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Chapter 12

Chapter 12. Intermolecular Forces:. Liquids, Solids, and Phase Changes. 12.1 물리적 상태와 상 변화의 개관 12.2 상 변화의 양적 측면 상 변화에 수반되는 열 상 변화의 평형 성질 상 그림 12.3 분자간 힘의 형태 이온－쌍극자 힘 쌍극자－쌍극자 힘 수소 결합 전하－유발쌍극자 힘 분산력 ( 런던 힘 ) 12.4 액체의 성질 표면장력 모세관 현상 점성 12.5 물의 독특한 성질 용매 성질 열 성질 표면 성질

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Chapter 12

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

  2. 12.1 물리적 상태와 상 변화의 개관 12.2 상 변화의 양적 측면 상 변화에 수반되는 열 상 변화의 평형 성질 상 그림 12.3 분자간 힘의 형태 이온－쌍극자 힘 쌍극자－쌍극자 힘 수소 결합 전하－유발쌍극자 힘 분산력(런던 힘) 12.4 액체의 성질 표면장력 모세관 현상 점성 12.5 물의 독특한 성질 용매 성질 열 성질 표면 성질 고체 물 및 액체 물의 밀도 12.6 고체 상태: 구조, 성질 및 결합 고체의 구조적 특성 결정성 고체 비결정성 고체 고체의 결합

  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. 상변화 gas vaporizing sublimination melting liquid condensing solid 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. 증발열 및 융해열

  7. Phase changes and their enthalpy changes. Figure 12.2

  8. 기체 내 입자들은 멀리 떨어져 무작위하게 움직이고 있고, 액체 내 입자들은 서로 접하고는 있으나 상대적으로 서로 움직이고 있고, 고체 내 입자들은 서로 접하고 경직된 구조에서 위치가 고정되어 있다. 이는 분자간 힘과 운동에너지의 상대적 크기에 기인한다. 물질의 각 상태에 따른 형태, 압축률, 유동성 등 거시적 차이는 분자수준에서 차이가 있기 때문이다. 고체, 액체,기체로 변하는 과정에서 에너지가 흡수된다. 주어진 조건하에서일어나는 각 상 변화에는 이에 해당하는 엔탈피 변화가 수반된다.

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

  10. 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)

  11. Liquid-gas equilibrium. Figure 12.4

  12. 액체 내 분자들의 속력분포와 온도 Figure 12.5

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

  14. The Clausius-Clapeyron Equation

  15. 1 760 torr 1 ln -40.5 x103 J/mol - T2 115 torr 308K 8.314 J/mol*K Using the Clausius-Clapeyron Equation 34.9℃에서 에탄올의 증기압은 115torr이다. 에탄올의 ΔH증발은40.5kJ/mol이다.증기압이 760torr인 온도(℃)를 계산하라. 주어진 것들 ΔH증발, P1, P2, T1을 식 12.1에 대입하고 T2에 대해 푼다. 여기서 R 값은 8.31 J/mol·K이므로 T1을 K 단위로 바꾸어 T2를 구한 다음 T2를 다시 ℃로 바꾼다. SOLUTION: 34.90C = 308.0K = T2 = 350K = 770C

  16. Phase diagrams(상도표) Figure 12.8 CO2 H2O

  17. van der Waal’s distance bond length covalent radius van der Waal’s radius Figure 12.9 Covalent and van der Waals radii.

  18. Figure 12.10 Periodic trends in covalent and van der Waals radii (in pm).

  19. 극성 분자들과 쌍극자－쌍극자 힘. Figure 12.11 solid liquid

  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. 쌍극자 모멘트와 끓는점 Figure 12.12

  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. 수소 결합과 끓는점 Figure 12.13

  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 12.14 instantaneous dipoles

  26. Figure 12.15 Molar mass and boiling point.

  27. Molecular shape and boiling point. Figure 12.16 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 12.17 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 12.18

  32. 표면장력과 입자들 간의 힘 Table 12.3 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 12.19 capillarity H2O Hg

  34. Table 12.4 Viscosity of Water (물의 점성도) 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 12.20 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 12.21

  38. The striking beauty of crystalline solids. Figure 12.22

  39. 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 12.23

  40. 1/8 atom at 8 corners The three cubic unit cells. Figure 12.24 (1 of 3) 단순입방 Atoms/unit cell = 1/8 * 8 = 1 배위수 = 6

  41. 1/8 atom at 8 corners 1 atom at center 배위수= 8 The three cubic unit cells. Figure 12.24 (2 of 3) 체심입방 Atoms/unit cell = (1/8*8) + 1 = 2

  42. 1/8 atom at 8 corners 1/2 atom at 6 faces 배위수 = 12 The three cubic unit cells. Figure 12.24 (3 of 3) 면심입방 Atoms/unit cell = (1/8*8)+(1/2*6) = 4

  43. Packing of spheres. Figure 12.26 simple cubic (52% packing efficiency) body-centered cubic (68% packing efficiency)

  44. 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 12.26 (continued) closest packing of first and second layers abab… (74%) abcabc… (74%)

  45. 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

  46. 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

  47. Figure 12.27 Diffraction of x-rays by crystal planes.

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