View
215
Download
2
Category
Preview:
Citation preview
TEMAS SELECTOS DE FISICOQUÍMICA
¡¡BIENVENIDOS!!
Dr. René D. Peralta.Dpto. de Procesos de Polimerización.Correo electrónico: rene@ciqa.mxTel. 01 844 438 9830 Ext. 1260.
Maestría en Ciencia e Ingeniería de Materiales.
PEÑOLES
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.
3
Chapter 7
One-component Phase
Equilibriumhttp://www.google.com/search?hl=es&rlz=1W1SNYX&q=phase+equilibria+in+one+component+ppt&btnG=Buscar&aq=f&aqi=&aql=&oq=&gs_rfai=
4
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 faseuna mezcla de hielo y agua – dos fases
Solido – un cristal es una sola faseuna aleación de dos metales – dos fases (inmiscible)- una fase (miscible)
Physical Chemistry
Chapter 7
5
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.
Physical Chemistry
Chapter 7
6
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 químicamente independiente de un sistema.
Número de componentes, c: el número mínimo de especies independientes necesario para definir la composición de todas las fases presentes en el sistema.
Physical Chemistry
Chapter 7
7
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.
Mezcla de etanol y agua:
sistema de dos componentes.
agua
etanol agua
Physical Chemistry
Chapter 7
8
La Regla de las Fases.Cuando ocurre una reacción,
CaCO3(s) CaO(s) + CO2(g)
Fase 1
Sistema de dos componentes
CaO CO2
Fase 2 Fase 3
CaO + CO2 CaCO3
Physical Chemistry
Chapter 7
El número de constituyentes el número de componentes
9
La Regla de las Fases.Counting components
NH4Cl(s) NH3(g) + HCl(g)
Phase 1
a one-component system
NH4Cl
Phase 2
NH4Cl NH3 + HCl
How many components are present in a system in which ammonium chloride undergoes thermal decomposition?
three constituents
additional NH3 or HCl
two-component
Physical Chemistry
Chapter 7
10
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
Physical Chemistry
Chapter 7
11
La Regla de las Fases.Two assumptions:
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.
(1) no chemical reactions occur
(2) every chemical species is present in every phase
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 intensive variables is p(c-1)+2.
Physical Chemistry
Chapter 7
12
La Regla de las Fases.At equilibrium, the chemical potential of a component j must be the same in every 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).
j, = j, =… for p phase
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
Physical Chemistry
Chapter 7
13
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
Equilibrio de fases de un componente.
For a one-component system (pure water)
Physical Chemistry
Chapter 7
14
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.
Physical Chemistry
Chapter 7
15
La Regla de las Fases.
Solid-liquid phase boundary: a plot of the freezing point at various P.
PTf Tb
T3 TcT
Gas stable
Liquid stable
Solid stable
T
Triple point
Critical point
vapor
liquid solid
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.
Physical Chemistry
Chapter 7
16
La Regla de las Fases.
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).
T3 Tc
P
T
Triple point
Critical point
vapor
liquid solid
Critical point: at which (critical P and critical T) the surface disappears.
Physical Chemistry
Chapter 7
17t/℃
A
DC
0.00611
0.01
solid
gas
liquid
O
P /
10 5
Pa
374.2
218 atm
Diagrama de fases del H2O : P — T
Line
Point
Region
99.974
1 atm
0.0024
I
R
S
Y
Tf TbT3
Physical Chemistry
Chapter 7
18
Diagrama de fases del H2O : P — T
Region (s, l, g):
f=2, one phase
t/℃
A
D C
0.00611
0.01
solid
gas
liquid
O
P /
10 5
Pa
374.2
218 atm
99.974
1 atm
0.0024
I
R
S
Y
Tf TbT3
Line (OA, AD, AC):
f=1, two phases in equilibrium
Point (A):
f=0, three phases in equilibrium
Tc
Physical Chemistry
Chapter 7
19
Pure water
vapor
P=611Pa
icet=0.01℃
Triple point
In a sealed vessel
(a) Triple point of H2O
Air and vapor
P=101.325 kPa
ice
Air-saturated water
t=0℃
Freezing point
In an open vessel
(b) Freezing point of H2O
Difference between triple point and freezing point
Physical Chemistry
Chapter 7
20
Difference between triple point and freezing 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
Why the freezing point is lower than the triple point?
Physical Chemistry
Chapter 7
21
La Ecuacion de Clapeyron.
Fig. 7.5 two neighboring points on a two-phase line of a one-component system.
dT
P
T
+
Phase
Phase
Phase equilibrium:
dP1
2
For a pure substance
mm GG
At point 1, 1,1, mm GG
At point 2, 2,2, mm GG
mmmm dGGdGG 1,1,
mm dGdG (7.13)
Physical Chemistry
Chapter 7
Ecuación de Clausius-Clapeyron
Rudolf Clausius1822 – 1888
Físico matemático alemán.
Emile Clapeyron1799 - 1864
Ingeniero francés.
(courtesy F. Remer)
23
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
24
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
25
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
26
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
27
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
28
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
29
La Ecuacion de Clapeyron.
For a single phase
i iidnVdPSdTdG
dnVdPSdTdG (7.14)pure phase
mm nGGnGG ,/
dnGndGdG mm
dnVdPSdTdnGndG mm
VdPSdTndGm
(7.15)dPVdTSdG mmm one-phase, one-component
Physical Chemistry
Chapter 7
30
La Ecuacion de Clapeyron.
For any point on the - equilibrium line
(7.15)dPVdTSdG mmm
mm dGdG (7.13)
dPVdTSdPVdTS mmmm (7.16)
dTSSdPVV mmmm )()(
V
S
V
S
VV
SS
dT
dP
m
m
mm
mm
(7.17)*
Physical Chemistry
Chapter 7
31
La Ecuacion de Clapeyron.
For a reversible (equilibrium) phase change
THS /
VT
H
VT
H
dT
dP
m
m
(7.18)*
V
S
V
S
VV
SS
dT
dP
m
m
mm
mm
(7.17)*
Clapeyron Equation (Clausius-Clapeyron equation)
one component two-phase equilibrium
Physical Chemistry
Chapter 7
32
La Ecuacion de Clapeyron.
Fig. 7.5: two neighboring points on a two-phase line of a one-component system.
dT
P
T
+
Phase
Phase
The slope of the phase boundaries
dP1
2
VT
H
VT
H
dT
dP
m
m
m
m
V
S
dT
dP
Any phase equilibrium of any pure substance
Physical Chemistry
Chapter 7
The Clausius-Clapeyron Equation
ln P =
-HvapR
1T
C
ln P2P1
= -Hvap
R1T2
1T1
SAMPLE PROBLEM 12.1 Using the Clausius-Clapeyron Equation
SOLUTION:
PROBLEM: 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.
PLAN: We are given 4 of the 5 variables in the Clausius-Clapeyron equation. Substitute and solve for T2.
ln
P2P1
= -Hvap
R1
T2
1T1
34.90C = 308.0K
ln760 torr115 torr
=-40.5 x103 J/mol8.314 J/mol*K
1T2
1308K
-
T2 = 350K = 770C
En este punto, continuar con este archivo, con ejemplos seleccionados: Sect. 5 Phase Equilibria in a One-Component System
36
The liquid-vapor boundary The solid-vapor boundary
PRTgVVPRTgVV msubmvap /)(,/)(
Solid-gas or liquid-gas equilibrium, not near Tc
VT
H
V
S
dT
dP
VT
H
V
S
dT
dP
sub
sub
sub
sub
vap
vap
vap
vap
,
2
ln
RT
H
dT
Pd m (7.19)*
(7.20)R
H
Td
Pd m
1ln
Physical Chemistry
Chapter 7
37
The liquid-vapor boundary The solid-vapor boundary
Solid-gas or liquid-gas equilibrium, not near Tc
2
1 2
2
1
1ln dT
RTHPd m
121
2 11ln
TTR
H
P
P m (7.21)
nbp
mm
RT
H
RT
HatmP
)/ln( (7.22)
liquid-gas equilibrium, not near Tc
Physical Chemistry
Chapter 7
38
The solid-liquid boundary
Solid-liquid equilibrium, small temperature range
VT
H
V
S
dT
dP
fus
fus
fus
fus
dTVT
HdT
V
SdP
fus
fus
fus
fus
2
1
2
1
2
1(7.23)
)()( 121
1212 TTVT
HTT
V
SPP
fus
fus
fus
fus
(7.24)
Physical Chemistry
Chapter 7
39
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:
dTVT
HdT
V
SdP
fus
fus
fus
fus
2
1
2
1
2
1(7.23)
K
TbarP
15.273ln
7.1
10008.60.1/
3
*
* lnT
T
V
HPP
fus
fus
Physical Chemistry
Chapter 7
40
The Phase Rule
The formula gives the following values:
K
TbarP
15.273ln)1053.3(0.1/ 4
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
41053.3
/0.1exp)15.273(
barPKT
Then, with P=1.5 kbar, T=262 K or –11oC.
Physical Chemistry
Chapter 7
41
The Phase RuleP1=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.
Physical Chemistry
Chapter 7
42
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. )(
mm GG
Physical Chemistry
Chapter 7
43
Solid-solid Phase Transitions
Fig. 7.9 (a)
Phase diagram of S (part)
E
t/℃80 120 160
P /
10 5
Pa
102
100
10-2
10-4
10-6
104
gasB
C
liquid
monoclinicorthorhombic
solid
95 119
151Three triple points:
B: 95 oC
C: 119 oC
E: 151 oC
Physical Chemistry
Chapter 7
44
Solid-solid Phase Transitions
There are many different types of Phase Transition.
Fusion, vaporization……
Ehrenfest Classification:
Changes of enthalpy and volume
VVVPP trsmm
TT
,,
T
HSSS
TTtrs
trsmmPP
,,
(7.15)dPVdTSdG mmm
Physical Chemistry
Chapter 7
45
Solid-solid Phase Transitions
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.
First-order phase transition
Tt
H
Tt
V
Tt
S
Tt
CP
Tt
T
Physical Chemistry
Chapter 7
46
Solid-solid Phase Transitions
CP is the slope of H-T. at Tt, the slope of H and Cp are infinite.
First-order phase transition
A first-order phase transition is also characterized by an infinite heat capacity at the transition temperature.
PP T
HC
Tt
H
Tt
V
Tt
S
Tt
CP
Tt
T
Physical Chemistry
Chapter 7
47
Solid-solid Phase Transitions
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.
Second-order phase 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.
Physical Chemistry
Chapter 7
48
Solid-solid Phase Transitions
First-order phase transition
Second-order phase transition
Tt
H
Tt
V
Tt
S
Tt
CP
Tt
T
Tt
H
Tt
V
Tt
S
Tt
CP
Tt
T
Physical Chemistry
Chapter 7
49
Solid-solid Phase Transitions
First-order Second-order
CP
Tt
T
CP
Tt
T
Lambda
CP
Tt
T
PP T
HC
dT
dqC P
TTP t tTTPC
not first-order
tTTPC
Physical Chemistry
Chapter 7
¡Atracciones futuras!Equilibrio químico en una mezcla
de gases ideales.
TEMAS SELECTOS DE FISICOQUÍMICA
Equilibrio de fases en sistemas de un componente.
Soluciones.
Recommended