Post on 04-Jun-2018
8/13/2019 B18PA1 NHN 03 Vision
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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
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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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5/32
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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8/13/2019 B18PA1 NHN 03 Vision
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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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14/32
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
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
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
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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16/32
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
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
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
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
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
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
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
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
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
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)
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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.
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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 (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,
constituent,component
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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8/13/2019 B18PA1 NHN 03 Vision
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8/13/2019 B18PA1 NHN 03 Vision
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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
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
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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
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,
constituent,component
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)
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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
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,
constituent,component
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)
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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
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,
constituent,component
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.
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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
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,
constituent,component
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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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 diagram of water
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 diagram of CO2 and He
CO24He
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