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Chemistry, Instructor Solutions Manual

Chapter 8

Chapter 8 Effects of Intermolecular Forces Solutions to Problems in Chapter 8 8.1 The condensation temperature of a substance reflects the strength of its interatomic attractions. Xe has the highest condensation temperature, indicating that it has the greatest interatomic attractions. Ar, with the lowest condensation temperature, has the least interatomic attractions. At 140 K, Xe is in a condensed phase because the kinetic energy of the atoms is not enough to overcome the interatomic attractions. Kr and Ar, however, are gases because the kinetic energy of their atoms is sufficient to overcome their interatomic attractions. At 100K, only argon atoms have kinetic energies sufficient to overcome its interatomic attractions, so it is the only gas. Xe and Kr are both in condensed phases.

8.2 Condensation and boiling temperatures refer to the same characteristic: both indicate the temperature for the gas–liquid phase change. The condensation temperature of a substance measures the strength of its intermolecular attractions. Thus, carbon tetrachloride has much larger intermolecular attractions than methane does. At room temperature (298 K) and standard pressure, methane is a gas because its molecules have enough kinetic energy to overcome its intermolecular forces. Carbon tetrachloride is a liquid because its molecules do not have enough kinetic energy to overcome its intermolecular forces.

8.3 For any given substance, intermolecular attractions are constant. Their relative importance depends on two features: how close together molecules are, on average; © John Wiley and Sons Canada, Ltd.

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and how much kinetic energy of motion molecules have, on average. For any given substance, molecular volume is constant, so its relative importance depends on how close together molecules are. (a) Molecules are farther apart, making intermolecular attractions less significant, and molecular size is less significant. (b) Molecules are closer together, so intermolecular attractions and molecular sizes are more significant. (c) Increasing the temperature at constant pressure leads to a volume increase, so molecules are farther apart, making intermolecular attractions and molecular size less significant. 8.4 For any given substance, intermolecular attractions are constant. Their relative importance depends on two features: how close together molecules are, on average; and how much kinetic energy of motion molecules have, on average. For any given substance, molecular volume is constant, so its relative importance depends on how close together molecules are. (a) Molecular kinetic energies are smaller, but the average distance between molecules does not change, so intermolecular attractions are more significant; molecular size is unchanged. (b) Molecules are closer together, so intermolecular attractions and molecular sizes are more significant. (c) Removing molecules at constant volume increases the average distance between molecules, so intermolecular attractions and molecular sizes become less significant. 8.5 Pictures of atomic arrangements should show a high degree of order for a solid, close packing but some disorder for a liquid, and large distances between atoms for a gas:

8.6 Pictures of atomic arrangements should show a high degree of order for a solid, close packing but some disorder for a liquid, and large distances between atoms for a gas:

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8.7 To determine deviation from ideal behaviour, calculate the pressure using the ideal gas equation and compare the calculated result with the experimental value:

ρideal

−1 −1 nRT (1.00 mol ) (0.08314 L bar mol K ) ( 40.0 + 273.15 K ) = = = 21.7 bar V 1.20 L

ρ − ρideal % Deviation = (100% )  actual ρideal 

  19.7 bar − 21.7 bar   = (100%)   = − 9.2% 21.7 bar   

8.8 To determine deviation from ideal behaviour, calculate the volume using the ideal gas equation and compare the calculated result with the experimental value: nH2 = V ideal

m 3.000 g = = 1.488 mol M 2.016 g/mol nRT (1.488 mol)(0.08314 L bar mol −1 K −1)(273.15 K) = = 0.1747 L 193.5 bar P  10− 3 L  = 189.18 cm3  = 0.18918 L 3   1 cm 

=

V actual

% Deviation = (100%)

0.18918 L − 0.1747 L Vactual − Videal = (100%) = 8.31% Videal 0.1747 L

8.9 To calculate pressure using the van der Waals equation, we must know n, V, T, a, and b: 2  nRT   n a  p= −   2  V − nb   V 

Table 2-3 in your textbook lists values for Cl 2 : a = 657.9 kPa L2 mol–2 , b = 0.0562 L/mol m nCl 2 =  M

 103 g   1 mol   = 1.25 kg   1 kg   70.906 g  = 17.63 mol     

 0.0562 L     = 14.01 L  1 mol  

(V − nb ) = 15.0 L − 17.63 mol  

2 −2 (17.63 mol)(8.314 L kPa mol −1 K −1 )(295 K) (17.63 mol) (657.9 kPa L mol ) − 14.01 L (15.0 L)2 p = 3086 kPa − 909 kPa = 2177 kPa = 21.8 bar 2

p=

nRT (17.63 mol)(0.08314 L bar mol− 1 K− 1 )(295 K) = = 29.1 bar V 15.0 L  p − pideal   21.8 bar − 29.1 bar  −25% % Deviation = ( 100%)  actual  = ( 100%)  = pideal 29.1 bar    

pideal =

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8.10 To calculate pressure using the van der Waals equation, we must know n, V, T, a, and b: 2 nRT   n a  p =  −   2   V − nb   V 

Table 2–3 in your textbook lists values for Cl2 : a = 657.9 kPa L2 mol–2 , b = 0.0562 L/mol nCl2 =

10.5 g = 0.1481 mol 70.906 g/mol

T = 145 + 273 = 418 K

  0.0562 L   (V − nb) = 5.00 L −  0.1481 mol   = 4.99 L  1 mol    (0.1481 mol)(8.314 L kPa mol −1 K −1)(418 K) (0.1481 mol) 2 (657.9 L2 kPa mol −2 ) p= − 4.99 L (4.99 L) 2 p = 103.1 kPa − 0.58 kPa = 102.5 kPa = 1.02 bar p ideal =

nRT (0.1481 mol)(0.08314 L bar mol −1 K −1)(418 K) = = 1.03 bar V 5.00 L

p = 1.02 bar, p ideal = 1.03 bar 8.11 Ease of liquefaction depends on the magnitude of intermolecular attractions: the larger the attractions, the easier the substance is to liquefy. All these molecules are symmetrical (tetrahedral), so the ranking depends entirely on dispersion forces. Ease of liquefaction will increase with molecular size: CH 4 (hardest to liquefy) < CF 4 < CCl4 (easiest to liquefy). 8.12 Boiling point depends on the magnitude of intermolecular attractions: the larger the attractions, the higher the boiling point. All these substances are symmetrical (atomic), so the ranking depends entirely on dispersion forces. Boiling point will increase with the total number of electrons (atomic number): He (lowest bp) < Ne < Ar < Xe (highest bp). 8.13 Polarizability increases with the size of the highest occupied orbitals. Draw the molecules with larger orbitals (CCl4 ) to show greater distortion than for the molecules with smaller orbitals (CH 4 ). These molecules are close to spherical in the absence of polarization:

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8.14 Polarizability increases with atomic size. Show the atoms that have larger orbitals (Xe) with greater distortion than the distortion for the atoms with smaller orbitals (Ne). These atoms are initially spherical:

8.15 Boiling point depends on the magnitude of intermolecular attractions: the larger the attractions, the higher the boiling point. The size of intermolecular attractions depends on the amount of dispersion forces (size of molecule and size of orbitals), the presence of dipolar forces, and hydrogen bonding. Among these substances, ethanol forms hydrogen bonds, so it has the highest boiling point. Propane, being smaller than n-pentane, has the lowest boiling point: Propane (lowest bp) < n-pentane < ethanol (highest bp). 8.16 Boiling point depends on the magnitude of intermolecular attractions: the larger the attractions, the higher the boiling point. The size of intermolecular attractions depends on the amount of dispersion forces (size of molecule and size of orbitals), presence of dipolar forces due to polar bonds, and hydrogen bonding. Among these substances, n-propanol forms hydrogen bonds, so it has the highest boiling point. Dimethyl ether, being smaller than diethyl ether, has the lowest bp: Dimethyl ether (lowest bp) < diethyl ether < n-propanol (highest bp). 8.17 Hydrogen bonding capability requires the presence of an electronegative atom with lone pairs (O, F, or N) and an H–X bond to a highly electronegative atom (X = O, N, or F): (b and d) hydrogen bonding; (a and c) no hydrogen bonding.

8.18 Hydrogen bonding capability requires the presence of an electronegative atom with lone pairs (O, F, or N) and an H–X bond to a highly electronegative atom (X = O, N, or F). Water meets both requirements, so any substance that meets either requirement will form hydrogen bonds with water: (a and b) No hydrogen bonding; (c and e) meet both requirements, form hydrogen © John Wiley and Sons Canada, Ltd.

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bonds with water; (d) has an O atom with lone pairs, forms hydrogen bonds with H atoms from water.

8.19 Hydrogen bonding capability requires the presence of an electronegative atom with lone pairs (O, F, or N) and an H–X bond to a highly electronegative atom (X = O, N, or F): (a) Ammonia has one N atom with a lone pair that can form a hydrogen bond to any H atom of another ammonia molecule:

(b) Two types of hydrogen bonds are possible, between an N atom of NH 3 and an H atom of H 2 O, and between an O atom of H 2 O and an H atom of NH 3 .

8.20 Hydrogen bonding capability requires the presence of an electronegative atom with lone pairs (O, F, or N) and an H–X bond to a highly electronegative atom (X = O, N, or F): © John Wiley and Sons Canada, Ltd.

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(a) Methanol has an O atom with two lone pairs that can form a hydrogen bond to the H–O atom of another methanol molecule:

(b) Acetone does not have an H–X bond to O, N, or F, but it has an electronegative O atom with lone pairs that can form hydrogen bonds to the H atom of HF:

8.21 The viscosities of a set of similar substances increases with increasing chain length, because of increased dispersion forces (stronger intermolecular attractions make it harder for molecules to slide by one another) and increased “tangling” (longer chains are more entangled). Thus, the order of increasing viscosity is pentane (C 5 ) < gasoline (C 8 ) < fuel oil (C 12 ). 8.22 Viscosity decreases as temperature increases. Thus, when the weather is cold, a high-viscosity oil will not flow well when the engine is cold, leading to high friction and possible engine damage. For this reason, a low-viscosity oil is preferred for winter use. 8.23 An oxide surface has the ability to form hydrogen bonds to water, so the meniscus of water in aluminum tubing will be concave, just like inside glass tubing. The nonmetallic nature of the oxide layer means that there will not be strong liquid-surface forces for mercury inside aluminum tubing, so the meniscus will be convex, just like inside glass tubing. 8.24 Platinum atoms on the surface of platinum tubing will not strongly attract water molecules, as there is no possibility of hydrogen bonding. Consequently, the meniscus of water inside platinum tubing will be convex. For mercury, in contrast, metal–metal interactions are strong, so mercury forms a concave meniscus inside platinum tubing. 8.25 Paper towels contain cellulose (polysaccharide) , which forms good hydrogenbonding interactions with water, so aqueous solutions wet paper towels well and are absorbed effectively. There are no strong intermolecular interactions between an oil and the fibres of paper towels, so salad dressing is not absorbed effectively. 8.26 Wax contains hydrocarbon molecules that cannot form hydrogen bonds to water molecules, so water forms beads on a freshly waxed surface. Windshield glass is a © John Wiley and Sons Canada, Ltd.

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silicate structure with many –OH and =O groups that can form hydrogen bonds with water molecules, so water spreads out as a film on a clean windshield. 8.27 The vapour pressure of a substance at any given temperature is determined by the strength of intermolecular forces: the stronger the forces, the lower the vapour pressure. (a) Benzene has a higher vapour pressure, because chlorobenzene has a polar C–Cl bond that gives it a dipole moment and generates dipolar intermolecular forces; (b) hexane has a higher vapour pressure, because there are hydrogen bonding interactions for 1-hexanol that increase its intermolecular forces; (c) heptane has a higher vapour pressure, because octane, being the larger molecule, has stronger dispersion forces. 8.28 The vapour pressure of a substance at any given temperature is determined by the strength of intermolecular forces: the stronger the forces, the lower the vapour pressure. (a) Water has a lower vapour pressure, because water molecules form two hydrogen bonds per molecule whereas methanol forms only one; (b) 1-hexanol has a lower vapour pressure, because the magnitude of dispersion forces increases as a molecule gets bigger; (c) chloroform has a lower vapour pressure, because both molecules have similar dipole moments but the three chlorine atoms of chloroform give it larger dispersion forces. 8.29 The position of an element in the periodic table, the chemical formula, and a knowledge of polyatomic ions all help in identifying types of solids: Sn, a metal, is a metallic solid; S 8 has a specific molecular formula, so it is a molecular solid; Se is not a metal, so it is a network solid; SiO 2 is described in your textbook as a network solid; and Na2 SO 4 contains Na+ cations and SO 4 2– polyatomic anions, so it is an ionic solid. 8.30 The position of an element in the periodic table, the chemical formula, and knowledge of polyatomic ions all help in identifying types of solids: Pt, a metal, is a metallic solid; P 4 has a specific molecular formula, so it is a molecular solid; Ge is not a metal, so it is a network solid; As 2 O 3 is not ionic, and we cannot write a satisfactory molecular Lewis structure, so it is a network solid; and (NH 4 ) 3 PO 4 contains NH 4 + cations and PO 4 3− polyatomic anions, so it is an ionic solid. 8.31 Consult your textbook for information about the different types of solids. (a) The bonding in metals comes from extended networks of delocalized electrons, whereas the bonding in network solids includes many individual covalent bonds. (b) Metals conduct electricity, are malleable and ductile, and are shiny. Network solids are non-conductors or semiconductors, are brittle, and often have dull appearances. 8.32 Consult your textbook for information about the different types of solids. (a) The particles in molecular solids are neutral molecules such as P 4 or S 8 , so the © John Wiley and Sons Canada, Ltd.

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forces are dispersion forces. The particles in ionic solids are cations and anions, so the forces are electrical. (b) Molecular solids tend to be soft and low-melting. Ionic solids tend to be brittle and high-melting. 8.33 The position of an element in the periodic table, the chemical formula, and knowledge of polyatomic ions all help in identifying types of solids: (a) Br 2 is a discrete neutral molecule. It forms a molecular solid. (b) KBr contains cations and anions. It forms an ionic solid. (c) Ba is an alkaline earth metal. It forms a metallic solid. (d) SiO 2 is described in your textbook as a network solid. (e) CO2 is a covalent molecule. It forms a molecular solid. 8.34 The position of an element in the periodic table, the chemical formula, and knowledge of polyatomic ions all help in identifying types of solids: (a) HCl is a discrete neutral molecule. It forms a molecular solid. (b) KCl contains cations and anions. It forms an ionic solid. (c) NH 4 NO 3 contains cations and anions. It forms an ionic solid. (d) Mn is a transition metal. It forms a metallic solid. (e) Si is a non-metal. It forms a network solid. 8.35 The density of a solid depends on the nature of the elements it contains and on how compact the bonding network is. A structure with a more open bonding pattern has a lower density than one with a more compact bonding pattern. Graphite uses sp2 hybrids and has an extended p-bonding network, making it planar with a relatively open structure between successive planes. Diamond uses sp3 hybrids and has a compact network of bonds:

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8.36 The density of a solid depends not only on the nature of the elements it contains but also on how compact the bonding network is. A structure with a more open bonding pattern has a lower density than one with a more compact bonding pattern. Amorphous silica has some irregularities in its bond network, giving it a relatively open structure compared with crystalline quartz, which is entirely regular and compact in bonding:

8.37 To determine a chemical formula from a unit cell, count the number of atoms of 1 (each is part each element contained within the unit cell. Atoms on a face count 2 1 of two unit cells), those on an edge count (each is part of four unit cells), and 4 1 those at a corner count (each is part of eight unit cells). There is one Ti atom 8 within the cell. Ca:

1 (8 corner atoms) = 1 atom 8

O:

1 (6 face atoms) = 3 atoms 2

The chemical formula is CaTiO 3 . 8.38 To determine a chemical formula from a unit cell, count the number of atoms of 1 (each is part each element contained within the unit cell. Atoms on a face count 2 1 of two unit cells), those on an edge count (each is part of four unit cells), and 4 1 those at a corner count (each is part of eight unit cells): 8 Na: 8 interior atoms = 8 atoms O:

1 1 (6 face atoms) + (8 corner atoms) = 4 atoms 2 8

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The Na:O ratio is 2:1, giving a chemical formula of Na2 O. 8.39 The problem states that carborundum is diamond-like. Diamond is composed entirely of tetrahedral carbon atoms (SN = 4), and the empirical formula of carborundum, SiC, indicates that alternate C atoms should be replaced with Si atoms. Each C atom bonds to four Si atoms and each Si atom bonds to four C atoms:

8.40 The problem states that quartz is a network solid. Quartz is composed of tetrahedral silicon atoms (SN = 4) and tetrahedral oxygen atoms, and the empirical formula of quartz, SiO 2 , indicates that there are twice as many O atoms as Si atoms. Each Si atom bonds to four O atoms and each O atom bonds to two Si atoms and has two lone pairs:

8.41 A unit cell has to be a shape that can be used over and over to build the entire pattern. The upper screen shows a unit cell containing two complete fish, and the lower screen shows a unit cell that contains no complete fish, but notice that it 1 1 contains (2 edge) + (4 corner) = 2 fish overall, just like the other unit cell: 2 4

8.42 A unit cell has to be a shape that can be used over and over to build the entire pattern. The left-hand screen shows a unit cell containing no complete stars. Notice 1 1 that it contains (2) = 1 eight-pointed...