1版权所有：华东理工大学物理化学教研室
Physical Chemistry
Peter Atkins
(Sixth edition)
Bilingual Program
2版权所有：华东理工大学物理化学教研室
Part 1,Equilibrium
Bilingual Program
4,The Second Law,the concepts
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This chapter will explain the origin of the
spontaneity of physical and chemical change.
Two simple processes are examined,It shows
that a property,the entropy can be defined,
measured,and used to discuss spontaneous
changes quantitatively,This chapter also
i n t r o d u c e s a m a j o r s u b s i d i a r y
thermodynamic property,the Gibbs energy.
4,The Second Law,the concepts
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The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
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The direction of spontaneous change
Some things happen naturally,A gas expands to
fill the available volume,a hot body cools to the
temperature of its surroundings,and a chemical
reaction runs in one direction rather than another.
But some things don't,We can confine a gas to a
smaller volume,we can cool an object with a
refrigerator,and we can force some reactions to
go in reverse,However,none of these processes
happens spontaneously.
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The direction of spontaneous change
The characteristic of these two processes,
spontaneous and non-spontaneous is
summarized by the Second Law of
thermodynamics,
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Sketch for the distribution of energy
4.1 The dispersal of energy
The direction of spontaneous change
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Sketch for the distribution of energy
4.1 The dispersal of energy
The spontaneous changes
are always accompanied by
a dispersal of energy into a
more disordered form.
The direction of spontaneous change
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(a) A ball resting on a warm
surface; the atoms are under-
going thermal motion;
(b) For the ball to fly upwards,
some of the random vibra-
tional motion would have to
change into coordinated,
directed motion.
4.1 The dispersal of energy
Sketch for the distribution of energy
(a) (b)
The direction of spontaneous change
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The spontaneous changes are always
accompanied by a dispersal of energy
into a more disordered form.
The direction of spontaneous change
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The entropy of an isolated system increases
in the course of a spontaneous change:
In the term of entropy
totS?
0
where Stot is the total entropy of the system and its
surroundings,Thermodynamically irreversible
processes are spontaneous processes,and must be
accompanied by an increase in entropy.
4.1 The dispersal of energy
The direction of spontaneous change
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The characteristic of these two process,
spontaneous and non-spontaneous is
summarized by the Second Law of
Thermodynamics,
The direction of spontaneous change
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The Second Law may be
expressed in terms of the
entropy,a state function
that lets us assess whether
one state is accessible from
another by a spontaneous
change,the Second Law
uses the entropy to identify
the spontaneous changes
among those permissible
changes.
The First Law led to the intro-
duction of the internal energy,
a state function that lets us
assess whether a change is
permissible; only those change
may occur for which the U of
an isolated system remains
constant,First Law uses the
internal energy to identify
permissible changes;
Statements on the Second Law of Thermodynamic
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No process is possible in which the sole result is the
absorption of heat from a reservoir and its complete
conversion into work.
The entropy of an isolated system increases in the
course of a spontaneous change:
totS?
0
The spontaneous changes are always accompanied
by a dispersal of energy into a more disordered
form.
Statements on the Second Law of Thermodynamic
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The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
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1),The thermodynamic definition of entropy
a),Heat stimulates disorderly motion in the
surroundings; work stimulates uniform motion
of atoms in the surroundings,does not change the
degree of disorder,and so does not change the
entropy.
b),A change in the extent to which energy is
dispersed in a disorderly manner depends on the
quantity of energy transferred as heat.
4.2 Entropy
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1),The thermodynamic definition of entropy
T
qS r e vdd?
The thermodynamic definition of entropy is based on
The units of entropy,J K -1,
The units of Molar entropy,J K -1 mol –1 (the same as that of R)
4.2 Entropy
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1),The thermodynamic definition of entropy
For a measurable change between two states i and f this
expression integrates to:
fi r e vd TqS?
To calculate the difference in entropy between any two
states of a system,integrate the heat supplied at each
stage of the path divided by the temperature at which
the heat is supplied.
4.2 Entropy
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Example
Calculate the change in entropy when
50kJ of energy is transferred reversibly
and isothermally as heat to a large block
of copper at (0) ℃,and (b) 70 ℃.
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Example
Answer:
fi r e vr e vd TqqTS 1?
Method:
12
3
r e v JK101,8
2 7 3 K
J1050Δ
T
qS( a )
12
3
r e v JK101,5
2 7 3 ) K( 7 0
J1050Δ
T
qS( b )
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1),The thermodynamic definition of entropy
sur
revs u r,
sur
d
d
T
q
S?
The change in entropy of the surroundings
sursur
sur
sur T
q
T
qS
Because the temperature and pressure of the surroundings
is commonly constant,whatever the change takes place,
reversibly or irreversibly,we have
s u r
s u r
s u r
dd
T
qS?
4.2 Entropy
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0S s u r
For any adiabatic change,
qsur = 0,so
The change of entropy of the surroundings can
be calculated by dividing the heat transferred by
T at which the transfer takes place(regardless of
how the change is brought about in the system).
4.2 Entropy
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Illustration
To calculate the entropy change in the surroundings when
1.00 mol H2O(l) is formed from its elements under
standard conditions at 298 K,we use = -286 kJ from
a table,The heat released is supplied to the surroundings,
now regarded as being at constant pressure,so qsur= +286
kJ,Therefore,
H?
1
s u r JKK
J 9 5 9
2 9 8
1086.2 5S
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2),The entropy as a state function
The entropy is one of the function of states,which
implies that:
0 Sd
A thermodynamic cycle; the overall
change in a state function is zero.
Hot source,Th
low source,Tc
4.2 Entropy
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The Carnot cycle
Carnot cycle –
four reversible stages:
The structure of a Carnot cycle.
1),Reversible isothermal expansion
at Th,?S1= qh/Th,qh>0.
4.2 Entropy - The entropy as a state function
2),Reversible adiabatic expansion
S2= 0.
3),Reversible isothermal compression
at Tc,?S3= qc/Tc,qc < 0.
4),Reversible adiabatic compression
S4= 0.
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2),The entropy as a state function
The total change in entropy around the closed cycle is:
c
c
h
h
c
c
h
h 00d
T
q
T
q
T
q
T
q
S
since
c
h
c
h
T
T
q
q
then
0d SThe structure of a Carnot cycle.
h
h
T
q
c
c
T
q
0
0
4.2 Entropy - The entropy as a state function
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A
B
hh V
Vn R Tq ln
B
A
cc V
VnR Tq ln
c
h
c
h
T
T
q
q
0
c
c
h
hd
T
q
T
qS
4.2 Entropy - The entropy as a state function
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In the limit of infinitesimal cycles,
the non-canceling edges of the
Carnot cycles match the overall
cycle exactly; the sum becomes
an integral,This result implies
that dS is an exact differential and
S is state function.
p e r i m e t e r
r e v
a l l
r e v 0
T
q
T
q
0d S
2),The entropy as a state function
4.2 Entropy
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The thermodynamic definition of entropy
T
qS r e vdd?
s u rs u r
revs u r,
s u r
ddd
T
q
T
qS
For the system
s u rt o t SSS
For the surroundings
For the total
4.2 Entropy
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The efficiency of a heat engine,ε,
The efficiency is the work
done divided by the heat
supplied from the hot source.
ha d s o r b e dh e a t
p e r f o r m e d w o r k
q
w
ε
The structure of a heat engine
4.2 Entropy
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h
c
h
ch
q
q
q
qq
ε
1
Note that qc?0.
The structure of a heat engine
The work performed by the engine
is the difference between the heat
supplied by hot source and that
returned to the cold sink,then:
The efficiency of a heat engine,ε,
4.2 Entropy
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For a Carnot cycle:
h
c
r e v T
T
ε 1
The structure of a heat engine
The efficiency of a heat engine,ε,
The second law of thermodynamics
implies that all reversible engines
have the same efficiency regardless
of their construction.
4.2 Entropy
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Suppose an engine that is working reversibly between
a hot source at a temperature Th and a cold sink at a
temperature T,then
3),The thermodynamic temperature
hTεT )1(
hh
c
T
T
T
T
ε 11
4.2 Entropy
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3),The thermodynamic temperature
hTεT )1(
1),The zero of the thermodynamic temperature scale occurs for
a Carnot efficiency of 1;
2),On the Kelvin scale it is defined by setting the temperature of
the triple point of water as 273.16 K exactly.
3),It is possible to measure temperature on a purely mechanical
basis.
The expression of the thermodynamic temperature scale
defined by Kelvin.
4.2 Entropy
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4),The Clausius inequality
s u rs u r ddo r 0dd SSSS
Consider a system in thermal and mechanical contact with
its surroundings at the same temperature,T,Any change of
state is accompanied by a change in entropy of the system,
dS,and of the surroundings,dSsur,Then
Since
dd s u r qq
dd sur TqSIt follows that dd TqS?
4.2 Entropy
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4),The Clausius inequality
dd T qS?
For any change,the Clausius inequality is expressed as,
For the isolated system:
0d?S
In an isolated system the entropy of the system alone
cannot decrease when a spontaneous change takes place.
4.2 Entropy
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4),The Clausius inequality
For the isolated system:
0d?S
> irreversible
< impossible
= reversible
0 S
4.2 Entropy
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4),The Clausius inequality
0s u r SS
> spontaneous
< impossible
= equilibrorus
4.2 Entropy
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The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
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4.3 Entropy changes accompanying specific processes
Because a change in the degree of molecular order
occurs when a substance freezes or boils,we should
expect the transition to be accompanied by a change
in entropy,
1),The entropy of phase transitions
In a reversible process
In an irreversible process
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The entropy of phase transition at Ttrs,Reversible
At the normal transition temperature,Ttrs,any transfer of
heat between the system and its surroundings is reversible
because the two phases in the system are in equilibrium,
Hq t r sSince at constant p,the change in molar
entropy of the system is
t r s
t r s
t r s T
HS
H2O(s)
1 mol
273K
1 atm
H2O(l)
1 mol
273K
1 atm
reversible
ΔS
t r s
t r s
t r s T
HS
At Ttrs =273K
4.3 Entropy changes accompanying specific processes
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The entropy of transition at T,Irreversible
If a process changes at which the temperature deviates the
normal transition temperature,Ttrs,the process is not a
reversible and the above equation can not be directly used
to calculate the change in entropy.In this case
At T=263K
(T≠Ttrs)
H2O(l)
1 mol
263K
1 atm
irreversible
H2O(l)
1 mol
273K
1 atm
H2O(s)
1 mol
273K
1 atm
reversible
ΔS
ΔS1 ΔS3
ΔS2
H2O(s)
1 mol
263K
1 atm
ΔS=ΔS1 +ΔS2 +ΔS3
4.3 Entropy changes accompanying specific processes
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At T=263K
(T≠Ttrs)
H2O(l)
1 mol
263K
1 atm
irreversible
H2O(l)
1 mol
273K
1 atm
H2O(s)
1 mol
273K
1 atm
reversible
ΔS
ΔS1 ΔS3
ΔS2
H2O(s)
1 mol
263K
1 atm
H2O(s)
1 mol
273K
1 atm
H2O(l)
1 mol
273K
1 atm
reversible
ΔS
t r s
t r s
t r s T
HS
ΔS=ΔS1 +ΔS2 +ΔS3
At Ttrs =273K
4.3 Entropy changes accompanying specific processes
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1),The entropy of phase transition at the Ttrs
If the phase transition is exothermic (Δ trsH < 0),
the entropy change is negative,Δ trsS < 0,
which is consistent with the system becoming
more ordered.
If the transition is endothermic(Δ trsH > 0),
the entropy change is positive,Δ trsS > 0,
which is consistent with the system becoming
more disordered,
4.3 Entropy changes accompanying specific processes
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2),The isothermal expansion of a perfect gas
The change in entropy of a perfect gas that expands
isothermally from Vi to Vf is
fi r e vr e vd TqqTS 1?
The heat absorbed during a reversible isothermal
expansion of a perfect gas can be calculated from:
△ U = q + w and △ U = 0,
For a reversible change qrev = -wrev
4.3 Entropy changes accompanying specific processes
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For a perfect gas in the isothermal process
i
f
r e v
f
i
d
V
VnR T
V
VnR Tw V
VT
ln,
It follows that
i
f
r e vr e v V
VnR Twq ln
i
f
V
VnRS ln
The change in entropy of a perfect gas that expands
isothermally from Vi to Vf is
4.3 Entropy changes accompanying specific processes
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i
f
V
VnRS ln
Because S is a state function,this expression applies
whether the change of state occurs reversibly or
irreversibly,If the change is reversible,the entropy
change in the surroundings must be such as to give
ΔStot= 0,If the expansion occurs freely (w = 0) and
irreversibly,and if T remains constant,then q = 0.
Consequently,ΔSsur=0,and the total entropy change
is given by this equation,And in this case,
ΔS? 0,spontaneous process
4.3 Entropy changes accompanying specific processes
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3),The variation of entropy with temperature
The entropy of a system at a temperature Tf from a knowle-
dge of its entropy at a temperature Ti and the heat supplied
to change its temperature from one value to the other:
)()( if TSTSS
fi r e vif d)) TqTSTS ((
From the definition of constant-pressure heat capacity,
so long as the system is doing no non-expansion work:
4.3 Entropy changes accompanying specific processes
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3),The variation of entropy with temperature
TCq p dd r e v?
f
i
rev
if
d
T
qTSTS
When Cp is independent of T in the temperature range of interest:
i
f
if T
T
CTSTS p ln
f
ii
d
T
TCTS p
4.3 Entropy changes accompanying specific processes
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Example - Calculating the entropy change
Calculate the entropy change when argon at 25℃ and
1.00 atm in a container of volume 500 cm3 is allowed
to expand to 1000 cm3 and is simultaneously heated to
100 ℃,
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Method,Because S is a state function,we are free to choose
the most convenient path from the initial state,
21 SSS
RCC pV mm,,
if TTCS mV ln,2
The second step,reversible heating at constant volume to
the final temperature,The entropy change in this step is
if VVnRS ln1
The first step,reversible isothermal expansion to the final
volume.The entropy change in this step is
Example - Calculating the entropy change
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11121 0,1 7 5 J K0,0 5 7 J K0,1 1 8 J K SSS
RCCTTCS Vpp m,m,if a n d lnΔ F r o m
1
if1
K0,1 1 8 J 5001000ln
ln
nR
VVnRS
Answer,The amount of Ar present is (from the perfect
gas equation and the data for the initial state):
n= pV/RT = 0.0204 mol
1
-1-1
0,0 5 7 J K
298K373Kln)m o l( 1 2,4 8 J K)( 0,0 2 0 4 m o l
2S
Example - Calculating the entropy change
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4),The measurement of entropy form heat capacity
The entropy of a system at a temperature T based on its
entropy at T = 0
f
f u s
0
f ( s ) d( 0))(
T
H
T
TCSTS T p
T pT p T TCT HT TCb bf T
b
v a p
T
d)g(d)l(
The entropy is equal to the area beneath the upper curve up to
the corresponding temperature,plus the entropy of each phase
transition passed.
4.3 Entropy changes accompanying specific processes
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Debye extrapolation
Theoretically,the heat
capacity is proportional
to T 3 when T is low,
Cp = aT 3The variation of C
p
/T with
the temperature
4.3 Entropy changes accompanying specific processes
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The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
版权所有：华东理工大学物理化学教研室 56
4.4 The Third Law of thermodynamics
The entropy change accompanying any physical or
chemical transformation approaches zero as the
temperature approaches zero:
1),The Nernst heat theorem
ΔS T? 0 K = 0
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2),The Third Law thermodynamics
If the entropy of every element in its most stable state atT=0
is taken as zero,then every substance has a positive entropy
which atT=0 may become zero,and which does become zero
for all perfect crystalline substances,including compounds.
S(0K) = 0,for all perfect crystalline substances
The Third Law does not state that entropies are zero at
T = 0; it merely implies that all perfect materials have the
same entropy at that temperature,Choosing this common
value as zero is then a matter of convenience,
4.4 The Third Law of thermodynamics
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3),Third-Law entropies
The Third-Law entropy,standard entropy,:
)( TS
On the basis of S(0K) = 0 for all perfect crystals,the
entropy of the substance in its standard state at the
Temperature T is denoted
)( TS
4.4 The Third Law of thermodynamics
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The standard reaction entropy,,the difference
between the molar entropies of the pure,separated
products and the pure,separated reactants,all
substances being in their standard states at the
specified temperature.
Sr?
m
R e a c t a n t s
m
P r o d u c t s
r SνSνS
( J ) m
J
Jr SνS
v-stoichiometric coefficient
4.4 The Third Law of thermodynamics
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Calculate the standard reaction entropy of
the following reaction at 25 ℃:
Illustration
O ( l )H ( g )O( g )H 22212
g,Og,HlO,HS 2Θm212Θm2ΘmΘr SSS
-1-1-1-121 m o l1 6 3,3 J Km o lJK)0.205(7.1309.69
m
R e a c t a n t s
m
P r o d u c t s
r SνSνS
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A sample consisting of 2.00 mol of a diatomic
perfect gas at 250K is compressed reversibly
and adiabatically until its temperature reaches
300K,Calculate q,w,ΔU,ΔH andΔS,Given
that CV,m=27.5 K-1 mol-1,
Example
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2.00mol (id.g)
250K,p1,V1
2.00mol (id.g)
300K,p2,V2ΔU,ΔH,ΔS
q,w
Answer,Because the compression process is
reversible and adiabatic,it follows that
0 rev qq
0dΔ f
i
r e v
T
qS
Example
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kJ 752,UqUw
an d S i n ce m,TnCH p
-1-1-1-1m,m,m o l3 5,8 1 4 J Km o l8,3 1 4 ) J K( 2 7,5 RCC Vp
3,58k J3581,4J
250) K( 300)m ol( 35,814 JK( 2,0m ol )
t h e n
-1-1
H
U? TnC
V m,
K)2 5 03 0 0()m o lJK5.27()m o l0.2( -1-1
kJ75.2J2750
Example
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A sample of copper of mass 2.75 kg is
cooled at constant pressure from 330K
to 275K,Calculate (a) the energy that
must be removed as heat,and (b) the
change in entropy of the sample.
2.75kg(Copper)
330K,p
2.75kg(Copper)
275K,p
q,ΔS
Example
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Answer,The heat released is
TnCTCq pp m,
3 3 0 ) K-( 2 7 5)m o l( 2 4,4 4 J K)
k g m o l106 3,5 4
2,7 5 k g
( 1-1-
13-
J105 8,2 3
Example
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ln
dd
( b )
m
r e v
i
f
p
p
T
T
nC
T
TC
T
q
S,
3 3 02 7 5ln)m o l( 2 4,4 4 J K)k g m o l106 3,5 4 2,7 5 k g( 1-1-13-
-1193 J K
Example
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The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
版权所有：华东理工大学物理化学教研室 68
Concentrating on the system
Entropy is the basic concept for discussing the direction
of natural change,but to use it we have to analyze
changes in both the system and its surroundings,We
have seen that it is always very simple to calculate the
entropy change in the surroundings,and we shall now
see that it is possible to devise a simple method for
taking that contribution into account automatically,This
approach focuses our attention on the system and
simplifies discussions.
4,The Second Law,the concepts
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4.6 The Helmholtz and Gibbs energies
This inequality can be developed in two ways according
to the conditions of (a) constant volume,and (b) constant
pressure under which the process occurs
0dd
T
qS
1) The Helmholtz and Gibbs energies
For a system in thermal equilibrium with its surroundings
at a temperature T,when a change in the system occurs and
there is a transfer of energy as heat between the system and
the surroundings,the Clausius inequality is
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(a) Heat transfer at constant volume,zero of non-
expansion work,dqV=dU
The criterion for spontaneous change is expressed solely
in terms of the state functions of the system.
At either constant internal energy (dU = 0) or constant
entropy (dS = 0):
0d,?VUS 0d,?VSU
0dd TUS
UST dd? w o r k ) e x p a n i s i o n-n o n no,c o n s t a n t ( p
4.6 The Helmholtz and Gibbs energies
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The first inequality states that,in a system at constant V
and constant U,the entropy increases in a spontaneous
change,The second inequality should be interpreted that,
if the entropy of the system is unchanged,then there must
be an increase in entropy of the surroundings,which can
be achieved only if the energy of the system decreases as
energy flows out as heat.
0d,?VUS 0d,?VSU
4.6 The Helmholtz and Gibbs energies
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(b) Heat transfer at constant pressure,zero of
non-expansion work,dqp=dH
At either constant enthalpy (dH = 0) or constant entropy
(dS = 0):
HST dd? w o r k ) e x p a n i s i o n-n o n no,c o n s t a n t ( p
0d,?pHS 0d,?pSH
The entropy of the system at constant p must increase if its
enthalpy remains constant; the enthalpy must decrease if the
entropy of the system is constant.
4.6 The Helmholtz and Gibbs energies
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The Helmholtz energy,A,TSUA
The Gibbs energy,G,TSHG
c) The definition of the Helmholtz and Gibbs energies
When the state of the system changes at constant T:
STUA ddd STHG ddd
4.6 The Helmholtz and Gibbs energies
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STUA ddd STHG ddd
0dd STU 0dd STHFrom and
The criterion for spontaneous change is then:
0d,?VTA 0d,?pTG
< 0 spontaneous
= 0 equilibrorus
1) The Helmholtz and Gibbs energies
UST dd? w o r k ) e x p a n i s i o n-n o n no,c o n s t a n t ( p
HST dd? w o r k ) e x p a n i s i o n-n o n no,c o n s t a n t ( p
at constant T,V
we=0
at constant T,p
we=0
4.6 The Helmholtz and Gibbs energies
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The Clausius inequality,TdS ≥ dq and
the First Law,dU = dq + dw,it then follows that
2) Maximum work,wmax
dU ≤ TdS+dw or dw ≥ dU -TdS
The maximum energy that can be obtained from the
system as work in a reversible change is then given by
dwmax = dU -TdS
STUA ddd S in c e
4.6 The Helmholtz and Gibbs energies
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2) Maximum work
Awm a x
Aw dd m a x?
The change in the Helmholtz energy is equal to the
maximum work accompanying a process.
A is also called the ‘maximum work function’,or the
‘work function’,
(A,Arbeit is the German word for work)
4.6 The Helmholtz and Gibbs energies
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Example - Calculating the maximum available work
When 1mol C6 H12O6 (glucose) is oxidized to carbon
dioxide and water at 25 ℃ according to the equation
C6 H12O6(s) + 6O2 (g)? 6C O2 (g) +? 6H2O (l)
How much of this energy change can be extracted as
(a) heat at constant pressure,(b) work? Given that
11r1r m o l1 8 2,4 J K2 8 0 8 k J m o l S,U?
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Method,We know that the heat released at constant
pressure is equal to the value of ΔH,so we need to
relate,which is given,Suppose that all
the gases involved are perfect and the equation is
RTnUH g rr
UH rr to
For the maximum available from the process,
Awma x
Example - Calculating the maximum available work
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Ar?(b) Because T = 298 K,the value of
Answer,(a) Because Δng = 0,so
Therefore,at constant pressure,the energy available as
heat is 2808 kJmol-1.
1rr 2 8 0 8 k J m o l UH
1rrr 2 8 6 2 k J m o l STUA
Therefore,the combustion of 1.000 mol C6 H12O6 can be
used to produce up to 2862 kJ of work,
Example - Calculating the maximum available work
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3),Maximum non-expansion work at constant T&p
a n d dd r e vww? STqq r e v ddd
For a reversible change
ST)pV(wST dddd r e v
STHG ddd so
revdw
consists of expansion work (-pdV)
and non-expansion work(dwe)
)(dd r e v pVw
)(dddd pVwqH
From the definition that H = U + PV
4.6 The Helmholtz and Gibbs energies
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4),Maximum non-expansion work at constant T&p
For the reversible change at constant p and T
)(ddd rev pVwG
At constant temperature and pressure the maximum non-
expansion work is equal to the change in Gibbs energy.
pVVpwVpG dd)dd(d reve,
pVw dd r e ve,
r e ve,dd wG?
4.6 The Helmholtz and Gibbs energies
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Example - Calculating the maximum non-expansion work of a reaction
How much energy is available for sustaining muscular
and nervous activity from the combustion of 1.00 mol
of glucose molecules under standard conditions at 37℃
(blood temperature)?
Given that The standard entropy of reaction is +182.4
JK-1 mol -1and the
Hr? is -2808 kJmol
-1
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Method,The non-expansion work available from
the reaction is equal to the change in standard
Gibbs energy for the reaction,
STHG rrr
The standard reaction Gibbs energy is then:
1k Jm ol2865
)m o lJK1 8 2 0 4()K3 1 0(k J m o I2 8 0 8G 111 r
Example - Calculating the maximum non-expansion work of a reaction
版权所有：华东理工大学物理化学教研室 84
we,max =-2865 kJ
for the combustion of 1 mol glucose molecules,and
the reaction can be used to do up to 2865 kJ of non-
expansion work.
A person of mass 70 kg would need to do 2.1 kJ of
work to climb vertical through 3.0 m; therefore,at
least 0.13 g of glucose is needed to complete the task.
Example - Calculating the maximum non-expansion work of a reaction
版权所有：华东理工大学物理化学教研室 85
The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
版权所有：华东理工大学物理化学教研室 86
4.7 Standard molar Gibbs energies
The standard Gibbs energy of reaction can be based on the
standard entropies and enthalpies of reaction:
The standard Gibbs energy of formation is the standard
reaction Gibbs energy for the formation of a compound
from its elements in their reference states.
R e ac t a n t s
ff
P r o d u c t s
r GvGvG
STHG rrr
The standard Gibbs energies of reaction
The standard Gibbs energies of formation
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Standard Gibbs energies o formation of the elements in
their reference states are zero,because their formation is
'null' reaction,
R e ac t a n t s
ff
P r o d u c t s
r GvGvG
Illustration,To calculate the standard Gibbs energy of the
reaction CO(g)+? O2 (g)? C O2 (g) at 25℃,
g),(Og)( C O,g),( C O 2ff2fr GGGG 21
12 5 7,2 k J mo l( 0 )( 1 3 7,2 )3 3 9 4,4 21
4.7 Standard molar Gibbs energies
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
Physical Chemistry
Peter Atkins
(Sixth edition)
Bilingual Program
2版权所有：华东理工大学物理化学教研室
Part 1,Equilibrium
Bilingual Program
4,The Second Law,the concepts
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This chapter will explain the origin of the
spontaneity of physical and chemical change.
Two simple processes are examined,It shows
that a property,the entropy can be defined,
measured,and used to discuss spontaneous
changes quantitatively,This chapter also
i n t r o d u c e s a m a j o r s u b s i d i a r y
thermodynamic property,the Gibbs energy.
4,The Second Law,the concepts
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The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
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The direction of spontaneous change
Some things happen naturally,A gas expands to
fill the available volume,a hot body cools to the
temperature of its surroundings,and a chemical
reaction runs in one direction rather than another.
But some things don't,We can confine a gas to a
smaller volume,we can cool an object with a
refrigerator,and we can force some reactions to
go in reverse,However,none of these processes
happens spontaneously.
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The direction of spontaneous change
The characteristic of these two processes,
spontaneous and non-spontaneous is
summarized by the Second Law of
thermodynamics,
版权所有：华东理工大学物理化学教研室 7
Sketch for the distribution of energy
4.1 The dispersal of energy
The direction of spontaneous change
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Sketch for the distribution of energy
4.1 The dispersal of energy
The spontaneous changes
are always accompanied by
a dispersal of energy into a
more disordered form.
The direction of spontaneous change
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(a) A ball resting on a warm
surface; the atoms are under-
going thermal motion;
(b) For the ball to fly upwards,
some of the random vibra-
tional motion would have to
change into coordinated,
directed motion.
4.1 The dispersal of energy
Sketch for the distribution of energy
(a) (b)
The direction of spontaneous change
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The spontaneous changes are always
accompanied by a dispersal of energy
into a more disordered form.
The direction of spontaneous change
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The entropy of an isolated system increases
in the course of a spontaneous change:
In the term of entropy
totS?
0
where Stot is the total entropy of the system and its
surroundings,Thermodynamically irreversible
processes are spontaneous processes,and must be
accompanied by an increase in entropy.
4.1 The dispersal of energy
The direction of spontaneous change
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The characteristic of these two process,
spontaneous and non-spontaneous is
summarized by the Second Law of
Thermodynamics,
The direction of spontaneous change
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The Second Law may be
expressed in terms of the
entropy,a state function
that lets us assess whether
one state is accessible from
another by a spontaneous
change,the Second Law
uses the entropy to identify
the spontaneous changes
among those permissible
changes.
The First Law led to the intro-
duction of the internal energy,
a state function that lets us
assess whether a change is
permissible; only those change
may occur for which the U of
an isolated system remains
constant,First Law uses the
internal energy to identify
permissible changes;
Statements on the Second Law of Thermodynamic
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No process is possible in which the sole result is the
absorption of heat from a reservoir and its complete
conversion into work.
The entropy of an isolated system increases in the
course of a spontaneous change:
totS?
0
The spontaneous changes are always accompanied
by a dispersal of energy into a more disordered
form.
Statements on the Second Law of Thermodynamic
版权所有：华东理工大学物理化学教研室 15
The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
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1),The thermodynamic definition of entropy
a),Heat stimulates disorderly motion in the
surroundings; work stimulates uniform motion
of atoms in the surroundings,does not change the
degree of disorder,and so does not change the
entropy.
b),A change in the extent to which energy is
dispersed in a disorderly manner depends on the
quantity of energy transferred as heat.
4.2 Entropy
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1),The thermodynamic definition of entropy
T
qS r e vdd?
The thermodynamic definition of entropy is based on
The units of entropy,J K -1,
The units of Molar entropy,J K -1 mol –1 (the same as that of R)
4.2 Entropy
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1),The thermodynamic definition of entropy
For a measurable change between two states i and f this
expression integrates to:
fi r e vd TqS?
To calculate the difference in entropy between any two
states of a system,integrate the heat supplied at each
stage of the path divided by the temperature at which
the heat is supplied.
4.2 Entropy
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Example
Calculate the change in entropy when
50kJ of energy is transferred reversibly
and isothermally as heat to a large block
of copper at (0) ℃,and (b) 70 ℃.
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Example
Answer:
fi r e vr e vd TqqTS 1?
Method:
12
3
r e v JK101,8
2 7 3 K
J1050Δ
T
qS( a )
12
3
r e v JK101,5
2 7 3 ) K( 7 0
J1050Δ
T
qS( b )
版权所有：华东理工大学物理化学教研室 21
1),The thermodynamic definition of entropy
sur
revs u r,
sur
d
d
T
q
S?
The change in entropy of the surroundings
sursur
sur
sur T
q
T
qS
Because the temperature and pressure of the surroundings
is commonly constant,whatever the change takes place,
reversibly or irreversibly,we have
s u r
s u r
s u r
dd
T
qS?
4.2 Entropy
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0S s u r
For any adiabatic change,
qsur = 0,so
The change of entropy of the surroundings can
be calculated by dividing the heat transferred by
T at which the transfer takes place(regardless of
how the change is brought about in the system).
4.2 Entropy
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Illustration
To calculate the entropy change in the surroundings when
1.00 mol H2O(l) is formed from its elements under
standard conditions at 298 K,we use = -286 kJ from
a table,The heat released is supplied to the surroundings,
now regarded as being at constant pressure,so qsur= +286
kJ,Therefore,
H?
1
s u r JKK
J 9 5 9
2 9 8
1086.2 5S
版权所有：华东理工大学物理化学教研室 24
2),The entropy as a state function
The entropy is one of the function of states,which
implies that:
0 Sd
A thermodynamic cycle; the overall
change in a state function is zero.
Hot source,Th
low source,Tc
4.2 Entropy
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The Carnot cycle
Carnot cycle –
four reversible stages:
The structure of a Carnot cycle.
1),Reversible isothermal expansion
at Th,?S1= qh/Th,qh>0.
4.2 Entropy - The entropy as a state function
2),Reversible adiabatic expansion
S2= 0.
3),Reversible isothermal compression
at Tc,?S3= qc/Tc,qc < 0.
4),Reversible adiabatic compression
S4= 0.
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2),The entropy as a state function
The total change in entropy around the closed cycle is:
c
c
h
h
c
c
h
h 00d
T
q
T
q
T
q
T
q
S
since
c
h
c
h
T
T
q
q
then
0d SThe structure of a Carnot cycle.
h
h
T
q
c
c
T
q
0
0
4.2 Entropy - The entropy as a state function
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A
B
hh V
Vn R Tq ln
B
A
cc V
VnR Tq ln
c
h
c
h
T
T
q
q
0
c
c
h
hd
T
q
T
qS
4.2 Entropy - The entropy as a state function
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In the limit of infinitesimal cycles,
the non-canceling edges of the
Carnot cycles match the overall
cycle exactly; the sum becomes
an integral,This result implies
that dS is an exact differential and
S is state function.
p e r i m e t e r
r e v
a l l
r e v 0
T
q
T
q
0d S
2),The entropy as a state function
4.2 Entropy
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The thermodynamic definition of entropy
T
qS r e vdd?
s u rs u r
revs u r,
s u r
ddd
T
q
T
qS
For the system
s u rt o t SSS
For the surroundings
For the total
4.2 Entropy
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The efficiency of a heat engine,ε,
The efficiency is the work
done divided by the heat
supplied from the hot source.
ha d s o r b e dh e a t
p e r f o r m e d w o r k
q
w
ε
The structure of a heat engine
4.2 Entropy
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h
c
h
ch
q
q
q
ε
1
Note that qc?0.
The structure of a heat engine
The work performed by the engine
is the difference between the heat
supplied by hot source and that
returned to the cold sink,then:
The efficiency of a heat engine,ε,
4.2 Entropy
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For a Carnot cycle:
h
c
r e v T
T
ε 1
The structure of a heat engine
The efficiency of a heat engine,ε,
The second law of thermodynamics
implies that all reversible engines
have the same efficiency regardless
of their construction.
4.2 Entropy
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Suppose an engine that is working reversibly between
a hot source at a temperature Th and a cold sink at a
temperature T,then
3),The thermodynamic temperature
hTεT )1(
hh
c
T
T
T
T
ε 11
4.2 Entropy
版权所有：华东理工大学物理化学教研室 34
3),The thermodynamic temperature
hTεT )1(
1),The zero of the thermodynamic temperature scale occurs for
a Carnot efficiency of 1;
2),On the Kelvin scale it is defined by setting the temperature of
the triple point of water as 273.16 K exactly.
3),It is possible to measure temperature on a purely mechanical
basis.
The expression of the thermodynamic temperature scale
defined by Kelvin.
4.2 Entropy
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4),The Clausius inequality
s u rs u r ddo r 0dd SSSS
Consider a system in thermal and mechanical contact with
its surroundings at the same temperature,T,Any change of
state is accompanied by a change in entropy of the system,
dS,and of the surroundings,dSsur,Then
Since
dd s u r qq
dd sur TqSIt follows that dd TqS?
4.2 Entropy
版权所有：华东理工大学物理化学教研室 36
4),The Clausius inequality
dd T qS?
For any change,the Clausius inequality is expressed as,
For the isolated system:
0d?S
In an isolated system the entropy of the system alone
cannot decrease when a spontaneous change takes place.
4.2 Entropy
版权所有：华东理工大学物理化学教研室 37
4),The Clausius inequality
For the isolated system:
0d?S
> irreversible
< impossible
= reversible
0 S
4.2 Entropy
版权所有：华东理工大学物理化学教研室 38
4),The Clausius inequality
0s u r SS
> spontaneous
< impossible
= equilibrorus
4.2 Entropy
版权所有：华东理工大学物理化学教研室 39
The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
版权所有：华东理工大学物理化学教研室 40
4.3 Entropy changes accompanying specific processes
Because a change in the degree of molecular order
occurs when a substance freezes or boils,we should
expect the transition to be accompanied by a change
in entropy,
1),The entropy of phase transitions
In a reversible process
In an irreversible process
版权所有：华东理工大学物理化学教研室 41
The entropy of phase transition at Ttrs,Reversible
At the normal transition temperature,Ttrs,any transfer of
heat between the system and its surroundings is reversible
because the two phases in the system are in equilibrium,
Hq t r sSince at constant p,the change in molar
entropy of the system is
t r s
t r s
t r s T
HS
H2O(s)
1 mol
273K
1 atm
H2O(l)
1 mol
273K
1 atm
reversible
ΔS
t r s
t r s
t r s T
HS
At Ttrs =273K
4.3 Entropy changes accompanying specific processes
版权所有：华东理工大学物理化学教研室 42
The entropy of transition at T,Irreversible
If a process changes at which the temperature deviates the
normal transition temperature,Ttrs,the process is not a
reversible and the above equation can not be directly used
to calculate the change in entropy.In this case
At T=263K
(T≠Ttrs)
H2O(l)
1 mol
263K
1 atm
irreversible
H2O(l)
1 mol
273K
1 atm
H2O(s)
1 mol
273K
1 atm
reversible
ΔS
ΔS1 ΔS3
ΔS2
H2O(s)
1 mol
263K
1 atm
ΔS=ΔS1 +ΔS2 +ΔS3
4.3 Entropy changes accompanying specific processes
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At T=263K
(T≠Ttrs)
H2O(l)
1 mol
263K
1 atm
irreversible
H2O(l)
1 mol
273K
1 atm
H2O(s)
1 mol
273K
1 atm
reversible
ΔS
ΔS1 ΔS3
ΔS2
H2O(s)
1 mol
263K
1 atm
H2O(s)
1 mol
273K
1 atm
H2O(l)
1 mol
273K
1 atm
reversible
ΔS
t r s
t r s
t r s T
HS
ΔS=ΔS1 +ΔS2 +ΔS3
At Ttrs =273K
4.3 Entropy changes accompanying specific processes
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1),The entropy of phase transition at the Ttrs
If the phase transition is exothermic (Δ trsH < 0),
the entropy change is negative,Δ trsS < 0,
which is consistent with the system becoming
more ordered.
If the transition is endothermic(Δ trsH > 0),
the entropy change is positive,Δ trsS > 0,
which is consistent with the system becoming
more disordered,
4.3 Entropy changes accompanying specific processes
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2),The isothermal expansion of a perfect gas
The change in entropy of a perfect gas that expands
isothermally from Vi to Vf is
fi r e vr e vd TqqTS 1?
The heat absorbed during a reversible isothermal
expansion of a perfect gas can be calculated from:
△ U = q + w and △ U = 0,
For a reversible change qrev = -wrev
4.3 Entropy changes accompanying specific processes
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For a perfect gas in the isothermal process
i
f
r e v
f
i
d
V
VnR T
V
VnR Tw V
VT
ln,
It follows that
i
f
r e vr e v V
VnR Twq ln
i
f
V
VnRS ln
The change in entropy of a perfect gas that expands
isothermally from Vi to Vf is
4.3 Entropy changes accompanying specific processes
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i
f
V
VnRS ln
Because S is a state function,this expression applies
whether the change of state occurs reversibly or
irreversibly,If the change is reversible,the entropy
change in the surroundings must be such as to give
ΔStot= 0,If the expansion occurs freely (w = 0) and
irreversibly,and if T remains constant,then q = 0.
Consequently,ΔSsur=0,and the total entropy change
is given by this equation,And in this case,
ΔS? 0,spontaneous process
4.3 Entropy changes accompanying specific processes
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3),The variation of entropy with temperature
The entropy of a system at a temperature Tf from a knowle-
dge of its entropy at a temperature Ti and the heat supplied
to change its temperature from one value to the other:
)()( if TSTSS
fi r e vif d)) TqTSTS ((
From the definition of constant-pressure heat capacity,
so long as the system is doing no non-expansion work:
4.3 Entropy changes accompanying specific processes
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3),The variation of entropy with temperature
TCq p dd r e v?
f
i
rev
if
d
T
qTSTS
When Cp is independent of T in the temperature range of interest:
i
f
if T
T
CTSTS p ln
f
ii
d
T
TCTS p
4.3 Entropy changes accompanying specific processes
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Example - Calculating the entropy change
Calculate the entropy change when argon at 25℃ and
1.00 atm in a container of volume 500 cm3 is allowed
to expand to 1000 cm3 and is simultaneously heated to
100 ℃,
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Method,Because S is a state function,we are free to choose
the most convenient path from the initial state,
21 SSS
RCC pV mm,,
if TTCS mV ln,2
The second step,reversible heating at constant volume to
the final temperature,The entropy change in this step is
if VVnRS ln1
The first step,reversible isothermal expansion to the final
volume.The entropy change in this step is
Example - Calculating the entropy change
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11121 0,1 7 5 J K0,0 5 7 J K0,1 1 8 J K SSS
RCCTTCS Vpp m,m,if a n d lnΔ F r o m
1
if1
K0,1 1 8 J 5001000ln
ln
nR
VVnRS
Answer,The amount of Ar present is (from the perfect
gas equation and the data for the initial state):
n= pV/RT = 0.0204 mol
1
-1-1
0,0 5 7 J K
298K373Kln)m o l( 1 2,4 8 J K)( 0,0 2 0 4 m o l
2S
Example - Calculating the entropy change
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4),The measurement of entropy form heat capacity
The entropy of a system at a temperature T based on its
entropy at T = 0
f
f u s
0
f ( s ) d( 0))(
T
H
T
TCSTS T p
T pT p T TCT HT TCb bf T
b
v a p
T
d)g(d)l(
The entropy is equal to the area beneath the upper curve up to
the corresponding temperature,plus the entropy of each phase
transition passed.
4.3 Entropy changes accompanying specific processes
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Debye extrapolation
Theoretically,the heat
capacity is proportional
to T 3 when T is low,
Cp = aT 3The variation of C
p
/T with
the temperature
4.3 Entropy changes accompanying specific processes
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The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
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4.4 The Third Law of thermodynamics
The entropy change accompanying any physical or
chemical transformation approaches zero as the
temperature approaches zero:
1),The Nernst heat theorem
ΔS T? 0 K = 0
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2),The Third Law thermodynamics
If the entropy of every element in its most stable state atT=0
is taken as zero,then every substance has a positive entropy
which atT=0 may become zero,and which does become zero
for all perfect crystalline substances,including compounds.
S(0K) = 0,for all perfect crystalline substances
The Third Law does not state that entropies are zero at
T = 0; it merely implies that all perfect materials have the
same entropy at that temperature,Choosing this common
value as zero is then a matter of convenience,
4.4 The Third Law of thermodynamics
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3),Third-Law entropies
The Third-Law entropy,standard entropy,:
)( TS
On the basis of S(0K) = 0 for all perfect crystals,the
entropy of the substance in its standard state at the
Temperature T is denoted
)( TS
4.4 The Third Law of thermodynamics
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The standard reaction entropy,,the difference
between the molar entropies of the pure,separated
products and the pure,separated reactants,all
substances being in their standard states at the
specified temperature.
Sr?
m
R e a c t a n t s
m
P r o d u c t s
r SνSνS
( J ) m
J
Jr SνS
v-stoichiometric coefficient
4.4 The Third Law of thermodynamics
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Calculate the standard reaction entropy of
the following reaction at 25 ℃:
Illustration
O ( l )H ( g )O( g )H 22212
g,Og,HlO,HS 2Θm212Θm2ΘmΘr SSS
-1-1-1-121 m o l1 6 3,3 J Km o lJK)0.205(7.1309.69
m
R e a c t a n t s
m
P r o d u c t s
r SνSνS
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A sample consisting of 2.00 mol of a diatomic
perfect gas at 250K is compressed reversibly
and adiabatically until its temperature reaches
300K,Calculate q,w,ΔU,ΔH andΔS,Given
that CV,m=27.5 K-1 mol-1,
Example
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2.00mol (id.g)
250K,p1,V1
2.00mol (id.g)
300K,p2,V2ΔU,ΔH,ΔS
q,w
Answer,Because the compression process is
reversible and adiabatic,it follows that
0 rev qq
0dΔ f
i
r e v
T
qS
Example
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kJ 752,UqUw
an d S i n ce m,TnCH p
-1-1-1-1m,m,m o l3 5,8 1 4 J Km o l8,3 1 4 ) J K( 2 7,5 RCC Vp
3,58k J3581,4J
250) K( 300)m ol( 35,814 JK( 2,0m ol )
t h e n
-1-1
H
U? TnC
V m,
K)2 5 03 0 0()m o lJK5.27()m o l0.2( -1-1
kJ75.2J2750
Example
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A sample of copper of mass 2.75 kg is
cooled at constant pressure from 330K
to 275K,Calculate (a) the energy that
must be removed as heat,and (b) the
change in entropy of the sample.
2.75kg(Copper)
330K,p
2.75kg(Copper)
275K,p
q,ΔS
Example
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Answer,The heat released is
TnCTCq pp m,
3 3 0 ) K-( 2 7 5)m o l( 2 4,4 4 J K)
k g m o l106 3,5 4
2,7 5 k g
( 1-1-
13-
J105 8,2 3
Example
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ln
dd
( b )
m
r e v
i
f
p
p
T
T
nC
T
TC
T
q
S,
3 3 02 7 5ln)m o l( 2 4,4 4 J K)k g m o l106 3,5 4 2,7 5 k g( 1-1-13-
-1193 J K
Example
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The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
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Concentrating on the system
Entropy is the basic concept for discussing the direction
of natural change,but to use it we have to analyze
changes in both the system and its surroundings,We
have seen that it is always very simple to calculate the
entropy change in the surroundings,and we shall now
see that it is possible to devise a simple method for
taking that contribution into account automatically,This
approach focuses our attention on the system and
simplifies discussions.
4,The Second Law,the concepts
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4.6 The Helmholtz and Gibbs energies
This inequality can be developed in two ways according
to the conditions of (a) constant volume,and (b) constant
pressure under which the process occurs
0dd
T
qS
1) The Helmholtz and Gibbs energies
For a system in thermal equilibrium with its surroundings
at a temperature T,when a change in the system occurs and
there is a transfer of energy as heat between the system and
the surroundings,the Clausius inequality is
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(a) Heat transfer at constant volume,zero of non-
expansion work,dqV=dU
The criterion for spontaneous change is expressed solely
in terms of the state functions of the system.
At either constant internal energy (dU = 0) or constant
entropy (dS = 0):
0d,?VUS 0d,?VSU
0dd TUS
UST dd? w o r k ) e x p a n i s i o n-n o n no,c o n s t a n t ( p
4.6 The Helmholtz and Gibbs energies
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The first inequality states that,in a system at constant V
and constant U,the entropy increases in a spontaneous
change,The second inequality should be interpreted that,
if the entropy of the system is unchanged,then there must
be an increase in entropy of the surroundings,which can
be achieved only if the energy of the system decreases as
energy flows out as heat.
0d,?VUS 0d,?VSU
4.6 The Helmholtz and Gibbs energies
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(b) Heat transfer at constant pressure,zero of
non-expansion work,dqp=dH
At either constant enthalpy (dH = 0) or constant entropy
(dS = 0):
HST dd? w o r k ) e x p a n i s i o n-n o n no,c o n s t a n t ( p
0d,?pHS 0d,?pSH
The entropy of the system at constant p must increase if its
enthalpy remains constant; the enthalpy must decrease if the
entropy of the system is constant.
4.6 The Helmholtz and Gibbs energies
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The Helmholtz energy,A,TSUA
The Gibbs energy,G,TSHG
c) The definition of the Helmholtz and Gibbs energies
When the state of the system changes at constant T:
STUA ddd STHG ddd
4.6 The Helmholtz and Gibbs energies
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STUA ddd STHG ddd
0dd STU 0dd STHFrom and
The criterion for spontaneous change is then:
0d,?VTA 0d,?pTG
< 0 spontaneous
= 0 equilibrorus
1) The Helmholtz and Gibbs energies
UST dd? w o r k ) e x p a n i s i o n-n o n no,c o n s t a n t ( p
HST dd? w o r k ) e x p a n i s i o n-n o n no,c o n s t a n t ( p
at constant T,V
we=0
at constant T,p
we=0
4.6 The Helmholtz and Gibbs energies
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The Clausius inequality,TdS ≥ dq and
the First Law,dU = dq + dw,it then follows that
2) Maximum work,wmax
dU ≤ TdS+dw or dw ≥ dU -TdS
The maximum energy that can be obtained from the
system as work in a reversible change is then given by
dwmax = dU -TdS
STUA ddd S in c e
4.6 The Helmholtz and Gibbs energies
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2) Maximum work
Awm a x
Aw dd m a x?
The change in the Helmholtz energy is equal to the
maximum work accompanying a process.
A is also called the ‘maximum work function’,or the
‘work function’,
(A,Arbeit is the German word for work)
4.6 The Helmholtz and Gibbs energies
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Example - Calculating the maximum available work
When 1mol C6 H12O6 (glucose) is oxidized to carbon
dioxide and water at 25 ℃ according to the equation
C6 H12O6(s) + 6O2 (g)? 6C O2 (g) +? 6H2O (l)
How much of this energy change can be extracted as
(a) heat at constant pressure,(b) work? Given that
11r1r m o l1 8 2,4 J K2 8 0 8 k J m o l S,U?
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Method,We know that the heat released at constant
pressure is equal to the value of ΔH,so we need to
relate,which is given,Suppose that all
the gases involved are perfect and the equation is
RTnUH g rr
UH rr to
For the maximum available from the process,
Awma x
Example - Calculating the maximum available work
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Ar?(b) Because T = 298 K,the value of
Answer,(a) Because Δng = 0,so
Therefore,at constant pressure,the energy available as
heat is 2808 kJmol-1.
1rr 2 8 0 8 k J m o l UH
1rrr 2 8 6 2 k J m o l STUA
Therefore,the combustion of 1.000 mol C6 H12O6 can be
used to produce up to 2862 kJ of work,
Example - Calculating the maximum available work
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3),Maximum non-expansion work at constant T&p
a n d dd r e vww? STqq r e v ddd
For a reversible change
ST)pV(wST dddd r e v
STHG ddd so
revdw
consists of expansion work (-pdV)
and non-expansion work(dwe)
)(dd r e v pVw
)(dddd pVwqH
From the definition that H = U + PV
4.6 The Helmholtz and Gibbs energies
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4),Maximum non-expansion work at constant T&p
For the reversible change at constant p and T
)(ddd rev pVwG
At constant temperature and pressure the maximum non-
expansion work is equal to the change in Gibbs energy.
pVVpwVpG dd)dd(d reve,
pVw dd r e ve,
r e ve,dd wG?
4.6 The Helmholtz and Gibbs energies
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Example - Calculating the maximum non-expansion work of a reaction
How much energy is available for sustaining muscular
and nervous activity from the combustion of 1.00 mol
of glucose molecules under standard conditions at 37℃
(blood temperature)?
Given that The standard entropy of reaction is +182.4
JK-1 mol -1and the
Hr? is -2808 kJmol
-1
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Method,The non-expansion work available from
the reaction is equal to the change in standard
Gibbs energy for the reaction,
STHG rrr
The standard reaction Gibbs energy is then:
1k Jm ol2865
)m o lJK1 8 2 0 4()K3 1 0(k J m o I2 8 0 8G 111 r
Example - Calculating the maximum non-expansion work of a reaction
版权所有：华东理工大学物理化学教研室 84
we,max =-2865 kJ
for the combustion of 1 mol glucose molecules,and
the reaction can be used to do up to 2865 kJ of non-
expansion work.
A person of mass 70 kg would need to do 2.1 kJ of
work to climb vertical through 3.0 m; therefore,at
least 0.13 g of glucose is needed to complete the task.
Example - Calculating the maximum non-expansion work of a reaction
版权所有：华东理工大学物理化学教研室 85
The direction of spontaneous change
4.1 The dispersal of energy
4.2 Entropy
4.3 Entropy changes accompanying specific
processes
4.4 The Third Law of thermodynamics
4.5 Reaching very low temperatures
Concentrating on the system
4.6 The Helmholtz and Gibbs energies
4.7 Standard molar Gibbs energies
4,The Second Law,the concepts
版权所有：华东理工大学物理化学教研室 86
4.7 Standard molar Gibbs energies
The standard Gibbs energy of reaction can be based on the
standard entropies and enthalpies of reaction:
The standard Gibbs energy of formation is the standard
reaction Gibbs energy for the formation of a compound
from its elements in their reference states.
R e ac t a n t s
ff
P r o d u c t s
r GvGvG
STHG rrr
The standard Gibbs energies of reaction
The standard Gibbs energies of formation
版权所有：华东理工大学物理化学教研室 87
Standard Gibbs energies o formation of the elements in
their reference states are zero,because their formation is
'null' reaction,
R e ac t a n t s
ff
P r o d u c t s
r GvGvG
Illustration,To calculate the standard Gibbs energy of the
reaction CO(g)+? O2 (g)? C O2 (g) at 25℃,
g),(Og)( C O,g),( C O 2ff2fr GGGG 21
12 5 7,2 k J mo l( 0 )( 1 3 7,2 )3 3 9 4,4 21
4.7 Standard molar Gibbs energies
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
