Presentation is loading. Please wait.

Presentation is loading. Please wait.

Liquids, Solids, and Phase Changes

Similar presentations


Presentation on theme: "Liquids, Solids, and Phase Changes"— Presentation transcript:

1 Liquids, Solids, and Phase Changes
Chapter 10

2 Gases, Liquids and Solids
Gases have little or no interactions. Liquids and solids have significant interactions.

3 Gases, Liquids and Solids
Liquids and solids have well-defined volume. Liquid molecules “flow,” while solids are held “rigid.” Gases, Liquids and Solids

4 Intermolecular Forces
Intermolecular forces are attractive forces between molecules. Intramolecular forces hold atoms together in a molecule. Intermolecular vs Intramolecular 41 kJ to vaporize 1 mole of water (inter) 930 kJ to break all O-H bonds in 1 mole of water (intra) “Measure” of intermolecular force boiling point melting point Generally, intermolecular forces are much weaker than intramolecular forces.

5 Intermolecular Forces
Dipole Moments 01 Polar covalent bonds form between atoms of different electronegativity. This is described as a bond dipole.

6 Dipole Moments 02 Dipole Moment (µ): The measure of net molecular polarity or charge separation. µ = Q ´ r r = distance between charges + = Q, – = –Q

7 Dipole Moments: µ = Q ´ r Q = Charge of electron: Q = 1.60 x 10­-19 C,
r = bond length, m μ , dipole moments are expressed in debyes (D) where 1 D = x 10–30 C·m What is the dipole moment if one proton separated from one electron by a distance of 100 PM μ = Q x r = (1.60 x 10­-19 C) (100 x 10­-12 m) (1D/3.336 x C . m) = 4.80 D

8 Dipole Moments 03 Polarity can be illustrated with an electrostatic potential map. These show electron-rich groups as red and electron-poor groups as blue-green.

9 Dipole Moments 04

10 % ionic character for HCl = (1.03/6.09) x 100% = 16.9%
The dipole moment of HCl is 1.03 D, and the distance between the atoms is 127 pm. Calculate the percent ionic character of the HCl bond μ = Q x r = (1.60 x 10­-19 C)(127 x 10­-12 m) (1D/3.336 x C . m)= 6.09 D % ionic character for HCl = (1.03/6.09) x 100% = 16.9%

11 Intermolecular Forces 01
Attractive forces between molecules and ions. Several types of forces: Dipole–dipole Instantaneous induced dipole (dispersion forces) Ion–dipole Hydrogen “bonds.”

12 Intermolecular Forces
Dipole-Dipole Forces Attractive forces between polar molecules Orientation of Polar Molecules in a Solid

13 Intermolecular Forces
Ion-Dipole Forces Attractive forces between an ion and a polar molecule Ion-Dipole Interaction

14 Intermolecular Forces 04
London Dispersion Forces: Attraction is due to instantaneous, temporary dipoles formed due to electron motions.

15 Intermolecular Forces
Dispersion Forces Continued Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted. Polarizability increases with: greater number of electrons more diffuse electron cloud Dispersion forces usually increase with molar mass. XXXXX XXXX 2

16 What type(s) of intermolecular forces exist between each of the following molecules?
HBr HBr is a polar molecule: dipole-dipole forces. There are also dispersion forces between HBr molecules. CH4 CH4 is nonpolar: dispersion forces. S O SO2 SO2 is a polar molecule: dipole-dipole forces. There are also dispersion forces between SO2 molecules.

17 Hydrogen Bond Hydrogen Bond: Molecules containing N–H, O–H, or F–H groups, and an electronegative O, N, or F.

18 Why is the hydrogen bond considered a “special” dipole-dipole interaction?
Decreasing molar mass Decreasing boiling point

19 Which of the following molecules can hydrogen bond with itself?
1, 2 2, 3 3, 4 1, 2, 3 1, 2, 3, 4

20 Which of the following molecules can hydrogen bond with itself?
1, 2 2, 3 3, 4 1, 2, 3 1, 2, 3, 4

21 Of the following substances, predict which has the lowest boiling point based on London dispersion forces. He Ne Ar Kr Xe

22 Correct Answer: He Ne Ar Kr Xe
More massive species have more polarizability and stronger London dispersion forces; consequently, amongst the noble gases He has the lowest boiling point.

23 Of the following substances, predict which has the highest boiling point based upon intermolecular forces? CH4 H2O H2S SiH4 H2Se NH ……. O=C

24 Correct Answer: CH4 H2O H2S SiH4 H2Se
Of these, only H2O has any hydrogen bonding. Hydrogen bonding substantially increases the intermolecular forces, and hence the boiling point.

25 With nerves as steady as a chemical bond.
© 2003 John Wiley and Sons Publishers Courtesy Ken Karp With nerves as steady as a chemical bond.

26 Figure 13.1: “Floating” a tack on water.
© 2003 John Wiley and Sons Publishers Figure 13.1: “Floating” a tack on water.

27 Figure 13.2: Place a tack on the surface of a glass of water.
© 2003 John Wiley and Sons Publishers Figure 13.2: Place a tack on the surface of a glass of water. The tack remains on the water’s surface as you gently push a toothpick into the water. The tack drops as soon as someone else pushes the toothpick through the water’s surface, no matter how gently. The secret lies in the chemistry of liquids, surfaces, and detergents. Courtesy Ken Karp Figure 13.2: Place a tack on the surface of a glass of water.

28 Surface Tension Water strider walks on a pond without penetrating the surface

29 Surface Tension

30 Strong intermolecular forces
Properties of Liquids Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area. Strong intermolecular forces High surface tension

31 Viscosity Viscosity is the measure of a liquid’s resistance to flow
and is related to the ease with which molecules move around, and thus to the intermolecular forces.

32 Surface Tension Tensiometer

33 Intermolecular Forces 09
Viscosity is the measure of a liquid’s resistance to flow and is related to the ease with which molecules move around, and thus to the intermolecular forces. Another unit of viscosity is kg/m.s

34 Phase Changes 01

35 Phase Changes Phase Change (State Change): A change in physical form but not the chemical identity of a substance. Fusion (melting): Vaporization : Sublimation: solid to liquid liquid to gas solid to gas Freezing: Condensation: Deposition: liquid to solid gas to liquid gas to solid

36 Phase Changes 02 (∆Hvap ) kJ/mol (∆Hvap ) = 40.67 KJ/mol (∆Hfus)
(∆Hfus) = 6.01 KJ/mol

37 Phase Changes Vapor Pressure: The pressure exerted by gaseous molecules above a liquid.

38 Clausius-Clapeyron Equation
Molar heat of vaporization (DHvap) is the energy required to vaporize 1 mole of a liquid. ln P = - DHvap RT + C Clausius-Clapeyron Equation P = (equilibrium) vapor pressure T = temperature (K) R = gas constant (8.314 J/K•mol)

39 Vapor Pressure The boiling point of a liquid is the temperature at which its vapor pressure equals atmospheric pressure. The normal boiling point is the temperature at which its vapor pressure is 760 torr.

40 Evaporation, Vapor Pressure, and Boiling Point
+ C 1 T - DHvap R ln Pvap = y = m x + b There is also a two-point form of the Clausius-Clapeyron equation.

41 Evaporation, Vapor Pressure, and Boiling Point

42 Vapor Pressure

43 By taking measurements at two temps, we get:
ln P = - DHvap RT + C Clausius-Clapeyron Equation By taking measurements at two temps, we get:

44 The normal boiling point of benzene is 80. 1 °C, and ΔHvap = 30
The normal boiling point of benzene is 80.1 °C, and ΔHvap = 30.8 kJ/mol, what is boiling point of benzene on top of Mount Everest, where P = 260 mm Hg P1 = 760 mm Hg; P2 = 260 mm Hg; t1 = 80.1oC, T2 = ? ΔHvap = 30.8 kJ/mol , R = J / K . mol Solve for T2 (the boiling point for benzene at 260 mm Hg). T2 = 320 K ; t = 47oC (boiling point is lower at lower pressure)

45 Which statement is true?
200 400 600 800 Vapor Pressure (mm Hg) 25 50 75 100 Temperature ( C) Boiling point ~120°C Boiling point ~95°C Boiling point ~75°C Melting point ~95°C Melting point ~75°C

46 Which statement is true?
200 400 600 800 Vapor Pressure (mm Hg) 25 50 75 100 Temperature ( C) Boiling point ~120°C Boiling point ~95°C Boiling point ~75°C Melting point ~95°C Melting point ~75°C

47 Solids An amorphous solid does not possess a well-defined arrangement and long-range molecular order they do not have a fixed sharp melting point. . A glass is an optically transparent fusion product of inorganic materials that has cooled to a rigid state without crystallizing Structure of a crystalline solid is based on the unit cell, a basic repeating structural unit. Crystalline quartz (SiO2) Non-crystalline quartz glass

48 Kinds of Solids Amorphous Solids: Particles are randomly arranged and have no ordered long-range structure. Example - rubber. Crystalline Solids: Particles have an ordered arrangement extending over a long range. ionic solids molecular solids covalent network solids metallic solids

49 Kinds of Solids Ionic Solids: Particles are ions ordered in a regular three-dimensional arrangement and held together by ionic bonds. Example - sodium chloride.

50 Unit cells in 3 dimensions
lattice point At lattice points: Atoms Molecules Ions Unit cells in 3 dimensions Unit Cell

51 Simple Cubic Packing Body-Centered Cubic Packing

52 Crystal Structure Simple Cube Body-Centered Cube:

53 Crystal Structure 04 Face-Centered Cube:

54 Crystalline Solids We can determine the empirical formula of an ionic solid by determining how many ions of each element fall within the unit cell.

55 Hexagonal Closest Pack
A-B-A-B- Space used 74%

56 Cubic Closes Pack Space used 74% A-B-C-A-B-C

57

58 Unit Cells and the Packing of Spheres in Crystalline Solids
Unit Cell: A small repeating unit that makes up a crystal.

59 Types of Crystals What are the empirical formulas for these compounds?
(a) Orange:chlorine; Gray:cesium (b) Blue:sulfur; Gray: zinc (c) Green:fluorine, Gray: calcium (c) (b) (a) CsCl ZnS CaF2

60 Types of Crystal 04 Carbon:

61 Carbon Allotropes

62 Types of Crystals lattice points: Atoms Molecules Ions
Covalent Crystals Lattice points occupied by atoms Held together by covalent bonds Hard, high melting point carbon atoms diamond graphite

63 Types of Crystals Molecular Crystals
Lattice points occupied by molecules Held together by intermolecular forces Soft, low melting point 11.6

64 Cross Section of a Metallic Crystal
Types of Crystals Metallic Crystals Lattice points occupied by metal atoms Held together by metallic bonds Soft to hard, low to high melting point Good conductors of heat and electricity Cross Section of a Metallic Crystal nucleus & inner shell e- mobile “sea” of e-

65 Ionic Crystals Covalent Crystals Molecular Crystals Metallic Crystals

66 Simple cubic cell Body-centered cubic Face-centered cubic

67 When silver crystallizes, it forms face-centered cubic cells
When silver crystallizes, it forms face-centered cubic cells. The unit cell edge length is 409 pm. Calculate the density of silver. d = m V V = a3 = (409 pm)3 = 6.83 x cm3 4 atoms/unit cell in a face-centered cubic cell 107.9 g mole Ag x 1 mole Ag 6.022 x 1023 atoms x m = 4 Ag atoms = 7.17 x g d = m V 7.17 x g 6.83 x cm3 = = 10.5 g/cm3

68 X-Ray Crystallography
Diffraction is the scattering of radiation by an object containing regularly spaced lines, with a spacing that is equivalent to the wavelength of radiation. Diffraction is due to interference between two waves passing through the same region of space at the same time.

69

70 Extra distance = BC + CD = 2d sinq = nl (Bragg Equation)

71 X rays of wavelength nm are diffracted from a crystal at an angle of Assuming that n = 1, what is the distance (in pm) between layers in the crystal? nl = 2d sin q n = 1 q = l = nm = 154 pm nl 2sinq = 1 x 154 pm 2 x sin14.17 d = = 315 pm

72 Titanium metal has a density of 4
Titanium metal has a density of 4.54 g/cm3, and an atomic radius of pm. What is the structure of cubic unit cell? mass of one Ti atom = x 10­-23 g/atom (problem 10.78) r = pm = x 10­-12 m r = x 10­-12 m = x 10­-8 cm Calculate the volume and then the density for Ti assuming it is Simple (primitive) cubic, body-centered cubic, and face-centered cubic. Compare the calculated density with the actual density to identify the unit cell. For primitive cubic: a = 2r; volume = a3 = [2(1.448 x 10­-8 cm)]3 = x 10­-23 cm3 density = m/v = g/cm3

73 For face-centered cubic:
a = √8. r; volume = a3 = [2√2 . (1.448 x 10-­8 cm)]3 = x 10­-23 cm3 density = = g/cm3 For body-centered cubic: a = 4r/√3; volume = a3 = x 10­23 cm3 density = = g/cm3 The calculated density for a face-centered cube (4.630 g/cm3) is closest to the actual density of 4.54 g/cm3. Ti crystallizes in the face-centered cubic unit cell.

74 Phase Diagrams 01

75 This phase diagram of water is just meant to introduce a phase diagram before defining all the pieces. Water

76 Phase Diagrams Normal BP: Occurs at 1 atm.
Critical Point: A combination of temperature and pressure beyond which a gas cannot be liquefied. Critical Temperature: The temperature beyond which a gas cannot be liquefied regardless of the pressure. Critical Pressure: The pressure beyond which a liquid cannot be vaporized regardless of the temperature. Supercritical Fluid: A state of matter beyond the critical point that is neither liquid nor gas. Triple Point: A point at which three phases coexist in equilibrium.

77 Supercritical CO2 Caffeine extraction from green coffee with supercritical CO2 Application of Supercritical CO2 in dry cleaning

78 Phase Diagrams Carbon Dioxide
Supercritical fluid extraction of caffeine from coffee beans is a fun example. The positive slope of the solid-liquid boundary line means solid carbon dioxide is more dense than liquid carbon dioxide. An isotherm can be drawn to show how the liquid changes to a solid when the pressure is increased. Also, notice that the triple point for solid-liquid-gas occurs at 5.11 atm. Thus, you cannot have liquid carbon dioxide below that pressure.

79 Phase Diagrams 05 Approximately, what is the normal boiling point and what is the normal melting point of the substance? What is the physical state when: T = 150 K, P = 0.5 atm T = 325 K, P = 0.9 atm T = 450 K, P = 265 atm

80 The End

81 Coordination Numbers

82


Download ppt "Liquids, Solids, and Phase Changes"

Similar presentations


Ads by Google