Intermolecular Forces ( 분자간힘 ) & Liquids and Solids ( 액체및고체 ) Chapter 11

Intermolecular Forces ( 분자간힘 ) & Liquids and Solids ( 액체및고체 ) Chapter 11

목차 1. 액체와고체의분자운동론 2. 분자간힘 3. 액체의특성 4. 결정구조 5. 결정에의한 X-선회절 6. 결정의형태 7. 비결정질고체 8. 상변화 9. 상평형그림

액체와고체의분자운동론 상 (phase) 은그계의다른부분과접해있지만분명한경계로구분되어있는균일한부분 Solid phase ( 고체상 ) ice ( 얼음 ) Liquid phase ( 액체상 ) water ( 물 ) } 2 Phases ( 상 ) ( 기체, 액체및고체의특성 ) ( 밀도 ) ( 압축성 ) ( 분자의운동 ) 11.1

Intermolecular forces ( 분자간힘 ) : 분자들사이에작용하는인력 Intramolecular forces ( 분자내힘 ) : 분자에서원자들을서로붙들고있는힘 Intermolecular vs Intramolecular ( 분자간 & 분자내힘의비교 ) 41 kj to vaporize 1 mole of water (inter) 930 kj to break all O-H bonds in 1 mole of water (intra) 일반적으로분자간힘이분자내힘보다작다. 분자간힘의척도 boiling point melting point h vap H fus H sub 11.2

Intermolecular Forces ( 분자간힘 ) Dipole-Dipole Forces ( 쌍극자-쌍극자힘 ) polar molecules ( 극성분자 ) 들사이의인력정전기적, Coulomb 법칙 고체내극성분자들의배열 11.2

Intermolecular Forces ( 분자간힘 ) Ion-Dipole Forces ( 이온-쌍극자힘 ) Ion ( 이온 ) 과 polar molecule ( 극성분자 ) 사이에작용하는인력정전기적, Coulomb 법칙 Ion-Dipole Interaction 11.2

Intermolecular Forces ( 분자간힘 ) Dispersion Forces ( 분산힘 ) 원자나분자에서순간적으로유도된유도쌍극자로인하여작용하는인력, 모든형태의화학종에존재함. ion-induced dipole ( 이온 - 유도쌍극자 ) interaction 유도쌍극자 - 유도쌍극자상호작용 dipole-induced dipole ( 쌍극자 - 유도쌍극자 ) interaction 11.2

분산힘은주변에있는이온의전하량, 쌍극자의세기에따라달라지고원자나분자의편극도에의해결정된다. Polarizability ( 편극도 ): 원자나분자의전자분포가고르지않게되기쉬운정도 편극도는전자수가많을수록, 전자구름이넓은영역에확산되어있을수록증가한다. 분산힘은분자량이커질수록증가한다. 11.2

연습문제 11.1 What type(s) of intermolecular forces exist between each of the following molecules? HBr HBr is a polar molecule( 극성분자 ). dipole-dipole forces & dispersion forces ( 쌍극자 - 쌍극자 ) ( 분산력 ) CH 4 SO 2 CH 4 is nonpolar( 비극성 ). dispersion forces O S O ( 분산력 ) SO 2 is a polar molecule( 극성분자 ). dipole-dipole forces & dispersion forces 11.2

Hydrogen Bond ( 수소결합 ) N-H, O-H, or F-H 등과같은극성결합에서수소원자와전기음성적인 O, N, F 원자사이에작용하는특별한형태의쌍극자 - 쌍극자상호작용. A H B or A H A A & B = N, O, or F 11.2

액체의특성 Surface tension ( 표면장력 ) : 액체의표면을잡아늘리거나증가시키는데필요한에너지 ( 단위면적당 ) Strong intermolecular forces High surface tension 11.3

Properties of Liquids Cohesion ( 응집력 ): 같은분자들사이의인력 Adhesion ( 부착력 ): 서로다른분자들사이의인력 Adhesion 모세관현상 Cohesion 부착력 > 응집력 부착력 < 응집력 11.3

Properties of Liquids Viscosity ( 점성도 ): 흐름에대한유체의저항정도 Strong intermolecular forces High viscosity 느리게흐름 11.3

Water is a Unique Substance 수소결합 - 좋은용매, 높은비열, 덜조밀한고체상태 Maximum Density 4 0 C 물의밀도 얼음의밀도 < 물의밀도 11.3

A crystalline solid ( 결정질고체 ): 원자, 분자, 이온이먼공간에걸쳐정해진측정자리에규칙적으로배열되어있는고체 An amorphous solid ( 비결정질고체 ): 먼공간에걸친규칙적배열이없는고체 A unit cell ( 단위세포 ): 결정질고체에서구조적으로반복되는기본단위 ( 격자점 ) lattice point At lattice points: Atoms Molecules Ions Unit Cell Unit cells in 3 dimensions 11.4

7 가지단위세포 단순입방정계정방정계사방정계 마름모정계 단사정계삼사정계육방정계

단순입방체 (simple cubic) 구조에서구의쌓임 X 구의배위수 (coordination number) = 6 11.4

입방체의 3 가지형태 단순입방체 Simple cubic 체심입방체 Body-centered cubic 면심입방체 Face-centered cubic

체심입방체 (body-centered cubic) 에서구의배열 Shared by 8 unit cells Shared by 2 unit cells

단순입방체 1 atom/unit cell (8 x 1/8 = 1) 체심입방체 2 atoms/unit cell (8 x 1/8 + 1 = 2) 면심입방체 4 atoms/unit cell (8 x 1/8 + 6 x 1/2 = 4) 11.4

Closest packing ( 최조밀쌓임 ) 배위수 = 12 ABA 배열 육방최조밀쌓임 (hexagonal close-packed) Mg, Ti, Zn hcp ccp 면심입방체 ABC 배열 입방최조밀쌓임 (cubic close-packed) Al, Ni, Ag

3 가지입방체에서모서리길이와원자반지름과의관계

예제 11.3 금 (Au) 은입방최조밀쌓임구조 ( 면심입방체 ) 결정을이루며, 19.3 g/cm 3 의밀도를갖는다. 금의원자반지름은 (pm)? 결정단위세포의부피와질량을이용한다. 단위세포내에있는원자의총개수 = (8x1/8) + (6x1/2)= 4 단위세포질량, m = 4 원자 1단위세포 1몰 6.022 10 23 원자 197.0g Au 1몰 Au = 1. 31 10-21 g/ 단위세포 V = m d = 1.31 10-21 19.3 g/cm g 3 = 6.79 10-23 cm 3 a = 3 V = 3 6.79 10-23 cm 3 = 4.08 10 8 cm = 8r r = a 8 = 4.08 10 8 8 cm = 1.44 10 8 cm = 144pm

연습문제 11.3 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 = a 3 = (409 pm) 3 = 6.83 x 10-23 cm 3 4 atoms/unit cell in a face-centered cubic cell m = 4 Ag atoms 107.9 g x mole Ag x 1 mole Ag 6.022 x 10 23 atoms = 7.17 x 10-22 g d = m V 7.17 x 10-22 g = = 10.5 g/cm 6.83 x 10-23 cm 3 3 11.4

X- 선회절 (X-ray diffraction) crystal Photographic plate X-ray tube 원자층에의한 x- 선반사 Extra distance = BC + CD = 2d sinθ = nλ (Bragg Equation) 11.5

연습문제 11.4 X rays of wavelength 0.154 nm are diffracted from a crystal at an angle of 14.17 0. Assuming that n = 1, what is the distance (in pm) between layers in the crystal? nλ = 2d sin θ n = 1 θ = 14.17 0 λ = 0.154 nm = 154 pm d = nλ 2sinθ = 1 x 154 pm 2 x sin14.17 = 77.0 pm 예제 11.4 풀어볼것 11.5

Types of Crystals ( 결정의형태 ) Ionic Crystals ( 이온결정 ) Lattice points occupied by cations and anions Held together by electrostatic attraction ( 격자에너지 ) Hard, brittle, high melting point Poor conductor of heat and electricity 물에녹으면전도성 CsCl ZnS CaF 2 11.6

Types of Crystals Covalent Crystals ( 공유결정 ) Lattice points occupied by atoms Held together by covalent bonds ( 공유결합 ) Hard, high melting point Poor conductor of heat and electricity carbon atoms diamond graphite 11.6

Types of Crystals Molecular Crystals ( 분자결정 ) Lattice points occupied by molecules Held together by intermolecular forces ( 분자간힘 ) Soft, low melting point Poor conductor of heat and electricity 예 ) SO 2, I 2, P 4, S 8 11.6

Types of Crystals Metallic Crystals ( 금속결정 ) 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 핵 & 내부전자 Cross Section of a Metallic Crystal mobile sea of e - 11.6

Amorphous solid ( 비결정질고체 ) Glass ( 유리 ): 무기물질을결정화시키지않고굳은상태가되도록냉각시켜서광학적으로투명하게만든용융화합물 Crystalline quartz (SiO 2 ) Non-crystalline quartz glass 11.7