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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 if

If 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 how

How 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
http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/


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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

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 dU

and dH

are 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 is 151000J.How big is the change in enthalpy, H?


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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

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 = +90kJmol

1

N2(g) + 2 O2(g) N2O4(l) H2 = 20kJmol1

2 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 a

system


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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

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=dqrevT

(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

Cp

T dT (20)

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

S=CplnT2

T1(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 JK

1mol1)

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

V2

V1(22)

For the change in entropy in this process follows:

S=nRlnV2

V1(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 JK1? (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 =

32R. (Atkins, 4.12)


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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: 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

(T

b)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 JK

1mol1.

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 Jmol1, Tf= 273K, fusS= 22 Jmol

1K1

2 Vaporization of water:

water373 K, 1 bar

steam (C 4)

vapH= 41000Jmol1, 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.


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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


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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= ve

Ph2CHCl+(in water) Ph2CH+(aq) + Cl (aq) S= ve

Would guess S= +veas number of particles increase but ions aresolvated, meaning an ordering of water molecules around the ions.


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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 = nJ

ntotalninumber of moles

i

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 xA

The 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= nARlnpA

p

nBRlnpB

p

(31)

= nR(xAln xA+xBln xB) (32)

mixS 0 (33)

Ideal gases always mix.


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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=kBlnW (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..


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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

Counting possibilities

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

Counting possibilities

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 configurationsNA molecules of A NA! identical configurations

W = N!

(N NA)!NA! (35)

What about the Jacks (B molecules)?

Does Wchange when they are included?


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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=kBlnW =kBln

N!

(NNA)!NA!

(37)

=kB lnN! ln(N NA)! lnNA! (38)=kB[NlnN N (N NA)ln(N NA) + (N NA) NAlnNA+ NA] (39)

= kBN

ln

N NA

N

+NA

N 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)


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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 JK

1mol1

Calculated entropy of CO(Boltzmann):Sm (298 K) = 198 JK

1mol1

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 JK1mol1

IfSm(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


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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 JK

1mol1 andrH= 572 kJmol

1

Standard reaction entropy, Sr

Sr =i

iSm,i(products)

j

jSm,j(reactants) (44)


23/32

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.


24/32

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 systemwhereC =. . . 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:

dSdQ

T and dH= dQ (p= const.)

TdS dH

TdS dH 0

dH TdS 0 (45)

Gibbs free energy

Definition: G=H TS

and consequently:

dG = dH TdS SdT

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

Spontaneous change/process: dGT,p 0

Equilibrium of a system at a fixed temperature and pressurecorresponds to the minimum of the Gibbs free energy.


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27/32

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: TTf liquid vs.solid.

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

T

p

= Sm


28/32

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 that

Vm() dp Sm() dT =Vm() dp Sm() dT

(Sm() Sm()) dT = (Vm() Vm()) dp

dp

dT =

trsS

trsV Clapeyron equation (56)


29/32

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 attemperatureT. 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)


30/32

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.


31/32

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 temperature Tc.

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

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.
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32/32

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

b18pa1 nhn 03 vision

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