (N 1) N

=N! (N NA)! N!

(N NA)! =W


18/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14


Now we have overcounted the possibilities WSeveral different ways lead to the same microstate. For example, inthe case of:

# 1 # 2 . . . . . . # 3# 1 # 3 . . . . . . # 2# 2 # 1 . . . . . . # 3

# 2 # 3 . . . . . . # 1# 3 # 1 . . . . . . # 2# 3 # 2 . . . . . . # 1

The three Jills (A molecules) are indistinguishable. The sixconfigurations constitute the same microstate.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14


To account for this overcounting, how many ways ofplacing molecules on a given set of sites?

1 molecules of A

1 identical configurations

2 molecules of A 2 identical configurations3 molecules of A 6 identical configurations... molecules of A ... identical configurations

NA molecules of A NA! identical configurations

W = N!(N NA)!NA! (35)

What about the Jacks (B molecules)?

DoesWchange when they are included?


19/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14


The remaining empty offices are all filled up withJacks (B molecules): N=NA+NB

In the same way as we are not able to tell the difference betweenempty and empty, we can not tell the difference between one Jackand another.

Adding the Jacks does not increase W any further.

W = N!(N NA)!NA! S=kBln

N!

(N NA)!NA! (36)

We now have a formula to calculate the entropy of any cubic latticewith N sites, NA molecules of A and NB molecules of B.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Stirlings approximation

The factorials are inconvenient for calculations.

Fortunately . . . ln N! Nln N NThis approximation becomes more exact for large N.

We typically deal with Nof the order of 1023.

Stirling's approximation

Y

0.1

1

10

100

1,000

0.1

1

10

100

1,000

N

1 10 100

1 10 100

Y = ln N!Y = NlnN- N


20/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy of mixing

Making use of the Stirling approximation

S=kBln W =kBln

N!

(NNA)!NA!

(37)

=kB ln N! ln(N NA)! ln NA! (38)=kB[Nln N N (N NA)ln(N NA) + (N NA) NAln NA+NA] (39)

= kBN

ln

N NA

N

+

NAN

ln

NA

N NA

(40)

Using mole fractions: xA=NA/N and xB= (N NA)/N .

= kBNln xB+xAlnxAxB = kBN[(1 xA) ln xB+xAln xA]= kBN(xAln xA+xBln xB) (41)

Using N=nNAvo. and R=kBNAvo.

S= nR(xAln xA+xBln xB) (42)

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Entropy of mixing: S= nR(xAln xA+ xBln xB)

Before mixing: Sinitial=S(xA= 1) + S(xB = 1) = 0

A A A

A A A

B B B

B B B

After mixing: Sfinal= nR(xAln xA+ xBln xB)

A B B

B A A

A B A

B A B

only one out of thepossible

arrangements

mixS=Sfinal Sinitial= nR(xAln xA+xBln xB)

We allow completely random mixing.

Macroscopic thermodynamics

mixS= nR(xAln xA+xBln xB)


21/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Third law of Thermodynamics

Third law

A perfect crystal at T= 0 K has zero entropy.

Alternative definitions

The entropies of all perfectly crystalline substances are thesame at T = 0.

AtT= 0 for a perfect crystal, atoms and ions are regularlyarranged with no disorder leading to S= 0.

The Third-Law entropy or absoluteentropy at any temperature, S(T)can be calculated using heat capacity

and phase-change data. In the figureon the right, the entropy follows fromintegrating the Cp/T curve. Thestandard molar entropy, Sm isreported at a pressure ofp= 1 barand at the temperature of interest(often 298.15 K for reference).

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Residual entropy

Measured entropy of CO (heatcapacity, boiling point):Sm (298 K) = 192 JK1mol1Calculated entropy of CO(Boltzmann):Sm (298 K) = 198 JK1mol1

Where does this discrepancycome from?

Orientation at T= 0 K means Nmolecules of CO can be arrangedin W = 2N ways leading toSm(0) =kln 2

NA =NAkln 2 =Rln 2 5.8 JK1mol1IfSm(0)> 0 this is called the

residual entropy.Water is another example withSm(0) =kln(3/2)

NA =NAkln(3/2) =Rln(3/2) 3.4 JK1mol1

CO

H2O


22/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction


Mixing

Possibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Equilibrium and direction of changeWhat do we know so far about entropy and the directionof change?

isolated system dSsystem>0 spontaneousnon-isolated system dSsystem+ dSsurroundings>0 spontaneous

Clausius inequalitydS dq

T (system) (43)

It is an inequality for spontaneous irreversible change but an equalityfor a reversible change where the equilibrium is maintained.Implications of entropyWhenever considering the implications of entropy for a non-isolated

system, we must always consider the total change of the system andthe surroundings.We need to deal with this in chemistry but there are two other statefunctions that will be able to help us finding the direction of changeby focussing on the system alone:G: Change in Gibbs free energyA: Change in Helmholtz energy (less common)

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Introduction

Irreversible change

Chemicalreactions

MixingPossibilities

Mixing functions

Third law

Direction ofchange

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Standard reaction entropy and directionExample

2 H2(g) + O2(g) H2O(l) (C 7)

Data at T= 298.15K: Sm (H2(g)) = 131 JK1mol1,Sm (O2(g)) = 205 JK1mol1, Sm (H2O(l)) = 70 JK1mol1 andrH=

572 kJ

mol1

Standard reaction entropy, Sr

Sr =i

iSm,i(products)

j

jSm,j(reactants) (44)


23/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,constituent,component

Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Free energy and physical equilibrium

Key concepts

Gibbs free energy change G

Helmholtz energy change A

Phase boundaries:temperature and pressure

Phase stability

triple and critical point

Goal: describe the phase of acomponent based on the Gibbsfree energy.

Phase diagram

Reading

Chapter 4 Thermodynamics: the Second LawChapter 5 Physical equilibria: pure substances

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,

constituent,component

Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Phase, constituent and component (1)

PhaseA phase is a state of matter that is uniform throughout, not only inchemical composition but also in physical state. (Gibbs)

P=. . . describes the number of phases present in a system.

P= 1single phase

solidliquid

gas

P= 2phase mixture

solid/solidsolid/liquid

solid/gas etc.


24/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Phase, constituent and component (2)

ConstituentA constituent is a chemical species (ion or molecule) that is presentin the system.

ComponentA component is a chemically independent constituent of a systemwhere C =. . . is the number of components in the system.

P= 2 P= 2 P= 1

C= 1 C = 2 C = 2

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Gibbs free energy

From the Clausius inequality:

dS dQT

and dH= dQ (p= const.)

TdS dHTdS dH 0dH TdS 0 (45)

Gibbs free energy

Definition: G=H TSand consequently:

dG = dH TdS SdT= dH TdS (T = const.) see equ.45 (46)

Spontaneous change/process: dGT,p 0Equilibrium of a system at a fixed temperature and pressurecorresponds to the minimum of the Gibbs free energy.


25/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Change in Gibbs free energyWe describe changes in an open systems where the pressure isconstant with the Gibbs free energy.

Combining 1st and 2nd lawdU= dQ pdV and dH= dQ+Vdp

dU=TdS pdVdG= dH SdT TdS

= dU+pdV+Vdp SdT TdS=TdS pdV+pdV+Vdp SdT TdS

dG=Vdp SdT (47)Temperature and pressure dependence ofG

The free energy is a state function:

Pressure dependence: G

p =V T = const. (48)

Temperature dependence: G

T = S p= const. (49)

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Helmholtz free energyConstant temperature and volumeTo describe a closed system (V= const.) in contact with a heatbath (T= const.) we cannot use the Gibbs free energy. Theadiabatic bomb calorimeter is such an example.

Helmholtz free energy

Definition: A= U TSand consequently:

dA= dU TdS SdT= dq pdV TdS SdT=TdS pdV TdS SdT

dA= pdV SdT (50)

Maximum workConsidering the total change in entropy one can show that theHelmholtz free energy is a measure for the maximum work that canbe performed by a system:

A w (51)


26/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Gibbs free energy: pressure dependence

The (molar) Gibbs free energy varies with pressure. In general:

Gm=Gm(pf) Gm(pi)withp=pf pi T= const.

Gm=Vmp (52)

Note: The molar volume of liquids and solids does change very littlewith pressure. However, the molar volume of gases strongly dependson pressure.

The pressure dependence ofG for an ideal gas

Gm(pf) =Gm(pi) +RTlnpfpi (53)

Gas always expands into a vacuum.

An empty gas cylinder has never been observed tospontaneously refill.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Gibbs free energy: temperature dependence

The (molar) Gibbs free energy varies with temperature:

Gm=Gm(Tf) Gm(Ti)withT =Tf Ti p= const.

Gm=

SmT (54)


27/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Chemical potential of pure substances

Spontaneous process:dG0 for all substances (T >0 K)

The phase with the lowest chemicalpotential is the most stable one.

Sm(s)< Sm(l)< Sm(v)

The phase with the lower molarentropy is the stable phase at lowertemperatures: T Tf liquid vs.solid.

Two phases have the same chemicalpotential at the transitiontemperature (equilibrium).

T

p

= Sm


28/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Chemical potential: variation with pressure

Vm>0 for all substances

The phase with the higher molar

volume (phase ) is the stable phaseat lower pressure.

The phase with the lower molarvolume (phase ) is stable at higherpressures.

High-pressure phases are denser thanlow-pressure phases.

For most substances Vm(l)> Vm(s).Important exception: for waterVm(l)< Vm(s).

pT =Vm

pressure,p

chemicalpotential,

phase

phase

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Phase boundaries

At a phase transition thechemical potentials of twocoexisting phases are equal:

(; p, T) =(; p, T)

d() = d()

Using d= Vmdp SmdT it follows thatVm() dp Sm() dT =Vm() dp Sm() dT

(Sm()

Sm()) dT = (Vm()

Vm()) dp

dp

dT =

trsS

trsV Clapeyron equation (56)


29/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The solid-liquid boundary

Melting/fusion goes with a molar enthalpy change fusH attemperature T. The molar entropy of melting is therefore fusH/T.

dpdT

= fusHTfusV

Slope of solid-liquid boundary (57)

Entropy of melting is positive.

Volume change upon melting ismostly positive and small(exception: water!).

pp

dp= fusHfusV

TT

1T

dT (58)

p=p +fusH

fusV ln

T

T (59)

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The liquid-vapour boundary

The entropy of vaporization at temperature T is equal to vapH/T.

dp

dT =

vapH

TvapV Slope of solid-liquid boundary (60)

Entropy of vaporization is positive.

Volume change upon vaporization islarge and positive.

dp/ dT >0 but smaller than for thesolid-liquid boundary.

dp

dT =

vapH

T(RT/p) (ideal gas) (61)

d ln p

dT =

vapH

RT2 (62)

p=pe ; = vapH

R

1

T 1

T

Clausius-Clapeyron (63)


30/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The gas-liquid critical point

Critical point ends the liquid-vapour equilibrium line.

Gas can only be liquefied below TC.

Clausius-Clapeyron equation breaks down near critical point

(ideal gas, vapH(T) = const., Vvap Vm(g)).

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The solid-vapour boundary and triple point

The Clausius-Clapeyron equation can also be used to calculate thesolid-vapour boundary.

dp

dT =

subH

TsubV Slope of solid-vapour boundary (64)

subH= fusH+ vapH

Triple point

At the triple point all threecoexistence curves intersect.

All three phases are inequilibrium at the triplepoint.


31/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Liquid phase

Limitations

For a substance with a volume increase from solid to liquid(most substances but not water), the liquid phase does onlyexist above the triple point temperature T

tand below the

critical temperatureTc.

For any substance the liquid phase does not exist below thetriple point pressurept.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The Gibbs phase rule

Phase rule

F =C P+ 2 (65)F: Number of degrees of freedom e.g. pressure, temperature, molefraction

C: Number of componentsP: Number of phases

Why can four phases not exist in equilibrium?As an example, tin can exist in four phases (solid white tin, solidgrey tin, liquid tin, vapour tin). But is there a point at which allfour phases exist in equilibrium?

Gm(1, T, p) =Gm(2, T, p) Gm(2, T, p) =Gm(3, T, p) Gm(3, T, p) =Gm(4, T, p)

System of three equations with two unknowns (T, p) does nothave a solution.

F =C P+ 2 = 1 4 + 2 = 1 does not make sense.It also shows for a triple point: F= 0. There is only a singletriple point for one value ofT, p.


32/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Phase diagram of water

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Introduction

Phase,


Free energy

Chemicalpotential

Phase boundaries

Phase rule

Phase diagrams

Lecture 7 & 8

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Phase diagram of CO2 and He

CO24He


33/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant

Le Chateliersprinciple

Partial molarquantities

Mixing of idealgases

Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Chemical equilibrium: reactions and mixtures

Chemical equilibriumDuring a chemical reaction dn is caused by the (dis-)appearance ofcomponents. Consider the gas phase reaction:

AA(g) +BB(g) CC(g) +DD(g) (C 8)

iare the stochiometric numbers of a balanced reaction.

Physical equilibriumSo far we observed mass flow dn as a consequence of a phasetransition.How do partial molar volumes and partial molar Gibbs free energiesin a mixture effect the equilibrium?

Reading

Chapter 7 Chemical equilibrium: The principlesChapter 6 The properties of mixtures, sections 6.1and 6.2, pp. 123127

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Extent of a chemical reaction

AA(g) +BB(g) CC(g) +DD(g)

The extent of a reaction is defined so that any change by dchanges the amount of any species J by:

d= 1

JdnJ Extent of reaction (66)

Note: the extent of a reaction is also referred to as the composition.

Gibbs energy

G =CCn+DDn AAn BBn (67)Remember that A, the chemical potentialof A is equivalent to thepartial molar Gibbs free energyof A, Gm(A) =G(A)/nA.


34/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Reaction Gibbs free energy rGIf in equation67we divide by n we obtain the reaction Gibbsenergy

rG = G

n = (CC+DD) (AA+BB) (68)

Or more general based on the extent of reaction when pressure andtemperature are kept constant:

rG =

G

p,T

(69)

rG0reverse reaction is spontaneousrG= 0equilibrium

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Example: ammonia synthesis (1)

Extent of reaction

N2(g) + 3 H2(g) 2 NH3(g) (C 9)

N2 = 1H2 =

3

NH3 = 2

d= 11dnN2 = 1

3dnH2 =1

2dnNH3

Reaction Gibbs free energy

dG=N2dnN2 +H2dnH2+NH3dnNH3

= N2d 3H2d+ 2NH3d=N2 3H2 + 2NH3

d

rG= N2 3H2 + 2NH3


35/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Variation of rG with In general the chemical potential of a species J can be expressedwith its activity aJ as:

J=J +RTln aJ (70)

where in the special case of mixtures of ideal gases aJ=pJ/p is

the partial pressure of J in respect to the standard pressure.

rG =

(C C+D D) (A A+B B)

+RT[Cln aC+Dln aD Aln aA Bln aB] (71)

Where the first line is the standard reaction Gibbs energy:

rG =

(C C+

D D) (A A+B B)

(72)

= Gm(C) C+Gm(D) D Gm(A) A+Gm(B) B(73)Combining equations71and73and using logarithm rules:n ln a= ln an and ln a+ ln b ln c= ln ab

c leads to:

rG = rG +RTln

aCC aDD

aAA aBB

(74)

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Example: ammonia synthesis (2)For ideal gases we can calculate the chemical potential directly.With J=

J +RTln pJ and pJ in bar follows:

rG = N2 3H2+ 2NH3=

N2 +RTln pN2

3

H2 +RTln pH2

+ 2 NH3 +RTln pNH3

= N2 3H2

+ 2NH3 + RTlnp2NH3

pN2 p3H2

rG = rG + RTln Q (75)

Standard reaction Gibbs energy rG

rG can be calculated from the standard Gibbs energies of

formation:

rG = Gm(products) Gm(reactants) (76)= rH

TrS (77)rG

= 2fGNH3

fGN2 3fGH2

= 2fGNH3


36/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Reaction quotient Qand equilibrium const. K

At standard conditions:Partial pressures are all 1 bar

rG = rG +RTln

12

1 13 = rG = 2fG

NH3

Reaction quotient,QQ=

aCC aDD

aAA aBB

(78)

At equilibriumAt equilibrium the reaction quotient Q is equivalent to theequilibrium constant Kand rG = 0.

Qequilibrium=K (79)

rG=

rG +RTln K= 0 (80)

ln K= rG

RT (81)

We can now calculate the equilibrium composition of a reactionmixture from tables of thermodynamic data.

The equilibrium constant is dimensionless.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Calculating an equilibrium compositionExample: Ammonia synthesis

K=p2NH3

pN2 p3H2

= 977

Species, J N2 H2 NH3

Initial pJ / bar 1.00 3.00 0Change / bar x 3x +2xEquilibriumpJ / bar 1.00 x 3.00 3x 2x


37/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Equilibrium constant in terms of concentration, Kc

Kc=[C]C [D]D

[A]A [B]B(82)

How are the equilibrium constant in terms of activity, Kand theequilibrium constant in terms of concentration, Kc related?

K =Kcc RTp

gas (83)=Kc

T

12.027K

gas(84)

where gas = (C+D) (A+B) with the standardconcentration c = 1 moldm3 and standard pressure p = 1 bar.Example: ammonia synthesisAtT = 298 K the equilibrium constant is K= 5.8 105.

Kc= ?

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Equilibria changes

How does the equilibrium respond to changes in . . .1 concentration?

2 pressure?

3 temperature?

Le Chateliers Principle

Perturbation of a system at equilibrium will cause the equilibriumposition to change in such a way as to tend to remove theperturbation.

Alternative wordings of Le Chateliers principle:

When a system at equilibrium is disturbed, the equilibriumposition will shift in the direction which tends to minimise, orcounteract, the effect of the disturbance.

If a dynamic equilibrium is disturbed by changing the conditions,the position of equilibrium moves to counteract the change.

Any change in status quo prompts an opposing reaction in theresponding system.


38/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Equilibria changes: concentration

Suppose we have an equilibrium established in the ammoniasynthesis:

N2(g) + 3 H2(g) 2 NH3(g)

What happens when we . . .1 increase the concentration of N2?2 decrease the concentration of N2?

Kc= [NH3]

2

[N2][H2]3

According to Le Chatelier, the position of equilibrium will move insuch a way as to counteract the change:

1 Equilibrium will move so that [N2] decreases again.

The position of equilibrium moves to the right ([NH3]increases).In general you might use this if, for example, one of thereactants was a relatively expensive material whereas the otherwas cheap and plentiful.

2 Equilibrium will move so that [N2] increases again.The position of equilibrium moves to the left ([NH3] decreases).

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Change in concentration

A + B AB Kc= [AB]

[A][B] =

5

1 1 = 5

C + D CD Kc= [CD]

[C][D]=

1

8=

2

4 4


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Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Equilibria changes: pressure

Consider a dissociative reaction where = extent of dissociation:

A(g) 2 B(g)

Start n 0Equ. n(1

) 2n

xin(1 )

n(1 ) + 2n2n

n(1 ) + 2n1 1 +

2

1 +

K =p2BpA

=x2Bp

2

xAp

=

(2)2(1 +)

(1 +)2(1 ) p

= 42

1 2 p

=

1 +

4p

K

12

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Equilibria changes: temperature

At temperatureT1 ln K1= rG

RT1= rH

RT1+

rS

R

At temperatureT2 ln K2= rG

RT2= rH

RT2+

rS

R

vant Hoff equation

ln K2 ln K1= rH

R

1

T1 1

T2

(85)

Alternative expression of the vant Hoffisochore:

d ln K

d(1/T) = rH

R (86)

Exothermic reactions: T favours the reactants: KEndothermic reactions: T favours the products: K


40/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Equilibrium response

2 NO2(g)brown

N2O4(g)colour-less

fH57.2 kJmol1

Change with pressurehttp://www.youtube.com/watch?v=PxJbp1SzGjY

Change with temperaturehttp://www.youtube.com/watch?v=tlGrBcgANSY

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Le Chateliers Principle: summary

Perturbation of a system at equilibrium will cause the equilibriumposition to change in such a way as to tend to remove theperturbation.

Concentration If the concentration of a reactant (solute or gas) is

increased, the equilibrium position shifts to use up theadded reactants resulting in more products.

Pressure If the pressure of an equilibrium system is increased,then the equilibrium position shifts toreduce thepressure.

Temperature If the temperature of an endothermic equilibriumsystem is increased, the equilibrium position shifts touse up the heat by producing more products.

If the temperature of an exothermic equilibriumsystem is increased, the equilibrium position shifts touse up the heat by producing more reactants.
http://www.youtube.com/watch?v=PxJbp1SzGjYhttp://www.youtube.com/watch?v=tlGrBcgANSYhttp://www.youtube.com/watch?v=PxJbp1SzGjYhttp://www.youtube.com/watch?v=tlGrBcgANSYhttp://www.youtube.com/watch?v=tlGrBcgANSYhttp://www.youtube.com/watch?v=PxJbp1SzGjY


41/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Thermodynamic description of mixtures

Motivation

How can we describe the physical properties of a mixture?

What is the chemical equilibrium of a mixture?

How do thermodynamical properties change in a mixture of

ideal gases?Mole fractions: System with J components

xJ mole fraction of component J

xJ= nJntotal

ninumber of molesi

xi=nA+nB+ +nN

ntotal= 1 (87)

Binary system with components A and B

xA= nA

nA+nBxB=

nBnA+nB

(88)

xA+xB= 1

xB= 1 xA

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Partial molar volume (1)

AimWe want to be able to calculate the volume of any liquid mixtureof two components A and B containing the amount nA of A and nBof B.

dV

dnA=

V

nA=Vm,A

Vm,A molar volume of component A

We can draw a similar plot for Vagainst nB with nA= 0.

V =VAnA+VBnB (89)


42/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Partial molar volume

Definition of partial molarvolumeThe partial molar volume VJ of asubstance J is defined as:

VJ=

V

nJ

p,T,n

VA depends on the composition of the mixture.Note that the partial molar volume of A is negative at b.

dV = V

nAp,T,nB dnA+ V

nBp,T,nA dnB=VAdnA+VBdnB

V =

nA0

VAdnA+

nB0

VBdnB=VA

nA0

dnA+VB

nB0

dnB

=VAnA+VBnB

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Partial molar volume

Remember that Vis a state function i.e. it does not matter ifwe first pour A into B or vice versa.

For example: Add A first and then B or the other way round.The final volume will always be the same.

We only have to know the partial volumes of the components in

the target mixture in order to calculate the total volume of themixture.

While molar volumes are always positive, partial molarquantities can be negative:

Example: limiting partial molar volume of MgSO4 in water is1.4 cm3mol1Addition of 1 mol of MgSO4 to a large quantity of waterdecreases the volume by 1.4 cm3.

The mixture contracts as salt breaks up , ions becomehydrated, slight collapse

The concept of partial molar quantities can be extended to anyextensive state function. (Partial molar Gibbs energy)


43/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Partial molar volume (2)

Example: water/ethanol mixture

H2O VH2O(xEtOH= 0) 18cm3mol1

Water is added to water. Thepartial molar volume of water is justthe molar volume of water.

VH2O(xEtOH= 1) 14cm3mol1Water is added to a huge amount ofethanol. Every water molecule willbe surrounded by ethanol molecules.

EtOH Density:

EtOH= 0.789gcm3

Molar mass:MEtOH= 46.07 gmol1

Molar volume:Vm,EtOH= 58.4 cm

3mol1

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Partial molar Gibbs energies

p,T, nB = const.

Chemical potentialThe chemical potential of acomponent J in a mixture isdefined as the partial molar Gibbsenergy:

J=

G

nJ

p,T,n

(90)

For a pure substance G =n Gm.It follows that: = Gm.

In the same way as we have derived an equation for the total volume

of a mixture, the total Gibbs energy of a binary mixture is:

G =nAA+nBB (91)


44/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Fundamental equation of chemicalthermodynamics

If the pressure, temperature and number of moles of eachcomponent change, then the Gibbs free energy changes by

dG =Vdp

SdT as before

+i

idni new term

(92)

The new termaccounts for changes in number of moles due to e.g.

phase changes

chemical reactions

The chemical potential is responsible for mass flow (dni) from one

phase to another, and the appearance or disappearance ofcomponents due to chemical reactions.

For binary systems the equation simplifies to:

dG =Vdp SdT+AdnA+BdnB (93)

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Gibbs-Duhem equation

Total Gibbs energy of a binary mixture

G =nAA+nBB (94)

Infinitesimal change:

dG = dnAA+ dnBB+nA dA+nB dB

From equation93with T, p= const. follows:

nAdA+nBdB= 0

From this special case of a binary mixture follows in general:

inidi = 0 Gibbs-Duhem equation (95)

The chemical potential of one component of a mixture cannotchange independently of the chemical potentials of the othercomponents.


45/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

The Gibbs energy of mixing of ideal gases (1)

Gi =. . .

Gf =. . .

G =AnA+BnB

mixG =Gf Gi

We need to know the chemicalpotentials of the gases beforeand after mixing.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14


The molar free energy of an ideal gas is:

Gm(p) =Gm +RTln

p

p

Since= Gm it follows that

= +RTln p

p

The pressure is always given in bar:

= +RTlnp

1 = +RTln p

Before mixing:

Gi=AnA+BnB=nA(A +RTln p) +nB(B +RTln p)

After mixing:

Gf =AnA+BnB=nA(A +RTln pA) +nB(

B +RTln pB)


46/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14


mixG=Gf Gi=nA(

A +RTln pA) +nB(

B +RTln pB)

(nA(

A +RTln p) +nB(

B +RTln p))

=nARTln pA

p +nBRTlnp

B

p

mixG is independent ofA and

B .

With equation30and the ideal gaslaw:

mixG=xAnRTln xA+xBnRTln xB

=nRT(xAln xA+xBln xB)(96)

mixG 0 from xJ 1 Ideal gases always mix.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Extent of areaction

Equilibriumconstant




Lecture 9 & 10

Lecture 11 & 12

Lecture 13 & 14

Other mixing functions

What happens to the entropy and enthalpy of an ideal gas mixture?

Entropy

mixS=

mixG

T

p,nA,nB

=

nR(xAln xA+xBln xB) (97)

mixS 0Note that we derived the entropy of mixing using statisticalthermodynamics (molecular view) in lectures 3 & 4 (see equ.42).

Enthalpy

From G = H TSmixH= mixG+TmixS

mixH= 0 (98)The enthalpy of mixing is zero since there are no interactionsbetween the gas molecules.The driving force comes entirely from the increase in entropy of thesystem.


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49/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10

Chemicalpotential of liquids

Ideal solutions

Ideal-dilutesolutions

Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Chemical potential of liquids

How to calculate the chemical potential of a liquid.

Ideal gas = +RTln p (single component)A =

A +RTln pA (ideal gas in a mixture)

Based on the equation of state: Vm=RT/p

Liquid We cannot calculate because we do not know theequation of state.

Reading

Chapter 6 The properties of mixtures

sections 6.3, 6.4, 6.5 and 6.8pp. 127 134 and pp. 140 142.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Vapour-liquid equilibrium

At equilibrium, the chemical potential of a substance in thevapour phase must be equal to its chemical potential in theliquid:

A(l) =A(g, p)

For the pure liquid A:

A=A(l) =

A(g) =

A +RTln p

A

(* denotes pure substance)

In the presence of a solute B:

A =A +RTln pA

Combining the last two equations:

A =A RTln pA (99)

A=A RTln pA+RTln pA

=A+RTlnpApA

(100)


50/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Ideal solutions

It has been found for closely related substances (e.g. benzene /toluene) that:

pA=xA

pA Raoults law (103)

The presence of a second component (B) reduces the rate at whichA molecules leave the surface of the liquid but does not inhibit therate at which they return.

rate of vaporization =k xArate of condensation =k pA

At equilibrium: k

xA=k

pA

pA= kk

xA

For the pure liquid: pA=pA and xA= 1

pA= k

k pA=xA pA

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Raoults law

Binary mixtureRaoults law is valid for both components in a binary mixture:

pA=xA pA and pB=xB pBWith Daltons law of partial pressures

ptotal =pA+pB

we can construct a vapour-pressure diagram.Vapour pressure diagram

Mixtures that follow Raoults lawthroughout the composition range areideal solutions or ideal mixtures.

The average energies of the A A, B B

and A B interactions are the same in anideal solution. They are NOT zero!

Remember that there are nointermolecular forces in an ideal gas.


51/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Chemical potential of liquids (2)

From equations100and103follows:

A=

A+RTln xA (104)The presence of a solute B lowers the chemical potential of thesolvent A.

Summary of chemical potentials

ideal gas (g) = (g) +RTln p

ideal gas in a mixture A(g) =A (g) +RTln pA

component in an ideal solution A(l) =A(l) +RTln xA

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Ideal solutions: Gibbs energy of mixing

The Gibbs energy of mixing two liquids to form an ideal solution iscalculated in the same way as for two ideal gases.

Gi =nAA+nB

B

Gf=nA(A+RTln xA)

+nB(B+RTln xB)

The averageenergies of theA A, B B and

AB inter-actions are thesame but notzero.

mixG =nRT(xAln xA+xBln xB) 0 (105)mixS=

nR(xAln xA+xBln xB)

0 (106)

mixH= 0 (107)

mixV =

mixG

p

T,nA,nB

= 0 (108)


52/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Real mixtures: deviations from Raoults law

If the average AA, BB and ABinteractions are notequal then theconcepts of the ideal mixture fail.

AB less attractive than AA,BBVapour pressure is greater thanideal case (positive deviation fromRaoults law). For example:Acetone / CS2

AB more attractive than AA,BB

Vapour pressure is less than idealcase (negative deviation fromRaoults law). For example:Acetone / Chloroform

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Behaviour close to pure substances

As xA 1, the partial pressureof A (solvent) approaches thevalue expected for an ideal

mixture.Raoults law still applicable forreal mixtures if the molefractions are close to one. Thered ellipse in the top figureshows the ideal solution regime.

The solvent molecules (A) arein an environment that differs

only slightly from that of thepure solvent.


53/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Behaviour at small mole fractions: Henrys law

For real mixtures the vapourpressure of the solute (B) isoften proportional to its mole

fraction.However, the constant ofproportionality is not thevapour pressure of the puresubstance but the Henrys lawconstant (KB):

pB=xBKB (109)

Mixtures for which the solute(B) obeys Henrys law and thesolvent (A) obeys Raoults laware called ideal-dilutesolutions.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Ideal-dilute solutions

Henrys law constantsin water at 298 K

K/ atm

O2 4.259 104

He 14.97 104Ne 12.30 104

Ar 3.955 104

1 The solvent molecules (A) are inan environment that differs onlyslightly from that of the puresolvent.

Raoults law2 The solute molecules (B) are in

an environment totally unlike thatin the pure substance (almost onlyBA interactions).Henrys law

The Henrys law constant KB dependson the nature of substance A.

Henrys law is used for calculating theamounts of gas dissolved in liquids(e.g. oxygen or anaesthetics in blood).


54/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Examples of real liquid mixtures

KB >pB

KA >pA

Unfavourable ABinteractions

KB


55/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Activity

Activity and pressure

aB=pBpB

(113)

Activity and mole fraction

aB=B xB (114)

Ideal mixtures: B = 1

Non-ideal mixtures: B 1

As xB 1, aB xB and hence B 1 (ideal solution regime).B=

B+RTln aB=

B+RTln xB

ideal mixture

+ RTln B correction term

(115)

The activity coefficient varies with p, T, and the mole fraction.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Vapour pressure diagrams

Phase rule

F =C P+ 2 Gibbs phase rule (116)F: Variance, C: Components, P: Phases in equilibrium

A binary mixture (C= 2) for which we keep the temperatureconstant: F = 3

P. Degrees of freedom are then the pressure and

the composition.ptotal =pA+pB=p

B+

pA pB

xA

Point a:1 phase present (liquid)

Point b (bubble-point line):Evaporation starts (2 phases).More volatile component dominates in thevapour phase.

Point c (dew-point line):1 phase present (vapour).The composition of the vapour phase isthe same as for the original liquid phase.


56/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Composition of the vapour

Question: At which pressure is the dew-line crossed? We need toknow ptotal =f(yA) where yA is the mole fraction of A in the vapourphase.

yA= pAptotalyB= 1 yA

ptotal =pB+ (p

A pB) xA

pA=xA pAIt follows:

yA= xA pA

pB

+ pA pB xA(117)

The vapour is richer in the morevolatile component.

ptotal = pA pB

pB+

pA pB

yA(118)

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

The lever rule

Vertical line: isoplethconstant composition

Horizontal lines in2-phase region: tie lines

nl =n l Lever rule

(119)


57/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Temperature-composition diagrams

A phase diagram can be used to discuss the process of fractionaldistillation. We change the y-axis in the phase diagram: p Twith p= const.

Fractional distillationIn a simple condensation thevapour is withdrawn andcondensed. In fractionaldistillation, the boiling andcondensation cycle is repeatedsuccessively.

Note that the liquid phase isnow in the lower part of thediagram.

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

Azeotropes

Lecture 11 & 12

Lecture 13 & 14

Vapour phase diagramExample: hexane/heptane mixtures (1)

1 Label the regions of thephase diagrams as to whichphase(s) is (are) present.

2 Most volatile component?hexane heptane

3 What are the boilingtemperatures of purehexane and heptane?hexane: 100 C; heptane:65 Chexane: 65 C; heptane:100 C

4 How would you obtain pureheptane?A few steps depending oninitial mixtureInfinite number of steps


58/77

Thermodynamics

N H Nahler

Lecture 1 & 2

Lecture 3 & 4

Lecture 5 & 6

Lecture 7 & 8

Lecture 9 & 10


Ideal solutions


Henrys law

Activity

Phase diagrams

b18pa1 nhn vision

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