Presentation on theme: "Chemistry: Atoms First"— Presentation transcript:
1 Chemistry: Atoms First Julia Burdge & Jason OverbyChapter 12Intermolecular Forces and the Physical Properties of Liquids and SolidsKent L. McCorkleCosumnes River CollegeSacramento, CA
2 Intermolecular Forces and the Physical properties of Liquids and Solids 1212.1 Intermolecular ForcesDipole-Dipole InteractionsHydrogen BondingDispersion ForcesIon-Dipole Interactions12.2 Properties of LiquidsSurface TensionViscosityVapor Pressure12.3 Crystal StructureUnit CellsPacking SpheresClosest Packing12.4 Types of CrystalsIonic CrystalsCovalent CrystalsMolecular CrystalsMetallic Crystals12.5 Amorphous Solids12.6 Phase ChangesLiquid-Vapor Phase TransitionSolid-Liquid Phase TransitionSolid-Vapor Phase Transition12.7 Phase Diagrams
3 12.1 Intermolecular Forces Intermolecular forces are attractive forces that hold particles together in the condensed phases.The magnitude (and type) of intermolecular forces is what determines whether the particles that make up a substance are a gas, liquid, or solid.GasLiquidSolid
4 Intermolecular Forces Attractive forces that act between atoms or molecules in a pure substance are collectively called van der Waals forces.Dipole-dipole interactions are attractive forces that act between polar molecules.The magnitude of the attractive forces depends on the magnitude of the dipole.
6 Hydrogen BondingHydrogen bonding is a special type of dipole-dipole interaction.Hydrogen bonding only occurs in molecules that contain H bonded to a small, highly electronegative atom such as N, O, or F.FHF
10 Worked Example 12.1What kind(s) of intermolecular forces exist in (a) CCl4(l), (b) CH3COOH(l), (c) CH3COCH3(l), and (d) H2S(l).Strategy Draw Lewis dot structures and apply VSEPR theory to determine whether each molecule is polar or nonpolar. Nonpolar molecules exhibit dispersion forces only. Polar molecules exhibit dipole-dipole interactions and dispersion forces. Polar molecules with N–H , F–H, or O–H bonds exhibit dipole-dipole interactions (including hydrogen bonding) and dispersion forces.(a) (b) (c) (d)
11 Worked Example 12.1 (cont.) (a) (b) (c) (d) Solution (a) CCl4 is nonpolar, so the only intermolecular forces are dispersion forces.(b) CH3COOH is polar and contains an O–H bond, so it exhibits dipole-dipole interactions (including hydrogen bonding) and dispersion forces.(c) CH3COCH3 is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits dipole-dipole interactions and dispersion forces.(d) H2S is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits dipole-dipole interactions and dispersion forces.Think About It Being able to draw correct Lewis structures is, once again, vitally important. Review, if you need to, the procedure for drawing them.
12 Ion-Dipole Interactions Ion-dipole interactions are Coulombic attractions between ions (either positive or negative) and polar molecules.
13 Properties of Liquids12.2Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area.The stronger the intermolecular forces, the higher the surface tension.
14 Capillary action is the movement of a liquid up a narrow tube. Properties of LiquidsCapillary action is the movement of a liquid up a narrow tube.Two types of forces bring about capillary action:cohesion is the attraction between like moleculesadhesion is the attraction between unlike moleculesAdhesive forces are greater than cohesive forcesCohesive forces aregreater than adhesive forces
15 Properties of LiquidsViscosity is a measure of a fluid’s resistance to flow.The higher the viscosity the more slowly a liquid flows.Liquids that have higher intermolecular forces have higher viscosities.
16 The number of molecules with enough kinetic energy to escape. Properties of LiquidsVapor pressure is also dependent on intermolecular forces.If a molecule at the surface of a liquid has enough kinetic energy, it can escape to the gas phase in a process called vaporization.T1 < T2The number of molecules with enough kinetic energy to escape.
17 Evaporation: H2O(l) → H2O(g) Condensation: H2O(l) ← H2O(g) Properties of LiquidsThe vapor pressure increases until the rate of evaporation equals the rate of condensation.H2O(l) ⇌ H2O(g)Evaporation: H2O(l) → H2O(g)Condensation: H2O(l) ← H2O(g)When the forward process and reverse process are occurring at the same rate, the system is in dynamic equilibrium.
18 Properties of LiquidsThe vapor pressure increases until the rate of evaporation equals the rate of condensation.H2O(l) ⇌ H2O(g)
19 Properties of LiquidsThe vapor pressure increases with temperature.
20 Properties of LiquidsThe Clausius-Clapeyron equation relates the natural log of vapor pressure and the reciprocal of absolute temperature.ln P = natural log of vapor pressureΔHvap = the molar heat of vaporizationR = the gas constant (8.314 J/K•mol)T = the kelvin temperatureC is an experimentally determined constant
21 Properties of LiquidsThe Clausius-Clapeyron equation:Plotting ln P versus 1/T is a line with a slope of −ΔH/R.ΔH is assumed to be independent of temperature.
22 Properties of LiquidsThe Clausius-Clapeyron equation can be rearranged into a two point form:
23 Worked Example 12.2Diethyl ether is a volatile, highly flammable organic liquid that today is used mainly as a solvent. (It was used as an anesthetic during the nineteenth century and as a recreational intoxicant early in the twentieth century during prohibition, when ethanol was difficult to obtain.) The vapor pressure of diethyl ether is 401 mmHg at 18°C, and its molar heat of vaporization is 26 kJ/mol. Calculate its vapor pressure at 32°C.Strategy Given the vapor pressure at one temperature, P1, use the equation below to calculate the vapor pressure at a second temperature, P2.Temperature must be expressed in kelvins, so T1 = K and T2 = K. Because the molar heat of vaporization is given in kJ/mol, we will have to convert it to J/mol for the units of R to cancel properly: ΔHvap = 2.6×104 J/mol. The inverse function of ln x is ex.ln = −P1P2ΔHvapR1T2T1
24 Worked Example 12.2 (cont.) Solution ln = P1 P2 2.6×104 J/mol 8.314 J/K∙mol1K−K= −0.4928P1P2= e− =P10.6109= P2401 mmHg0.6109P2 == 6.6×102 mmHgThink About It It is easy to switch P1 and P2 or T1 and T2 accidentally and get the wrong answer to a problem such as this. One way to help safeguard against this common error is to verify that the vapor pressure is higher at the higher temperature.
25 Crystal Structure12.3A crystalline solid possess rigid and long-range order; its atoms, molecules, or ions occupy specific positions.A unit cell is the basic repeating structural unit of a crystalline solid.
26 Crystal StructureThere are seven types of unit cells.
27 Crystal StructureThe coordination number is defined as the number of atoms surrounding an atom in a crystal lattice.The value of the coordination number indicates how tightly the atoms are packed together.The basic repeating unit in the array of atoms is called a simple cubic cell.
28 Crystal StructureThere are three types of cubic cells.
29 Crystal StructureIn a body-centered cubic cell (bcc) the spheres in each layer rest in the depressions between spheres in the previous layer.The coordination number is 8.
30 Crystal StructureIn a face-centered cubic cell (fcc) the coordination number is 12.
31 Most of a cell’s atoms are shared by neighboring cells. Crystal StructureMost of a cell’s atoms are shared by neighboring cells.A corner atom is shared by eight unit cells.An edge atom is shared by four unit cells.A face-centered atom is shared by two unit cells.
32 Crystal StructureA simple cubic cell has the equivalent of only one complete atom contained within the cell.
33 Crystal StructureA body-centered cubic cell has two equivalent atoms:A face-centered cubic cell contains four complete atoms:
34 Hexagonal close-packed (hcp) structure: Crystal StructureHexagonal close-packed (hcp) structure:Site directly over an atom in layer AClose packing starts with a layer of atoms (A)Atoms in the second layer (B) fit into the depressions of the first layerHexagonal close-packed structure.
35 Cubic close-packed (ccp) structure: Crystal StructureCubic close-packed (ccp) structure:Site directly over an atom in layer A(hcp)Site NOT directly over an atom in layer A (ccp)Cubic close-packed structure35
36 Crystal Structure Closest packing: Hexagonal close-packing (hcp) Cubic close-packing (ccp) corresponds to a face-centered cubit cell.36
37 Edge length (a) and radius (r) are related: Crystal StructureEdge length (a) and radius (r) are related:Simple cubicBody-centered cubicFace-centered cubic37
38 Worked Example 12.3Gold crystallizes in a cubic close-packed structure (face-centered cubic unit cell) and has a density of 19.3 g/cm3. Calculate the atomic radius of an Au atom in angstroms (Å).Strategy Using the given density and the mass of gold continued within a face-centered cubic unit cell, determine the volumes of the unit cell. Then, use the volume to determine the value of a, and use the equation a = √8r to find r. Be sure to use consistent units for mass, length, and volume. The face-centered cubic unit cell contains a total of four atoms of gold [six faces, each shared by two unit cells, and eight corners, each shared by eight unit cells]. d = m/V and V = a3.Solution First, we determine the mass of gold (in grams) contained within a unit cell:m = × × = 1.31×10-21 g/unit cell4 atomsunit cell1 mol6.022×1023 atoms197.0 g Au1 mol Au
39 Worked Example 12.3 (cont.) Solution Then we calculate the volume of the unit cell in cm3:V = = = 6.78×10-23 cm3Using the calculated volume and the relationship V = a3 (rearranged to solve for a), we determine the length of a side of a unit cell:a = = √6.78×10-23 cm3 = 4.08×10-8 cmUsing the relationship provided a = √8r (rearranged to solve for r), we determine the radius of a gold atom in centimeters.r = = = 1.44×10-8 cmFinally, we convert centimeters to angstroms:1.44×10-8 cm × × = 1.44 Åmd1.31×10-21 g19.3 g/cm3Think About It Atomic radii tend to be on the order of 1 Å, so this answer is reasonable.3a√84.08×10-8 cm√81×10-2 m1 cm1 Å1×10-10 m
40 Types of Crystals12.4Ionic crystals are composed of charged ions that are held together by Coulombic attraction.The unit cell of an ionic compound can be defined be either the positions of the anions or the positions of the cations.
41 Crystal structures of three ionic compounds: Types of CrystalsCrystal structures of three ionic compounds:CsClSimple cubic latticeZnSZincblende structure(based on FCC)CaF2fluorite structure(based on FCC)41
42 Worked Example 12.4How many of each ion are contained within a unit cell of ZnS?Strategy Determine the contribution of each ion in the unit cell based on its position. Referring to the figure, the unit cell has four Zn2+ ions completely contained within the unit cell, and S2- ions at each of the eight corners and at each of the six faces. Interior ions (those completely contained within the unit cell) contribute one, those at the corners each contribute one-eighth, and those on the faces contribute one-half.Think About It Make sure that the ratio of cations to anions that you determine for a unit cell matches the ratio expressed in the compound’s empirical formula.Solution The ZnS unit cell contains four Zn2+ ions (interior) and four S2- ions [8 × (corners) and 6 × (faces)]1812
43 Worked Example 12.5The edge length of the NaCl unit cell is 564 pm. Determine the density of NaCl in g/cm3.Strategy Use the number of Na+ and Cl- ions in a unit cell (four of each) to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is mass divided by volume (d = m/V). Be careful to use units consistently.The masses of Na+ and Cl- ions are amu and amu, respectively. The conversion factor from amu to grams isso the masses of the Na+ and Cl- ions are 3.818×10-23 g and 5.887×10-23 g, respectively. The unit cell length is564 pm × × = 5.64×10-8 cm1 g6.022×1023 amu1×10-12 m1 pm1 cm1×10-2 m
44 Worked Example 12.5 (cont.)Solution The mass of the unit cell is 3.882×10-22 g (4 × 3.818×10-23 g + 4 × 5.887×10-23 g). The volume of a unit cell is 1.794×10-22 cm3 [(5.64×10-8 cm)3]. Therefore, the density is given byd = = 2.16 g/cm33.882×10-22 g1.794×10-22 cm3Think About It If you were to hold a cubic centimeter (1 cm3) of salt in your hand, how heavy would you expect it to be? Common errors in this type of problem include errors of unit conversion–especially with regard to length and volume. Such errors can lead to results that are off by many orders of magnitude. Often you can use common sense to gauge whether or not a calculated answer is reasonable. For instance, simply getting the centimeter-meter conversion upside down would result in a calculated density of 2.16×1012 g/cm3! You know that a cubic centimeter of salt doesn’t have a mass that large. (That’s billions of kilograms!) If the magnitude of a result is not reasonable, go back and check your work.
45 Types of CrystalsIn covalent crystals, atoms are held together in an extensive three-dimensional network entirely by covalent bonds.45
46 Types of CrystalsIn molecular crystals, the lattice points are occupied by molecules; the attractive forces between them are van der Waals forces and/or hydrogen bonding.46
47 Worked Example 12.6The metal iridium (Ir) crystallizes with a face-centered cubic unit cell. Given that the length of the edge of a unit cell is 383 pm, determine the density of iridium in g/cm3.Strategy A face-centered metallic crystal contains four atoms per unit cell [8 × (corners) and 6 × (faces)]. Use the number of atoms per cell and the atomic mass to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is then mass divided by volume (d = m/V). Be sure to make all necessary unit conversions.The mass of an Ir atom is amu. The conversion factor from amu to grams isso the mass of an Ir atom is 3.192×10-22 g. The unit cell length is383 pm × × = 3.83×10-8 cm18121 g6.022×1023 amu1×10-12 m1 pm1 cm1×10-2 m
48 Worked Example 12.6 (cont.)Solution The mass of the unit cell is 1.277×10-21 g (4 × 3.192×10-22 g). The volume of a unit cell is 5.618×10-23 cm3 [(3.83×10-8 cm)3]. Therefore, the density is given byd = = 22.7 g/cm31.277×10-21 g5.62×10-23 cm3Think About It Metals typically have high densities, so common sense can help you decide whether or not your calculated answer is reasonable.
49 Electrons are delocalized over the entire crystal. Types of CrystalsIn metallic crystals, every lattice point is occupied by an atom of the same metal.Electrons are delocalized overthe entire crystal.Delocalized electrons make metalsgood conductors.Large cohesive force resulting fromdelocalization makes metals strong.49
51 Amorphous Solids12.5Amorphous solids lack a regular three-dimensional arrangement of atoms.Glass is an amorphous solid.Glass is a fusion product.SiO2 is the chief component.Na2O and B2O3 are typically fused with molten SiO2 and allowed to cool without crystallizing.
54 Phase Changes12.6A phase is a homogeneous part of a system that is separated from the rest of the system by a well defined boundary.When a substance goes from one phase to another phase, it has undergone a phase change.Example Phase ChangeFreezing of water H2O(l) → H2O(s)Evaporation (or vaporization) of water H2O(l) → H2O(g)Melting (fusion) of ice H2O(s) → H2O(l)Condensation of water vapor H2O(g) → H2O(l)Sublimation of dry ice CO2(s) → CO2(g)Deposition of iodine I2(g) → I2(s)
56 Phase ChangesThe boiling point of a substance is defined as the temperature at which its vapor pressure equals the external atmospheric pressure.The molar heat of vaporization (ΔHvap) is the amount of heat required to vaporize a mole of substance at its boiling point.56
57 ice ⇌ water H2O(s) ⇌ H2O(l) Phase ChangesThe transformation of a liquid to a solid is called freezing.The reverse process is called melting, or fusion.The melting point (freezing point) of a solid (or liquid) is the temperature at which the solid and liquid phases coexist in equilibrium.ice ⇌ waterH2O(s) ⇌ H2O(l)In dynamic equilibrium, the forward and reverse process are occurring at the same rate.57
58 Phase ChangesThe molar heat of fusion (ΔHfus) is the energy required to melt 1 mol of a solid.58
59 Phase Changes Heating curves: Liquid and vapor in equilibrium TimeLiquid and vapor in equilibriumBoiling pointVaporMelting pointLiquidSolid and liquid in equilibriumSolid59
60 Solid I2 in equilibrium with its vapor Phase ChangesSublimation is the process by which molecules go directly from the solid phase to the vapor phase.Deposition is reverse process of sublimation.The molar enthalpy of sublimation (ΔHsub) of a substance is the energy required to sublime 1 mole of a solid.ΔHsub = ΔHfus + ΔHvapSolid I2 in equilibrium with its vapor60
61 Worked Example 12.7(a) Calculate the amount of heat deposited on the skin of a person burned by g of liquid water at 100.0°C and (b) the amount of heat deposited by 1.00 g of steam at 100.0°C. (c) Calculate the amount of energy necessary to warm g of water from 0.0°C to body temperature and (d) the amount of heat required to melt g of ice 0.0°C and then warm it to body temperature. (Assume that body temperature is 37.0°C.)
62 Worked Example 12.7 (cont.)Strategy For the purpose of following the sign conventions, we can designate the water as the system and the body as the surroundings. (a) Heat is transferred from hot water to the skin in a single step: a temperature change. (b) The transfer of heat from steam to the skin takes place in two steps: a phase change and a temperature change. (c) Cold water is warmed to body temperature in a single step: a temperature change. (d) The melting of ice and the subsequent warming of the resulting liquid water takes place in two steps: a phase change and a temperature change. In each case, the heat transferred during a temperature change depends on the mass of the water, the specific heat of the water, and the change in temperature. For the phase changes, the heat transferred depends on the amount of water (in moles) and the molar heat of vaporization (ΔHvap) or molar heat of fusion (ΔHfus). In each case, the total energy transferred or required is the sum of the energy changes for the individual steps.The specific heat is J/g∙°C for water and 1.99 J/g∙°C for steam. ΔHvap is kJ/mol and ΔHfus is 6.01 kJ/mol. Note: The ΔHvap of water is the amount of heat required to vaporize a mole of water. However, we want to know how much heat is deposited when water vapor condenses, so we use −40.79 kJ/mol.
63 Worked Example 12.7 (cont.)Solution (a) ΔT = 37.0°C – 100.0°C = –63.0°Cq = msΔT = 1.00 g × ×–63.0°CThus, 1.00 g of water at 100.0°C deposits kJ of heat on the skin. (The negative sign indicates that heat is given off by the system and absorbed by the surroundings.)(b)q1 = nΔHvap = mol ×q2 = msΔT = 1.00 g × ×–63.0°CThe overall energy deposited on the skin by 1.00 g of steam is the sum of q1 and q2:–2.26 kJ + (–0.264 kJ) = –2.53 kJ4.184 Jg∙°C= –2.64×102 J = –0.264 kJ1.00 g18.02 g/mol= mol water−40.79 kJmol= –2.26 kJ4.184 Jg∙°C= –2.64×102 J = –0.264 kJ
64 Worked Example 12.7 (cont.) Solution (c) ΔT = 37.0°C – 0.0°C = 37.0°C q = msΔT = 1.00 g × ×37.0°CThe energy required to warm g of water from 0.0°C to 37.0°C is 15.5 kJ.(d)q1 = nΔHfus = 5.55 mol ×q2 = msΔT = g × ×37.0°CThe energy rquired to melt g of ice at 0.0°C and warm it to 37.0°C is the sum of q1 and q2:33.4 kJ kJ = 48.9 kJ4.184 Jg∙°C= 1.55×104 J = 15.5 kJThink About It In problems that include phase changes, the q values corresponding to the phase-change steps will be the largest contributions to the total. If you find that this is not the case in your solution, check to see if you have made the common error of neglecting to convert the q values corresponding to temperature changes from J to kJ.100.0 g18.02 g/mol= 5.55 mol water6.01 kJmol= 33.4 kJ4.184 Jg∙°C= 1.55×104 J = 15.5 kJ
65 Phase Diagrams12.6A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas.The triple point is theonly combination ofpressure and temperaturewhere three phases of asubstance exist inequilibrium.triple point
66 The phase diagram of water: Phase DiagramsThe phase diagram of water:66
67 Worked Example 12.8Using the following phase diagram, (a) determine the normal boiling point and the normal melting point of the substance, (b) determine the physical state of the substance at 2 atm and 110°C, and (c) determine the pressure and temperature that correspond to the triple point of the substance.Strategy Each point on the phase diagram corresponds to a pressure-temperature combination. The normal boiling and melting points are the temperatures at which the substance undergoes phase changes. These points fall on the phase boundary lines. The triple point is where the three phase boundaries meet.
68 Worked Example 12.8 (cont.)Solution By drawing lines corresponding to a given pressure and/or temperature, we can determine the temperature at which a phase change occurs, or the physical state of the substance under specified conditions.(a)The normal boiling and melting points are ~140°C and ~205°C, respectively.
69 Worked Example 12.8 (cont.) Solution (b) At 2 atm and 110°C the substance is a solid.
70 Worked Example 12.8 (cont.) Solution (c) The triple point occurs at ~0.8 atm and ~115°C.Think About It The triple point of this substance occurs at a pressure below atmospheric pressure. Therefore, it will melt rather than sublime when it is heated under ordinary conditions.
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