1. Chem 106, Prof. T. L. Heise 1 CHE 106: General Chemistry  CHAPTER ELEVEN Copyright © Tyna L. Heise 2001 All Rights Reserved

2. Chem 106, Prof. T. L. Heise 2  All elements can exist in three different phases of matter –Chemical properties remain the same –Physical properties are different –Many of the substances to be considered in solid and liquid phase are molecular »Virtually all liquids are molecular »Physical properties examined are due to the forces of attractions NOT within a molecule BUT between molecules Molecular Comparison of Solids and Liquids Chapt. 11.1

3. Chem 106, Prof. T. L. Heise 3 Solids, Liquids, and Gases Chapt. 11.1

4. Chem 106, Prof. T. L. Heise 4 Solids, Liquids, and Gases Chapt. 11.1 Substances can be changed from one phase to another by -heating and cooling to change the average kinetic energy -adding and decreasing pressure changes the distance between molecules which leads to an alteration in intermolecular forces - increasing pressure pushes molecules closer together, increasing molecular attraction - decreasing pressure allows molecules to spread farther apart, decreasing molecular attraction

5. Chem 106, Prof. T. L. Heise 5 Intermolecular Forces The strengths of intermolecular forces of different substances vary greatly, however they are much weaker than ionic or covalent bonds: You are not actually breaking apart a molecule, just separating molecules from each other as you move from solid to liquid and the to a gas phase. Physical Properties reflect the IM forces - boiling points - melting points Chapt. 11.2

6. Chem 106, Prof. T. L. Heise 6 Intermolecular Forces Three types of IM forces: dipole - dipole forces ion - dipole forces London dispersion forces hydrogen bonding Chapt. 11.2

7. Chem 106, Prof. T. L. Heise 7 Ion - Dipole forces Important in aqueous solutions - important between an ion and the partial charge on the end of a polar compound negative ion attracts to + positive ion attracts to - Chapt. 11.2

8. Chem 106, Prof. T. L. Heise 8 Dipole - Dipole forces Important in aqueous solutions - important between two dipoles negative end attracts to + positive end attracts to - **for molecules of equal size, the strengths of dipole forces increases with electronegativity Chapt. 11.2

9. Chem 106, Prof. T. L. Heise 9 London Dispersion forces Important between two nonpolar molecules - important to look at instantaneous electron arrangements - at an instant, the position of an electron can force a temporary dipole moment. - the temporary dipole moment can induce its’ neighbor to experience a temporary dipole moment, causing an attraction - this force is only significant when the molecules are very close to each other Chapt. 11.2

10. Chem 106, Prof. T. L. Heise 10 London Dispersion forces Chapt. 11.2

11. Chem 106, Prof. T. L. Heise 11 London Dispersion forces The amount of polarizability increases with size of molecule. - boiling point increases with molecular weight The more contact a molecule has with another also increases the polarizability - n-pentane has higher boiling point than neopentane due to larger available surface area Chapt. 11.2

12. Chem 106, Prof. T. L. Heise 12 Hydrogen Bonding A special intermolecular force which exists between the H of one molecule and the O, F, or N of a neighboring polar molecule Chapt. 11.2

13. Chem 106, Prof. T. L. Heise 13 Hydrogen Bonding Chapt. 11.2

14. Chem 106, Prof. T. L. Heise 14 Intermolecular Forces Chapt. 11.2 Summary: Intermolecular Forces can be determined using1. Composition 2. Structure Dispersion forces are in ALL molecules Strengths increase with increased weight, and depend on shape Dipole-dipole forces add to dispersion forces and are found ONLY in polar molecules Hydrogen bonds are found between the H of one molecule and the F, O, or N of another None of these are as strong as ionic or covalent bonds, but of these, hydrogen bonding is the strongest

15. Chem 106, Prof. T. L. Heise 15 Properties of Liquids Chapt. 11.3 Two important properties of Liquids: Viscosity - the resistance of a liquid to flow low viscosity flows easily high viscosity flows slowly Surface Tension - the energy required to increase the surface area of a liquid by a unit amount. Molecules at surface all are attracted inwards instead of in all directions, this packs surface molecules closer together The measurement of the inward forces that must be overcome is the surface tension

16. Chem 106, Prof. T. L. Heise 16 Properties of Liquids Chapt. 11.3 Viscosity

17. Chem 106, Prof. T. L. Heise 17 Properties of Liquids Chapt. 11.3 Surface Tension

18. Chem 106, Prof. T. L. Heise 18 Phase Changes Chapt. 11.4 Many important properties of liquids and solids relate to the ease in which they change from one state to another:

19. Chem 106, Prof. T. L. Heise 19 Phase Changes Chapt. 11.4 Each phase change is accompanied by a change in energy of the system. Energy must be added to overcome intermolecular forces and achieve a less ordered state. Energy must be released as intermolecular forces begin to form and a molecule achieves a more ordered state

20. Chem 106, Prof. T. L. Heise 20 Phase Changes Chapt. 11.4 Heating Curves: plot of the process of changing phase of substance Heat of Fusion: the energy needed to change a solid to a liquid Heat of Vaporization: the amount of energy needed to change a liquid to a gas

21. Chem 106, Prof. T. L. Heise 21 Phase Changes Chapt. 11.4 Special Notes: Supercooling: reducing the heat of the liquid so fast it’s temperature falls below its freezing point, but it remains a liquid. This occurs because the liquid never has a chance to form an ordered solid. Critical Temperature: Highest temperature at which a substance can exist as a liquid Critical Pressure: The pressure needed to bring about liquefaction at critical temp.

22. Chem 106, Prof. T. L. Heise 22 Vapor Pressure Chapt. 11.5 Molecules can escape from the surface of liquid into the gas phase by vaporization or evaporation.

23. Chem 106, Prof. T. L. Heise 23 Vapor Pressure Chapt. 11.5 Why? Molecules of liquids move at various speeds At any instant a molecule on the surface may possess enough energy to overcome its intermolecular forces and escape into the gas phase The movement of molecules between liquid and gas phases goes on continuously, eventually only so many molecules can escape in a closed container and a dynamic equilibrium will be attained

24. Chem 106, Prof. T. L. Heise 24 Vapor Pressure Chapt. 11.5 Open Containers As liquid evaporates, the gas molecules move away from surface of liquid, causing the possibility of recapture by the liquid to be very small Equilibrium never occurs, and the vapor continues to form until the liquid is gone Liquids that easily evaporate due to low intermolecular forces, creating a high vapor pressure in a closed container, are considered volatile

25. Chem 106, Prof. T. L. Heise 25 Vapor Pressure Chapt. 11.5 Boiling Point: the temperature at which the vapor pressure equals the external pressure acting on the surface of the liquid. Normal Boiling Point: the boiling point of a liquid at 1 atm

26. Chem 106, Prof. T. L. Heise 26 Phase Diagrams Chapt. 11.6

27. Chem 106, Prof. T. L. Heise 27 Structures of Solids Chapt. 11.7 Crystalline Solid: particles are in a well ordered arrangement - flat surfaces or surfaces at definite angles to each other - orderly stacking of particles causes them to have orderly shapes - examples - quartz and diamonds - melts at specific temperature

28. Chem 106, Prof. T. L. Heise 28 Structures of Solids Chapt. 11.7 Amorphous Solid: particles have no orderly structure - mixtures of molecules that do not stack well together. - composed of large complicated molecules - examples: rubber and glass - intermolecular forces vary in strength from one part of solid to another due to irregularities in solid, melting point varies

29. Chem 106, Prof. T. L. Heise 29 Structures of Solids Chapt. 11.7 Unit Cells: the repeating unit of a crystalline solid that builds the definite patterns - three dimensional array called a lattice - generally parallelpipeds, can be described using two terms, length and angle between edges

30. Chem 106, Prof. T. L. Heise 30 Structures of Solids Chapt. 11.7 Unit Cells

31. Chem 106, Prof. T. L. Heise 31 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.

32. Chem 106, Prof. T. L. Heise 32 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron. 1) D = m V

33. Chem 106, Prof. T. L. Heise 33 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron. 1) D = m V Mass = 55.845 amu

34. Chem 106, Prof. T. L. Heise 34 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron. 1) D = m V Mass = 55.845 amu 1 g 6.02 x 10 23 amu

35. Chem 106, Prof. T. L. Heise 35 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron. 1) D = m V Mass = 2(55.8) amu 1 g = 1.85 x 10 -22 g 6.02 x 10 23 amu

36. Chem 106, Prof. T. L. Heise 36 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron. 1) D = m V 3 Volume = (2.8664 angstroms) (10 -10 m) 1 angstroms

37. Chem 106, Prof. T. L. Heise 37 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron. 1) D = m V 3 Volume = (2.8664 angstroms) (10 -10 m) = 2.36 x 10 -29 1 angstroms

38. Chem 106, Prof. T. L. Heise 38 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron. 1) D = m V Volume = 2.36 x 10 -29 m 3 1 L 10 -3 m 3

39. Chem 106, Prof. T. L. Heise 39 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron. 1) D = m V Volume = 2.36 x 10 -29 m 3 1 L = 2.36 x 10 -26 L 10 -3 m 3

40. Chem 106, Prof. T. L. Heise 40 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 ≈ on each side. Calculate the density of this form of iron. 1) D = m = 1.85 x 10 -22 g V 2.36 x 10 -26 L

41. Chem 106, Prof. T. L. Heise 41 Structures of Solids Chapt. 11.7 Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 ≈ on each side. Calculate the density of this form of iron. 1) D = m = 1.85 x 10 -22 g = 7,838 g V 2.36 x 10 -26 L L

42. Chem 106, Prof. T. L. Heise 42 Structures of Solids Chapt. 11.7 Close packing of spheres: structures adopted by crystalline solids are those that bring particles in closest contact to maximize intermolecular attractions * Each sphere has 12 neighbors, known as the coordination number

43. Chem 106, Prof. T. L. Heise 43 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (a) How many aluminum atoms are in the unit cell

44. Chem 106, Prof. T. L. Heise 44 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (a) How many aluminum atoms are in the unit cell 1/2 atom at each of 6 faces = 3 atoms 1/8 atom at each of 8 corners = 1 atom 4 atoms total

45. Chem 106, Prof. T. L. Heise 45 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (b) What is the coordination number of each aluminum atom?

46. Chem 106, Prof. T. L. Heise 46 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (b) What is the coordination number of each aluminum atom? 12

47. Chem 106, Prof. T. L. Heise 47 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (c) If each aluminum atom has a radius of 1.43 angstroms, what is the length of each side?

48. Chem 106, Prof. T. L. Heise 48 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (c) If each aluminum atom has a radius of 1.43 angstroms, what is the length of each side? 4 radius on diagonal=4(1.43 angstroms)= 5.72 angstroms

49. Chem 106, Prof. T. L. Heise 49 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (c) If each aluminum atom has a radius of 1.43 A, what is the length of each side? 4 radius on diagonal = 4(1.43 A) = 5.72 A a 2 + b 2 = c 2 2a 2 = c 2 2a 2 = (5.72 A) 2 2a 2 = 32.72 A a 2 = 16.36 A a = 4.04 A

50. Chem 106, Prof. T. L. Heise 50 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal

51. Chem 106, Prof. T. L. Heise 51 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal 1) D = m V

52. Chem 106, Prof. T. L. Heise 52 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal 1) D = m V Mass = 26.98 amu

53. Chem 106, Prof. T. L. Heise 53 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal 1) D = m V Mass = 26.98 amu 1 g 6.02 x 10 23 amu

54. Chem 106, Prof. T. L. Heise 54 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal 1) D = m V Mass = 4(26.98) amu 1 g = 1.79 x 10 -22 g 6.02 x 10 23 amu

55. Chem 106, Prof. T. L. Heise 55 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal 1) D = m V 3 Volume = (4.04 A) (10 -10 m) 1 L 1 A 10 -3 m 3

56. Chem 106, Prof. T. L. Heise 56 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal 1) D = m V Volume = 6.59 x 10 -26 L

57. Chem 106, Prof. T. L. Heise 57 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal 1) D = m = 1.79 x 10 -22 g V 6.59 x 10 -26 L

58. Chem 106, Prof. T. L. Heise 58 Structures of Solids Chapt. 11.7 Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell) (d) Calculate the density of the of aluminum metal 1) D = m = 1.79 x 10 -22 g = 2,720 g/L V 6.59 x 10 -26 L

59. Chem 106, Prof. T. L. Heise 59 Bonding in Solids Chapt. 11.7 The physical properties of crystalline solids, such as melting point and hardness, depend both on the arrangements off particles and on the attractive forces between particles in solids. 4 Types : Molecular Covalent & Network Ionic Metallic

60. Chem 106, Prof. T. L. Heise 60 Bonding in Solids Chapt. 11.7 Molecular - made up of atoms or molecules - London dispersion forces, dipole-dipole forces, hydrogen bonding - soft, low to moderately high melting point, poor thermal and electrical conductivity Network Covalent - atoms connected in networks - covalent bonding - very hard, very high melting point, often poor thermal and electrical conductivity

61. Chem 106, Prof. T. L. Heise 61 Bonding in Solids Chapt. 11.7 Ionic - made up of positively and negatively charged ions - electrostatic attractions - hard, high melting point, poor thermal and electrical conductivity Metallic - atoms - metallic bonding - soft to very hard, low to very high melting point, excellent thermal and electrical conductivity, malleable and ductile

62. Chem 106, Prof. T. L. Heise 62  Three types of Intermolecular Forces  Viscosity  Surface Tension  Phase changes  Heats of Fusion and Vaporization  Vapor Pressure  Phase Diagrams, triple point  Crystalline solids  Amorphous solids  Unit Cells  Bonding in Solids Chapter Eleven; Review