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TEMAS SELECTOS DE FISICOQUÍMICA

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Dr. René D. Peralta.Dpto. de Procesos de Polimerización.Correo electrónico: [email protected]. 01 844 438 9830 Ext. 1260.

Maestría en Ciencia e Ingeniería de Materiales.

PEÑOLES

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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=

http://www.google.com/search?hl=es&rlz=1W1SNYX&q=phase+equilibria+in+one+component+ppt&btnG=Buscar&aq=f&aqi=&aql=&oq=&gs_rfai

http://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 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


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


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


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


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


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


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


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


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


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


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


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


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.

PP T

HC

Tt

H

Tt

V

Tt

S

Tt

CP

Tt

T

Physical Chemistry

Chapter 7

47


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



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


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.

temas selectos de fisicoquÍmica ¡¡bienvenidos!! dr. rené d. peralta. dpto. de procesos de...

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