TEMAS SELECTOS DE FISICOQUÍMICA

1. Maestría en Ciencia e Ingeniería de Materiales. PEÑOLES TEMAS SELECTOS DE FISICOQUÍMICA Dr. René D. Peralta. Dpto. de Procesos de Polimerización. Correo electrónico: rene@ciqa.mx Tel. 01 844 438 9830 Ext. 1260. ¡¡BIENVENIDOS!!

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. Chapter 7 One-component Phase Equilibrium 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. 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)

5. Physical Chemistry Chapter 7 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.

6. 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 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.

7. Physical Chemistry Chapter 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. agua Mezcla de etanol y agua: sistema de dos componentes. etanol agua

8. Physical Chemistry Chapter 7 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

9. Physical Chemistry Chapter 7 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 additional NH3 or HCl NH4Cl NH4Cl  NH3 + HCl

10. f = c – p + 2 (7.7) Physical Chemistry Chapter 7 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. no reactions

11. 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 intensive variables is p(c-1)+2.

12. 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

13. Physical Chemistry Chapter 7 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

14. 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.

15. Physical Chemistry Chapter 7 La Regla de las Fases. Tf Tb  P solid Critical point liquid Triple point vapor Solid stable Liquid stable Gas stable T3 Tc T 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.

16. Physical Chemistry Chapter 7 La Regla de las Fases. P solid Critical point liquid 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.

17. Physical Chemistry Chapter 7 Diagrama de fases del H2O : P — T D C 218 atm Y I solid liquid Line Point Region S P / 10 5 Pa 1 atm R gas 0.00611 A O 0.01 99.974 374.2 0.0024 T3 Tb Tf t/℃

18. Physical Chemistry Chapter 7 Diagrama de fases del H2O : P — T Region (s, l, g): f=2, one phase D C 218 atm Y I Line (OA, AD, AC): f=1, two phases in equilibrium solid liquid S P / 10 5 Pa 1 atm R gas 0.00611 A Point (A): f=0, three phases in equilibrium O 0.01 99.974 374.2 0.0024 Tc T3 Tb Tf t/℃

19. vapor Air and vapor P=611Pa P=101.325 kPa ice ice t=0.01℃ t=0℃ Pure water Air-saturated water Triple point In a sealed vessel Freezing point In an open vessel (a) Triple point of H2O (b) Freezing point of H2O Physical Chemistry Chapter 7 Difference between triple point and freezing point

20. 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

21. (7.13) Physical Chemistry Chapter 7 La Ecuacion de Clapeyron. P Phase equilibrium:  + Phase  dP 2 For a pure substance 1 Phase  At point 1, T dT At point 2, Fig. 7.5 two neighboring points on a two-phase line of a one-component system.

22. Ecuación de Clausius-Clapeyron Rudolf Clausius 1822 – 1888 Físico matemático alemán. Emile Clapeyron 1799 - 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. (7.15) Physical Chemistry Chapter 7 La Ecuacion de Clapeyron. For a single phase pure phase (7.14) one-phase, one-component

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

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

32. Physical Chemistry Chapter 7 La Ecuacion de Clapeyron. P  + Phase  dP 2 1 Phase  T The slope of the phase boundaries dT Fig. 7.5: two neighboring points on a two-phase line of a one-component system. Any phase equilibrium of any pure substance

33. The Clausius-Clapeyron Equation

34. PROBLEM: The vapor pressure of ethanol is 115 torr at 34.90C. If DHvap 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. 1 760 torr 1 ln -40.5 x103 J/mol - T2 115 torr 308K 8.314 J/mol*K SAMPLE PROBLEM 12.1 Using the Clausius-Clapeyron Equation SOLUTION: 34.90C = 308.0K = T2 = 350K = 770C

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

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

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

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

39. (7.23) Physical Chemistry Chapter 7 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:

40. Physical Chemistry Chapter 7 The Phase Rule 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 Then, with P=1.5 kbar, T=262 K or –11oC.

41. Physical Chemistry Chapter 7 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.

42. 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.

43. 104 102 orthorhombic 100 P / 10 5 Pa monoclinic E 10-2 151 solid 10-4 liquid 10-6 C t/℃ B 119 gas 95 80 120 160 Physical Chemistry Chapter 7 Solid-solid Phase Transitions Phase diagram of S (part) Three triple points: B: 95 oC C: 119 oC E: 151 oC Fig. 7.9 (a)

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

45. V H  S CP T Tt Tt Tt Tt Tt Physical Chemistry Chapter 7 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.

46. V H  S CP T Tt Tt Tt Tt Tt Physical Chemistry Chapter 7 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.

47. 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.

48. V H  S CP T Tt Tt Tt Tt Tt V H  S CP T Tt Tt Tt Tt Tt Physical Chemistry Chapter 7 Solid-solid Phase Transitions First-order phase transition Second-order phase transition

49. CP  CP CP  T T T Tt Tt Tt Physical Chemistry Chapter 7 Solid-solid Phase Transitions First-order Second-order Lambda not first-order

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