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TEMAS SELECTOS DE
FISICOQUÍMICA
Maestría en Ciencia e Ingeniería de Materiales.
PEÑOLES
¡¡BIENVENIDOS!!
Dr. René D. Peralta.
Dpto. de Procesos de Polimerización.
Correo electrónico: [email protected]
Tel. 01 844 438 9830 Ext. 1260.
CONTENIDO DEL CURSO
8. Principios extremos y relaciones termodinámicas. 
9. Equilibrio químico en una mezcla de gases ideales. 
10. Equilibrio de fases en sistemas de un componente.
11. Soluciones.
Chapter 7
One-component
Phase Equilibrium
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3
Physical Chemistry
Chapter 7
La Regla de las Fases.
Fase: un estado de la materia que es uniforme en
composición química y estado físico. (Gibbs)
Numero de fases (p):
Gas o mezcla gaseosa – una sola fase.
Líquido – una, dos y tres fases
dos líquidos totalmente miscibles – una sola fase
una mezcla de hielo y agua – dos fases
Solido – un cristal es una sola fase
una aleación de dos metales – dos fases (inmiscible)
- una fase (miscible)
4
Chapter 7
Physical Chemistry
La Regla de las Fases.
(a)
(b)
La diferencia entre (a) una solución de una sola-fase,
en la cual la composición es uniforme en una escala
microscópica, y (b) una dispersión, en la cual, regiones
de un componente están embebidas en una matriz de
un segundo componente.
5
Physical Chemistry
Chapter 7
La Regla de las Fases.
La diferencia entre (a) constituyente y (b) componente.
(a) Constituyente: una especie química (un ion o una
molécula) que está presente en un sistema.
(b) Componente: un constituyente
independiente de un sistema.
químicamente
Número de componentes, c: el número mínimo de
especies independientes necesario para definir la
composición de todas las fases presentes en el sistema.
6
Chapter 7
Physical Chemistry
La Regla de las Fases.
Cuando no tiene lugar una reacción,
El número de constituyentes = el número de
componentes.
Agua pura:
sistema de un componente.
agua
Mezcla de etanol y agua: sistema de dos componentes.
etanol
agua
7
Chapter 7
Physical Chemistry
La Regla de las Fases.
Cuando ocurre una reacción,
CaCO3(s)  CaO(s) + CO2(g)
Fase 1
Fase 2
Fase 3
Sistema de dos componentes
CaO
CO2
CaO + CO2  CaCO3
El número de constituyentes  el número de
componentes
8
Chapter 7
Physical Chemistry
La Regla de las Fases.
Counting components
How many components are present in a system in
which ammonium chloride undergoes thermal
decomposition?
NH4Cl(s)  NH3(g) + HCl(g)
Phase 1
Phase 2
three constituents
two-component
a one-component system
NH4Cl
NH4Cl  NH3 + HCl
additional NH3 or
HCl
9
Chapter 7
Physical Chemistry
La Regla de las Fases.
Degree of freedom or Variance (f): the number of
intensive variables that can be changed independently
without disturbing the number of phases in
equilibrium.
The phase rule: a general relation among the variance
f, the number of components c and the number of
phases p at equilibrium for a system of any
composition.
f=c–p+2
(7.7)
no reactions
10
Physical Chemistry
Chapter 7
La Regla de las Fases.
Two assumptions:
(1) no chemical reactions occur
(2) every chemical species is present in every phase
Counting the total number of intensive variables
(properties that do not depend on the size of the
system). The pressure P and temperature T count as 2.
Specify the composition of a phase by giving the mole
fractions of c-1 components (because x1+x2+…+xc=1, and
all mole fractions are known if all except one are
specified.)
There are p phases, the total number of composition
variables is p(c-1). At this stage, the total number of
11
intensive variables is p(c-1)+2.
Physical Chemistry
Chapter 7
La Regla de las Fases.
At equilibrium, the chemical potential of a component
j must be the same in every phase:
j, = j, =… for p phase
That is, there are p-1 equations to be satisfied for each
component j. as there are c components, the total
number of equations is c(p-1).
Each equation reduces the freedom to vary one of the
p(c-1)+2 intensive variables. It follows that the total
variance is
f = p(c-1) + 2 - c(p-1) = c – p + 2
12
Chapter 7
Physical Chemistry
Equilibrio de fases de un componente.
For a one-component system (pure water)
f＝1－p＋2＝3－p，(C＝1）
f ≥0, p ≥1, 3≥p≥1
p＝1，f＝2
p＝2，f＝1
p＝3，f＝0
13
Physical Chemistry
Chapter 7
La Regla de las Fases.
Phase diagram: shows the regions of pressure and
temperature at which its various phases are
thermodynamically stable.
Phase boundary: a boundary between regions, shows
the values of P and T at which two phases coexist in
equilibrium.
14
Chapter 7
Physical Chemistry
La Regla de las Fases.
Tf

Tb
P
solid
liquid
Critical
point
Triple
point
Solid
stable
Liquid
stable
vapor
Gas
stable
T
T3
Tc
T
Solid-liquid phase boundary: a plot of the freezing point at various P.
Liquid-vapor phase boundary: a plot of the vapor P of liquid against T.
Solid-vapor phase boundary: a plot of the sublimation vapor P against T.
15
Chapter 7
Physical Chemistry
La Regla de las Fases.
P
solid
liquid
Critical
point
Triple
point
vapor
T3
Tc
T
Triple point: at which three different phases (s, l, g) all simultaneously
coexist in equilibrium. It occurs at a single definite pressure and
temperature characteristic of the substance (outside our control).
Critical point: at which (critical P and critical T) the surface disappears.
16
Chapter 7
Physical Chemistry
Diagrama de fases del H2O : P — T
D
218 atm
C
Y
P / 10 5 Pa
I
solid
liquid
Line
S
Point
1 atm
Region
R
0.00611
gas
A
O
0.0024 0.01
T3
Tf
99.974
Tb
374.2
t/℃
17
Chapter 7
Physical Chemistry
Diagrama de fases del H2O : P — T
Region (s, l, g):
D
P / 10 5 Pa
218 atm
C
f=2, one phase
Y
I
solid
Line (OA, AD, AC):
liquid
S
f=1, two phases in
equilibrium
1 atm
R
0.00611
gas
A
Point (A):
O
0.0024 0.01
T3
Tf
99.974
Tb
t/℃
374.2
Tc
f=0, three phases in
equilibrium
18
Chapter 7
Physical Chemistry
Difference between triple point and freezing point
vapor
Air and vapor
P=611Pa
P=101.325 kPa
ice
Pure water
t=0.01℃
ice
Air-saturated
water
Triple point
Freezing point
In a sealed vessel
In an open vessel
(a) Triple point of H2O
t=0℃
(b) Freezing point of H2O
19
Physical Chemistry
Chapter 7
Difference between triple point and freezing point
Why the freezing point is lower than the triple point?
The higher pressure lowers the freezing point
compared with that of pure water
The dissolved air (i.e. N2 and O2) lowers the freezing
point compared with that of pure water
20
Chapter 7
Physical Chemistry
La Ecuacion de Clapeyron.
Phase equilibrium:
P
   
For a pure substance

m

m
G G
At point 1,
At point 2,
 +
Phase 
dP
2
1
Gm ,1  Gm,1
Gm , 2  Gm, 2
Gm,1  dGm  Gm,1  dGm
dGm  dGm
dT
Phase 
T
Fig. 7.5 two neighboring
points on a two-phase line of a
one-component system.
(7.13)
21
Ecuación de Clausius-Clapeyron
Rudolf Clausius
1822 – 1888
Físico matemático alemán.
Emile Clapeyron
1799 - 1864
Ingeniero francés.
(courtesy F. Remer)
The Clapeyron Equation
• Assume two phases (  and ) of a pure substance are at equilibrium at a
certain p and T
– (p,T) = (p,T)
• p and T can be changed infinitesimally (by dp and dT) in such a way that
the two phases remain at equilibrium
• d = Vmdp –SmdT
–  = Gm
• Change in chemical potential for each phase must be the same
– d = d
• dP/dT = Sm/Vm
The Clapeyron equation
– Vm = V,m - V,m
– Sm = S,m - S,m
23
The Solid-Liquid Phase Boundary
• Melting (fusion) is accompanied by a molar enthalpy change, fusH at a
temperature T
• T is the melting point temperature
• fusS = fusH / T
– Reversible phase transition
• dp/dT = fusH / T fusV
– fusV = Vm(liquid) - Vm(solid)
• dp/dT is large and generally positive
–
–
–
–
fusV is very small and generally positive
fusH is positive (melting is an endothermic process)
dp/dT is the slope of the phase boundary
For water, the slope is negative because molar volume of ice is greater than
molar volume of liquid water
24
The Liquid-Vapor Phase Boundary
• dp/dT = vapH / T vapV
– vapH is the enthalpy of vaporization
– vapV = Vm(gas) - Vm(liquid)
• dp/dT is positive
• The magnitude of dp/dT (the slope) for the liquid-vapor phase boundary
is much smaller than the magnitude of the slope of the solid-liquid phase
boundary
– vapV  fusV
25
The Clausius-Clapeyron Equation
• dp/dT = vapH /T vapV
– vapV = Vm(gas) - Vm(liquid)  Vm(gas)
• dp/dT = vapH / T Vm
– Vm is the molar volume of the gas
– Vm = RT/P (assuming ideal gas behavior)
• dp/dT = p vapH / RT2
• dlnp/dT = vapH / RT2
– The Clausius-Clapeyron equation
• dlnp/d(1/T) = - vapH / R
– A plot of lnp versus 1/T yields a graph with slope = -vapH / R
– Linear relation at least over moderate temperature interval because vapH
varies only slightly with temperature
26
Clausius Clapeyron equation – The Two-Point Form
• Integration of Clausius-Clapeyron equation yields:
• ln(P2/P1) = - (vapH /R) (1/T2 –1/T1)
– vapH assumed independent of T
– P1 is the vapor pressure at temperature T1
– P2 is the vapor pressure at temperature T2
• The heat of vaporization of a liquid can be calculated if the vapor
pressure of the liquid is known at two temperatures
• From the vapor pressure at a given temperature and the heat of
vaporization, one can estimate the vapor pressure at a different
temperature
• The vapor pressure of a liquid is 1atm (760 torr) at the normal boiling
point
27
The Solid-Vapor Phase Boundary
• dp/dT = subH / T subV
– subV = Vm,g – Vm,s  Vm,g
• The sublimation curve is steeper than the boiling point curve because
subH > vapH
• The two-point form of the Clausius-Clapeyron equation can be used to
calculate the heat of sublimation of a solid from its sublimation pressure
at two temperatures
28
Chapter 7
Physical Chemistry
La Ecuacion de Clapeyron.
For a single phase
dG  SdT  VdP  i i dni
dG   SdT  VdP  dn
Gm  G / n,
pure phase
(7.14)
G  nGm
dG  ndGm  Gm dn
ndGm  Gm dn  SdT  VdP  dn
ndGm  SdT  VdP
dGm  Sm dT  Vm dP
one-phase, onecomponent
(7.15)
29
Chapter 7
Physical Chemistry
La Ecuacion de Clapeyron.
For any point on the - equilibrium line
dGm  dGm
dGm  Sm dT  Vm dP
 S m dT  Vm dP   S m dT  Vm dP
(7.13)
(7.15)
(7.16)
(Vm  Vm )dP  ( S m  S m )dT
dP S m  S m S m S
 



dT Vm  Vm Vm V
(7.17)*
30
Chapter 7
Physical Chemistry
La Ecuacion de Clapeyron.
For a reversible (equilibrium) phase change
S  H / T
dP S m  S m S m S
 



dT Vm  Vm Vm V
(7.17)*
dP H m
H


dT TVm TV
(7.18)*
one component two-phase equilibrium
Clapeyron Equation (Clausius-Clapeyron equation)
31
Chapter 7
Physical Chemistry
La Ecuacion de Clapeyron.
dP S m

dT Vm
P
 +
Phase 
dP
dP H m
H


dT TVm TV
The slope of the phase
boundaries
Any phase equilibrium of any pure
substance
2
1
dT
Phase 
T
Fig. 7.5: two neighboring points on a
two-phase line of a one-component
system.
32
The Clausius-Clapeyron Equation
ln P =
-Hvap 
1 
   C
R
T 
P2
-Hvap  1
1 
ln
=
  
R
P1
T2 T1
SAMPLE PROBLEM 12.1
The vapor pressure of ethanol is 115 torr at 34.90C. If Hvap of
ethanol is 40.5 kJ/mol, calculate the temperature (in 0C) when
the vapor pressure is 760 torr.
PROBLEM:
PLAN:
Using the Clausius-Clapeyron Equation
We are given 4 of the 5 variables in the Clausius-Clapeyron
equation. Substitute and solve for T2.
SOLUTION:
P2
-Hvap  1
1 
ln
=
  
P1
R T2 T1
760 torr
ln
115 torr
=
-40.5 x103 J/mol
8.314 J/mol*K
T2 = 350K = 770C
34.90C = 308.0K
1
1
T2
308K
En este punto, continuar con este archivo, con ejemplos seleccionados:
Sect. 5 Phase Equilibria in a One-Component System
Chapter 7
Physical Chemistry
The liquid-vapor boundary
The solid-vapor boundary
dP  vap S  vap H


,
dT  vapV T vapV
 vapV  Vm ( g )  RT / P,
dP  subS  sub H


dT  subV T subV
 subV  Vm ( g )  RT / P
d ln P H m

dT
RT 2
d ln P  H m

R
1
d 
T 
(7.19)*
(7.20)
Solid-gas or liquid-gas equilibrium, not near Tc
36
Chapter 7
Physical Chemistry
The liquid-vapor boundary
The solid-vapor boundary

2
1
d ln P  H m 
2
1
1
dT
2
RT
H m  1 1 
P2
  
ln

P1
R  T2 T1 
(7.21)
Solid-gas or liquid-gas equilibrium, not near Tc
ln( P / atm)  
H m H m

RT
RTnbp
(7.22)
liquid-gas equilibrium, not near Tc
37
Chapter 7
Physical Chemistry
The solid-liquid boundary
dP  fus S  fus H


dT  fusV T fusV

2
1
P2  P1 
dP  
 fus S
 fusV
2
1
 fus S
 fusV
dT  
(T2  T1 ) 
2
1
 fus H
T1 fusV
 fus H
T fusV
dT
(T2  T1 )
(7.23)
(7.24)
Solid-liquid equilibrium, small temperature range
38
Chapter 7
Physical Chemistry
Constructing a solid-liquid phase
boundary
Example: construct the ice-liquid phase boundary for water at
temperature between –1oC and 0oC. What is the melting temperature of
ice under a pressure of 1.5 kbar? fusH = +6.008 kJ/mol, fusV = -1.7
cm3/mol.
Answer:

2
dP  
2
 fus S
dT  
2
 fus H
dT
1 T V
 fusV
fus
 fus H  T 
*
PP 
ln  * 
 fusV  T 
  6.008 103  
T



P / bar  1.0  
ln 


 1.7

  273.15K 
1
1
(7.23)
39
Chapter 7
Physical Chemistry
The Phase Rule
T


P / bar  1.0  (3.53 10 ) ln 

 273.15K 
4
The formula gives the following values:
T/oC
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
P/bar
130
105
79
53
27
1.0
What is the melting temperature of ice under a pressure of 1.5 kbar?
Rearrange the formula into
 1.0  P / bar 
T  (273.15K )  exp 
4 
 3.53 10 
Then, with P=1.5 kbar, T=262 K or –11oC.
40
Chapter 7
Physical Chemistry
The Phase Rule
P1=1.0 bar, T1=273 K
P2=1.5 kbar, T2=262 K
Comment: notice the decrease in melting temperature with increasing
pressure: water is denser than ice, so ice responds to pressure by
tending to melt.
41
Physical Chemistry
Chapter 7
Solid-solid Phase Transitions
Polymorphism:
Many substances have more than one solid form which has a
different crystal structure and is thermodynamically stable over
certain ranges of T and P.
Allotropy:
Polymorphism in elements.
Metastable:
The rate of conversion of  to  is slow enough to allow  to


exist for a significant period of time. (Gm  Gm )
42
Chapter 7
Physical Chemistry
Solid-solid Phase Transitions
Phase diagram of S (part)
104
102
B: 95
oC
C: 119
oC
E: 151 oC
P / 10 5 Pa
Three triple points:
100
orthorhombic
monoclinic
E
151
solid
liquid
10-2
10-4
10-6
C
B
95
119
gas
80
120
160
t/℃
Fig. 7.9 (a)
43
Chapter 7
Physical Chemistry
Solid-solid Phase Transitions
There are many different types of Phase Transition.
Fusion, vaporization……
Ehrenfest Classification:
Changes of enthalpy and volume
dGm  Sm dT  Vm dP
  

 P
  

 T
(7.15)
   
  
  V ,m  V ,m   trsV
T  P T
   
 trs H
  
   S  ,m  S ,m   trs S  
T
 P  T  P
44
Chapter 7
Physical Chemistry
Solid-solid Phase Transitions
First-order phase transition
Because trsV and trsS are non-zero for melting and vaporization for
such transitions, the slopes of the chemical potential plotted against
either pressure or temperature are different on either side of the
transition. The first derivatives of the chemical potentials with
respect to pressure and temperature are discontinuous at the
transition.
V

H
Tt
Tt
S
Tt
CP
Tt
Tt
T
45
Chapter 7
Physical Chemistry
Solid-solid Phase Transitions
First-order phase transition
CP is the slope of H-T. at Tt, the slope of H and Cp are infinite.
A first-order phase transition is also characterized by an infinite heat
capacity at the transition temperature.
V

H
Tt
Tt
 H 
CP  

 T  P
CP
S
Tt
Tt
Tt
T
46
Physical Chemistry
Chapter 7
Solid-solid Phase Transitions
Second-order phase transition
The first derivative of the chemical potential with respect to
temperature is continuous but its second derivative with respect to
temperature is discontinuous at the transition.
A continuous slope of  (a graph with the same slope on either side of
the transition) implies that the volume and entropy (and hence the
enthalpy) do not change at the transition.
The heat capacity is discontinuous at the transition but does not
become infinite.
47
Chapter 7
Physical Chemistry
Solid-solid Phase Transitions
V

H
Tt
Tt
S
Tt
CP
Tt
Tt
T
First-order phase transition
V

H
Tt
Tt
S
Tt
CP
Tt
Second-order phase transition
Tt
T
48
Chapter 7
Physical Chemistry
Solid-solid Phase Transitions
CP

Tt
CP
T
First-order
Tt
T
CP
T Tt
T
Tt
Second-order
 H 
CP  

 T  P
dqP


dT

CP
Lambda
not first-order
CP
T Tt

CP
T Tt

49
TEMAS SELECTOS DE FISICOQUÍMICA
¡Atracciones futuras!
Equilibrio químico en una mezcla de
gases ideales. 
Equilibrio de fases en sistemas de un
componente. 
Soluciones.