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PHYSICAL
CHEMISTRY
CHAPTER 4
MATERIAL EQUILIBRIUM
1
ANIS ATIKAH BINTI AHMAD
[email protected]
SUBTOPIC
Introduction to Material Equilibrium
 Entropy and Equilibrium
 The Gibbs and Helmholtz Energies
 Thermodynamic Relations for a System
Equilibrium
 Calculation of Changes in State Function
 Phase Equilibrium
 Reaction Equilibrium

2
WHAT IS MATERIAL EQUILIBRIUM?
In each phase of the closed system, the number
of moles of each substances present remains
constant in time
 No net chemical reactions are occurring in the
system
 No net transfer of matter from one part of the
system to another
 Concentration of chemical species in the
various part of the system are constant

3
Reaction equilibrium
Material
equilibrium
Phase equilibrium
4
ENTROPY AND EQUILIBRIUM



Entropy, is a measure
of the "disorder" of a
system. What "disorder
refers to is really the
number of different
microscopic states a
system can be in, given
that the system has a
particular fixed
composition, volume,
energy, pressure, and
temperature.
While energy strives to
be minimal, entropy
strives to be maximal
Entropy wants to grow.
Energy wants to shrink.
Together, they make a
compromise.
5
ENTROPY AND EQUILIBRIUM
Example: In isolated system (not in material
equilibrium)
 The spontaneous chemical reaction or transport
of matter are irreversible process that increase
the ENTROPY
 The process was continued until the system’s
entropy is maximized.
 Once it is maximized, any further process can
only decrease entropy –(violate the second law)

6
•

isolated systems:
is one with rigid
walls that has no
communication (i.e.,
no heat, mass, or
work transfer) with
its surroundings. An
example of an
isolated system
would be an
insulated container,
such as an insulated
gas cylinder
isolated
(Insulated)
System:
U = constant
Q=0
7
Consider a system at T;


The system is not in material equilibrium
but is in mechanical and thermal
equilibrium

The surroundings are in material,
mechanical and thermal equilibrium

System and surroundings can exchange
energy (as heat and work) but not matter
Since system and surroundings are isolated , we have
dqsurr= -dqsyst (1)

Since, the chemical reaction or matter transport within the non
equilibrium system is irreversible, dSuniv must be positive:
dSuniv = dSsyst + dSsurr > 0 (2)
8
The surroundings are in thermodynamic
equilibrium throughout the process.
 Therefore, the heat transfer is reversible, and
dSsurr= dqsurr/T (3)


The systems is not in thermodynamic
equilibrium, and the process involves an
irreversible change in the system, therefore
DSsyst ≠dqsyst/T (4)
9


Equation (1) to (3) give
dSsyst > -dSsurr = -dqsurr/T = dqsyst/T (5)
dSuniv = dSsyst + dSsurr >0 (2)
Therefore
dSsyst > dqsyst/T
dS > dqirrev/T
dqsurr= -dqsyst (1)
dSsurr= dqsurr/T (3)
(6)
closed syst. in them. and mech. equilib.
10
When the system has reached material
equilibrium, any infinitesimal process is a change
from a system at equilibrium to one infinitesimally
close to equilibrium and hence is a reversible
process.
 Thus, at material equilibrium we have,
ds = dqrev/T (7)


Combining (6) and (7):
ds ≥dq/T (8) material change, closed syst. in
them & mech. Equilib
11
The first law for a closed system is
dq = dU – dw (9)
 Eq 8 gives dq≤ TdS
 Hence for a closed system in mechanical and
thermal equilibrium we have
dU – dw ≤ TdS

ds ≥ dq/T (8)
Or
dU ≤ TdS + dw (10)
12
THE GIBSS & HELMHOLTZ ENERGIES
A spontaneous process at constant-T-and-V
is accompanied by a decrease in the
Helmholtz energy, A.
A spontaneous process at constant-T-and-P is
accompanied by a decrease in the Gibbs
energy, G.
dA = 0
at equilibrium, const. T, V
dG = 0
at equilibrium, const. T, P
13
HELMHOLTZ FREE ENERGY
A  U - TS
Consider material equilibrium at constant T and V
dU  TdS + dw
dU  TdS + SdT – SdT + dw
dU  d(TS) – SdT + dw
d(U – TS)  – SdT + dw
d(U – TS) – SdT - PdV
dw = -P dV
for P-V work
only
at constant T and V, dT=0, dV=0
d(U – TS)  0
Equality sign holds at material equilibrium
14
HELMHOLTZ FREE ENERGY
For a closed system (T & V constant), the state
function U-TS, continually decrease during the
spontaneous, irreversible process of chemical
reaction and matter transport until material
equilibrium is reached
 d(U-TS)=0
at equilibrium

15
HELMHOLTZ FREE ENERGY
d (U  TS )   SdT  dw
dA  SdT  dw
dA  dw
A  w
Closed system, in thermal &mechanic.
equilibrium
const. T
wby   w
wby   A
const. T, closed syst.
It turns out that A carries a greater significance than being simply a
signpost of spontaneous change:
The negative change in the Helmholtz energy is
equal to the maximum work the system can do:
wby,max   A
16
GIBBS FREE ENERGY
G  H – TS  U + PV – TS
Consider material equilibrium for constant T & P, into
dU  T dS + dw with dw = -P dV
dU  T dS + S dT – S dT - P dV + V dP – V dP
dU  d(TS) – SdT – d(PV) + VdP
d(U + PV – TS)  – SdT + VdP
d(H – TS) – SdT + VdP
at constant T and P, dT=0, dP=0
17
d(H – TS)  0
GIBBS FREE ENERGY
the state function H-TS, continually decrease
during material changes (constant T and P) ,
until material equilibrium is reached.
 This is the minimisation of Gibbs free energy.

d(H – TS)  0
GIBBS FREE
ENERGY,
G=H-TS
G = H – TS = U + PV - TS
18
GIBBS FREE ENERGY
G  H – TS  U + PV – TS
G
Constant T, P
G decreases during the
approach to equilibrium,
reaching minimum at
equilibrium
Equilibrium reached
Time
dGT,P  0
19
GIBBS FREE ENERGY
As G of the system decrease at constant T & P,
Suniv increases.
WHY?
Consider a system in mechanical and thermal equilibrium which
undergoes an irreversible chemical reaction or phase change at constant
T and P.
Suniv  S surr  S syst  H syst / T  S syst
 (H syst  TS syst ) / T  Gsyst / T
Suniv  Gsyst / T
closed syst., const. T, V, P-V work only
20
The decrease in Gsyst as the system proceeds to equilibrium at constant T
and P corresponds to a proportional increase in S univ
G  H – TS  U + PV – TS
G  U– TS + PV A + PV
dG  dA  PdV  VdP
dG   SdT  dw  PdV  VdP
dG  SdT  dw  PdV  VdP
const. T and P, closed syst.
dG  dw  PdV
If the P-V work is done in a mechanically reversible manner, then
dw   PdV  dwnon P V
dG  dwnon P V
or
wby,non P V  G
G  wnon P V   wby,non P V21
const. T and P, closed
syst.
For a reversible change
wby,non PV  G
The maximum non-expansion work from a process
at constant P and T is given by the value of -G
w nonPV , max  G
(const. T, P)
22
Thermodynamic Relations for a
System in Equilibrium
6 Basic Equations:
dU = TdS - PdV
closed syst., rev. proc.,
P-V work only
H  U + PV
A  U – TS
G  H - TS
 U 
CV  

 T V
closed syst., in equilib.,
P-V work only
 H 
CP  

 T  P
closed syst., in equilib.,
P-V work only
23
Basic
Equations
 S 
CV  T  
 T V
closed syst., in equilib.
 S 
CP  T  
 T  P
The rates of change of U, H, and S with respect to T can be determined
from the heat capacities CP and CV.
Heat
capacities
(CP
CV )
Key properties
24
The Gibbs Equations
dU = TdS - PdV
dH = TdS + VdP
dA = -SdT - PdV
closed syst., rev.
proc., P-V work
only
dG = -SdT + VdP
How to derive dH, dA and dG?
25
The Gibbs Equations
dH = ?
H  U + PV
dH = d(U + PV)
= dU + d(PV)
dU = TdS - PdV
= dU + PdV + VdP
= (TdS - PdV) + PdV + VdP
dH = TdS + VdP
26
dA = ?
A  U - TS
dA = d(U - TS)
dU = TdS - PdV
= dU - d(TS)
= dU - TdS - SdT
= (TdS - PdV) - TdS - SdT
dA = -SdT - PdV
dG = ?
G  H - TS
dG = d(H - TS)
dH = TdS+VdP
= dH - d(TS)
= dH - TdS - SdT
= (TdS + VdP) - TdS - SdT
dG = -SdT + VdP
27
The Gibbs equation dU= T dS – P dV implies that U is being considered
a function of the variables S and V. From U= U (S,V) we have
 U 
 U 
dU  
 dS  
 dV
 S V
 V  S
 U 
 U 

 T 
  P
 S V
 V  S
 G 

  S
 T  P
 G 

 V
 P T
(dU  TdS  PdV )
(dG = -SdT + VdP)
The Power of thermodynamics:
Difficultly measured properties to be expressed in terms of easily 28
measured properties.
The Euler Reciprocity
Relations
If Z＝f(x，y)，and Z has continuous second partial derivatives,
then
dz  Mdx  Ndy
 z 
M  
 x  y
 z 

N  
 y  x
  z    z 
    
y  x  x  y 
That is
 M

 y

 N 
  

 x  x  y
29
The Maxwell Relations
(Application of Euler relation to Gibss equations)
dU = TdS - PdV
The Gibbs equation (4.33) for dU is
dU＝TdS－PdV
dS＝0
dV＝0
 U 

  P
 V  S
 U 

 T
 S V
Applying Euler Reciprocity,

V
  U 
 U 

 


S  V 
 S 

T S    P V
V
S
 P 
 T 

 

 S V  V  S
30
These are the Maxwell Relations
 P 
 T 

 

 S V  V  S
 T 
 V 





 P  S  S  P
The first two are little used.
 S 
 P 

 

 V T  T V
 S 
 V 
   

 P T
 T  P
The last two are extremely valuable.
The equations relate the isothermal pressure and volume variations of
entropy to measurable properties.
31
DEPENDENCE OF STATE FUNCTIONS ON T,
P, AND V
We now find the dependence of U, H, S and G on
the variables of the system.
 The most common independent variables are T
and P.
 We can relate the temperature and pressure
variations of H, S, and G to the measurable Cp,α,
and κ

32
Volume dependence of U
The Gibbs equation gives dU＝TdS－PdV
For an isothermal process dUT＝TdST－PdVT
Divided above equation by dVT, the infinitesimal volume change
at constant T, to give
dU T
dST
T
P
dVT
dVT
 U 
 S 

  T
 P
 V T
 V T
 S   P 
From Maxwell Relations

  
 V T  T V
T
 U 
 P 
P

  T   P 

 V T
 T V
T subscripts
indicate that the
infinitesimal
changes dU, dS,
and dV are for a
constant-T process
33
 U 
Temperature dependence of U  T   CV

V
Temperature dependence of H  H   CP
 T  P
Pressure dependence of H
From Basic Equations
from Gibbs equations, dH＝TdS＋VdP
 H 
 S 

  T  V
 P T
 P T
 S 
 V 
   

 P T
 T  P
From Maxwell Relations
 H 
 V 

  T 
  V  TV  V
 P T
 T  P
34
Temperature dependence of S
From Basic Equations
CP
 S 

 
T
 T  P
Pressure dependence of S
From Maxwell Relations
The equations of this
section apply to a
closed system of fixed
composition and also to
a closed system where
the composition
changes reversibly
 S 
 V 

  
  V
 P T
 T  P
Temperature and Pressure dependence of G
The Gibbs equation (4.36) for dG is
dG ＝ -SdT + VdP
dT＝0
 G 

 V
 P T
dP＝0
 G 

  S
 T  P
35
Joule-Thomson Coefficient
(easily measured quantities)
 JT
 T 
 
 P  H
 JT
 H 
 
 / CP
 P T
from Chapter 2
From
pressure
dependence
of H
 H 
 V 

  T 
  V  TV  V
 P T
 T  P
 JT
( 1
 V
)[TV  V ]  
CP
 CP

(T  1)

36
Heat-Capacity Difference
(easily measured quantities)
 U 
 V 
CP  CV  
  P 

 V T
 T  P
From
volume
dependence
of U
T
 U 
 P 
P

  T   P 

 V T
 T V
 T  V 
CP  CV  


   T  P
CP  CV 
TV

2
 V 


T

P
  V 1 
 V 

  V
 T  P
37
Heat-Capacity Difference
CP  CV 
TV 2

1. As T  0, CP  CV
2. CP  CV (since  > 0)
3. CP = CV (if  = 0)
38
EXAMPLE 1
CP  CV 
TV 2

÷n
CP ,m  CV ,m 
TVm 2

39
 U 


 V T
Internal Pressure
Ideal gases
 U 

 0
 V T
Solids, Liquids, & Non-ideal Gases
T
 U 
 P 
P

  T   P 

 V T
 T V
Solids
300 J/cm3 (25 oC, 1 atm)
Liquids
300 J/cm3 (25 oC, 1 atm)
Strong intermolecular forces in solids and liquids.
40
CALCULATION OF CHANGES IN STATE FUNCTION
1.
Calculation of ΔS
Suppose a closed system of constant composition goes from
state (P1,T1) to state (P2,T2), the system’s entropy is a
function of T and P
 S 
 S 
dS  
 dT  
 dP
 T  P
 P T
CP
dS 
dT  VdP
T
41
Integration gives:
S  S 2  S1 

2
1
2
CP
dT  V dP
1
T

Since S is a state function, ΔS is independent of the path used to
connect states 1 and 2. A convenient path (Figure 4.3) is first to
hold P constant at P1 and change T from T1 to T2. Then T is held
constant at T2, and P is changed from P1 to P2.
For step (a), dP=0 and gives
S a 
T2

T1
CP
dT
T
const P  P1
For step (b), dT=0 and gives
Sb  

P2
P1
V dP
const T  T2
42
EXAMPLE 2
43
2. Calculation of ΔH
 H 
 H 
dH  
 dT  
 dP
 T  P
 P T
 CP dT  (TV  V )dP
2
2

 V  TV  dP
1
1
H  C P dT 
ΔU can be easily found from ΔH using :
ΔU = ΔH – Δ (PV)
Alternatively
we can write
down the
equation for ΔU
similar to:
44
3. Calculation of ΔG
For isothermal process:
G  H  TS const T
Alternatively, ΔG for an isothermal process that does not
involve an irreversible composition change can be found as:
P2

G  VdP const T
from slide 28
 G 

 V
 P T
P1
A special case:
G  0
rev
[Since H  q, S  q / T ]
process at const T
and P
45
Phase Equilibrium
A phase equilibrium involves the same chemical species
present in different phase. [ eg:C6H12O6(s) C6H12O6(g) ]
46
Phase equilib, in closed syst, P-V
work only
For the spontaneous flow of
-
moles of j from phase
to phase
Closed syst that has not
yet reached phase
equilibrium
47
Suppose that substance j is initially absent in phase
.
If initially
>
, then j flows from phase β to phase δ until the
equilibrium is reached.
However, if
Then, j cannot flow out of δ (since it is absent from δ ). The system will
therefore unchanged with time and hence in equilibrium. So, when a
substance is absent from a phase, the equilibrium condition becomes:
Phase equilib, j absent from
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Reaction Equilibrium
A reaction equilibrium involves different chemical species
present in the same phase.
During a chemical reaction, the change Δn in the no. of moles of
each substance is proportional to its stoichometric coefficient v,
where the proportionality constant is the same for all species.
This proportionality constant is called the extent of reaction
For general chemical reaction undergoing a definite amount of
reaction, the change in moles of species i, ,
, equals
multiplied by the proportionality constant
:
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The condition for chemical-reaction
equilibrium in a closed system is
Reaction equilib, in closed
system., P-V work only
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EXAMPLE 3
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