CHAPTER 11: INTERMOLECULAR FORCES AND LIQUIDS AND SOLIDS. Chemistry 1411 Joanna Sabey

Transcription

1 CHAPTER 11: INTERMOLECULAR FORCES AND LIQUIDS AND SOLIDS Chemistry 1411 Joanna Sabey

2 Forces Phase: homogeneous part of the system in contact with other parts of the system but separated from them by a well-defined boundary. State volume/shape Density Compressibility Motion of molecules Gas Assumes the volume and shape of its container Liquid Has a definite volume but assumes the shape of its container Solid Has a definite shape and volume Low High High Very compressible Only slightly compressible Virtually Incompressible Very free motion Slide past one another freely Vibrate about fixed positions

3 Forces Intermolecular Forces: attractive forces between molecules Measure of Intermolecular forces: melting point Boiling point ΔH vaporization ΔH fusion ΔH sublimation Intramolecular Forces: holds atoms together in a molecule, chemical bondings. In general intermolecular forces are weaker than intramolecular forces.

4 Intermolecular Forces Dipole-Dipole Forces: attractive forces between polar molecules.

5 Intermolecular Forces Ion-Dipole Forces: attract and ion and a polar molecule to each other.

6 Intermolecular Forces Dispersion Forces: attractive forces that arise as a result of temporary dipoles induced in atoms or molecules. Induced dipole: the separation of positive and negative charges in the atom, non-polar molecule, is due to the proximity of an ion or polar molecule.

7 Intermolecular Forces Polarizability: the ease with which the electron distribution in the atom or molecule can be distorted. Increases with greater number of electrons and a more diffuse electron cloud. Dispersion forces usually increase with molar mass.

8 Intermolecular Forces What type(s) of intermolecular forces exist between the following pairs? HBr and H 2 S Dipole-dipole and dispersion forces Cl 2 and CBr 4 Dispersion forces I 2 and NO 3 - Ion-induced dipole and dispersion forces NH 3 and C 6 H 6 Dipole induced dipole forces and dispersion forces

9 Intermolecular Forces Hydrogen Bond: a special dipole-dipole interaction between the hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom.

10 Hydrogen Bonding

11 Hydrogen Bonding Which of the following can form hydrogen bonds with water? CH 3 OCH 3 Yes CH 4 No F 2 Yes HCOOH Yes Na + No

12 Properties of Liquids Surface tension: the amount of energy required to stretch or increase the surface of a liquid by a unit area. Strong intermolecular forces make strong surface tension Cohesion: the intermolecular attraction between like molecules. Adhesion: an attraction between unlike molecules. Viscosity: a measure of a fluid s resistance to flow. Strong Intermolecular forces make high viscosity solution.

13 Crystal Structure Crystalline solid: posses rigid and long-range order; its atoms, molecules, or ions occupy specific positions. Amorphous solids: do not possess a well-defined arrangement and long-range molecular order. Unit Cell: the basic repeating structural unit of a crystalline solid lattice point At lattice points: Atoms Molecules Ions

14 Seven Basic Unit Cells

15 Cubic Unit Cells

16 Atoms Shared by 8 unit cells Shared by 4 unit cells Shared by 2 unit cells

17 Closet Packing Closet Packing: the most efficient arrangement of spheres. hexagonal cubic

18 Structure Gold (Au) crystallizes in a cubic close-packed structure (the face-centered cubic unit cell) and has a density of 19.3 g/cm 3. Calculate the atomic radius of gold in picometers. Calculate the Mass of Gold Au using the structure: Mass=(4 atoms/1unit cell)(1 mol/ 6.02X10 23 atoms)(197.0 g/mol) =1.31 X g/unit cell Calculate the volume of unit cell: D=m/V V=m/D V = (1.31 X g/unit cell)/ (19.3 g/cm 3 ) = 6.97X10-23 cm 3 Determine the edge length(a) of the cell: Determine the radius: R = 144 pm

19 X-Ray Diffraction X-ray diffraction: the scattering of X rays by the units of a crystalline solid.

20 Bragg Equation: X-Ray Diffraction 2d sinθ = nλ

21 X-Ray Diffraction X rays of wavelength nm strike an aluminum crystal; the rays are reflected at an angle of Assuming that n = 1, calculate the spacing between the planes of aluminum atoms (in pm) that is responsible for this angle of reflection. The conversion factor is obtained from 1 nm = 1000 pm. Convert the wavelength to pm: 0.154nm X (1000pm/1nm) = 154 pm Rearrange Equation: 2d sinθ = nλ d= n λ / 2 sin θ Plug in values to solve for the spacing: d= (1)( 154 pm) / 2 sin (19.3 ) d= 233 pm

22 Crystals Ionic Crystals: composed of charged species and the anions and cations are generally quite different in size. Held together by electrostatic attraction Hard, brittle, high melting point Poor conductor of heat and electricity CsCl ZnS CaF 2

23 Crystals Covalent Crystals: atoms are held together in an extensive three-dimensional network entirely by covalent bonds. Hard, high melting point Poor conductor of heat and electricity diamond graphite

24 Crystals Molecular Crystals: the lattice points are occupied by molecules and the attractive forces between them are van der Waals forces and/or hydrogen bonding. Held together by intermolecular forces Soft, low melting point Poor conductor of heat and electricity

25 Crystals Metallic Crystals: every lattice point in a crystal is occupied by an atom of the same metal. Soft to hard, low to high melting point Good conductors of heat and electricity nucleus & inner shell e - mobile sea of e -

26 Metallic Crystals

27 Amorphous Solids Amorphous Solids: lack a regular three-dimensional arrangement of atoms. Glass: An optically transparent fusion product of inorganic materials that has cooled to a rigid state without crystallizing. Crystalline quartz (SiO 2 ) Non-crystalline quartz glass

28 Crystals

29 Phase Changes Least Order Greatest Order

30 Liquid-Vapor Equilibrium Equilibrium vapor pressure: the vapor pressure measured when a dynamic equilibrium exists between condensation and evaporation. Evaporation, vaporization: process in which a liquid is transformed into a gas Condensation: the change from the gas phase to the liquid phase. Dynamic Equilibrium: the rate of a forward process is exactly balance by the rate of the reverse process.

31 Vaporization Molar heat of vaporization ( H vap ): the energy required to vaporize 1 mole of a liquid at its boiling point. Clausius-Clapeyron Equation ln P = - H vap RT P = (equilibrium) vapor pressure T = temperature (K) R = gas constant (8.314 J/K mol) + C Vapor Pressure Versus Temperature

32 Vaporization

33 Vaporization Diethyl ether is a volatile, highly flammable organic liquid that is used mainly as a solvent. The vapor pressure of diethyl ether is 401 mmhg at 18 C. Calculate its vapor pressure at 32 C. Plug in given values into the second equation, convert Celsius to Kelvin. P= 656 mmhg

34 Critical Temperature and Pressure Critical Temperature (T c ): the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure Critical pressure (P c ): the minimum pressure that must be applied to bring about liquefaction at the critical temperature. T < T c T > T c T ~ T c T < T c

35 Liquid-Solid Equilibrium Melting point of solid /freezing point of a liquid: A liquid is the temperature at which solid and liquid phases coexist at equilibrium. Molar heat of fusion ( H fus ): the energy required to melt 1 mole of a solid substance at its freezing point.

36 Heating Curve

37 Solid-Gas Equilibrium Molar heat of sublimation ( H sub ): the energy required to sublime 1 mole of a solid. H sub = H fus + h vap Sublimation: the process in which molecules go directly from the solid into the vapor phase Deposition: molecules make the transition from vapor to solid directly.

38 Solid-Gas Equilibrium Calculate the amount of energy (in kilojoules) needed to heat 346 g of liquid water from 0 C to 182 C. Assume that the specific heat of water is J/g C over the entire liquid range and that the specific heat of steam is 1.99 J/g C. Find the heat change(q) at each stage: Heating of water: q 1 =msδt =(346g)(4.184 J/ g X C)(182 C-0 C) = 145 kj Evaporating of water: q 2 = (346g) (1 mol /18.02 g) (40.97 kj / mol) = 783 kj Heating steam from 100 C to 182 C: q 3 = (346g)(1.99J/ J/g C) (182 C- 100 C) = 56.5 kj Add the heat changes together: q = 145 kj kj kj = 985 kj

39 Phase Diagram Phase Diagram: summarizes the conditions at which a substance exists as a solid, liquid, or gas. Phase Diagram of Water Triple Point: the only condition under which all three phases can be in equilibrium with each other.