1. 1 States of Matter: Liquids and Solids

2. 2 States of Matter  Comparison of gases, liquids, and solids. (See Figure 11.2)(See Figure 11.2) –Gases are compressible fluids. Their molecules are widely separated. –Liquids are relatively incompressible fluids. Their molecules are more tightly packed. –Solids are nearly incompressible and rigid. Their molecules or ions are in close contact and do not move.

3. 3 Changes of State  A change of state or phase transition is a change of a substance from one state to another. solid liquid gas elt melting ez freezing en condensation ili boiling m sublimation (see Figure 11.3) ns condensation or pos deposition

4. 4 Vapor Pressure  Liquids are continuously vaporizing. –If a liquid is in a closed vessel with space above it, a partial pressure of the vapor state builds up in this space. –The vapor pressure of a liquid is the partial pressure of the vapor over the liquid, measured at equilibrium at a given temperature. (See Figure 11.4)(See Figure 11.4)

5. 5 Vapor Pressure  The vapor pressure of a liquid depends on its temperature. (See Figure 11.7)(See Figure 11.7) –As the temperature increases, the kinetic energy of the molecular motion becomes greater, and vapor pressure increases. –Liquids and solids with relatively high vapor pressures at normal temperatures are said to be volatile.

6. Henry’s Law  Look up Henry’s Law in your textbook.  Complete the activity keeping this law in mind. 6

7. 7 Boiling Point  The temperature at which the vapor pressure of a liquid equals the pressure exerted on the liquid is called the boiling point. –As the temperature of a liquid increases, the vapor pressure increases until it reaches atmospheric pressure. –At this point, stable bubbles of vapor form within the liquid. This is called boiling. –The normal boiling point is the boiling point at 1 atm.

8. 8 Freezing Point  The temperature at which a pure liquid changes to a crystalline solid, or freezes, is called the freezing point. –The melting point is identical to the freezing point and is defined as the temperature at which a solid becomes a liquid. –Unlike boiling points, melting points are affected significantly by only large pressure changes.

9. 9 Heat of Phase Transition  To melt a pure substance at its melting point requires an extra boost of energy to overcome lattice energies. –The heat needed to melt 1 mol of a pure substance is called the heat of fusion and denoted  H fus. –For ice, the heat of fusion is 6.01 kJ/mol.

10. 10 Heat of Phase Transition  To boil a pure substance at its melting point requires an extra boost of energy to overcome intermolecular forces. –The heat needed to boil 1 mol of a pure substance is called the heat of vaporization and denoted  H vap. (see Figure 11.9) (see Figure 11.9) –For ice, the heat of vaporization is 40.66 kJ/mol.

11. 11 A Problem to Consider  The heat of vaporization of ammonia is 23.4 kJ/mol. How much heat is required to vaporize 1.00 kg of ammonia? –First, we must determine the number of moles of ammonia in 1.00 kg (1000 g).

12. 12 A Problem to Consider  The heat of vaporization of ammonia is 23.4 kJ/mol. How much heat is required to vaporize 1.00 kg of ammonia? –Then we can determine the heat required for vaporization.

13. 13 Clausius-Clapeyron Equation  We noted earlier that vapor pressure was a function of temperature. –It has been demonstrated that the logarithm of the vapor pressure of a liquid varies linearly with absolute temperature. –Consequently, the vapor pressure of a liquid at two different temperatures is described by:

14. 14 A Problem to Consider  Carbon disulfide, CS 2, has a normal boiling point of 46 ° C (vapor pressure = 760 mmHg) and a heat of vaporization of 26.8 kJ/mol. What is the vapor pressure of carbon disulfide at 35 ° C? –Substituting into the Clausius-Clapeyron equation, we obtain:

15. 15 A Problem to Consider  Carbon disulfide, CS 2, has a normal boiling point of 46 ° C (vapor pressure = 760 mmHg) and a heat of vaporization of 26.8 kJ/mol. What is the vapor pressure of carbon disulfide at 35 ° C? – Taking the antiln we obtain:

16. 16 Phase Diagrams  A phase diagram is a graphical way to summarize the conditions under which the different states of a substance are stable. –The diagram is divided into three areas representing each state of the substance. –The curves separating each area represent the boundaries of phase changes.

17. 17 Phase Diagrams  Below is a typical phase diagram. It consists of three curves that divide the diagram into regions labeled “solid, liquid, and gas”. B temperature pressure A C D solidliquid gas..

18. 18 Phase Diagrams  Curve AB, dividing the solid region from the liquid region, represents the conditions under which the solid and liquid are in equilibrium. B temperature pressure A C D solidliquid gas..

19. 19 Phase Diagrams  Usually, the melting point is only slightly affected by pressure. For this reason, the melting point curve, AB, is nearly vertical. B temperature pressure A C D solidliquid gas..

20. 20 Phase Diagrams  If a liquid is more dense than its solid, the curve leans slightly to the left, causing the melting point to decrease with pressure. B temperature pressure A C D solidliquid gas..

21. 21 Phase Diagrams  If a liquid is less dense than its solid, the curve leans slightly to the right, causing the melting point to increase with pressure. B temperature pressure A C D solidliquid gas..

22. 22 Phase Diagrams  Curve AC, which divides the liquid region from the gaseous region, represents the boiling points of the liquid for various pressures. B temperature pressure A C D solidliquid gas..

23. 23 Phase Diagrams  Curve AD, which divides the solid region from the gaseous region, represents the vapor pressures of the solid at various temperatures. B temperature pressure A C D solidliquid gas..

24. 24 Phase Diagrams  The curves intersect at A, the triple point, which is the temperature and pressure where three phases of a substance exist in equilibrium. B temperature pressure A C D solidliquid gas..

25. 25 Phase Diagrams  The curves intersect at A, the triple point, which is the temperature and pressure where three phases of a substance exist in equilibrium. B temperature pressure A C D solidliquid gas.. (see Figures 11.11 11.11 and 11.12)11.12)

26. Rhombic v. Monoclinic 26

27. 27 Phase Diagrams  The temperature above which the liquid state of a substance no longer exists regardless of pressure is called the critical temperature. B temperature pressure A C D solidliquid gas.. T crit

28. 28 Phase Diagrams  The vapor pressure at the critical temperature is called the critical pressure. Note that curve AC ends at the critical point, C. B temperature pressure A C D solidliquid gas.. T crit P crit (see Figure 11.13)

29. 29 Properties of Liquids; Surface Tension and Viscosity  The molecular structure of a substance defines the intermolecular forces holding it together. –Many physical properties of substances are attributed to their intermolecular forces. –These properties include vapor pressure and boiling point. –Two additional properties shown in Table 11.2 are surface tension and viscosity.

30. 30 Properties of Liquids; Surface Tension and Viscosity  Surface tension is the energy required to increase the surface area of a liquid by a unit amount. –A molecule within a liquid is pulled in all directions, whereas a molecule on the surface is only pulled to the interior. (See Figure 11.16).(See Figure 11.16). –As a result, there is a tendency for the surface area of the liquid to be minimized (See Figure 11.18 ).(See Figure 11.18 ).

31. 31 Properties of Liquids; Surface Tension and Viscosity  Surface tension is the energy required to increase the surface area of a liquid by a unit amount. –This explains why falling raindrops are nearly spherical, minimizing surface area. –In comparisons of substances, as intermolecular forces increase, the apparent surface tension also increases. –intermolecular forces surface tension

32. 32 Intermolecular Forces; Explaining Liquid Properties  Viscosity is the resistance to flow exhibited by all liquids and gases. –Viscosity can be illustrated by measuring the time required for a steel ball to fall through a column of the liquid. (see Figure 11.20)(see Figure 11.20) –Even without such measurements, you know that syrup has a greater viscosity than water. –In comparisons of substances, as intermolecular forces increase, viscosity usually increases. –intermolecular forces viscosity

33. 33 Intermolecular Forces; Explaining Liquid Properties  Many of the physical properties of liquids (and certain solids) can be explained in terms of intermolecular forces, the forces of attraction between molecules. –Three types of forces are known to exist between neutral molecules. 1.Dipole-dipole forces 2.London (or dispersion) forces 3.Hydrogen bonding

34. 34 Intermolecular Forces; Explaining Liquid Properties  The term van der Waals forces is a general term including dipole-dipole and London forces. –Van der Waals forces are the weak attractive forces in a large number of substances. –Hydrogen bonding occurs in substances containing hydrogen atoms bonded to certain very electronegative atoms. –Van der Waals forces0.1 to 10 kJ/mol –Hyderogen bonding10 to 40 kJ/mol

35. 35 Dipole-Dipole Forces  Polar molecules can attract one another through dipole-dipole forces. –The dipole-dipole force is an attractive intermolecular force resulting from the tendency of polar molecules to align themselves positive end to negative end. H Cl   H  

36. 36 London Forces  London forces are the weak attractive forces resulting from instantaneous dipoles that occur due to the distortion of the electron cloud surrounding a molecule. –London forces increase with molecular weight. The larger a molecule, the more easily it can be distorted to give an instantaneous dipole. –All covalent molecules exhibit some London force.

37. 37

38. 38 Van der Waals Forces and the Properties of Liquids  In summary, intermolecular forces play a large role in many of the physical properties of liquids and gases. These include: –vapor pressure –boiling point –surface tension –viscosity

39. 39 Van der Waals Forces and the Properties of Liquids  The vapor pressure of a liquid depends on intermolecular forces. When the intermolecular forces in a liquid are strong, you expect the vapor pressure to be low. –As intermolecular forces increase, vapor pressures decrease.

40. 40 Van der Waals Forces and the Properties of Liquids  The normal boiling point is related to vapor pressure and is lowest for liquids with the weakest intermolecular forces. –When intermolecular forces are weak, little energy is required to overcome them. –Consequently, boiling points are low for such compounds.

41. 41 Van der Waals Forces and the Properties of Liquids  Surface tension increases with increasing intermolecular forces. –Surface tension is the energy needed to reduce the surface area of a liquid. –To increase surface area, it is necessary to pull molecules apart against the intermolecular forces of attraction.

42. 42 Van der Waals Forces and the Properties of Liquids  Viscosity increases with increasing intermolecular forces because increasing these forces increases the resistance to flow. –Other factors, such as the possibility of molecules tangling together, affect viscosity. –Liquids with long molecules that tangle together are expected to have high viscosities.

43. 43 Hydrogen Bonding  Hydrogen bonding is a force that exists between a hydrogen atom covalently bonded to a very electronegative atom, X, and a lone pair of electrons on a very electronegative atom, Y. –To exhibit hydrogen bonding, one of the following three structures must be present. H NOHFH ::: –Only N, O, and F are electronegative enough to leave the hydrogen nucleus exposed.

44. 44 Hydrogen Bonding  Molecules exhibiting hydrogen bonding have abnormally high boiling points compared to molecules with similar Van der Waals forces. –For example, water has the highest boiling point of the Group VI hydrides. (see Figure 11.24A)(see Figure 11.24A) –Similar trends are seen in the Group V and VII hydrides. (see Figure 11.24B)(see Figure 11.24B)

45. 45 Hydrogen Bonding  A hydrogen atom bonded to an electronegative atom appears to be special. –The electrons in the O-H bond are drawn to the O atom, leaving the dense positive charge of the hydrogen nucleus exposed. –It’s the strong attraction of this exposed nucleus for the lone pair on an adjacent molecule that accounts for the strong attraction. –A similar mechanism explains the attractions in HF and NH 3.

46. 46 Hydrogen Bonding H H O : : H H O : : H H O : : H H O : :

47. 47 Solid State  A solid is a nearly incompressible state of matter with a well-defined shape. The units making up the solid are in close contact and in fixed positions. –Solids are characterized by the type of force holding the structural units together. –In some cases, these forces are intermolecular, but in others they are chemical bonds (metallic, ionic, or covalent).

48. 48 Solid State  From this point of view, there are four types of solids. –Molecular (Van der Waals forces) –Metallic (Metallic bond) –Ionic (Ionic bond) –Covalent (Covalent bond)

49. 49 Types of Solids  A molecular solid is a solid that consists of atoms or molecules held together by intermolecular forces. –Many solids are of this type. –Examples include solid neon, solid water (ice), and solid carbon dioxide (dry ice).

50. 50 Types of Solids  A metallic solid is a solid that consists of positive cores of atoms held together by a surrounding “sea” of electrons (metallic bonding). –In this kind of bonding, positively charged atomic cores are surrounded by delocalized electrons. –Examples include iron, copper, and silver.

51. 51 Types of Solids  An ionic solid is a solid that consists of cations and anions held together by electrical attraction of opposite charges (ionic bond). –Examples include cesium chloride, sodium chloride, and zinc sulfide (but ZnS has considerable covalent character).

52. 52 Types of Solids  A covalent network solid is a solid that consists of atoms held together in large networks or chains by covalent bonds. –Examples include carbon, in its forms as diamond or graphite (see Figure 11.27), asbestos, and silicon carbide.(see Figure 11.27) –Table 11.5 summarizes these four types of solids.

53. 53 Physical Properties  Many physical properties of a solid can be attributed to its structure. –For a solid to melt, the forces holding the structural units together must be overcome. –For a molecular solid, these are weak intermolecular attractions. –Thus, molecular solids tend to have low melting points (below 300 o C). Melting Point and Structure

54. 54 Physical Properties  Many physical properties of a solid can be attributed to its structure. –For ionic solids and covalent network solids to melt, chemical bonds must be broken. –For that reason, their melting points are relatively high. –See Table 11.1.See Table 11.1. Melting Point and Structure

55. 55 Physical Properties  Many physical properties of a solid can be attributed to its structure. –Note that for ionic solids, melting points increase with the strength of the ionic bond. –Ionic bonds are stronger when: 1.The magnitude of charge is high. 2.The ions are small (higher charge density). Melting Point and Structure

56. 56 Physical Properties  Many physical properties of a solid can be attributed to its structure. –Metals often have high melting points, but there is considerable variability. –Melting points are low for Groups IA and IIA but increase as you move into the transition metals. –The elements in the middle of the transition metals have the highest melting points. Melting Point and Structure

57. 57 Physical Properties  Many physical properties of a solid can be attributed to its structure. –Hardness depends on how easily structural units can be moved relative to one another. –Molecular solids with weak intermolecular attractions are rather soft compared with ionic compounds, where forces are much stronger. Hardness and Structure

58. 58 Physical Properties  Many physical properties of a solid can be attributed to its structure. –Covalent network solids are quite hard because of the rigidity of the covalent network structure. –Diamond and silicon carbide (SiC), three- dimensional covalent network solids, are among the hardest substances known. Hardness and Structure

59. 59 Physical Properties  Many physical properties of a solid can be attributed to its structure. –Molecular and ionic crystals are generally brittle because they fracture easily along crystal planes. –Metallic solids, by contrast, are malleable. Hardness and Structure

60. 60 Physical Properties  Many physical properties of a solid can be attributed to its structure. –Molecular and ionic solids are generally considered nonconductors. –Ionic compounds conduct in their molten state, as ions are then free to move. –Metals are all considered conductors. Electrical Conductivity and Structure

61. 61 Physical Properties  Many physical properties of a solid can be attributed to its structure. –Of the covalent network solids, only graphite conducts electricity. –This is due to the delocalization of the resonant  electrons in graphite’s sp 2 hybridization. Electrical Conductivity and Structure

62. 62 Crystalline Solids; Crystal Lattices and Unit Cells  Solids can be crystalline or amorphous. –A crystalline solid is composed of one or more crystals; each crystal has a well-defined, ordered structure in three dimensions. Examples include sodium chloride and sucrose. –An amorphous solid has a disordered structure. It lacks the well-defined arrangement of basic units found in a crystal. Glass is an amorphous solid.

63. 63 Shapes of Crystals  Straw Activity

64. 64 Crystal Lattices  A crystal lattice is the geometric arrangement of lattice points in a crystal. –A unit cell is the smallest boxlike unit from which you can construct a crystal by stacking the units in three dimensions (see Figure 11.29).(see Figure 11.29). –There are seven basic shapes possible for unit cells, which give rise to seven crystal systems used to classify crystals (see Figure 11.31 and Table 11.6).(see Figure 11.31 Table 11.6).

65. 65 Crystal Lattices  A crystal lattice is the geometric arrangement of lattice points in a crystal. –These seven systems can have more than one possible crystal lattice. –A “primitive” lattice has lattice points only at the corners of each cell.

66. 66 Crystal Lattices  A crystal lattice is the geometric arrangement of lattice points in a crystal. –Other lattices in the same crystal may have lattice points on the “faces” of the unit cell. –Following is a description of the cubic crystal system.

67. 67 Cubic Unit Cells A simple cubic unit cell is a cubic cell in which the lattice points are situated only at the corners (see Figure 11.30).(see Figure 11.30) –A body-centered cubic unit cell is one in which there is a lattice point in the center of the cell as well as at the corners. –A face-centered cubic unit cell is one in which there are lattice points at the center of each face of the cell as well as at the corners, (see Figures 11.32 and 11.33).(see Figures 11.3211.33).

68. 68 Crystal Defects  There are principally two kinds of defects that occur in crystalline substances. –Chemical impurities, such as in rubies, where the crystal is mainly aluminum oxide with an occasional Al 3+ ion replaced with Cr 3+, which gives a red color. –Defects in the formation of the lattice. Crystal planes may be misaligned, or sites in the crystal lattice may remain vacant.

69. Crystal Defects  Point Defects Vacancies / Schottky defects Interstitial Antisite Topological Impurity / Substitution  Line Defects  Planar Defects 69

70. Point Defects  Vacancies / Schottky defects

71. Point Defects Interstitial

72. Point Defects  Antisite Na Cl Na Cl Na Cl Na Cl Cl Cl Na Cl Na Cl Na Cl Na Cl  Atom belongs in solid, but not in that place.

73. Point Defects Topological chemical bonding is topologically different from the surroundings Ex: 6 carbons in ring → rings of 5 and 7, shile total number of atoms remains constant.

74. Point Defects  Impurity / Substitution  The technique of purposefully substituting, or doping, a solid is used to produce: microchips, lasers, and in the amplification of light signals through fiberopitic cable, to name a few.

75. Line Defects  Dislocations are linear defects around which some of the atoms of the crystal lattice are misaligned.  Edge dislocations are caused by the termination of a plane of atoms in the middle of a crystal

76. Line Defects  a line defect  a slip of the part of crystal over an atomic plane relative to another part  A screw dislocation results when atomic planes form a spiral ramp winding around the line of the dislocation

77. Planar Defects  Grain boundaries  Usually result when one crystal grows into another

78. Planar Defects  Antiphase  Each side of the boundary has an opposite phase: For example if the ordering is usually ABABABAB, an anti phase boundary takes the form of ABABBABA.

79. Planar Defects  Stacking Fault  a one or two layer interruption in the stacking sequence; when stacking one of the layers on top of another, the atoms are not directly on top of one another

80. 80 Calculations Involving Unit Cell Dimensions  X-ray diffraction is a method for determining the structure and dimensions of a unit cell in a crystalline compound. –Once the dimensions and structure are known, the volume and mass of a single atom in the crystal can be calculated. –The determination of the mass of a single atom gave us one of the first accurate determinations of Avogadro’s number.

81. 81 Determination of Crystal Lattice by X-Ray Diffraction  When x-rays are reflected from the planes of a crystal, they show a diffraction pattern that can be recorded on photographic film (see Figure 11.47).(see Figure 11.47). –Analysis of these diffraction patterns allows the determination of the positions of the atoms in the unit cell of the solid.

82. 82 Sample Calculation  Silver crystals are face-centered cubics, with a cell edges of 4.086 angstroms. What is the distance between center of the two closest Ag atoms? What is the atomic radius of silver in this crystal? How many nearest neighbors does each atom have?

83. 83 What is the distance between center of the two closest Ag atoms?  Atoms are assumed to touch along face diagonals  The hypotenuse is equal to twice the center to center distance.  c = 2(4.086)  a = b = 4.086 angstroms  (4.086) 2 + (4.086) 2 =c 2  c 2 = √33.3908  c = 5.7785 angstroms  d = c/2  d = 2.889 angstroms 4.086 angstroms c a b

84. 84 What is the atomic radius of silver in this crystal?  The hypotenuse is of the unit cell face is four times the radius of the atom.  r = d = 2.889 angstroms 2 2  d = 1.445 angstroms 4.086 angstroms ca b

85. 85 How many nearest neighbors does each atom have?  The central atom has  4 nearest neighbors in the x-y plane  4 nearest neighbors in the x-z plane, and  4 nearest neighbors in the y-z plane  a total of 12 nearest neighbors.

86. 86  Nickel has a face-centered unit cell with an edge length of 352.4 pm. The density of nickel is 8.91 g/cm 3. From the atomic weight, calculate Avagadro’s number. Sample Calculation

87. 87 What is the mass of a single nickel atom?  Change pm to cm  pm to cm  352.4pm 1 x 10 -12 m 100 cm 1 pm 1m  = 3.524 x 10 -8 cm c a b 352.4 pm

88. What is the mass of a single nickel atom?  Change pm to cm  Use the density to calculate mass  3.524 x 10 -8 cm  D = m v  8.91 g = m cm 3 (3.524 x 10 -8 cm ) 3  m = 3.90 x 10 -22 g/ unit cell 88

89. 89  Determine atoms per unit cell to calculate g / atom  8 vertices 1/8 atom = 1 atom vertex  6 faces 1/2 atom = 3 atoms face  4 total atoms  3.90 x 10 -22 g/ unit cell = 4 atoms  9.75 x 10 -23 g / atom

90. 90  Use atomic weight to calculate Avagadro’s number  Atomic wt of Nickel is 59 g/ mol  59 g / mol = 9.75 x 10 -23 g / atom  6.05 x 10 23 atoms / mol

91. 91 Operational Skills  Calculating the heat required for a phase change of a given mass of substance.  Calculating vapor pressures and heats of vaporization.  Relating the conditions for the liquification of a gas to the critical temperature.  Identifying intermolecular forces.  Determining relative vapor pressure on the basis of intermolecular attraction.  Identifying types of solids.

92. 92 Operational Skills  Determining the relative melting points based on types of solids.  Determining the number of atoms per unit cell.  Calculating atomic mass from unit-cell dimension and density.  Calculating unit-cell dimensions from unit-cell type and density.

93. 93 Figure 11.2: Representation of the States of Matter Return to Slide 2

94. 94 Figure 11.4: Measurement of the vapor pressure of water. Return to Slide 4

95. 95 Figure 11.7: Variation of vapor pressure with temperature. Return to Slide 5

96. 96 Figure 11.9: Heating curve for water. Return to Slide 9

97. 97 Figure 11.11: Phase diagram for water (not to scale). Return to Slide 24

98. 98 Figure 11.12: Phase diagrams for carbon dioxide and sulfur (not to scale). Return to Slide 24

99. 99 Figure 11.13: Observing the critical phenomenon. Return to Slide 26

100. 100 Figure 11.16: Explaining Surface Tension Return to Slide 28

101. 101 Figure 11.18: Demonstration of Surface Tension of Water Return to Slide 28

102. 102 Figure 11.20: Comparison of the viscosities of two liquids. Photo courtesy of James Scherer. Return to Slide 30

103. 103 Figure 11.24: Boiling point versus molecular weight for hydrides. Return to Slide 41

104. 104 Figure 11.24: Boiling point versus molecular weight for hydrides. Return to Slide 41

105. 105 Figure 11.27: Structures of diamond and graphite. Return to Slide 49

106. 106 Return to Slide 51

107. 107 Figure 11.29: A two-dimensional pattern. Return to Slide 60

108. 108 Figure 11.31: Unit-cell shapes of the different crystal systems. Return to Slide 60

109. 109 Return to Slide 60

110. 110 Figure 11.30: Crystal structure and crystal lattice of copper. Return to Slide 63

111. 111 Figure 11.32: Cubic unit cells. Return to Slide 63

112. 112 Figure 11.33: Space-filling representation of cubic unit cells. Return to Slide 63

113. 113 Figure 11.47: A crystal diffraction pattern. From Preston, Proceedings of the Royal Society, A, Volume 172, plate 4, figure 5A Return to Slide 66