P a g e 1 Unit 3: Chapter 11/12: Liquids, Solids and Phase Changes Homework: Read Chapter 11 and 12 Keep up with assignments Liquids and solids are quite different from gases due to their attractive forces between the particles. Intermolecular dispersion forces increase attractions as the surface area increases. The foot pads of a gecko demonstrate this and allow a gecko to climb on vertical glass or even ceilings. Millions of microhairs (setae) line the toes. Each seta branches to hundred flattened tips called spatula. The unusually close contact with a surface allows for the short distance intermolecular forces to attract. Geckos can adhere to almost any surface Review Lewis Structures/how to determine polarity of substance: ion, polar, nonpolar. Polar verses Nonpolar particles: Polar molecules have an overall dipole moment ( that we learned about earlier. This dipole moment occurs when polar covalent bonds within a molecule do not cancel each other out. Nonpolar molecules or particles like the noble gases have zero dipole moment. In general, like will dissolve in like. Two polar liquids are generally miscible with each other. Two nonpolar liquids are miscible as well. But, when you take a polar liquid (vinegar) and a nonpolar liquid (vegetable oil) the two will be immiscible, and due to different densities one floats on the other.
P a g e 2 Intramolecular Forces: Intramolecular forces occur within a substance (molecule, ionic compound or network solid). These very strong forces include ionic bonding and polar and nonpolar covalent bonding. It takes high energy and/or temperatures to break apart intramolecular bonds. Bond energies vary around 150 to 1000kJ/mol. Ionic lattice energies vary from around 600 to16000 kj/mol. When water is vaporized from a liquid to a solid the intramolecular bonding between H-O-H does not break. H 2 O (l) H 2 O (g) When water undergoes electrolysis: 2 H 2 O (l) 2 H 2 (g) + O 2 (g), the intramolecular bonds between H-O-H break apart while new intramolecular bonds H-H and O=O form. Intermolecular Forces: Intermolecular forces are weak attractions that hold molecules and noble gas particles close together in a liquid or solid form. An alternative name for intermolecular forces is the van der Waals forces. They include London Dispersion Forces, dipole-dipole forces, and hydrogen bonds. An additional attraction is the ion-dipole forces that occur between polar molecules and ions, as in saltwater. Ion-Dipole Forces: Fairly strong intermolecular attractions. Energies range 10-50 kj/mol Electrical interactions attract between the charge of an ion and the opposite partial charge on the polar molecule solvent. Solvent cage (polar water) surrounds an ion (Na +1 or Cl -1 ) in a way that opposite charges attract. (NaCl and water) Interaction energy, E = z /r 2, where z is the ion charge, is the dipole moment on the ion and r is the distance between.
P a g e 3 London Dispersion Forces: Weaker of the intermolecular attractions. Energies range 1-10 kj/mol LD Forces exist in all molecules (polar and nonpolar) and in noble gases. Atoms develop temporary dipole arrangement of charges as an electron moves around the nucleus. This instantaneous dipole induces a similar dipole on neighbors Interatomic interaction is weak and short-lived The magnitude of the induced dipole depends on several factors larger the molecule, greater volume of electron cloud greater the molecular weight, more electrons, larger e -1 cloud more spread out the shape, maximizing the surface area, surface to surface contact the GREATER the attraction.
P a g e 4 Dipole - Dipole Forces: Intermolecular attractions found in polar molecules. Typical energy range 3-4 kj/mol DD forces are stronger as the distance between the opposite partial charges are closer. DD forces are stronger as the charge differentials between the opposite partial charges (higher dipole moment, ) are greater. Hydrogen Bonds: Stronger of the weak intermolecular attractions. Energies 10-40 kj/mol H-bonding is a special case of DD forces in which the distance is small and the dipole moment is large H-bonding occurs between molecules which have H directly bonded to F, O, or N within the molecule. F, O and N all have nonbonding electrons that play a role in creating the H-bond. Water has especially strong attractions since it can create two H-bonds from each O and an H bond from each H in a water molecule.
P a g e 5 Example 2: Identify the expected type of intermolecular attractions in the following. a) NH 4 Br (aq) b) C 2 H 4 c) HF d) Ne e) C 6 H 5 NH 2 f) H 2 S
P a g e 6 Example 3: Example 4: Choose the substance that has the stronger intermolecular attraction in each, explain your choice. a) C 4 H 10 vs C 3 H 7 OH b) C 2 H 6 vs Ne c) H 2 S vs H 2 O d) HCl vs HI e) Normal pentane, C 5 H 12 vs 2,2-dimethylpropane, C 5 H 12 Identify the type of attractions (intermolecular for molecules or noble gas, or intramolecular for ionic and network solids) Order to the best of your ability the lowest to highest expected melting point. NaCl, C (diamond), He, C 2 H 6, H 2 O, CH 2 O, KI, MgS Properties of Liquids: Surface Tension: Resistance to spreading out and increasing surface area. Molecules at the surface feel attractive forces on only one side and are pulled toward the liquid causing the liquid to bead up (spherical droplets) Small, nonpolar molecules with weak intermolecular attractions have low surface tension Viscosity: Resistance to flow. Small, nonpolar molecules with weak intermolecular attractions have low viscosity (CCl 4 ) Large, spread out, molecules with stronger intermolecular forces have higher viscosity (maple syrup)
P a g e 7 Capillary Action: ability of liquid to rise up inside a narrow tube. For liquid to rise in a capillary, both adhesive (between the liquid and the tube surface) attraction and cohesive (liquid to itself) attraction must be present. Narrower tubes will have a greater rise. Water will rise inside a glass tube well since it has both strong adhesive and cohesive forces. Mercury lacks adhesive attraction and will not rise in a capillary tube. Evaporation Rate: Rate in which liquid will vaporize in an open container. Vapor Pressure: The vapor pressure above a liquid inside a closed container Normal Melting Point: the temperature at 1 atmosphere pressure that the solid changes to liquid. Normal Boiling Point: the temperature at 1 atmosphere pressure that the liquid converts to gas. Enthalpy of Fusion: Energy required converting 1 mole of solid to liquid at its melting point. Enthalpy of Vaporization: Energy required converting 1 mole of liquid to gas at its melting point. The Clausius Clapeyron Equation: Vapor pressure exponentially rises as the temperature increases. The Clausius- Clapeyron equation below shows the relationship between vapor pressure, temperature and enthalpy of vaporization. ln(p 2 /P 1 ) = ( H vap /R)(1/T 1-1/T 2 ) Example 6: Chloroform (CHCl 3 ) has a normal boiling point of 62.0 C. Its enthalpy of vaporization is 29.2 kj/mol. Determine the vapor pressure of chloroform in a closed container at 24 C.
P a g e 8 Solution miscibility: Solubility depends, in part, on the attractive forces of the solute and solvent molecules, like dissolves like, miscible liquids will always dissolve in each other. Polar substances dissolve in polar solvents hydrophilic groups = OH, CHO, C=O, COOH, NH2, Cl Nonpolar molecules dissolve in nonpolar solvents hydrophobic groups = C-H, C-C Many molecules have both hydrophilic and hydrophobic parts solubility in water becomes a competition between the attraction of the polar groups for the water and the attraction of the nonpolar groups for their own kind
P a g e 9 Phase Changes: Melting (fusion), freezing, vaporization, condensation, sublimation, deposition. Heating or Cooling Curves and calculating the energy involved. Q = ms T for heating or cooling a constant phase. (s in J/mol C) Q = n H for phase changes (n for moles, H is generally kj/mol)
P a g e 10 Example 5: How much energy is required to change 9.00 grams of ice (H 2 O solid) at -25.0 C to steam (H 2 O gas) at 125.0 C? Given: specific heat capacities: 2.03 J/g C for H 2 O solid, 4.184 J/g C for H 2 O liquid, 2.0 J/g C for H 2 O gas H fus = 6.01 kj/mol; H vap = 40.67 kj/mol
P a g e 11 Phase Diagrams: Pressure (y axis) verses Temperature (x axis) Solid, liquid, gas, supercritical fluid, triple point normal melting and boiling points.
P a g e 12 Phase diagram for CO2 Try This: Draw a phase diagram with Pressure on the y axis and Temperature on the x axis that corresponds to the following information The triple point is 0.50 atm, 40 C The normal melting point is 42 C and the normal boiling point is 170 C The supercritical point is 12 atm and 220 C
P a g e 13 Solids: Amorphous: no long range order (obsidian, glass) Crystalline: has a repeating pattern of angles, distances between atoms. Unit Cells, the Structure of Crystalline Solids: Simple cells, Coordination number Simple cubic, 6 Body centered cubic, 8 Closest packing: With spheres it is more efficient to offset rows of atoms
P a g e 14 Hexagonal closest packed, 12, abababab Cubic closest packed, 12, abcabcabc, face centered cubic Counting atoms in a unit cell: body = 1 face = ½ edge = ¼ corner = 1 8
P a g e 15 Example 7: Types of Solids: Ionic Draw a NaCl cell. Each corner and face has a sodium ion. Chloride ions are placed interstitially on 12 edges and in the center. How many Na and Cl atoms make up one unit cell? Lattice sites occupied by ions Held together by attractions between oppositely charged ions, every cation attracts all anions around it, and vice-versa The coordination number represents the number of close cation anion interactions in the crystal The higher the coordination number, the more stable the solid, lowers the potential energy of the solid The coordination number depends on the relative sizes of the cations and anions that maintains charge balance, generally, anions are larger than
P a g e 16 cations. The number of anions that can surround the cation is limited by the size of the cation. The closer in size the ions are, the higher the coordination number is Molecular (polar and nonpolar) The lattice sites are occupied by molecules: CO 2, H 2 O, C 12 H 22 O 11 The molecules are held together by intermolecular attractive forces: dispersion forces, dipole dipole attractions, and H-bonds Because the attractive forces are weak, they tend to have low melting points, generally < 300 C Covalent Network Atoms attached to their nearest neighbors by covalent bonds Because of the directionality of the covalent bonds, these do not tend to form closest-packed arrangements in the crystal Because of the strength of the covalent bonds, these have very high melting points, generally > 1000 C Dimensionality of the network affects other physical properties Quartz (SiO 2 ) is shown. Melts at ~1600 C, Very hard Nonbonding Atomic Solids Noble gases in solid form Solid held together by weak dispersion forces, very low melting Tend to arrange atoms in closest-packed structure, maximizes attractive forces and minimizes energy Metallic Solid held together by metallic bonds, strength varies with sizes and charges of cations; coulombic attractions Melting point varies Mostly closest-packed arrangements of the lattice points, cations and sea of electrons
P a g e 17 Sea of electrons band theory Metal Alloys Band Theory: substitutional alloys brass: 2/3 Cu, 1/3 Zn sterling silver: 93% Ag, 7% Cu pewter: 96% Sn, 4% Cu interstitial alloys steel: Fe with 0-1.5% C The structures of metals and covalent network solids result in every atom s orbitals being shared by the entire structure For large numbers of atoms, this results in a large number of molecular orbitals that have approximately the same energy; we call this an energy band When two atomic orbitals combine they produce both a bonding and an antibonding molecular orbital When many atomic orbitals combine they produce a band of bonding molecular orbitals and a band of antibonding molecular orbitals The band of bonding molecular orbitals is called the valence band The band of antibonding molecular orbitals is called the conduction band HOMO, highest occupied molecular orbital LUMO, lowest unoccupied molecular orbital
P a g e 18 Band Gap At absolute zero, all the electrons will occupy the valence band As the temperature rises, some of the electrons may acquire enough energy to jump to the conduction band The difference in energy between the valence band and conduction band is called the band gap The larger the band gap, the fewer electrons there are with enough energy to make the jump Doping Semiconductors : Doping is adding impurities to the semiconductor s crystal to increase its conductivity Goal is to increase the number of electrons in the conduction band
P a g e 19 n-type semiconductors do not have enough electrons themselves to add to the conduction band, so they are doped by adding electron-rich impurities p-type semiconductors are doped with an electron-deficient impurity, resulting in electron holes in the valence band. Electrons can jump between these holes in the valence band, allowing conduction of electricity. When a p-type semiconductor adjoins an n-type semiconductor, the result is an p-n junction. Electricity can flow across the p-n junction in only one direction This is called a diode
P a g e 20 Solids and Liquids: Example Problems: 1. Identify for the following if there is a dipole moment or not. CO 2 SF 4 BrF 5 BeH 2 H 2 O NH 3 C 3 H 8 2. The dipole moment of ClF is observed to be 0.88 D. Its bond length is 163 pm. Solve for the calculated dipole moment if it was ionic [Cl] +1 [F] -1 using = Q x r. Solve for the percent ionic character in the actual Cl-F bond. (1 D = 3.336 x 10-30 C m. The charge of a single electron is 1.60 x 10-19 C. 3. Identify all of the following which exhibit ion-dipole forces. NaCl (s) NaCl (aq) Na (s) Cl 2 (g) K 2 CO 3 (aq) 4. Which is expected to have the largest London Dispersion Forces? a) C 2 H 6 b) C 8 H 18 c) N 2 d) CO 2 e) C 3 H 8 5. Describe the essential nature of the following and include example molecules. a) London Dispersion Forces b) Dipole-dipole attractions c) H- Bonding d) Ion-dipole attractions 6. Identify the types of intermolecular attractions (London Dispersion, Dipole-dipole, Hydrogen bonding) and/or intramolecular attractions (Covalent or Ionic bonding) that are present in each of the following. NaBr CH 2 Br 2 BeH 2 H 2 O Al 2 S 3 C 3 H 8 SiC 7. Arrange the following in order of expected increasing boiling point. CH 3 CH 2 OH CH 3 CH 2 CH 3 H 3 C O CH 3 CH 3 CH 2 NH 2 8. Briefly describe each of the following terms and predict what is expected for the following as liquids as the attraction forces get stronger. a) Will viscosity increase or decrease? b) Will surface tension increase or decrease? c) Will normal melting point increase or decrease? d) Will normal boiling point increase or decrease? e) Will enthalpy of vaporization ( Hvap) increase or decrease? f) Will vapor pressure increase or decrease? g) Will the evaporation rate increase or decrease? 9. Ether has a normal boiling point of 34.6 C and the enthalpy of vaporization for ether is 28.6kJ/mol. Calculate the vapor pressure of ether at 20.0 C?
P a g e 21 10. A kitchen pressure cooker operates at 1.50 atm. The H vap of water is 40.7 kj/mole. What is the boiling point of water in the pressure cooker in C? The normal boiling point of water is 100 C. 11. When a substance melts from solid to liquid at the normal melting point, will the sign for H fus be positive or negative? And will the S fus be positive or negative? 12. A 250-gram sample of liquid mercury at 30 C is added to a large quantity of liquid nitrogen kept at its normal boiling point of -196 C in a thermally insulated container. Information: Normal freezing point for Hg = -39 C Specific heat for liquid Hg = 0.138 J/g C Specific heat of solid Hg = 0.126 J/g C Enthalpy of fusion for Hg = 11.5 J/g Normal boiling point liquid N 2= -196 C Enthalpy of Vaporization N 2 = 199 J/g a) Solve for the total joules of energy lost by the mercury as it goes from a liquid at 30 C to a solid at -196 C b) Remembering the law of conservation of energy states that the energy lost is equal but opposite in sign as the energy gained. What mass of liquid nitrogen must be vaporized as Hg is brought to the temperature of liquid N 2 (-196 C)? 13. How many atoms are inside a body centered cubic cell made up of atoms in each corner and one in the center. 14. How many atoms are inside a face centered cubic cell made up of atoms in each corner and one on each face. 15. A binary ionic compound M x A y, crystallizes in a cubic structure that contains eight anions entirely within the unit cell and the metal cations are on each corner and on each face. What is the empirical formula? 16. Describe the essential nature/characteristics of the following types of solids and include example molecules. a) Polar molecule b) Nonpolar molecule c) Ionic d) Covalent network e) Metallic
P a g e 22 17. Use the Phase Diagram for questions below. atm. D 2.0 C 1.8 1.6 F 1.4 H P E 1.2 1.0 0.8 I 0.6 0.4 0.2 A J T G 0.0 0 10 20 30 40 50 60 70 80 90 100 110 T C a) What phase is represented by E F G b) What is T and why is it significant? c) What is C and why is it significant? d) What two phases are in equilibrium at H I J e) Name the phase changes occurring on the lines TD TC TA f) Estimate the normal melting point g) Estimate the normal boiling point h) What phase exists at 0.80 atm and 40 C? i) Starting at point J if the temperature increases 50 C the phase exists. By increasing the pressure at this new point by 1 atm the new phase is