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Chapter 12

Chapter 12. Intermolecular Forces:. Liquids, Solids, and Phase Changes. Intermolecular Forces: Liquids, Solids, and Phase Changes. 12.1 An Overview of Physical States and Phase Changes. 12.2 Quantitative Aspects of Phase Changes. 12.3 Types of Intermolecular Forces.

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Chapter 12

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  1. Chapter 12 Intermolecular Forces: Liquids, Solids, and Phase Changes

  2. Intermolecular Forces: Liquids, Solids, and Phase Changes 12.1 An Overview of Physical States and Phase Changes 12.2 Quantitative Aspects of Phase Changes 12.3 Types of Intermolecular Forces 12.4 Properties of the Liquid State 12.5 The Uniqueness of Water 12.6 The Solid State: Structure, Properties, and Bonding 12.7 Advanced Materials

  3. ATTRACTIVE FORCES electrostatic in nature Intramolecular forces bonding forces These forces exist within each molecule. They influence the chemical properties of the substance. Intermolecular forces nonbonding forces These forces exist between molecules. They influence the physical properties of the substance.

  4. exothermic endothermic Phase Changes sublimination vaporizing melting solid liquid gas condensing freezing

  5. Gas Conforms to shape and volume of container high high Liquid Conforms to shape of container; volume limited by surface very low moderate Solid Maintains its own shape and volume almost none almost none Table 12.1 A Macroscopic Comparison of Gases, Liquids, and Solids State Shape and Volume Compressibility Ability to Flow

  6. Figure 12.1 Heats of vaporization and fusion for several common substances.

  7. Phase changes and their enthalpy changes. Figure 12.2

  8. Figure 12.3 A cooling curve for the conversion of gaseous water to ice.

  9. Quantitative Aspects of Phase Changes Within a phase, a change in heat is accompanied by a change in temperature which is associated with a change in average Ek as the most probable speed of the molecules changes. q = (amount)(molar heat capacity)(T) During a phase change, a change in heat occurs at a constant temperature, which is associated with a change in Ep, as the average distance between molecules changes. q = (amount)(enthalpy of phase change)

  10. Liquid-gas equilibrium. Figure 12.4

  11. The effect of temperature on the distribution of molecular speed in a liquid. Figure 12.5

  12. Figure 12.6 Figure 12.7 A linear plot of vapor pressure- temperature relationship. Vapor pressure as a function of temperature and intermolecular forces.

  13. The Clausius-Clapeyron Equation

  14. 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

  15. test tube with ice iodine solid iodine vapor iodine solid Iodine subliming. Figure 12.8

  16. CO2 H2O Phase diagrams for CO2 and H2O. Figure 12.9

  17. van der Waal’s distance bond length covalent radius van der Waal’s radius Figure 12.10 Covalent and van der Waals radii.

  18. Figure 12.11 Periodic trends in covalent and van der Waals radii (in pm).

  19. liquid Polar molecules and dipole-dipole forces. Figure 12.12 solid

  20. .. .. .. O H F .. .. H N .. .. .. N O .. .. H F .. .. THE HYDROGEN BOND a dipole-dipole intermolecular force A hydrogen bond may occur when an H atom in a molecule, bound to small highly electronegative atom with lone pairs of electrons, is attracted to the lone pairs in another molecule. The elements which are so electronegative are N, O, and F. hydrogen bond donor hydrogen bond acceptor hydrogen bond acceptor hydrogen bond donor hydrogen bond acceptor hydrogen bond donor

  21. Dipole moment and boiling point. Figure 12.13

  22. PROBLEM: Which of the following substances exhibits H bonding? For those that do, draw two molecules of the substance with the H bonds between them. (b) (a) (c) PLAN: Find molecules in which H is bonded to N, O or F. Draw H bonds in the format -B: H-A-. (b) SAMPLE PROBLEM 12.2 Drawing Hydrogen Bonds Between Molecules of a Substance SOLUTION: (a) C2H6 has no H bonding sites. (c)

  23. Hydrogen bonding and boiling point. Figure 12.14

  24. Polarizability and Charged-Induced Dipole Forces distortion of an electron cloud • Polarizability increases down a group size increases and the larger electron clouds are further from the nucleus • Polarizability decreases left to right across a period increasing Zeff shrinks atomic size and holds the electrons more tightly • Cations are less polarizable than their parent atom because they are smaller. • Anions are more polarizable than their parent atom because they are larger.

  25. separated Cl2 molecules Dispersion forces among nonpolar molecules. Figure 12.15 instantaneous dipoles

  26. Figure 12.16 Molar mass and boiling point.

  27. Molecular shape and boiling point. Figure 12.17 fewer points for dispersion forces to act more points for dispersion forces to act

  28. PROBLEM: For each pair of substances, identify the dominant intermolecular forces in each substance, and select the substance with the higher boiling point. (d) Hexane (CH3CH2CH2CH2CH2CH3) or 2,2-dimethylbutane PLAN: Use the formula, structure and Table 2.2 (button). SAMPLE PROBLEM 12.3 Predicting the Type and Relative Strength of Intermolecular Forces (a) MgCl2 or PCl3 (b) CH3NH2 or CH3F (c) CH3OH or CH3CH2OH • Bonding forces are stronger than nonbonding(intermolecular) forces. • Hydrogen bonding is a strong type of dipole-dipole force. • Dispersion forces are decisive when the difference is molar mass or molecular shape.

  29. SAMPLE PROBLEM 12.3 Predicting the Type and Relative Strength of Intermolecular Forces continued SOLUTION: (a) Mg2+ and Cl- are held together by ionic bonds while PCl3 is covalently bonded and the molecules are held together by dipole-dipole interactions. Ionic bonds are stronger than dipole interactions and so MgCl2 has the higher boiling point. (b) CH3NH2 and CH3F are both covalent compounds and have bonds which are polar. The dipole in CH3NH2 can H bond while that in CH3F cannot. Therefore CH3NH2 has the stronger interactions and the higher boiling point. (c) Both CH3OH and CH3CH2OH can H bond but CH3CH2OH has more CH for more dispersion force interaction. Therefore CH3CH2OH has the higher boiling point. (d) Hexane and 2,2-dimethylbutane are both nonpolar with only dispersion forces to hold the molecules together. Hexane has the larger surface area, thereby the greater dispersion forces and the higher boiling point.

  30. H bonded to N, O, or F Figure 12.18 Summary diagram for analyzing the intermolecular forces in a sample. INTERACTING PARTICLES (atoms, molecules, ions) ions present ions not present ions only IONIC BONDING (Section 9.2) nonpolar molecules only DISPERSION FORCES only polar molecules only DIPOLE-DIPOLE FORCES ion + polar molecule ION-DIPOLE FORCES polar + nonpolar molecules DIPOLE- INDUCED DIPOLE FORCES HYDROGEN BONDING DISPERSION FORCES ALSO PRESENT

  31. hydrogen bonding occurs across the surface and below the surface the net vector for attractive forces is downward hydrogen bonding occurs in three dimensions The molecular basis of surface tension. Figure 12.19

  32. Table 12.3 Surface Tension and Forces Between Particles Surface Tension (J/m2) at 200C Substance Formula Major Force(s) diethyl ether CH3CH2OCH2CH3 dipole-dipole; dispersion 1.7x10-2 ethanol CH3CH2OH 2.3x10-2 H bonding butanol CH3CH2CH2CH2OH 2.5x10-2 H bonding; dispersion water H2O 7.3x10-2 H bonding mercury Hg 48x10-2 metallic bonding

  33. stronger cohesive forces adhesive forces Shape of water or mercury meniscus in glass. Figure 12.20 capillarity H2O Hg

  34. Table 12.4 Viscosity of Water at Several Temperatures viscosity - resistance to flow Viscosity (N*s/m2)* Temperature(0C) 20 1.00x10-3 40 0.65x10-3 0.47x10-3 60 80 0.35x10-3 *The units of viscosity are newton-seconds per square meter.

  35. The H-bonding ability of the water molecule. Figure 12.21 hydrogen bond donor hydrogen bond acceptor

  36. The Unique Nature of Water • great solvent properties due to polarity and • hydrogen bonding ability • exceptional high specific heat capacity • high surface tension and capillarity • density differences of liquid and solid states

  37. The hexagonal structure of ice. Figure 12.22

  38. The expansion and contraction of water. Figure 12.23

  39. The macroscopic properties of water and their atomic and molecular “roots”. Figure 12.24

  40. The striking beauty of crystalline solids. Figure 12.25 celestite pyrite amethyst halite

  41. lattice point unit cell unit cell portion of a 3-D lattice portion of a 2-D lattice The crystal lattice and the unit cell. Figure 12.26

  42. 1/8 atom at 8 corners The three cubic unit cells. Figure 12.27 (1 of 3) Simple Cubic Atoms/unit cell = 1/8 * 8 = 1 coordination number = 6

  43. 1/8 atom at 8 corners 1 atom at center coordination number = 8 The three cubic unit cells. Figure 12.27 (2 of 3) Body-centered Cubic Atoms/unit cell = (1/8*8) + 1 = 2

  44. 1/8 atom at 8 corners 1/2 atom at 6 faces coordination number = 12 The three cubic unit cells. Figure 12.27 (3 of 3) Face-centered Cubic Atoms/unit cell = (1/8*8)+(1/2*6) = 4

  45. Packing of spheres. Figure 12.28 simple cubic (52% packing efficiency) body-centered cubic (68% packing efficiency)

  46. layer a layer b hexagonal closest packing cubic closest packing layer c layer a hexagonal unit cell face-centered unit cell expanded side views Figure 12.28 (continued) closest packing of first and second layers abab… (74%) abcabc… (74%)

  47. PROBLEM: Barium is the largest nonradioactive alkaline earth metal. It has a body-centered cubic unit cell and a density of 3.62 g/cm3. What is the atomic radius of barium? (Volume of a sphere: V = 4/3pr3) reciprocal divided by M V = 4/3pr3 multiply by packing efficiency divide by Avogadro’s number volume of 1 mol Ba atoms SAMPLE PROBLEM 12.4 Determining Atomic Radius from Crystal Structure PLAN: We can use the density and molar mass to find the volume of 1 mol of Ba. Since 68%(for a body-centered cubic) of the unit cell contains atomic material, dividing by Avogadro’s number will give us the volume of one atom of Ba. Using the volume of a sphere, the radius can be calculated. density of Ba (g/cm3) radius of a Ba atom volume of 1 mol Ba metal volume of 1 Ba atom

  48. 137.3 g Ba 1 cm3 x mol Ba 3.62 g 26 cm3 mol Ba atoms x mol Ba atoms 6.022x1023 atoms SAMPLE PROBLEM 12.4 Determining Atomic Radius from Crystal Structure continued SOLUTION: Volume of Ba metal = = 37.9 cm3/mol Ba 37.9 cm3/mol Ba x 0.68 = 26 cm3/mol Ba atoms = 4.3x10-23 cm3/atom r3 = 3V/4p = 2.2 x 10-8cm

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