1. Intermolecular Forces, Liquid, and Solid Kinetic-Molecular View of Liquid and Solid Intermolecular Forces Liquid Properties Crystal Structure Phase Changes and Phase Diagram

2. I. Kinetic–Molecular View Gas Gas – Assume volume and shape of the container – Low density, very compressible, free motion Liquid Liquid – Has definite volume, assume shape of the container – High density, slightly compressible, flow readily Solid Solid – Definite volume, definite shape – High density, incompressible TPTP T

3. II. Intermolecular Forces Intermolecular force: attractive force between molecules Intramolecular force: hold atom together in a molecule Ex: 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 forces –Boiling point, melting point,  H vap  H sub,  H fus intramolecular forces > intermolecular forces

4. Types of Intermolecular Forces Ion-Dipole Forces – Attractive forces between an ion and a polar molecule Dipole-Dipole Forces – Attractive forces between polar molecules Dispersion Forces – Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules Hydrogen Bond – Special dipole-dipole forces involving H and F, O, N van der Waals Forces – includes dipole-dipole, dipole-induced dipole and dispersion forces

5. II. Intermolecular Forces A. Dipole-Dipole Forces Attractive forces between polar molecules Orientation of Polar Molecules in a Solid

6. II. Intermolecular Forces Attractive forces between an ion and a polar molecule Ion-Dipole Interaction B. Ion-Dipole Forces

7. II. Intermolecular Forces (1)Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules (exist in all molecules) (2)Resulted from electron cloud distortion ion-induced dipole interaction dipole-induced dipole interaction C. Dispersion Forces

8. II. Intermolecular Forces (3) Related to the polarizability of the molecule Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted. C. Dispersion Forces Polarizability generally increases with: greater number of electrons more diffuse electron cloud larger molar mass Ex: Arrange boiling point CH 3 CH 3 CH 3 CH 2 CH 3 He Ne Kr < <<

9. II. Intermolecular Forces (4) Depend on the molecular shape C. Dispersion Forces Ex: C 5 H 12 two isomers H H H CH 3 H 3 C - C - C - C - C H 3 H 3 C - C - CH 3 H H H CH 3 >

10. II. Intermolecular Forces D. Hydrogen Bond The hydrogen bond is a special dipole-dipole interaction between they hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom. A H … B A H … A or A & B are N, O, or F

11. II. Intermolecular Forces D. Hydrogen Bond Decreasing molar mass Decreasing boiling point

12. S O O (1) 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. CH 4 CH 4 is nonpolar: dispersion forces. SO 2 SO 2 is a polar molecule: dipole-dipole forces and dispersion forces between SO 2 molecules. Lets try these: (2) Which molecule can form H bond with each other? (a) CH 3 OH(b) CH 3 OCH 3 (c) F 2 Only (a) CH 3 OH CH 3 O H …. O CH 3 H

13. Predict the trend in boiling point for the following species: BaCl 2, H 2, CO, HF, Ne One more example: BaCl 2 > HF > CO > Ne > H 2 BaCl 2 : ionic bonding HF: HB + dispersion CO: dipole-dipole + dispersion Ne: dispersion H 2 : dispersion Stronger intermolecular forces, higher BP, MP

14. III. Properties of Liquids A. Surface Tension (1)The amount of energy required to stretch or increase the surface of a liquid by a unit area. Unit: J/m 2 Stronger intermolecular forces, higher surface tension

15. III. Properties of Liquids A. Surface Tension (2) Capillary action Cohesion is the attraction between like molecules Adhesion is an attraction between unlike molecules Capilliary rise: adh > coh coh > adh

16. III. Properties of Liquids B. Viscosity - fluid’s resistance to flow (1)Viscosity increases as interaction increases (2)Viscosity decreases as T increases (3)Viscosity depends on molecular shape larger for long flexible molecule

17. III. Properties of Liquids C. Water: a unique substance due to HB Maximum Density 4 0 C Density of Water Each H 2 O has 4 HB, hexagonal arrangement

18. Unit Cell Unit cells in 3 dimensions IV. Crystal Structure A crystalline solid possesses rigid and long-range order. In a crystalline solid, atoms, molecules or ions occupy specific (predictable) positions. An amorphous solid does not possess a well-defined arrangement and long-range molecular order (no sharp MP) A. Unit Cell the basic repeating structural unit of a crystalline solid. lattice point At lattice points: Atoms Molecules Ions

20. B. Cubic Cells IV. Crystal Structure

21. a. Simple cubic lattice (scc) B. Cubic Cell 1.No. of atom per unit cell 8 x 1/8 =1 2.No. of nearest neighbors (coordination number): 6 3.In a unit cell, 52% space is occupied by atom 4.Size of unit cell to size of atom r = radius of atom a = edge length of unit cell 2 r = a

22. b. body-centered cubic (bcc) B. Cubic Cell 1.No. of atom per unit cell 8 x 1/8 +1 =2 2.No. of nearest neighbors (coordination number): 8 3.In a unit cell, 68% space is occupied by atom 4.Size of unit cell to size of atom r = radius of atom a = edge length of unit cell b 2 = a 2 + a 2 C 2 = a 2 +b 2 = 3 a 2

23. c. face-centered cubic (fcc) B. Cubic Cell 1.No. of atoms per unit cell 8 x 1/8 + 6 x 1/2 =4 2.No. of nearest neighbors (coordination number): 12 3.In a unit cell, 74 % space is occupied by atom 4.Size of unit cell to size of atom r = radius of atom a = edge length of unit cell b 2 = a 2 + a 2 b = 4 r 2 a 2 = 16 r 2

24. Question: 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 mole Ag x 1 mole Ag 6.022 x 10 23 atoms x = 7.17 x 10 -22 g d = m V 7.17 x 10 -22 g 6.83 x 10 -23 cm 3 = = 10.5 g/cm 3

25. V. Bonding in Solids Amorphous solid Crystalline solid Crystalline quartz (SiO 2 ) Non-crystalline quartz glass

26. A. Ionic Crystals Lattice points occupied by cations and anions Held together by electrostatic attraction Hard, brittle, high melting point Poor conductor of heat and electricity in solid form Conduct electricity when melted CsClZnSCaF 2

27. B. Molecular Crystals Held together by intermolecular forces Soft, low melting point Poor conductor of heat and electricity Examples: CH 4, Ar, P 4, S 8

28. C. Covalent Crystals Form 3-D network Held together by covalent bonds Hard, high melting point Poor conductor of heat and electricity Examples include C, B, Si, SiO 2 (quartz) diamond graphite

29. D. Metallic Crystal 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 -

30. VI. Phase Changes A substance can be converted into different phases by adding Or removing heat Gas Liquid Solid Temperature Endothermic Vaporization Melting Condensation Freezing Sublimation Deposition

31. Before Evaporation At Equilibrium A. Liquid-Vapor Equilibrium

32. The equilibrium vapor pressure is the vapor pressure measured when a dynamic equilibrium exists between condensation and evaporation Rate of condensation Rate of evaporation = 1. Dynamic Equilibrium 2. Equilibrium vapor pressure a. as T b. as intermolecular forces

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

34. Question: The vapor pressure (VP) of ethanol is 100 mmHg at 34.9 o C. What is the VP a 52.6 o C? (  H vap = 39.3 kJ/mol) P 1 =100 mmHg T 1 =34.9 o C = 308.1 K T 2 =52.6 o C = 325.8 K P 2 =? P 2 =230 mmHg

35. (a)The boiling point is the temperature when the (equilibrium) vapor pressure = the external pressure. (b)The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm. 4. Boiling point

36. 5. Critical point The critical temperature (T c ) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure. The critical pressure (P c ) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature.

37. B. Liquid-Solid Equilibrium Molar heat of fusion (  H fus ) is the energy required to melt 1 mole of a solid substance.

38. C. Solid-Vapor Equilibrium Molar heat of sublimation (  H sub ) is the energy required to sublime 1 mole of a solid. Solid Sublimation, Deposition Gas  H sub =  H fus +  H vap  H sub  H vap  H fus Liquid

39. Heating (cooling curve)

40. Question: How much heat (in kJ) is needed to convert 80.0 g of ethanol at -130 o C to vapor at 96 o C? The specific heat of solid, liquid, and vapor ethanol are 0.97, 2.4, and 1.2 J/g.o C, respectively.  H fus = 5.02 kJ/mol,  H vap = 39.3 kJ/mol. The mp is -114 o C and bp is 78 o C. Step 1: heat solid up to mp (-114 o C), remember q=sm  T q 1 =0.97 J/g.o C x 80.0 g x (-114 o C -(-130 o C))x 1 kJ/1000J=1.24 kJ Step 2: melt solid to liquid q 2 = 80.0g x 1 mol C 2 H 5 OH/46.07 g x 5.02 kJ/mol=8.72 KJ Step 4: convert liquid to vapor q 4 = 80.0g x 1 mol C 2 H 5 OH/46.07 g x 39.3 kJ/mol=68.2 KJ Step 3: heat liquid from mp (-114 o C) to bp (78 o C) q 3 =2.4 J/g.o C x 80.0 g x (78 o C -(-114 o C))x 1 kJ/1000J=36.9 kJ Step 5: heat vapor from 78 o C to 96 o C q 5 =1.2 J/g.o C x 80.0 g x (96 o C -78 o C)x 1 kJ/1000J=1.73 kJ Total heat=q 1 +q 2 + q 3 + q 4 + q 5 =117 kJ

41. VII. Phase diagram Solid Liquid Gas sublimation deposition melting freezing vaporization condensation Critical point Triple point Pressure, P Temperature, T – summarizes the conditions at which a substance exists as a solid, liquid, or gas.

42. Phase Diagram of WaterPhase Diagram of CO 2