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TEMAS SELECTOS DE FISICOQUÍMICA ¡¡BIENVENIDOS!! Dr. René D. Peralta. Dpto. de Procesos de Polimerización. Correo electrónico:

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Presentation on theme: "TEMAS SELECTOS DE FISICOQUÍMICA ¡¡BIENVENIDOS!! Dr. René D. Peralta. Dpto. de Procesos de Polimerización. Correo electrónico:"— Presentation transcript:

1 TEMAS SELECTOS DE FISICOQUÍMICA ¡¡BIENVENIDOS!! Dr. René D. Peralta. Dpto. de Procesos de Polimerización. Correo electrónico: Tel Ext Maestría en Ciencia e Ingeniería de Materiales. PEÑOLES

2 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 3 Chapter 7 One-component Phase Equilibrium e+component+ppt&btnG=Buscar&aq=f&aqi=&aql=&oq=&gs_rfaihttp://www.google.com/search?hl=es&rlz=1W1SNYX&q=phase+equilibria+in+on e+component+ppt&btnG=Buscar&aq=f&aqi=&aql=&oq=&gs_rfai=

4 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 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) Physical Chemistry Chapter 7

5 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 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 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 etanolagua Physical Chemistry Chapter 7

8 8 La Regla de las Fases. Cuando ocurre una reacción, CaCO 3 (s)  CaO(s) + CO 2 (g) Fase 1 Sistema de dos componentes CaOCO 2 Fase 2Fase 3 CaO + CO 2  CaCO 3 Physical Chemistry Chapter 7 El número de constituyentes  el número de componentes

9 9 La Regla de las Fases. Counting components NH 4 Cl(s)  NH 3 (g) + HCl(g) Phase 1 a one-component system NH 4 Cl Phase 2 NH 4 Cl  NH 3 + HCl How many components are present in a system in which ammonium chloride undergoes thermal decomposition? three constituents additional NH 3 or HCl two-component Physical Chemistry Chapter 7

10 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 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 x 1 +x 2 +…+x c =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 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) c(p-1) = c – p + 2 Physical Chemistry Chapter 7

13 13 f = 1 - p + 2 = 3 - p , (C = 1 ) f ≥0, p ≥1, 3≥p≥1 p=1,f=2p=1,f=2 p=2,f=1p=2,f=1 p=3,f=0p=3,f=0 Equilibrio de fases de un componente. For a one-component system (pure water) Physical Chemistry Chapter 7

14 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 15 La Regla de las Fases. Solid-liquid phase boundary: a plot of the freezing point at various P. P TfTf TbTb  T3T3 TcTc T 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 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). T3T3 TcTc 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

17 17 t/℃t/℃ A D C solid gas liquid O P / 10 5 Pa atm Diagrama de fases del H 2 O : P — T Line Point Region atm I R S Y TfTf TbTb T3T3 Physical Chemistry Chapter 7

18 18 Diagrama de fases del H 2 O : P — T Region (s, l, g): f=2, one phase t/℃t/℃ A D C solid gas liquid O P / 10 5 Pa atm atm I R S Y TfTf TbTb T3T3 Line (OA, AD, AC): f=1, two phases in equilibrium Point (A): f=0, three phases in equilibrium TcTc Physical Chemistry Chapter 7

19 19 Pure water vapor P=611Pa ice t=0.01 ℃ Triple point In a sealed vessel (a) Triple point of H 2 O Air and vapor P= kPa ice Air-saturated water t=0 ℃ Freezing point In an open vessel (b) Freezing point of H 2 O Difference between triple point and freezing point Physical Chemistry Chapter 7

20 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 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: dP 1 2 For a pure substance At point 1, At point 2, (7.13) Physical Chemistry Chapter 7

22 Ecuación de Clausius-Clapeyron Rudolf Clausius 1822 – 1888 Físico matemático alemán. Emile Clapeyron Ingeniero francés. (courtesy F. Remer)

23 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  = V m dp –S m dT –  = G m Change in chemical potential for each phase must be the same – d   = d   dP/dT =  S m /  V m The Clapeyron equation –  V m = V ,m - V ,m –  S m = S ,m - S ,m

24 24 The Solid-Liquid Phase Boundary Melting (fusion) is accompanied by a molar enthalpy change,  fus H at a temperature T T is the melting point temperature  fus S =  fus H / T – Reversible phase transition dp/dT =  fus H / T  fus V –  fus V = V m (liquid) - V m (solid) dp/dT is large and generally positive –  fus V is very small and generally positive –  fus H 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 25 The Liquid-Vapor Phase Boundary dp/dT =  vap H / T  vap V –  vap H is the enthalpy of vaporization –  vap V = V m (gas) - V m (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 –  vap V   fus V

26 26 The Clausius-Clapeyron Equation dp/dT =  vap H /T  vap V –  vap V = V m (gas) - V m (liquid)  V m (gas) dp/dT =  vap H / T V m – V m is the molar volume of the gas – V m = RT/P (assuming ideal gas behavior) dp/dT = p  vap H / RT 2 dlnp/dT =  vap H / RT 2 – The Clausius-Clapeyron equation dlnp/d(1/T) = -  vap H / R – A plot of lnp versus 1/T yields a graph with slope = -  vap H / R – Linear relation at least over moderate temperature interval because  vap H varies only slightly with temperature

27 27 Clausius Clapeyron equation – The Two-Point Form Integration of Clausius-Clapeyron equation yields: ln(P 2 /P 1 ) = - (  vap H /R) (1/T 2 –1/T 1 ) –  vap H assumed independent of T – P 1 is the vapor pressure at temperature T 1 – P 2 is the vapor pressure at temperature T 2 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 28 The Solid-Vapor Phase Boundary dp/dT =  sub H / T  sub V –  sub V = V m,g – V m,s  V m,g The sublimation curve is steeper than the boiling point curve because  sub H >  vap H 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 29 La Ecuacion de Clapeyron. For a single phase (7.14) pure phase (7.15) one-phase, one- component Physical Chemistry Chapter 7

30 30 La Ecuacion de Clapeyron. For any point on the  -  equilibrium line (7.15) (7.13) (7.16) (7.17)* Physical Chemistry Chapter 7

31 31 La Ecuacion de Clapeyron. For a reversible (equilibrium) phase change (7.18)* (7.17)* Clapeyron Equation (Clausius-Clapeyron equation) one component two-phase equilibrium Physical Chemistry Chapter 7

32 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 dP 1 2 Any phase equilibrium of any pure substance Physical Chemistry Chapter 7

33 The Clausius-Clapeyron Equation

34 SAMPLE PROBLEM 12.1Using the Clausius-Clapeyron Equation SOLUTION: PROBLEM: The vapor pressure of ethanol is 115 torr at C. If  H vap of ethanol is 40.5 kJ/mol, calculate the temperature (in 0 C) 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 T C = 308.0K ln 760 torr 115 torr = x10 3 J/mol J/mol*K 1 T2T K - T 2 = 350K = 77 0 C

35 En este punto, continuar con este archivo, con ejemplos seleccionados: Sect. 5 Phase Equilibria in a One-Component System

36 36 The liquid-vapor boundary The solid-vapor boundary Solid-gas or liquid-gas equilibrium, not near T c (7.19)* (7.20) Physical Chemistry Chapter 7

37 37 The liquid-vapor boundary The solid-vapor boundary Solid-gas or liquid-gas equilibrium, not near T c (7.21) (7.22) liquid-gas equilibrium, not near T c Physical Chemistry Chapter 7

38 38 The solid-liquid boundary Solid-liquid equilibrium, small temperature range (7.23) (7.24) Physical Chemistry Chapter 7

39 39 Constructing a solid-liquid phase boundary Example : construct the ice-liquid phase boundary for water at temperature between –1 o C and 0 o C. What is the melting temperature of ice under a pressure of 1.5 kbar?  fus H = kJ/mol,  fus V = -1.7 cm 3 /mol. Answer : (7.23) Physical Chemistry Chapter 7

40 40 The Phase Rule The formula gives the following values: T/ o C P/bar What is the melting temperature of ice under a pressure of 1.5 kbar? Rearrange the formula into Then, with P=1.5 kbar, T=262 K or –11 o C. Physical Chemistry Chapter 7

41 41 The Phase Rule P 1 =1.0 bar, T 1 =273 K P 2 =1.5 kbar, T 2 =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 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. Physical Chemistry Chapter 7

43 43 Solid-solid Phase Transitions Fig. 7.9 (a) Phase diagram of S (part) E t/℃t/℃ P / 10 5 Pa gas B C liquid monoclinic orthorhombic solid Three triple points: B: 95 o C C: 119 o C E: 151 o C Physical Chemistry Chapter 7

44 44 Solid-solid Phase Transitions There are many different types of Phase Transition. Fusion, vaporization…… Ehrenfest Classification : Changes of enthalpy and volume (7.15) Physical Chemistry Chapter 7

45 45 Solid-solid Phase Transitions Because  trs V and  trs S 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  TtTt H TtTt V TtTt S TtTt CPCP TtTt T Physical Chemistry Chapter 7

46 46 Solid-solid Phase Transitions C P is the slope of H-T. at T t, the slope of H and C p are infinite. First-order phase transition A first-order phase transition is also characterized by an infinite heat capacity at the transition temperature.  TtTt H TtTt V TtTt S TtTt CPCP TtTt T Physical Chemistry Chapter 7

47 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 48 Solid-solid Phase Transitions First-order phase transition Second-order phase transition  TtTt H TtTt V TtTt S TtTt CPCP TtTt T  TtTt H TtTt V TtTt S TtTt CPCP TtTt T Physical Chemistry Chapter 7

49 49 Solid-solid Phase Transitions First-orderSecond-order CPCP TtTt T  CPCP TtTt T Lambda CPCP TtTt T  not first-order Physical Chemistry Chapter 7

50 ¡Atracciones futuras! Equilibrio químico en una mezcla de gases ideales.  TEMAS SELECTOS DE FISICOQUÍMICA Equilibrio de fases en sistemas de un componente.  Soluciones.


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