APPENDIX G BRIEF ANSWERS TO SELECTED PROBLEMS
A-18
Potential Energy
1.2 Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. (a) gas (b) liquid (c) liquid 1.4 Physical property: a characteristic shown by a substance itself, without any interaction with or change into other substances. Chemical property: a characteristic of a substance that appears as it interacts with, or transforms into, other substances. (a) Colour (yellow-green and silvery to white) and physical state (gas and metal to crystals) are physical properties. The interaction between chlorine gas and sodium metal is a chemical property. (b) Colour and magnetism are physical properties. No chemical changes. 1.6(a) Physical change; there is only a temperature change. (b) Chemical change; the change in appearance indicates an irreversible chemical change. (c) Physical change; there is only a change in size, not composition. (d) Chemical change; the wood (and air) become different substances with different compositions. 1.8(a) fuel (b) wood 1.13 Lavoisier measured the total mass of the reactants and products, not just the mass of the solids. The total mass of the reactants and products remained constant. His measurements showed that a gas was involved in the reaction. He called this gas oxygen (one of his key discoveries). 1.16 A well-designed experiment must have the following essential features: (1) There must be at least two variables that are expected to be related; (2) there must be a way to control all the variables, so that only one at a time may be changed; (3) the results must be reproducible. 1.19(a) (1 m)2/(100 cm)2 (b) (1000 m)2/(1 km)2 and (100 cm)2/(1 m)2 (c) (1000 m/1 km) and (1 h/3600 s) (d) (1000 g/1 kg) and (1m)3/(100 cm)3 1.21 An extensive property depends on the amount of material present. An intensive property is the same regardless of how much material is present. (a) extensive property (b) intensive property (c) extensive property (d) intensive property 1.23(a) increases (b) remains the same (c) decreases (d) increases (e) remains the same 1.26 1.43 nm 1.28 1 1011 nm 1.30(a) 2.07 109 km2 (b) $6.73 103 1.34(a) 5.52 103 kg/m3 (b) 5.52 mg/mm3 1.36(a) 2.56 109 mm3/cell (b) 1010 L 1.38(a) 9.626 cm3 (b) 64.92 g 1.40 2.70 g/cm3 1.42(a) 291 K (b) 109 K (c) 273 C 1.45(a) 2.47 107 m (b) 6.76 nm 1.52(a) none (b) none (c) 0.0410 (d) 4.0100 104 1.54(a) 0.00036 (b) 35.83 (c) 22.5 1.56 6x102 1.58(a) 134 m (b) 21,621 mm3 (c) 443 cm 1.60(a) 1.310000 105 (b) 4.7 104 (c) 2.10006 105 (d) 2.1605 103 1.62(a) 5550 (b) 10,070. (c) 0.000000885 (d) 0.003004 1.64(a) 8.025 104 (b) 1.0098 103 (c) 7.7 1011 1.66(a) 4.06 1019 J (b) 1.61 1024 molecules (c) 1.82 105 J/mol 1.68(a) Height measured, not exact. (b) Planets counted, exact. (c) Number of grams in a pound is not a unit
definition, not exact. (d) Definition of “millimetre,” exact. 1.70 7.50 0.05 cm 1.72(a) Iavg 8.72 g; IIavg 8.72 g; IIIavg 8.50 g; IVavg 8.56 g; sets I and II are most accurate. (b) Set III is the most precise, but is the least accurate. (c) Set I has the best combination of high accuracy and high precision. (d) Set IV has both low accuracy and low precision. 1.74(a)
compressed spring less stable⎯ energy stored in spring
(b)
Potential Energy
Chapter 1
two charges near each other less stable⎯ repulsion of like charges
1.76 7.7/1 1.78(a) density 0.21 g/L, will float (b) CO2 is denser than air, will sink (c) density 0.30 g/L, will float (d) O2 is denser than air, will sink (e) density 1.38 g/L, will sink (f) 0.55 g for empty ball; 0.50 g for ball filled with hydrogen 1.80(a) 8.0 1012 g (b) 4.1 105 m3 (c) $4.1 1014 1.82(a) 195.79C (b) 5.05 L 1.83(a) 2.6 m/s (b) 15 km (c) 12:45 pm 1.85 freezing point 3.7X; boiling point 63.3X 1.86 2.3 1025 g oxygen; 1.4 1025 g silicon; 5 1015 g each of ruthenium and rhodium
Chapter 2 Answers to Boxed Reading Problems: B2.1 (a) 5 peaks (b) m/e ratio of heaviest particle 74; m/e ratio of lightest particle35 B2.3(a) Since salt dissolves in water and pepper does not, add water to mixture and filter to remove solid pepper. Evaporate water to recover solid salt. (d) Heat the
Appendix G • Brief Answers to Selected Problems
mixture; the ethanol will boil off (distill), while the sugar will remain behind. • 2.1 Compounds contain different types of atoms; there is only one type of atom in an element. 2.4(a) The presence of more than one element makes pure calcium chloride a compound. (b) There is only one kind of atom, so sulfur is an element. (c) The presence of more than one compound makes baking powder a mixture. (d) The presence of more than one type of atom means cytosine cannot be an element. The specific, not variable, arrangement means it is a compound. 2.12(a) elements, compounds, and mixtures (b) compounds (c) compounds 2.14(a) Law of definite composition: the composition is the same regardless of its source. (b) Law of mass conservation: the total quantity of matter does not change. (c) Law of multiple proportions: two elements can combine to form two different compounds that have different proportions of those elements. 2.16(a) No, the percent by mass of each element in a compound is fixed. (b) Yes, the mass of each element in a compound depends on the amount of compound. 2.18 The two experiments demonstrate the law of definite composition. The unknown compound decomposes the same way both times. The experiments also demonstrate the law of conservation of mass since the total mass before reaction equals the total mass after reaction. 2.20(a) 1.34 g F (b) 0.514 Ca; 0.486 F (c) 51.4 mass % Ca; 48.6 mass % F 2.22(a) 0.603 (b) 322 g Mg 2.24 3.498 106 g Cu; 1.766 106 g S 2.26 compound 1: 0.905 S/Cl; compound 2: 0.451 S/Cl; ratio: 2.00/1.00 2.29 Coal A 2.31 Dalton postulated that atoms of an element are identical and that compounds result from the chemical combination of specific ratios of different elements. 2.32 If you know the ratio of any two quantities and the value of one of them, the other can always be calculated; in this case, the charge and the mass/charge ratio were known. 2.36 The atomic number is the number of protons in an atom’s nucleus. When the atomic number changes, the identity of the element changes. The mass number is the total number of protons and neutrons in the nucleus. The identity of an element is based on the number of protons, not the number of neutrons. The mass number can vary (by a change in number of neutrons) without changing the identity of the element. 2.39 All three isotopes have 18 protons and 18 electrons. Their respective mass numbers are 36, 38, and 40, with the respective numbers of neutrons being 18, 20, and 22. 2.41(a) These have the same number of protons and electrons, but different numbers of neutrons; same Z. (b) These have the same number of neutrons, but different numbers of protons and electrons; same N. (c) These have different numbers of protons, neutrons, and electrons; same A. 55 109 2.43(a) 38 18Ar (b) 25Mn (c) 47Ag 48 (b) 79 (c) 115B 2.45(a) 22Ti 34Se 22eⴚ
34eⴚ
5eⴚ
22pⴙ 26n0
34pⴙ 45n0
5pⴙ 6n0
A-19
2.47 69.72 u 2.49% abundance(35Cl) 75.774%, % abundance (37Cl) 24.226% 2.52(a) In the modern periodic table, the elements are arranged in order of increasing atomic number. (b) Elements in a group (or family) have similar chemical properties. (c) Elements can be classified as metals, metalloids, or nonmetals. 2.55 The alkali metals [Group 1] are metals and readily lose one electron to form cations; the halogens [Group 17] are nonmetals and readily gain one electron to form anions. 2.56(a) germanium; Ge; 14; metalloid (b) phosphorus; P; 15; nonmetal (c) helium; He; 18; nonmetal (d) lithium; Li; 1; metal (e) molybdenum; Mo; 6; metal 2.58(a) Ra; 88 (b) Si; 14 (c) Cu; 63.55 u (d) Br; 79.90 u 2.60 Atoms of these two kinds of substances will form ionic bonds, in which one or more electrons are transferred from the metal atom to the nonmetal atom to form a cation and an anion, respectively. 2.63 Coulomb’s law states the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. The product of charges in MgO [(2) (2)] is greater than that in LiF [(1) (1)]. Thus, MgO has stronger ionic bonding. 2.66 The Group 1 elements form cations, and the Group 17 elements form anions. 2.68 Each potassium atom loses one electron to form an ion with a 1 charge. Each sulfur atom gains two electrons to form an ion with a 2 charge. Two potassiums, losing one electron each, are required for each sulfur, which gains two electrons. The oppositely charged ions attract each other to form an ionic solid, K2S. 2.70 K; I 2.72(a) oxygen; 17; 16; 2 (b) fluorine; 19; 17; 2 (c) calcium; 40; 2; 4 2.74 Lithium forms the Liion; oxygen forms the O2 ion. Number of O2 ions 4.2 1021 O2 ions. 2.76 NaCl 2.78 The subscripts in a formula give the numbers of ions in a formula unit of the compound. The subscripts indicate that there are two Fions for each Mg1 ion. Using this information and the mass of each element, we can calculate percent mass of each element in the compound. 2.80 The two samples are similar in that both contain 20 billion oxygen atoms and 20 billion hydrogen atoms. They differ in that they contain different types of molecules: H2O2 molecules in the hydrogen peroxide sample, and H2 and O2 molecules in the mixture. In addition, the mixture contains 20 billion molecules (10 billion H2 and 10 billion O2), while the hydrogen peroxide sample contains 10 billion molecules. 2.84(a) Na3N, sodium nitride (b) SrO, strontium oxide (c) AlCl3, aluminum chloride 2.86(a) MgF2, magnesium fluoride (b) ZnS, zinc sulfide (c) SrCl2, strontium chloride 2.88(a) SnCl4 (b) iron(III) bromide (c) CuBr (d) manganese(III) oxide 2.90(a) cobalt(II) oxide (b) Hg2Cl2 (c) lead(II) acetate trihydrate (d) Cr2O3 2.92(a) BaO (b) Fe(NO3)2 (c) MgS 2.94(a) H2SO4; sulfuric acid (b) HIO3; iodic acid (c) HCN; hydrocyanic acid (d) H2S; hydrosulfuric acid 2.96(a) ammonium ion, NH4; ammonia, NH3 (b) magnesium sulfide, MgS; magnesium sulfite, MgSO3; magnesium sulfate, MgSO4 (c) hydrochloric acid, HCl; chloric acid, HClO3; chlorous acid, HClO2 (d) cuprous bromide, CuBr; cupric bromide, CuBr2 2.98 Disulfur tetrafluoride, S2F4 2.100(a) calcium chloride (b) copper(I) oxide (c) stannic fluoride (d) hydrochloric acid 2.102(a) 12 oxygen atoms; 342.2 u (b) 9 hydrogen atoms; 132.06 u (c) 8 oxygen atoms; 344.6 u 2.104(a) (NH4)2SO4; 132.15 u
A-20
Appendix G • Brief Answers to Selected Problems
divide by ᏹ (g/mol) Amount (mol) of each element
use amount of moles as subscripts Preliminary empirical formula
change to integer subscripts Empirical formula divide total number of atoms in molecule by the number of atoms in the empirical formula and multiply the empirical formula by that factor Molecular formula
(c) Find the empirical formula from the mass percents. Compare the number of atoms given for the one element to the number in the empirical formula. Multiply the empirical formula by the factor that is needed to obtain the given number of atoms for that element. Road Map (Same first three steps as in part (b).) Empirical formula divide the number of atoms of the one element in the molecule by the number of atoms of that element in the empirical formula and multiply the empirical formula by that factor
£
Molecular formula
(e) Count the numbers of the various types of atoms in the structural formula and put these into a molecular formula. Road Map Structural formula count the number of atoms of each element and use these numbers as subscripts
£
3.2(a) 12 mol C atoms (b) 1.445 1025 C atoms 3.7(a) left (b) left (c) left (d) neither 3.8(a) 121.64 g/mol (b) 76.02 g/mol (c) 106.44 g/mol (d) 152.00 g/mol 3.10(a) 134.7 g/mol (b) 175.3 g/mol (c) 342.17 g/mol (d) 125.84 g/mol 3.12(a) 1.1 102 g KMnO4 (b) 0.188 mol O atoms (c) 1.5 1020 O atoms 3.14(a) 9.73 g MnSO4 (b) 44.6 mol Fe(ClO4)3 (c) 1.74 1021 N atoms 3.16(a) 1.56 103 g Cu2CO3 (b) 0.0725 g N2O5 (c) 0.644 mol NaClO4; 3.88 1023 formula units NaClO4 (d) 3.88 1023 Na ions; 3.88 1023 ClO4 ions; 3.88 1023 Cl atoms; 1.55 1024 O atoms 3.18(a) 6.375 mass % H (b) 71.52 mass % O 3.20(a) 0.1252 mass fraction C (b) 0.3428 mass fraction O 3.23(a) 0.9507 mol cisplatin (b) 3.5 1024 H atoms 3.25(a) 195 mol rust (b) 195 mol Fe2O3 (c) 2.18 104 g Fe 3.27 CO(NH2)2 NH4NO3 (NH4)2SO4 KNO3 3.28(a) 3.12 104 mol PbS (b) 1.88 1025 Pb atoms 3.32(b) From the mass percent, determine the empirical formula. Add up the total number of atoms in the empirical formula, and divide that number into the total number of atoms in the molecule. The result is the multiplier that converts the empirical formula into the molecular formula.
Mass (g) of each element (express mass percent directly as grams)
£
Chapter 3
Road Map
£ £ £
(b) NaH2PO4; 119.98 u (c) KHCO3; 100.12 u 2.106(a) 108.02 u (b) 331.2 u (c) 72.08 u 2.108(a) SO3; sulfur trioxide; 80.07 u (b) C3H8; propane; 44.09 u 2.112 Separating the components of a mixture requires physical methods only; that is, no chemical changes (no changes in composition) take place, and the components maintain their chemical identities and properties throughout. Separating the components of a compound requires a chemical change (change in composition). 2.115(a) compound (b) homogeneous mixture (c) heterogeneous mixture (d) homogeneous mixture (e) homogeneous mixture 2.117(a) filtration (b) extraction or chromatography 2.119(a) fraction of volume 5.2 1013 (b) mass of nucleus 6.64466 1024 g; fraction of mass 0.999726 2.120 strongest ionic bonding: MgO; weakest ionic bonding: RbI 2.124(a) I NO; II N2O3; III N2O5 (b) I has 1.14 g O per 1.00 g N; II, 1.71 g O; III, 2.86 g O 2.128(a) Cl, 1.898 mass %; Na, 1.056 mass %; SO42, 0.265 mass %; Mg2, 0.127 mass %; Ca2, 0.04 mass %; K, 0.038 mass %; HCO3, 0.014 mass % (b) 30.72% (c) Alkaline earth metal ions, total mass % 0.167%; alkali metal ions, total mass % 1.094% (d) Anions (2.177 mass %) make up a larger mass fraction than cations (1.26 mass %). 2.130 Molecular formula, C4H6O4; molecular mass, 118.09 u; 40.68% by mass C; 5.122% by mass H; 54.20% by mass O 2.133(a) Formulas and masses in u: 15 N218O, 48; 15N216O, 46; 14N218O, 46; 14N216O, 44; 15N14N18O, 47; 15N14N16O, 45 (b) 15N218O, least common; 14N216O, most common 2.135 58.091 u 2.137 nitroglycerin, 39.64 mass % NO; isoamyl nitrate, 22.54 mass % NO 2.138 0.370 kg C; 0.0222 kg H; 0.423 kg O; 0.185 kg N 2.143 (1) chemical change (2) physical change (3) chemical change (4) chemical change (5) physical change
Molecular formula
3.34(a) CH2; 14.03 g/mol (b) CH3O; 31.03 g/mol (c) N2O5; 108.02 g/mol (d) Ba3(PO4)2; 601.8 g/mol (e) TeI4; 635.2 g/ mol 3.36 Disulfur dichloride; SCl; 135.04 g/mol 3.38(a) C3H6 (b) N2H4 (c) N2O4 (d) C5H5N5 3.40(a) Cl2O7 (b) SiCl4 (c) CO2 3.42(a) NO2 (b) N2O4 3.44(a) 1.20 mol F (b) 24.0 g M (c) calcium 3.47 C21H30O5 3.49 C10H20O 3.50 A balanced equation provides information on the identities of reactants and products, the physical states of reactants and products, and the molar ratios by which reactants form products. 3.53 b 3.54(a) 16Cu(s) S8(s) ¡ 8Cu2S(s) (b) P4O10(s) 6H2O(l) ¡ 4H3PO4(l) (c) B2O3(s) 6NaOH(aq) ¡ 2Na3BO3(aq) 3H2O(l) (d) 4CH3NH2(g) 9O2(g) ¡ 4CO2(g) 10H2O(g) 2N2(g)
Appendix G • Brief Answers to Selected Problems
2SO2(g) O2(g) ¡ 2SO3(g) Sc2O3(s) 3H2O(l) ¡ 2Sc(OH)3(s) H3PO4(aq) 2NaOH(aq) ¡ Na2HPO4(aq) 2H2O(l) C6H10O5(s) 6O2(g) ¡ 6CO2(g) 5H2O(g) 4Ga(s) 3O2(g) ¡ 2Ga2O3(s) 2C6H14(l) 19O2(g) ¡ 12CO2(g) 14H2O(g) 3CaCl2(aq) 2Na3PO4(aq) ¡ Ca3(PO4)2(s) 6NaCl(aq) 3.64 Balance the equation for the reaction: aAbB ¡ cC. Since A is the limiting reactant, A is used to determine the amount of C. Divide the mass of A by its molar mass to obtain the amount (mol) of A. Use the molar ratio from the balanced equation to find the amount (mol) of C. Multiply the amount (mol) of C by its molar mass to obtain the mass of C. Road Map 3.56(a) (b) (c) (d) 3.58(a) (b) (c)
Mass (g) of A
£ £ £
divide by ᏹ (g/mol) Amount (mol) of A
molar ratio between A and C
Amount (moles) of C multiply by ᏹ (g/mol) Mass (g) of C
3.66(a) 0.455 mol Cl2 (b) 32.3 g Cl2 3.68(a) 1.42 103 mol KNO3 (b) 1.43 105 g KNO3 3.70 195.8 g H3BO3; 19.16 g H2 3.72 2.60 103 g Cl2 3.74(a) I2(s) Cl2(g) ¡ 2ICl(s) ICl(s) Cl2(g) ¡ ICl3(s) (b) I2(s) 3Cl2(g) ¡ 2ICl3(s) (c) 1.33 103 gI2 3.76(a) 0.105 mol CaO (b) 0.175 mol CaO (c) calcium (d) 5.88 g CaO 3.78 1.36 mol HIO3, 239 g HIO3; 44.9 g H2O in excess 3.80 4.40 g CO2; 4.80 g O2 in excess 3.82 12.2 g Al(NO2)3, no NH4Cl, 48.7 g AlCl3, 30.7 g N2, 39.5 g H2O 3.84 50.% 3.86 90.5% 3.88 24.0 g CH3Cl 3.90 39.7 g CF4 3.91 A 3.95(a) C (b) B (c) C (d) B 3.98 No, instructions should read: “Take 100.0 mL of the 10.0 mol/L solution and, with stirring, add water until the total volume is 1000. mL.” 3.99(a) 7.85 g Ca(C2H3O2)2 (b) 0.254 mol/LKI (c) 124 mol NaCN 3.101(a) 4.65 g K2SO4 (b) 0.0653 mol/LCaCl2 (c) 1.11 1020 Mg2 ions 3.103(a) 0.0617 mol/L KCl (b) 0.00363 mol/L (NH4)2SO4 (c) 0.138 mol/L Na 3.105(a) 987 g HNO3/L (b) 15.7 mol/L HNO3 3.107 845 mL 3.109 0.88 g BaSO4 3.112(a) Instructions: Be sure to wear goggles to protect your eyes! Pour approximately 2.0 L of water into the container. Add to the water, slowly and with mixing, 0.90 L of concentrated HCl. Dilute to 3.0 L with more water. (b) 22.6 mL 3.115 Ionic or polar covalent compounds 3.116 Ions must be present, and they come from ionic compounds or from electrolytes such as acids and bases. 3.119 B 3.123(a) Benzene is likely to be insoluble in water because it is nonpolar and water
A-21
is polar. (b) Sodium hydroxide, an ionic compound, is likely to be very soluble in water. (c) Ethanol (CH3CH2OH) is likely to be soluble in water because the alcohol group (OH) is very polar, like the water molecule. (d) Potassium acetate, an ionic compound, is likely to be very soluble in water. 3.125(a) Yes, CsBr is a soluble salt. (b) Yes, HI is a strong acid. 3.127(a) 0.64 mol (b) 0.242 mol (c) 1.18 104 mol 3.129(a) 3.0 mol (b) 7.57 105 mol (c) 0.148 mol 3.131(a) 0.058 mol Al3; 3.5 1022 Al3 ions; 0.18 mol Cl; 1.1 1023 Cl ions (b) 4.62 104 mol Li; 2.78 1020 Li ions; 2.31 104 mol SO42; 1.39 1020 SO42 ions (c) 1.50 102 mol K; 9.02 1021 K ions; 1.50 102 mol Br; 9.02 1021 Br ions 3.133(a) 0.35 mol H (b) 6.3 103 mol H (c) 0.22 mol H 3.137 Spectator ions do not appear in a net ionic equation because they are not involved in the reaction and serve only to balance charges. 3.142 x 3 3.143 ethane propane cetyl palmitate ethanol benzene 3.148(a) Fe2O3(s) 3CO(g) ¡ 2Fe(s) 3CO2(g) (b) 3.39 107 g CO 3.150 89.8% 3.152(a) 2AB2 B2 ¡ 2AB3 (b) AB2 (c) 5.0 mol AB3 (d) 0.5 mol B2 3.154 B, C, and D have the same empirical formula, C2H4O; 44.05 g/mol 3.155 44.3% 3.158(a) C (b) B (c) D 3.165(a) 586 g CO2 (b) 10.5% CH4 by mass 3.167 10/0.66/1.0 3.172(a) 192.12 g/mol; C6H8O7 (b) 0.580 mol 3.173(a) N2(g) O2(g) ¡ 2NO(g) 2NO(g) O2(g) ¡ 2NO2(g) 3NO2(g) H2O(g) ¡ 2HNO3(aq) NO(g) (b) 2N2(g) 5O2(g) 2H2O(g) ¡ 4HNO3(aq) (c) 6.07 103 t HNO3 3.174 A 3.176 (a) 0.039 g heme (b) 6.3 105 mol heme (c) 3.5 103 g Fe (d) 4.1 102 g hemin 3.178 (a) 46.65 mass % N in urea; 31.98 mass % N in arginine; 21.04 mass % N in ornithine (b) 28.45 g N 3.180 29.54% 3.182(a) 89.3% (b) 1.47 g ethylene 3.184(a) 125 g salt (b) 65.6 L H2O
Chapter 4 Answers to Boxed Reading Problems: B4.1 The density of the atmosphere decreases with increasing altitude. High density causes more drag on the aircraft. At high altitudes, low density means that there are relatively few gas particles present to collide with the aircraft. B4.3 0.934%, 946 kPa • 4.1(a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger container. (b) The volume of the container holding the gas sample increases when heated, but the volume of the container holding the liquid sample remains essentially constant when heated. (c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced. 4.6 990 cm H2O 4.8(a) 75.5 kPa (b) 1.32 bar (c) 3.65 104 Pa (d) 107 kPa 4.10 0.953 bar 4.12 0.979 bar 4.18 At constant temperature and volume, the pressure of a gas is directly proportional to the amount (mol) of the gas. 4.20(a) Volume decreases to one-third of the original volume. (b) Volume increases by a factor of 3.0. (c) Volume increases by a factor of 4. 4.22(a) Volume decreases by a factor of 2. (b) Volume increases by a factor of 1.48. (c) Volume
Appendix G • Brief Answers to Selected Problems
Chapter 5 Answers to Boxed Reading Problem: B5.2 (a) 2C(s, coal) 2H2O(g) ¡ CH4(g) CO2(g) (b) 12 kJ/mol (c) 3.30 104 kJ
Increasing, H
• 5.4 Increase: eating food, lying in the sun, taking a hot bath. Decrease: exercising, taking a cold bath, going outside on a cold day. 5.6 The amount of the change in internal energy is the same for heater and air conditioner. Since both devices consume the same amount of electrical energy, the change in energy of the heater equals that of the air conditioner. 5.8 0 J 5.10 1.54 103 J/mol 5.12(a) 6.6 107 kJ (b) 1.6 107 kcal 5.15 8.8 h 5.17 Measuring the heat transfer at constant pressure is more convenient than measuring at constant volume. 5.19(a) exothermic (b) endothermic (c) exothermic (d) exothermic (e) endothermic (f) endothermic (g) exothermic 5.22 Reactants ⌬H ⫽ (⫺), (exothermic)
Products
Increasing, H
5.24(a) Combustion of ethane: 2C2H6(g) 7O2(g) ¡ 4CO2(g) 6H2O(g) heat 2C2H6 ⫹ 7O2
(initial)
⌬H ⫽ (⫺), (exothermic) 4CO2 ⫹ 6H2O
(final)
Increasing, H
(b) Freezing of water: H2O(l) ¡ H2O(s) heat H2O(l)
(initial) ⌬H ⫽ (⫺), (exothermic)
H2O(s)
(final)
5.26(a) 2CH3OH(l) 3O2(g) ¡ 2CO2(g) 4H2O(g) heat Increasing, H
decreases by a factor of 3. 4.24 144C 4.26 35.8 L 4.28 0.14 mol Cl2 4.30 0.873 g ClF3 4.33 yes 4.35 Beaker is inverted for H2 and upright for CO2. The molar mass of CO2 is greater than the molar mass of air, which, in turn, has a greater molar mass than H2. 4.39 5.78 g/L 4.41 1.76 103 mol AsH3; 3.43 g/L 4.43 51.1 g/mol 4.45 1.35 bar 4.47 38.7 g P4 4.49 41.2 g PH3 4.51 0.0249 g Al 4.55 C5H12 4.57(a) 0.90 mol (b) 0.00898 bar 4.58 286 mL SO2 4.60 10.1 kPa SiF4 4.65 At STP, the volume occupied by a mole of any gas is the same. At the same temperature, all gases have the same average kinetic energy, resulting in the same pressure. 4.68(a) PA PB PC (b) EA EB EC (c) rateA rateB rateC (d) total EA total EB total EC (e) dA dB dC (f) collision frequency in A collision frequency in B collision frequency in C 4.69 13.21 4.71(a) curve 1 (b) curve 1 (c) curve 1; fluorine and argon have about the same molar mass 4.73 14.9 min 4.75 4 atoms per molecule 4.78 negative deviations; N2 Kr CO2 4.80 At 1 bar; at lower pressures, the gas molecules are farther apart and intermolecular forces are less important. 4.83 6.89 104 g/mol 4.86(a) 2.24 103 kPa (b) 2124 kPa 4.90(a) 79.4 kPa N2; 21.1 kPa O2; 0.04 kPa CO2; 0.46 kPa H2O (b) 74.2 mol % N2; 13.6 mol % O2; 5.2 mol % CO2; 6.1 mol % H2O (c) 1.6 1021 molecules O2 4.92(a) 4 102 mL (b) 0.013 mol N2 4.93 36.7 L NO2 4.98 Al2Cl6 4.100 1.52 102 mol SO3 4.104(a) 1.95 103 g Ni (b) 3.5 104 g Ni (c) 62 m3 CO 4.106(a) 9 volumes of O2(g) (b) CH5N 4.109 The lungs would expand by a factor of 4.86; the diver can safely ascend 15.99 m to a depth of 22 m. 4.111 6.00 g H2O2 4.116 6.53 103 g N2 4.120(a) xenon (b) water vapour (c) mercury vapour (d) water vapour 4.124 17.0 g CO2; 18.0 g Kr 4.130 Ne, 676 m/s; Ar, 481 m/s; He, 1.52 103 m/s 4.132(a) 0.052 g (b) 1.1 mL 4.139(a) 16.5 L CO2 (b) PH20 6.51 kPa; PO2 PCO2 49 kPa 4.144 332 steps 4.146 1.4 4.150 Ptotal 0.327 bar; PI233.4 103 bar
2CH3OH ⫹ 3O2
(initial)
⌬H ⫽ (⫺), (exothermic) 2CO2 ⫹ 4H2O
(final)
(b) 12 N2(g) O2(g) heat ¡ NO2(g) Increasing, H
A-22
NO2
(final)
⌬H ⫽ (⫹), (endothermic) 1 2 N2
⫹ O2
(initial)
5.28(a) This is a phase change from the solid phase to the gas phase. Heat is absorbed by the system so qsys is positive (). (b) The volume of the system is expanding as more moles of gas are present after the phase change than were present before the phase change. So the system has done work of expansion, and w is negative. Since Usys q w, q is positive, and w is negative, the sign of Usys cannot be predicted. It will be positive if q w and negative if q w. (c) Uuniv 0. If the system loses energy, the surroundings gain an equal amount of energy. The sum of the energy of the system and the energy of the surroundings remains constant. 5.31 To determine the specific heat capacity of a substance, you need its mass, the heat added (or lost), and the change in temperature. 5.33 Heat capacity is the quantity of heat required to raise the temperature of an object 1 K. Specific heat capacity is the quantity of heat required to raise 1 g of a material by 1 K. Molar heat capacity is the quantity of heat required to raise the temperature of 1 mol of a substance by 1 K. 5.35 6.9 103 J 5.37 295C 5.39 77.5C 5.41 45C 5.43 36.6C 5.50 The reaction has a positive H, because it requires the input of energy to break the oxygen-oxygen bond. 5.51 H is negative; it is opposite in sign and half of the value for the vapourization of 2 mol of H2O. 5.52(a) exothermic (b) 20.2 kJ per 18 mol of S8 produced. (c) q 4.2 102 kJ (d) q 15.7 kJ 5.54(a) 12 N2(g) 12 O2(g) ¡ NO(g), H 90.29 kJ/mol (b) q 10.5 kJ 5.56 q 1.88 106 kJ
Appendix G • Brief Answers to Selected Problems
5.60(a) C2H4(g) 3O2(g) ¡ 2CO2(g) 2H2O(g);
rH 1411 kJ/mol (b) 1.39 g C2H4 5.64 110.5 kJ/mol 5.65 813.4 kJ/mol 5.67 N2(g) 2O2(g) ¡ 2NO2(g); rH 66.4 kJ/mol; equation 1 is A, equation 2 is B, and equation 3 is C. 5.69 44.0 kJ/mol 5.72 The standard enthalpy of reaction, rH, is the enthalpy change for a reaction where all substances are in their standard states. The standard enthalpy of formation, fH, is the enthalpy change that accompanies the formation of one mole of a compound in its standard state from elements in their standard states. 5.74(a) 12 Cl2(g) Na(s) ¡ NaCl(s) (b) H2(g) 12 O2(g) ¡ H2O(g) (c) no changes 5.75(a) Ca(s) Cl2(g) ¡ CaCl2(s) (b) Na(s) 12 H2(g) C(s, graphite) 32 O2(g) ¡ NaHCO3(s) (c) C(s, graphite) 2Cl2(g) ¡ CCl4(l) (d) 12 H2(g) 12 N2(g) 32 O2(g) ¡ HNO3(l) 5.77(a) 1036.9 kJ/mol (b) 433 kJ/mol 5.79 157.3 kJ/mol 5.81(a) 503.9 kJ/mol (b) 1H 2 2H 504 kJ/mol 5.82(a) C18H36O2(s) 26O2(g) ¡ 18CO2(g) 18H2O(g) (b) 10,488 kJ/mol (c) 36.9 kJ; 8.81 kcal (d) 8.81 kcal/g 11.0 g 96.9 kcal of energy released 5.84(a) initial 23.7 L/mol; final 24.9 L/mol (b) 187 J (c) 1.2 102 J (d) 3.1 102 J (e) 310 J (f) H U P V U w (q w) w qP 5.93 (a) 1.2 102 mol CH4 (b) $0.0053/mol (c) $0.90 5.98(a) r1H 657.0 kJ/mol;
r2H 32.9 kJ/mol (b) 106.6 kJ/mol 5.99(a) 6.81 103 J (b) 243C 5.100 22.2 kJ/mol 5.101(a) 34 kJ/mol (b) 757 kJ 5.103(a) 1.25 103 kJ (b) 2.24 103C
Chapter 6 Answers to Boxed Reading Problem: B6.1 (a)slope 1.3 104 L/mol; y-intercept 0.00 (b) diluted solution 1.8 105 mol/L; original solution 1.4 104 mol/L • 6.2(a) x-ray ultraviolet visible infrared microwave radio waves (b) radio microwave infrared visible ultraviolet x-ray (c) radio microwave infrared visible ultraviolet x-ray 6.7 316 m; 3.16 1011 nm 6.9 2.5 1023 J 6.11 red yellow blue 6.13 1.3483 107 nm 6.16(a) 1.24 1015 s1; 8.21 1019 J (b) 1.4 1015 s1; 9.0 1019 J 6.18 Bohr’s key assumption was that the electron in an atom does not radiate energy while in a stationary state, and it can move to a different orbit only by absorbing or emitting a photon whose energy is equal to the difference in energy between two states. These differences in energy correspond to the wavelengths in the known line spectra for the hydrogen atom. A solar system model does not allow for the movement of electrons between levels. 6.20(a) absorption (b) emission (c) emission (d) absorption 6.22 Yes, the predicted line spectra are accurate. The energies (Z 2)(2.18 10 18 J) , where Z could be predicted from En 2 is the atomic number for the atom or ion.n The energy levels for Be3 will be greater by a factor of 16 (Z 4) than those for the hydrogen atom. This means that the pattern of lines will be similar, but the lines will be at different wavelengths.
A-23
6.23 434.17 nm 6.25 1875.6 nm 6.27 2.76 105 J/mol 6.29 d a c b 6.31 n 4 6.37 Macroscopic objects do exhibit a wavelike motion, but the wavelength is too small for humans to perceive. 6.39(a) 1.15 1036 m (b) 2 1033 m 6.41 2.2 1026 m/s 6.43 3.75 1036 kg/photon 6.47 The total probability of finding an electron at 52.9 pm is much greater for the 1s orbital than for the 2s orbital. 6.48(a) principal determinant of the electron’s energy or distance from the nucleus (b) determines the shape of the orbital (c) determines the orientation of the orbital in three-dimensional space 6.49(a) one (b) five (c) three (d) nine 6.51(a) ml 2, 1, 0, 1, 2 (b) ml 0 (if n 1, then l 0) (c) ml 3, 2, 1, 0, 1, 2, 3 6.53(a)
(b) z
z x
y
x y
6.55 Subshell
Allowable mI values
(a) d (l 2) (b) p (l 1) (c) f (l 3)
2, 1, 0, 1, 2 1, 0, 1 3, 2, 1, 0, 1, 2, 3
# of orbitals
5 3 7
6.57(a) n 5 and l 0; one orbital (b) n 3 and l 1; three orbitals (c) n 4 and l 3; seven orbitals 6.59(a) no; correct: n 2, l 1, ml 1 or: n 2, l 0, ml 0 (b) allowed (c) allowed (d) no; correct: n 5, l 3, ml 3 or n 5, l 2, ml 0 6.62(a) E (2.180 1018 J) (1/n2). This is identical to the expression from Bohr’s theory. (b) 3.028 1019 J (c) 656.1 nm 6.63(a) The attraction of the nucleus for the electrons must be overcome. (b) The electrons in silver are more tightly held by the nucleus. (c) silver (d) Once the electron is freed from the atom, its energy increases in proportion to the frequency of the light. 6.66 Li2 6.68(a) 2 ¡ 1 (b) 5 ¡ 2 (c) 4 ¡ 2 (d) 3 ¡ 2 (e) 6 ¡ 3 6.72(a) l 1 or 2 (b) l 1 or 2 (c) l 3, 4, 5, or 6 (d) l 2 or 3 6.74(a) 1 6.022 1023 1 2 bZ2 a b 2 1 mol ninitial (b) 3.28 107 J/mol (c) 205 nm (d) 22.8 nm 6.76(a) 5.293 1011 m (b) 5.293 109 m 6.78 6.4 1027 photons 6.80(a) no overlap (b) overlap (c) two lines (d) At longer wavelengths, the hydrogen spectrum begins to become a continuous band. 6.82(a) 7.56 1018 J; 2.63 108 m (b) 5.122 1017 J; 3.881 109 m (c) 1.2 1018 J; 1.66 107 m 6.84(a) 1.87 1019 J (b) 3.58 1019 J 6.86(a) red (Sr); green (Ba) (b) 5.89 kJ (Sr); 5.83 kJ (Ba) 6.88(a) This is the wavelength of maximum absorbance, so it gives the highest sensitivity. (b) ultraviolet region (c) 1.93 102 g vitamin A/g oil 6.92 1.0 1018 photons/s 6.95 3s ¡ 2p; 3d ¡ 2p; 4s ¡ 2p; 3p ¡ 2s ¢E (2.18 10 18 J)a
A-24
Appendix G • Brief Answers to Selected Problems
7.31(a) O; Group 16; Period 2
Chapter 7 7.1 Elements are listed in the periodic table in an ordered, systematic way that correlates with a periodicity of their chemical and physical properties. The theoretical basis for the table in terms of atomic number and electron configuration does not allow for a “new element” between Sn and Sb. 7.3(a) predicted atomic mass 54.23 u (b) predicted melting point 6.3C 7.6 The quantum number ms relates to just the electron; all the others describe the orbital. 7.9 Shielding occurs when electrons protect or shield other electrons from the full nuclear attraction. The effective nuclear charge is the nuclear charge an electron actually experiences. As the number of electrons, especially core electrons, increases, the effective nuclear charge decreases. 7.11(a) 6 (b) 10 (c) 2 7.13(a) 6 (b) 2 (c) 14 7.16 Hund’s rule states that electrons will occupy empty orbitals in a given subshell(with parallel spins) before filling half-filled orbitals. The lowest energy arrangement has the maximum number of unpaired electrons with parallel spins. (a) correct (b) incorrect 1s
2s
2p
1s
2s
2p
7.18 Main-group elements from the same group have similar valence electron configurations. Valence electron configurations in a period (row) vary, with each succeeding element having an additional electron. 7.20 The maximum number of electrons in any energy level n is 2n2, so the n 4 energy level holds a maximum of 2(42) 32 electrons. 7.21(a) n 5, l 0, ml 0, and ms 12 or 12 (b) n 3, l 1, ml 1, 0 or 1 and ms 12 or 12 (c) n 5, l 0, ml 0, and ms 12 or 12 (d) n 2, l 1, ml 1, and ms 12 or 12 7.23(a) Rb: 1s22s22p63s23p64s23d104p65s1 (b) Ge: 1s22s22p63s23p64s23d104p2 (c) Ar: 1s22s22p63s23p6 7.25(a) Cl: 1s22s22p63s23p5 (b) Si: 1s22s22p63s23p2 (c) Sr: 1s22s22p63s23p64s23d104p65s2 7.27(a) Ti: [Ar] 4s23d2 4s
3d
4p
(b) Cl: [Ne] 3s23p5 3s
3p
(c) V: [Ar] 4s23d3 4s 3d 7.29(a) Mn: [Ar] 4s23d5 4s 3d (b) P: [Ne] 3s23p3 3p 3s (c) Fe: [Ar] 4s23d6 4s
3d
4p
[He] 2s 2p (b) P; Group 15; Period 3 [Ne] 3s 3p 7.33(a) Cl; Group 17; Period 3 [Ne] 3s 3p (b) As; Group 15; Period 4 [Ar] 4s 3d 4p 7.35(a) [Ar] 4s23d104p1; Group 13 (b) [He] 2s22p6; Group 18 7.37 core Electrons
(a) O (b) Sn (c) Ca (d) Fe (e) Se
Valence Electrons
2 46 18 18 28
6 4 2 8 6
7.39(a) B; Al, Ga, In, and Tl (b) S; O, Se, Te, and Po (c) La; Sc, Y, and Ac 7.41(a) C; Si, Ge, Sn, and Pb (b) V; Nb, Ta, and Db (c) P; N, As, Sb, and Bi 7.43 Na (first excited state): 1s22s22p63p1 1s 2s 2p 3s 3p 7.50 A high IE1 and a very negative EA1 suggest that the elements are halogens, in Group 17, which form 1 ions. 7.53(a) K Rb Cs (b) O C Be (c) Cl S K (d) Mg Ca K 7.55(a) Ba Sr Ca (b) B N Ne (c) Rb Se Br (d) Sn Sb As 7.57 1s22s22p1 (boron, B) 7.59(a) Na (b) Na (c) Be 7.61(1) Metals conduct electricity; nonmetals do not. (2) Metal ions have a positive charge; nonmetal ions have a negative charge. (3) Metal oxides are mostly ionic and act as bases; nonmetal oxides are mostly covalent and act as acids. 7.62 Metallic character increases down a group and decreases to the right across a period. These trends are the same as those for atomic size and opposite those for ionization energy. 7.63 Possible ions are 2 and 4. 7.67(a) Rb (b) Ra (c) I 7.69(a) As (b) P (c) Be 7.71(a) Cl: 1s22s22p63s23p6 (b) Na: 1s22s22p6 (c) Ca2: 1s22s22p63s23p6 7.73(a) Al3: 1s22s22p6 (b) S2: 1s22s22p63s23p6 (c) Sr2: 1s22s22p63s23p64s2 3d104p6 7.75(a) 0 (b) 3 (c) 0 (d) 1 7.77 a, b, and d are paramegnetic 7.79(a) V3: [Ar] 3d 2, paramagnetic (b) Cd2: [Kr] 4d10, diamagnetic (c) Co3: [Ar] 3d 6, paramagnetic (d) Ag: [Kr] 4d10, diamagnetic 7.81 For palladium to be diamagnetic, all of its electrons must be paired. (a) You might first write the condensed electron configuration for Pd as [Kr] 5s24d 8. However, the partial orbital diagram is not consistent with diamagnetism. 5s
4d
5p
Appendix G • Brief Answers to Selected Problems
(b) This is the only configuration that supports diamagnetism, [Kr] 4d10. 5s 4d 5p (c) Promoting an s electron into the d subshell still leaves two electrons unpaired. 5s 4d 5p 7.83(a) Li Na K (b) Rb Br Se2 (c) F O2 N3 7.86 Ce: [Xe] 6s24f 15d1; Ce4: [Xe]; Eu: [Xe] 6s24f 7; Eu2: [Xe] 4f 7. Ce4 has a noble-gas configuration; Eu2 has a half-filled f subshell. 7.89 (a) SrBr2, strontium bromide (b) CaS, calcium sulfide (c) ZnF2, zinc fluoride (d) LiF, lithium fluoride 7.90 (a) 2009 kJ/mol (b) 549 kJ/mol 7.91 All ions except Fe8 and Fe14 are paramagnetic; Fe and Fe3 would be most attracted to a magnetic field.
Chapter 8 • 8.1(a) Greater ionization energy decreases metallic character. (b) Larger atomic radius increases metallic character. (c) Higher number of outer electrons decreases metallic character. (d) Larger effective nuclear charge decreases metallic character. 8.4(a) Cs (b) Rb (c) As 8.6(a) ionic (b) covalent (c) metallic 8.8(a) covalent (b) ionic (c) covalent 8.10(a) Rb (b) Si (c) I 8.12(a) Sr (b) P (c) S 8.14(a) 16; [noble gas] ns2np4 (b) 13; [noble gas] ns2np1 8.20(a) Ba2, [Xe]; Cl, [Ne]3s23p6, ⫺ 2⫺ 2 2 2 6 Cl ; BaCl2 (b) Sr , [Kr]; O , [He]2s 2p , O ; SrO ⫺ 3 2 6 (c) Al , [Ne]; F , [He]2s 2p , F ; AlF3 (d) Rb, [Kr]; O2, [He]2s22p6, O 2⫺; Rb2O. 8.22(a) 2 (b) 16 (c) 1 8.24(a) 13 (b) 2 (c) 16 8.26(a) BaS; the charge on each ion is twice the charge on the ions in CsCl. (b) LiCl; Li is smaller than Cs. 8.28(a) BaS; Ba2 is larger than Ca2. (b) NaF; the charge on each ion is half the charge on the ions in MgO. 8.30 788 kJ/ mol; the lattice energy for NaCl is less than that for LiF, because the Na and Cl ions are larger than Li and F ions. 8.33 336 kJ/mol 8.34 When two chlorine atoms are far apart, there is no interaction between them. As the atoms move closer together, the nucleus of each atom attracts the electrons of the other atom. The closer the atoms, the greater this attraction; however, the repulsions of the two nuclei and the electrons also increase at the same time. The final internuclear distance is the distance at which maximum attraction is achieved in spite of the repulsion. 8.35 The bond energy is the energy required to break the bond between H atoms and Cl atoms in one mole of HCl molecules in the gaseous state. Energy is needed to break bonds, so bond breaking is always endothermic and bond breakingH is positive. The quantity of energy needed to break the bond is released upon its formation, so bond formingH has the same magnitude as bond breakingH but is opposite in sign (always exothermic and negative). 8.39(a) I¬I Br¬Br Cl¬Cl (b) S¬Br S¬Cl S¬H (c) C¬N C“N C‚N 8.41(a) C¬O C“O; the C“O bond (bond order 2) is stronger than the C¬O bond (bond order 1). (b) C¬H O¬H; O is smaller than C so the O¬H bond is shorter and stronger than the C¬H bond. 8.43 Less energy is required to break weak bonds. 8.45 Both
A-25
are one-carbon molecules. Since methane contains fewer carbonoxygen bonds, it will have the greater heat of reaction per mole for combustion. 8.47 168 kJ/mol 8.49 22 kJ/mol 8.50 59 kJ/mol 8.51 Electronegativity increases from left to right and increases from bottom to top within a group. Fluorine and oxygen are the two most electronegative elements. Cesium and francium are the two least electronegative elements. 8.53 The H¬O bond in water is polar covalent. A nonpolar covalent bond occurs between two atoms with identical electronegativities. A polar covalent bond occurs when the atoms have differing electronegativities. Ionic bonds result from electron transfer between atoms. 8.56(a) Si S O (b) Mg As P 8.58(a) N P Si (b) As Ga Ca none 8.60 (a) N B (b) N O (c) C S (d) S O (e) N H (f) Cl O 8.62 a, d, and e 8.64(a) nonpolar covalent (b) ionic (c) polar covalent (d) polar covalent (e) nonpolar covalent (f) polar covalent; SCl2 SF2 PF3 ⬍ 8.66 (a) H I ⬍ H Br H Cl (b) H
⬍
C
H
O
⬍
H
F
⬍ (c) S Cl ⬍ P Cl Si Cl 8.69 He cannot serve as a central atom because it does not bond. H cannot because it forms only one bond. Fluorine cannot because it needs only one electron to complete its valence level, and it does not have d orbitals available to expand its valence level. Thus, it can bond to only one other atom. 8.71 All the structures obey the octet rule except c and g. F (b) Cl Se Cl (c) F C F 8.73(a) F
F
Si
O
F
8.75(a)
P
F
(b)
F
H
O
F
8.77(a)
N
O
¢£ N
O
N
O
C
S
⫺
¢£ ¢£
O N O
N N
N O
⫺ ⫺
¢£
N
N
N
⫺
formal charges: FCI 0, FCF 0
F F
I
F
F ⫺
H Al H
8.83(a) (b) 8.85(a)
⫺
N
H
S
N O
O
N
(b)
(c)
F
N
F
H
⫹
F O
8.79(a) (b) 8.81(a)
O
O
O
(b)
C
C
N
Cl
O
O
Br O
formal charges: FCH 0, FCAl 1
H ⫺
formal charges: FCC 1, FCN 0 formal charges: FCCl 0, FCO 1 ⫺ formal charges: FCBr 0, doubly O bonded FCO 0, singly bonded FCO 1; oxidation numbers: oxidation numberBr 5; oxidation numberO 2
⫺
A-26
Appendix G • Brief Answers to Selected Problems 2⫺
formal charges: FSS 0, singly bonded FSO 1, doubly bonded FSO 0; oxidation numbers: oxidation numberS 4; oxidation numberO 2 8.87(a) B has 6 valence electrons in BH3, so the molecule is electron deficient. (b) As has an expanded valence level with 10 electrons. (c) Se has an expanded valence level with 10 electrons. ⫺ (b) (c) (a) H Cl F O
S
O
(b)
O
B
F
H
As
F
Se
Cl
8.124
O
C
C
O
O
O
C
Cl
HC2O4⫺:
Cl
Be
⫹
Cl
C
C2O42⫺:
⫺
±£
Cl
Be
O
H
H
¢£
C
H
N
C
2–
¢£
C O
2–
O
¢£
C O
O C
2–
C O
O
Chapter 9 Answers to Boxed Reading Problem: B9.1 resonance form on the left: trigonal planar around C, trigonal pyramidal around N; resonance form on the right: trigonal planar around both C and N. • 9.2 The molecular shape and the electron-group arrangement are the same when no lone pairs are present on the central atom. 9.4 tetrahedral, AX4; trigonal pyramidal, AX3E; bent or V shaped, AX2E2 X X 9.6 X A
X
X X
X
A X
X
(a)
X
A X (d)
A
(e)
X
X
A X
X
X
(c) X
X
A X
X (b)
X
N
rH 367 kJ/mol 8.122(a) 1267 kJ/mol (b) 1226 kJ/mol (c) 1234.8 kJ/mol. The two answers differ by less than 10 kJ/mol. This is very good agreement since average bond energies were used to calculate answers a and b. (d) 37 kJ/mol
O
O O
–
C
O C
O
H
In H2C2O4, there are two shorter and stronger C“O bonds and two longer and weaker C¬O bonds. In HC2O42, the carbonoxygen bonds on the side retaining the H remain as one long, weak C¬O and one short, strong C“O. The carbon-oxygen bonds on the other side of the molecule have resonance forms with a bond order of 1.5, so they are intermediate in length and strength. In C2O42, all the carbon-oxygen bonds have a bond order of 1.5. 8.135 22 kJ/mol
A
⫹
C
O
O
Nitrogen
H
2–
¢£
H
±£
¢£
O
The single N¬N bond (bond order 1) is weaker and longer than the others. The triple bond (bond order 3) is stronger and shorter than the others. The double bond (bond order 2) has an intermediate strength and length. (b) H N N N N H H N N H
O
O O
O
Cl
Cl
Diazene
–
O
O
8.94 structure A 8.95(a) Shiny, conducts heat, conducts electricity, and is malleable. (b) Metals lose electrons to form positive ions 8.99(a) 800. kJ/mol, which is lower than the value in Table 8.2 (b) 2.417 104 kJ (c) 1690. g CO2 (d) 65.2 L O2 8.101(a) 125 kJ/mol (b) yes, since fH is negative (c) 392 kJ/mol (d) No, fH for MgCl2 is much more negative than that for MgCl. 8.103(a) 406 nm (b) 2.93 1019 J (c) 1.87 104 m/s 8.106 C¬Cl: 3.53 107 m; bond in O2: 2.40 107 m 8.107 XeF2: 132 kJ/mol; XeF4: 150. kJ/mol; XeF6: 146 kJ/mol 8.109(a) The presence of the very electronegative fluorine atoms bonded to one of the carbons makes the C¬C bond polar. This polar bond will tend to undergo heterolytic rather than homolytic cleavage. More energy is required to achieve heterolytic cleavage. (b) 1420 kJ/mol 8.112 13,286 kJ/mol 8.114 8.70 1014 s1; 3.45 107 m; the ultraviolet region of the electromagnetic spectrum 8.116(a) CH3OCH3(g): 326 kJ/ mol; CH3CH2OH(g): 369 kJ/mol (b) The formation of gaseous ethanol is more exothermic. (c) 43 kJ/mol H N N H N N 8.118(a) H N N H Hydrazine
H
C
O
2⫺
Cl
Cl
H
O
C ⫺
Cl
O
O
F Cl
C
O
8.89(a) Br expands its valence level to 10 electrons. (b) I has an expanded valence level of 10 electrons. (c) Be has only 4 valence electrons in BeF2, so the molecule is electron deficient. ⫺ (a) F Br F (b) Cl I Cl (c) F Be F 8.91
H
8.127 CH4: 409 kJ/mol O2; H2S: 398 kJ/mol O2 8.129(a) The O in the OH species has only 7 valence electrons, which is less than an octet, and 1 electron is unpaired. (b) 426 kJ/mol (c) 508 kJ/mol 8.132 H2C2O4: H O O H
H F
H
X
X (f)
Appendix G • Brief Answers to Selected Problems
9.8(a) trigonal planar, bent, 120 (b) tetrahedral, trigonal pyramidal, 109.5 (c) tetrahedral, trigonal pyramidal, 109.5 9.10(a) trigonal planar, trigonal planar, 120 (b) trigonal planar, bent, 120 (c) tetrahedral, tetrahedral, 109.5 9.12(a) trigonal planar, AX3, 120 (b) trigonal pyramidal, AX3E, 109.5 (c) trigonal bipyramidal, AX5, 90 and 120 9.14(a) bent, 109.5, less than 109.5 (b) trigonal bipyramidal, 90 and 120, angles are ideal (c) seesaw, 90 and 120, less than ideal (d) linear, 180, angle is ideal 9.16(a) C: tetrahedral, 109.5; O: bent, 109.5 (b) N: trigonal planar, 120 9.18(a) C in CH3: tetrahedral, 109.5; C with C“O: trigonal planar, 120; O with H: bent, 109.5 (b) O: bent, 109.5 9.20 OF2 NF3 CF4 BF3 BeF2 9.22(a) The C and N each have three groups, so the ideal angles are 120; the O has four groups, so the ideal angle is 109.5. The N and O have lone pairs, so the angles are less than ideal. (b) All central atoms have four pairs, so the ideal angles are 109.5. The lone pairs on the O reduce this value. (c) The B has three groups and an ideal bond angle of 120. All the oxygen atoms have four groups (ideal bond angles of 109.5), two of which are lone pairs that reduce the angle. ⫺ ⫹ 9.25 Cl Cl Cl Cl
HCl
P
Cl
Cl
Cl
P Cl
Cl
Cl Cl
Cl
Cl
C
Cl
C H
X
H
C
C
C
H
H
Y
Cl
Z
Yes, compound Y has a dipole moment. 9.37(a) formal charges for Al2Cl6: FCAl 1, FCend Cl 0, FCbridging Cl 1; formal charges for I2Cl6: FCI 1, FCend Cl 0, FCbridging Cl 1 (b) The iodine atoms are each AX4E2, and the shape around each is square planar. These square planar portions are adjacent, giving a planar molecule. 9.42 H
H C H3C
CH2
⫹
H2O2
±£
H3C
CH2
C
Chapter 10 10.1(a) sp2 (b) sp3d2 (c) sp (d) sp3 (e) sp3d 10.3 C has only 2s and 2p atomic orbitals, allowing for a maximum of four hybrid orbitals. Si has 3s, 3p, and 3d atomic orbitals, allowing it to form more than four hybrid orbitals. 10.5(a) six, sp3d2 (b) four, sp3 10.7(a) sp2 (b) sp2 (c) sp2 10.9(a) sp3 (b) sp3 (c) sp3 10.11(a) Si: one s and three p atomic orbitals form four sp3 hybrid orbitals. (b) C: one s and one p atomic orbital forms two sp hybrid orbitals. (c) S: one s, three p, and one d atomic orbital mix to form five sp3d hybrid orbitals. (d) N: one s and three p atomic orbitals mix to form four sp3 hybrid orbitals. 10.13(a) B (sp3 ¡ sp3) (b) A (sp2 ¡ sp3) ±£ 10.15 (a) s
p
s
p
⫹
H2O
O
(a) In epoxypropane, the shape around each C is tetrahedral, with ideal angles of 109.5. (b) The C that is not part of the three-membered ring should have close to the ideal angle. The atoms in the ring form an equilateral triangle, so the angles
sp3
±£
(b)
In the gas phase, PCl5 is AX5, so the shape is trigonal bipyramidal, and the bond angles are 120 and 90. The PCl4 ion is AX4, so the shape is tetrahedral, and the bond angles are 109.5. The PCl6 ion is AX6, so the shape is octahedral, and the bond angles are 90. 9.26 Molecules are polar if they have polar bonds that are not arranged to cancel each other. A polar bond is present any time there is a bond between elements with differing electronegativities. 9.29(a) CF4 (b) BrCl and SCl2 9.31(a) SO2, because it is polar and SO3 is not. (b) IF has a greater electronegativity difference between its atoms. (c) SF4, because it is polar and SiF4 is not. (d) H2O has a greater electronegativity difference between its atoms. H Cl Cl Cl Cl 9.33 H C
around the two carbon atoms in the ring are reduced from the ideal 109.5 to 60. 9.45(a) The F atoms will substitute at the axial positions first. (b) PF5 and PCl3F2 9.48 Trigonal planar molecules are nonpolar, so AY3 cannot be that shape. Trigonal pyramidal molecules and T-shaped molecules are polar, so AY3 could have either of these shapes. 9.51(a) 339 pm (b) 316 pm and 223 pm (c) 270 pm
Cl
P
A-27
sp2
p
⫹ e⫺
±£
(c) s
sp2
p
p
±£
10.17 (a) s
sp3
p
⫹ e⫺
±£
(b) s
p
sp3
⫹ e⫺ ±£
(c) s
p
d
sp3d 2 d3 10.20(a) False. A double bond is one and one bond. (b) False. A triple bond consists of one and two bonds. (c) True (d) True (e) False. A bond consists of a second pair of electrons after a bond has been previously formed. (f) False. End-to-end overlap results in a bond with electron density along the bond axis. 10.21(a) Nitrogen uses sp2 to form three bonds and one bond. (b) Carbon uses sp to form two bonds and two bonds. (c) Carbon uses sp2 to form three bonds and one bond. 10.23(a) N: sp2, forming 2 bonds and 1 bond F
N
O
(b) C: sp , forming 3 bonds and 1 bond 2
F
F C
C
F
F
(c) C: sp, forming 2 bonds and 2 bonds N
C
C
N
C CH3
CH2
H
CH2
C
C
H
cis
CH3
trans
The single bonds are all bonds. The double bond is one bond and one bond. 10.26 Four MOs form from the four p atomic orbitals. The total number of MOs must equal the number of atomic orbitals. 10.28(a) Bonding MOs have lower energy than antibonding MOs. Lower energy means more stable. (b) Bonding MOs do not have a nodal plane perpendicular to the bond. (c) Bonding MOs have higher electron density between nuclei than antibonding MOs. 10.30(a) two (b) two (c) four 10.32(a) A is *2p, B is 2p, C is 2p, and D is *2p. (b) *2p (A), 2p (B), and 2p (C) have at least one electron. (c) *2p (A) has only one electron. 10.34(a) stable (b) paramagnetic (c) ( 2s)2( *2s)1 10.36(a) C2 C2 C2 (b) C2 C2 C2 10.40(a) C (ring): sp2; C (all others): sp3; O (all): sp3; N: sp3 (b) 26 bonds (c) 6 electrons 10.42(a) 17 bonds (b) All carbons are sp2; the ring N is sp2, the other N atoms are sp3. 10.44(a) B changes from sp2 to sp3. (b) P changes from sp3 to sp3d. (c) C changes from sp to sp2. Two electron groups surround C in C2H2, and three electron groups surround C in C2H4. (d) Si changes from sp3 to sp3d2. (e) no change for S 10.46 P: tetrahedral, sp3; N: trigonal pyramid, sp3; C1 and C2: tetrahedral, sp3; C3: trigonal planar, sp2 10.51(a) B and D are present. (b) Yes, sp hybrid orbitals. (c) Two sets of sp orbitals, four sets of sp2 orbitals, and three sets of sp3 orbitals. 10.52 Through resonance, the C¬N bond gains some double-bond character, which hinders rotation about that bond. O C
O N
¢£
H
C
⫺ ⫹
“stick” more frequently, so the condensation rate increases. When the vapourization and condensation rates become equal, the vapour pressure becomes constant. 11.14 As intermolecular forces increase, (a) critical temperature increases, (b) boiling point increases, (c) vapour pressure decreases, and (d) heat of vapourization increases. 11.18 because the condensation of the vapour supplies an additional 41 kJ/mol 11.19 7.67 103 J 11.21 78.7 kPa 11.23 21.3 kJ/mol 11.25 51.2
p (bar)
H C
Answer to Boxed Reading Problem: B11.1 1.76 1010 m • 11.1 In a solid, the energy of attraction of the particles is greater than their energy of motion; in a gas, it is less. Gases have high compressibility and the ability to flow, while solids have neither. 11.4(a) Because the intermolecular forces are only partially overcome when fusion occurs but need to be totally overcome in vapourization. (b) Because solids have greater intermolecular forces than liquids do. (c) vap H condH 11.5(a) intermolecular (b) intermolecular (c) intramolecular (d) intramolecular 11.7(a) condensation (b) fusion (c) evaporation 11.9 The gas molecules slow down as the gas is compressed. Therefore, much of the kinetic energy lost by the propane molecules is released to the surroundings. 11.13 At first, the vapourization of liquid molecules from the surface predominates, which increases the number of gas molecules and hence the vapour pressure. As more molecules enter the gas phase, more gas molecules hit the surface of the liquid and
G
1
⫺200
⫺100 T (°C)
0
Solid ethene is denser than liquid ethene. 11.28 3280 kPa 11.32 O is smaller and more electronegative than Se; so the electron density on O is greater, which attracts H more strongly. 11.34 All particles (atoms and molecules) exhibit dispersion forces, but the total force is weak for small molecules. Dipole-dipole forces between small polar molecules dominate the dispersion forces. 11.37(a) hydrogen bonding (b) dispersion forces (c) dispersion forces 11.39(a) dipole-dipole forces (b) dispersion forces (c) hydrogen bonding 11.41 (a) H H H H
H
Chapter 11
L
10⫺3
N
10.55(a) C in ¬CH3: sp3; all other C atoms: sp2; O in two C¬O bonds: sp3; O in two C“O bonds: sp2 (b) two (c) eight; one 10.56(a) four (b) eight
S
H
H
H
C
H
10.25
Appendix G • Brief Answers to Selected Problems
H
A-28
H
C
C
H F
H H
H
O
H H
C O
H
C
C H
F H F (b) H 11.43(a) dispersion forces (b) hydrogen bonding (c) dispersion forces 11.45(a) I (b) CH2“CH2 (c) H2Se. In (a) and (c) the larger particle has the higher polarizability. In (b), the less tightly held electron clouds are more easily distorted. 11.47(a) C2H6; it is a smaller molecule exhibiting weaker dispersion forces than C4H10. (b) CH3CH2F; it has no H¬F bonds, so it exhibits only dipole-dipole forces, which are weaker than the hydrogen bonds of CH3CH2OH. (c) PH3; it has weaker intermolecular forces (dipole-dipole) than NH3 (hydrogen bonding). 11.49(a) HCl; it has dipole-dipole forces, and there are stronger ionic bonds in LiCl. (b) PH3; it has dipole-dipole forces, and there is stronger hydrogen bonding in NH3. (c) Xe; it exhibits weaker dispersion forces since its smaller size results in lower polarizability than the larger I2 molecules. 11.51(a) C4H8 (cyclobutane), because it is more compact than C4H10. (b) PBr3; the dipole-dipole forces in PBr3 are weaker than the ionic bonds in NaBr. (c) HBr; the dipoledipole forces in HBr are weaker than the hydrogen bonds in water. 11.53 As atomic size decreases and electronegativity increases, the electron density of an atom increases. Thus, the attraction to an H atom on another molecule increases while its
Appendix G • Brief Answers to Selected Problems
all directions is isotropic; otherwise, the substance is anisotropic. Liquid crystals have a degree of order only in certain directions, so they are anisotropic. 11.107(a) n-type semiconductor (b) p-type semiconductor 11.110(a) 2.6453 kPa (b) 0.0486 g 11.115 259 K 11.116(a) 2.26 kPa (b) 6.24 L 11.119(a) 2-furoic acid H O H
H C
O C
C
C
C
O
C
H
O
O
O
C
C C
C
H
H
H
H furfuryl alcohol H C
O
O C
C H
H
C
H
O C
C H
O
H
H
C C
H
(b) 2-furoic acid
H
H
H
O
O
C C
C
H
furfuryl alcohol H
C
C
H
H
H
H C
C
C H
O
O C
O
C H
C
C H
H
H
11.120(a) 50.1 metric tonne H2O (b) 1.12 108 kJ 11.121 2.9 g/m3 11.126(a) 1.1 min (b) 10. min (c) 100 80 60 40 20 0 20 40
T (°C)
bonded H atom becomes more positive. Fluorine is the smallest of the three and the most electronegative, so the hydrogen bonds in hydrogen fluoride are the strongest. Oxygen is smaller and more electronegative than nitrogen, so hydrogen bonds in water are stronger than hydrogen bonds in ammonia. 11.57 The cohesive forces in water and mercury are stronger than the adhesive forces to the nonpolar wax on the floor. Weak adhesive forces result in spherical drops. The adhesive forces overcome the even weaker cohesive forces in the oil, and so the oil drop spreads out. 11.59 Surface tension is defined as the energy needed to increase the surface area by a given amount, so units of energy per area are appropriate. 11.61 CH3CH2CH2OH HOCH2CH2OH HOCH2CH(OH) CH2OH. More hydrogen bonding means more attraction between molecules, so more energy is needed to increase surface area. 11.63 HOCH2CH(OH)CH2OH HOCH2CH2OH CH3CH2CH2OH. More hydrogen bonding means more attraction between neighbouring molecules, so they flow less easily. 11.68 Water is a good solvent for polar and ionic substances and a poor solvent for nonpolar substances. Water is a polar molecule and dissolves polar substances because their intermolecular forces are of similar strength. 11.69 A single water molecule can form four hydrogen bonds. The two hydrogen atoms each form one hydrogen bond to oxygen atoms on neighbouring water molecules. The two lone pairs on the oxygen atom form hydrogen bonds with hydrogen atoms on two neighbouring molecules. 11.72 Water exhibits strong capillary action, which allows it to be easily absorbed by the plant’s roots and transported upward to the leaves. 11.78 simple cubic 11.81 The energy gap is the energy difference between the highest filled energy level (valence band) and the lowest unfilled energy level (conduction band). In conductors and superconductors, the energy gap is zero because the valence band overlaps the conduction band. In semiconductors, the energy gap is small. In insulators, the gap is large. 11.83 atomic mass and atomic radius 11.84(a) face-centred cubic (b) bodycentred cubic (c) face-centred cubic 11.86(a) The change in unit cell is from a sodium chloride structure in CdO to a zinc blende structure in CdSe. (b) Yes, the coordination number of Cd changes from 6 in CdO to 4 in CdSe. 11.88(a) Nickel forms a metallic solid since it is a metal whose atoms are held together by metallic bonds. (b) Fluorine forms a molecular solid since the F2 molecules are held together by dispersion forces. (c) Methanol forms a molecular solid since the CH3OH molecules are held together by hydrogen bonds. (d) Tin forms a metallic solid since it is a metal whose atoms are held together by metallic bonds. (e) Silicon is in the same group as carbon, so it exhibits similar bonding properties. Since diamond and graphite are both network covalent solids, it makes sense that Si forms a network covalent solid as well. (f) Xe is an atomic solid since its individual atoms are held together by dispersion forces. 11.90 four 11.92(a) four Se2 ions, four Zn2 ions (b) 577.48 u (c) 1.77 1022 cm3 (d) 5.61 108 cm 11.94(a) insulator (b) conductor (c) semiconductor 11.96(a) Conductivity increases. (b) Conductivity increases. (c) Conductivity decreases. 11.98 1.68 108 cm 11.105 A substance whose properties are the same in
A-29
0
10
20
30 40 Time (min)
50
60
11.127 2.98 105 g BN 11.130 45.98 u
Chapter 12 Answers to Boxed Reading Problem: B12.1 (a) The colloidal particles in water generally have negatively charged surfaces and so repel each other, slowing the settling process. Cake alum, Al2(SO4)3, is added to coagulate the colloids. The Al3 ions neutralize the negative surface charges and allow the particles to aggregate and settle. (b) Water that contains large amounts of divalent cations (such as Ca2, and Mg2) is called hard water. During cleaning, these ions combine with the fatty-acid anions in soaps to produce insoluble deposits. (c) In reverse osmosis, a pressure greater than the osmotic pressure is applied to the solution, forcing the water back through the membrane, leaving the ions behind. (d) Chlorine may give the water an unpleasant odor, and can form carcinogenic chlorinated compounds. (e) The high concentration of NaCl displaces the divalent and polyvalent ions from the ion-exchange resin.
Appendix G • Brief Answers to Selected Problems
HO
C
C
C
C
H
H
H
H
OH
Does not form H bonds: H H
C H
H
Enthalpy
12.25(a) The volume of Na is smaller. (b) Sr2 has a larger ionic charge and a smaller volume. (c) Na is smaller than Cl. (d) O2 has a larger ionic charge with a similar ion volume. (e) OH has a smaller volume than SH. (f) Mg2 has a smaller volume. (g) Mg2 has both a smaller volume and a larger ionic charge. (h) CO32 has both a smaller volume and a larger ionic charge. 12.27(a) Na (b) Sr2 (c) Na (d) O2 (e) OH (f) Mg2 (g) Mg2 (h) CO32 12.29(a) 704 kJ/mol (b) The K ion contributes more because it is smaller and, therefore, has a greater charge density. 12.31(a) increases (b) decreases (c) increases 12.34 Add a pinch of X to each solution. Addition of a “seed” crystal of solute to a supersaturated solution causes the excess solute to crystallize immediately, leaving behind a saturated solution. The solution in which the added X dissolves is the unsaturated solution. The solution in which the added X remains undissolved is the saturated solution. 12.37(a) increase (b) decrease 12.39(a) 0.102 g O2 (b) 0.0214 g O2 12.42 0.20 mol/L 12.45(a) concentration (mol/L) and parts-by-volume (% w/v or
O
C H
O
H
Δ solutionH KCl(s)
H
K(aq) Cl(aq)
H
Δ hydrationH
H
Δ latticeH
% v/v) (b) parts-by-mass (% w/w) (c) molality 12.47 With just this information, you can convert between molality and concentration (mol/L), but you need to know the molar mass of the solvent to convert to mole fraction. 12.49(a) 0.944 mol/L C12H22O11 (b) 0.167 mol/L LiNO3 12.51(a) 0.0749 mol/L NaOH (b) 0.36 mol/L HNO3 12.53(a) Add enough distilled water to 4.25 g KH2PO4 to make 365 mL of aqueous solution. (b) Add enough distilled water to 125 mL of 1.25 mol/L NaOH to make 465 mL of solution. 12.55(a) Weigh out 48.0 g KBr, dissolve it in about 1 L distilled water, and then dilute to 1.40 L with distilled water. (b) Measure 82.7 mL of the 0.264 mol/L LiNO3 solution and add distilled water to make a total of 255 mL. 12.57(a) 0.896 mol/kg glycine (b) 1.21 mol/kg glycerol 12.59 4.48 mol/kg C6H6 12.61(a) Add 2.39 g C2H6O2 to 308 g H2O. (b) Add 0.0508 kg of 52.0 mass % HNO3 to 1.15 kg H2O to make 1.20 kg of 2.20 mass % HNO3. 12.63(a) 0.29 (b) 58 mass % (c) 23 mol/kg C3H7OH 12.65 42.6 g CsBr; mole fraction 7.16 103; 7.84 mass % 12.67 5.11 mol/kg NH3; 4.53 mol/L NH3; mole fraction 0.0843 12.69 2.5 ppm Ca2; 0.56 ppm Mg2 12.73 It conducts a large current. A strong electrolyte dissociates completely into ions in solution. 12.75 The boiling point is higher and the freezing point is lower for the solution than for the solvent. 12.78 A dilute solution of an electrolyte behaves more ideally than a concentrated one. With increasing concentration, the effective concentration deviates from the molar concentration because of ionic attractions. Thus, 0.050 mol/kg NaF has a boiling point closer to its predicted value. 12.81(a) strong electrolyte (b) strong electrolyte (c) nonelectrolyte (d) weak electrolyte 12.83(a) 0.6 mol of solute particles (b) 0.13 mol (c) 2 104 mol (d) 0.06 mol 12.85(a) CH3OH in H2O (b) H2O in CH3OH solution 12.87(a) II I III (b) bpII bpI bpIII (c) fpIII fpI fpII (d) vpIII vpI vpII 12.89 3.13 kPa 12.91 0.467C 12.93 79.5C 12.95 1.18 104 g C2H6O2 12.97(a) NaCl: 0.173 mol/kg and i 1.84 (b) CH3COOH: 0.0837 mol/kg and i 1.02 12.100 27.8 kPa for CH2Cl2; 6.40 kPa for CCl4 12.101 The fluid inside a bacterial cell is both a solution and a colloid. It is a solution of ions and small molecules and a colloid of large molecules (proteins and nucleic acids). 12.105 Soap micelles have nonpolar tails pointed inward and anionic heads pointed outward. The like charges on the heads of one micelle repel those on the heads of a neighbouring micelle. This repulsion between micelles keeps them from coagulating. Soap is more effective in freshwater than in seawater because the divalent cations in seawater combine with the anionic heads to form a precipitate. 12.109 3.4 109 L 12.113 0.0C: 4.53 104 mol/L O2; 20.0C: 2.83 104 mol/L O2; 40.0C: 2.00 104 mol/L O2 12.115(a) 89.9 g/mol (b) C2H5O; C4H10O2 (c) Forms H bonds: H
• 12.2 When a salt such as NaCl dissolves, ion-dipole forces cause the ions to separate, and many water molecules cluster around each ion in hydration shells. Ion-dipole forces bind the outermost shell to an ion. The water molecules in that shell form hydrogen bonds to others to create the next shell, and so on. 12.4 Sodium stearate is a more effective soap because the hydrocarbon chain in the stearate ion is longer than that in the ethanoate (acetate) ion. The longer chain allows for more dispersion forces with grease molecules. 12.7 KNO3 is an ionic compound and is therefore more soluble in water. 12.9(a) iondipole forces (b) hydrogen bonding (c) dipoleinduced dipole forces 12.11(a) hydrogen bonding (b) dipoleinduced dipole forces (c) dispersion forces 12.13(a) HCl(g), because the molecular interactions (dipole-dipole forces) in ethoxyethane are like those in HCl but not like the ionic bonding in NaCl. (b) CH3CHO(l), because the molecular interactions with ethoxyethane (dipole-dipole) are like those between CH3CHO, but not like the hydrogen bonds in water. (c) CH3CH2MgBr(s), because the molecular interactions (dipole-dipole and dispersion forces) are greater than between ethoxyethane and the ions in MgBr2. 12.16 Gluconic acid is soluble in water due to extensive hydrogen bonding by the OH groups attached to five of its carbons. The dispersion forces in the nonpolar tail of caproic acid are more similar to the dispersion forces in hexane; thus, caproic acid is soluble in hexane. 12.18 The energy changes needed to separate solvent into particles ( solventH), and that needed to mix the solvent and solute particles ( mixH) would be combined to obtain
solutionH. 12.22 Very soluble, because a decrease in enthalpy and an increase in entropy both favor the formation of a solution. 12.23 K(g) Cl(g)
H
A-30
C
C
H
H
H
Appendix G • Brief Answers to Selected Problems
12.119(a) 9.45 g NaF (b) 0.0017 g F 12.122(a) 68 g/mol (b) 2.1 102 g/mol (c) The molar mass of CaN2O6 is 164.10 g/mol. This value is less than the 2.1 102 g/mol calculated when the compound is assumed to be a strong electrolyte and is greater than the 68 g/mol calculated when the compound is assumed to be a nonelectrolyte. Thus, the compound does not dissociate completely in solution. (d) i 2.4 12.126(a) 1.82 104 g/mol (b) 3.41 105C 12.127 8.2 105 ng/L 12.131(a) 0.02 (b) 5 101 kPaL /mol (c) yes 12.133 Weigh 3.11 g of NaHCO3 and dissolve in 247 g of water. 12.137 c (mol solute/L solution) m (kg solvent/L solution) m dsolution. Thus, for very dilute solutions, molality density concentration (mol/L). For an aqueous solution, the number of litres of solution is approximately the same as the kg of solvent because the density of water is close to 1 kg/L, so m c. 12.139(a) 2.5 103 mol/L SO2 (b) The base reacts with the sulfur dioxide to produce calcium sulfite. The reaction of sulfur dioxide makes “room” for more sulfur dioxide to dissolve. 12.144(a) 7.74 103 mol /LkPa (b) 4 105 mol/L (c) 3 106 (d) 1 ppm 12.148(a)2.09 104 mol/LkPa (b) 8.84 ppm (c) kH: C6F14 C6H14 ethanol water. To dissolve oxygen in a solvent, the solvent molecules must be moved apart to make room for the gas. The stronger the intermolecular forces in the solvent, the more difficult it is to separate solvent particles and the lower the solubility of the gas. Both C6F14 and C6H14 have weak dispersion forces, with C6F14 having the weaker forces due to the electronegative fluorine atoms repelling each other. In both ethanol and water, the molecules are held together by strong hydrogen bonds with those bonds being stronger in water, as the boiling point indicates. 12.150(a) Yes, the phases of water can still coexist at some temperature and can therefore establish equilibrium. (b) The triple point would occur at a lower pressure and lower temperature because the dissolved air would lower the vapour pressure of the solvent. (c) Yes, this is possible because the gas-solid phase boundary exists below the new triple point. (d) No; at both temperatures, the presence of the solute lowers the vapour pressure of the liquid. 12.152(a) 2.8 104 g/mL (b) 81 mL
Chapter 13 13.1 The outermost electron is attracted by a smaller effective nuclear charge in Li because of shielding by the inner electrons, and it is farther from the nucleus in Li than in H. Both of these factors lead to a lower ionization energy for Li. 13.3(a) NH3 will form hydrogen bonds. F
N
F
H
N
H
H
F
(b) CH3CH2OH will form hydrogen bonds. H
H H
O
C H
H
H
H
C
H
H
C
C
H
H
O
H
13.5(a) 2Al(s) 6HCl(aq) ¡ 2AlCl3(aq) 3H2(g) (b) LiH(s) H2O(l) ¡ LiOH(aq) H2(g)
A-31
13.7(a) NaBH4: 1 for Na, 3 for B, 1 for H Al(BH4)3: 3 for Al, 3 for B, 1 for H LiAlH4: 1 for Li, 3 for Al, 1 for H (b) tetrahedral
H H
B
H
H
13.12(a) Group 13 or 3 (b) If E is in Group 3, the oxide will be more basic and the fluoride will be more ionic; if E is in Group 13, the oxide will be more acidic and the fluoride will be more covalent. 13.15(a) reducing agent (b) Alkali metals have relatively low ionization energies, which means they easily lose the outermost electron. (c) 2Na(s) 2H2O(l) ¡ 2Na(aq) 2OH2(aq) H2(g) and 2Na(s) Cl2(g) ¡ 2NaCl(s) 13.17 Density and ionic size increase down a group; the other three properties decrease down a group. 13.19 2Na(s) O2(g) ¡ Na2O2(s) 13.21 K2CO3(s) 2HI(aq) ¡ 2KI(aq) H2O(l) CO2(g) 13.25 Group 2 metals have an additional valence electron. The greater amount of shared electrons increases the strength of metallic bonding, which leads to higher melting points, higher boiling points, greater hardness, and greater density. 13.26(a) CaO(s) H2O(l) ¡ Ca(OH)2(s) (b) 2Ca(s) O2(g) ¡ 2CaO(s) 13.29(a) BeO(s) H2O(l) ¡ no reaction (b) BeCl2(l) 2Cl(solvated) ¡ BeCl42(solvated). Be behaves like other Group 2 elements in reaction (b). 13.32 The electron removed from Group 2 atoms occupies the outer s orbital, whereas in Group 13 atoms, the electron occupies the outer p orbital. For example, the electron configuration for Be is 1s22s2 and for B it is 1s22s22p1. It is easier to remove the p electron of B than an s electron of Be, because the energy of a p orbital is higher than that of the s orbital of the same level. Even though atomic size decreases because of increasing Zeff, IE decreases from 2 to 13. 13.33(a) Most atoms form stable compounds when they complete their valence shell (octet). Some compounds of Group 13 elements, like boron, have only six electrons around the central atom. Having fewer than eight electrons is called electron deficiency. (b) BF3(g) NH3(g) ¡ F3BNH3(g) B(OH)3(aq) OH(aq) ¡ B(OH)4(aq) 13.35 In2O3 Ga2O3 Al2O3 13.37 Apparent O.N., 3; The anion I3 has the general actual O.N., 1. I I I formula AX2E3 and bond angles of 180. (Tl3)(I)3 does not exist because of the low strength of the Tl¬I bond. 13.42(a) B2O3(s) 2NH3(g) ¡ 2BN(s) 3H2O(g) (b) 1.30 102 kJ/mol (c) 5.3 kg borax 13.43 Basicity in water is greater for the oxide of a metal. Tin(IV) oxide is more basic in water than carbon dioxide because tin has more metallic character than carbon. 13.45(a) Ionization energy generally decreases down a group. (b) The deviations (increases) from the expected trend are due to the presence of the first transition series between Si and Ge and of the lanthanides between Sn and Pb. (c) Group 13 13.48 Atomic size increases down a group. As atomic size increases, ionization energy decreases and so it is easier to form a positive ion. An atom that is easier to ionize exhibits greater metallic character.
A-32
Appendix G • Brief Answers to Selected Problems
13.50 a)
O O Si O
O
b)
8
O
O O
Si
Si
H
H
H
C
C
H
H
C
C
H
H
H
O Si
O O
O
O
13.53(a) diamond, C (b) calcium carbonate, CaCO3 (c) carbon dioxide, CO2 (d) carbon monoxide, CO (e) silicon, Si 13.57(a) 3 to 5 (b) For a group of nonmetals, the oxidation states range from the lowest, group number 8, or 15 18 3 for Group 15, to the highest, group number 10, or 15 10 5 for Group 15. 13.60 H3AsO4 H3PO4 HNO3 13.62(a) 4As(s) 5O2(g) ¡ 2As2O5(s) (b) 2Bi(s) 3F2(g) ¡ 2BiF3(s) (c) Ca3As2(s) 6H2O(l) ¡ 3Ca(OH)2(s) 2AsH3(g) ¢
13.64(a) N2(g) 2Al(s) ¡ 2AlN(s) (b) PF5(g) 4H2O(l) ¡ H3PO4(aq) 5HF(g) 13.66 Trigonal bipyramidal, with axial F atoms and equatorial Cl atoms; F Cl P F
13.70(a) (c)
O
Cl
O N
N
O
O O
N
O
(b) (d)
N
O
N
O
N
O
⫹
N
and O
O
⫺
N O
¢
N
O
N O
Cl
ger than Br¬Br bond. (b) Br¬Br bond is stronger than I¬I bond. (c) Cl¬Cl bond is stronger than F¬F bond. The fluorine atoms are so small that electron-electron repulsion of the lone pairs decreases the strength of the bond. 13.91(a) 3Br2(l) 6OH(aq) ¡ 5Br(aq) BrO3(aq) 3H2O(l) (b) ClF5(l) 6OH(aq) ¡ 5F(aq) ClO3(aq) 3H2O(l) 13.92(a) 2Rb(s) Br2(l) ¡ 2RbBr(s) (b) I2(s) H2O(l) ¡ HI(aq) HIO(aq) (c) Br2(l) 2I(aq) ¡ I2(s) 2Br(aq) (d) CaF2(s) H2SO4(l) ¡ CaSO4(s) 2HF(g) 13.94 HIO HBrO HClO HClO2 13.98 I2 Br2 Cl2, because Cl2 is able to oxidize Re to the 6 oxidation state, Br2 only to 5, and I2 only to 4 13.99(a) helium; (b) argon 13.101 Only dispersion forces hold atoms of noble gases together. 13.105 solnH 411 kJ/mol 13.109(a) Second ionization energies for alkali metals are so high because the electron being removed is from the next lower energy level and these are very tightly held by the nucleus. Also, the alkali metal would lose its noble gas electron configuration. (b) 2CsF2(s) ¡ 2CsF(s) F2(g); 405 kJ/mol 13.111(a) hyponitrous acid, H2N2O2; nitroxyl, HNO H N O (b) H O N N O H (c) In both species the shape is bent about the N atoms. 2⫺ 2⫺ (d) O N N
13.72(a) 2KNO3(s) ¡ 2KNO2(s) O2(g) (b) 4KNO3(s) ¡ 2K2O(s) 2N2(g) 5O2(g) 13.74(a) Boiling point and conductivity vary in similar ways down both groups. (b) Degree of metallic character and types of bonding vary in similar ways down both groups. (c) Both P and S have allotropes, and both bond covalently with almost every other nonmetal. (d) Both nitrogen and oxygen are diatomic gases at normal temperature and pressure. (e) O2 is a reactive gas, whereas N2 is not. Nitrogen can have any of six oxidation states, whereas oxygen has two. 13.76(a) NaHSO4(aq) NaOH(aq) ¡ Na2SO4(aq) H2O(l) (b) S8(s) 24F2(g) ¡ 8SF6(g) (c) FeS(s) 2HCl(aq) ¡ H2S(g) FeCl2(aq) (d) Te(s) 2I2(s) ¡ TeI4(s) 13.78(a) acidic (b) acidic (c) basic (d) amphoteric (e) basic 13.80 H2O H2S H2Te 13.83(a) O3, ozone (b) SO3, sulfur trioxide (c) SO2, sulfur dioxide (d) H2SO4, sulfuric acid (e) Na2S2O3.5H2O, sodium thiosulfate pentahydrate 13.85 S2F10(g) ¡ SF4(g) SF6(g); O.N. of S in S2F10 is 5; O.N. of S in SF4 is 4; O.N. of S in SF6 is 6. 13.87(a) 1, 1, 3, 5, 7 (b) The electron configuration for Cl is [Ne] 3s23p5. By gaining one electron, Cl achieves an octet. By forming covalent bonds, Cl completes or expands its valence level by maintaining electron pairs in bonds or as lone pairs. (c) Fluorine has only the 1 oxidation state because its small size and absence of d orbitals prevent it from forming more than one covalent bond. 13.89(a) Cl¬Cl bond is stron-
cis
trans
13.115 13 t 13.117 In a disproportionation reaction, a substance acts as both a reducing agent and an oxidizing agent because atoms of an element within the substance attain both higher and lower oxidation states in the products. The disproportionation reactions are b, c, d, e, and f. 13.119(a) Group 15 (b) Group 17 (c) Group 16 (d) Group 1 (e) Group 13 (f) Group 18 13.121 117.2 kJ/mol ⫺ ⫺ 13.126 O N O ¢¢ O N O O O
N N
O O
¢¢
O
N
O
⫹
The nitronium ion (NO2) has a linear shape because the central N atom has two surrounding electron groups, which achieve maximum repulsion at 180. The nitrite ion (NO2) bond angle is more compressed than the nitrogen dioxide (NO2) bond angle because the lone pair of electrons takes up more space than the lone electron. 13.128 2.29 104 g UF6 13.132 O, O, and O2 13.133(a) 39.96 mass % of As in CuHAsO3; 62.42 mass % of As in (CH3)3As (b) 0.35 g CuHAsO3
Chapter 14 • 14.2 Reaction rate is proportional to concentration. An increase in pressure will increase the concentration, resulting in an increased rate. 14.3 The addition of water will lower the concentrations of all dissolved solutes, and the rate will decrease. 14.5 An increase in temperature increases the rate by increasing the number of collisions between particles, but more
Appendix G • Brief Answers to Selected Problems
importantly, it increases the energy of collisions. 14.8(a) The instantaneous rate is the rate at one point in time during the reaction. The average rate is the average over a period of time. On a graph of reactant concentration vs. time, the instantaneous rate is the slope of the tangent to the curve at any point. The average rate is the slope of the line connecting two points on the curve. The closer together the two points (the shorter the time interval), the closer the average rate is to the instantaneous rate. (b) The initial rate is the instantaneous rate at the point on the graph where t 0, that is, when reactants are mixed. 14.10
Concentration
B(g)
A(g)
Time
Concentration
A(g)
B(g)
The initial rate is higher than the average rate because the rate decreases as reactant concentration decreases. ¢[B] ¢[C] 1 ¢[A] 4 mol/Ls 14.14 rate 2 ¢t ¢t ¢t ¢[A] ¢[C] 1 ¢[B] 14.16 rate 0.2 mol/Ls ¢t 2 ¢t ¢t 14.18 2N2O5(g) ¡ 4NO2(g) O2(g) ¢[N2] 1 ¢[H2] 1 ¢[NH3] 14.21 rate ¢t 3 ¢t 2 ¢t 1 ¢[O2] 1 ¢[O3] 14.22(a) rate 3 ¢t 2 ¢t (b) 1.45105mol/Ls 14.23(a) k is the rate constant, the proportionality constant in the rate law; it is reaction and temperature specific. (b) m represents the order of the reaction with respect to [A], and n represents the order of the reaction with respect to [B]. The order of a reactant does not necessarily equal its stoichiometric coefficient in the balanced equation. (c) L2/mol2min 14.25(a) Rate doubles. (b) Rate decreases by a factor of 4. (c) Rate increases by a factor of 9. 14.26 first order in BrO3; first order in Br; second order in H; fourth order overall 14.28(a) Rate doubles. (b) Rate is halved. (c) The rate increases by a factor of 16. 14.30 second order in NO2; first order in Cl2; third order overall 14.32(a) Rate increases by a factor of 9. (b) Rate increases by a factor of 8. (c) Rate is halved. 14.34(a) second order in A; first order in B (b) rate k[A]2[B] (c) 5.00 103 L2/mol2min 14.36(a) time1 (b) L/moltime (c) L2/mol2time (d) L3/2/mol3/2time 14.39(a) first order (b) second order (c) zero order 14.41 7 s 14.43(a) k 0.0660 min1 (b) 21.0 min [NH3] y-axis (ln [NH3]) 14.45(a) x-axis (time, s) 0 4.000 mol/L 1.38629 1.000 3.986 mol/L 1.38279 2.000 3.974 mol/L 1.37977
Time
1 ¢[AX2] 2 ¢t 1 (0.0088 mol/L 0.0500 mol/L) 2 (20.0 s 0 s) 0.00103 mol/Ls 0.0010 mol/Ls
ln [NH3]
14.12(a) Rate
(b) 0.06 0.05 0.04 [AX2]
1.387 1.386 1.385 1.384 1.383 1.382 1.381 1.380 1.379 0
[AX2] vs Time
0.03 0.02 0.01 0 0
20
10 Time (s)
30
A-33
0.5
1 1.5 Time (s)
2
k 3 103 s1 (b) t1/2 2 102 s 14.47 No, other factors that affect the rate are the energy and orientation of the collisions. 14.50 Measure the rate constant at a series of temperatures and plot ln k versus 1/T. The slope of the line equals Ea/R. 14.53 No, reaction is reversible and will eventually reach a state where the forward and reverse rates are equal. When this occurs, there are no concentrations equal to zero. Since some reactants are reformed from EF, the amount of EF will be less than 4 105 mol. 14.54 At the same temperature, both reaction mixtures have the same average kinetic energy, but the reactant molecules do not have the same average velocity. The trimethylamine molecule has greater mass than the ammonia molecule, so trimethylamine molecules will collide less often with HCl. Moreover, the
A-34
Appendix G • Brief Answers to Selected Problems
bulky groups bonded to nitrogen in trimethylamine mean that collisions with HCl having the correct orientation occur less frequently. Therefore, the rate of the reaction between NH3 and HCl is higher. 14.55 12 unique collisions 14.57 2.96 1018 14.59 0.033 s1 14.61(a) Ea(fwd) 215 kJ/mol Energy
Ea(rev) ABC ⫹ D
⌬rH ⫺55 kJ/mol
AB ⫹ CD
Reaction coordinate
(b) 2.70 102 kJ/mol (c) bond forming B C
A
D
bond weakening leading to bond breakage
14.64(a) Because the enthalpy change is positive, the reaction is endothermic. NOCl Cl Ea(rev)
Energy
Ea(fwd) H
Ea (fwd) 86 kJ rH 83 kJ
NO Cl2
activation energy, but increases the fraction of collisions with sufficient energy to equal or exceed the activation energy. 14.78(a) No. The spark provides energy that is absorbed by the H2 and O2 molecules to achieve the activation energy. (b) Yes. The powdered metal acts as a heterogeneous catalyst, providing a surface on which the oxygen-hydrogen reaction can proceed with a lower activation energy. 14.83(a) Water does not appear as a reactant in the ratedetermining step. (b) Step 1: rate1 k1[(CH3)3CBr] Step 2: rate2 k2[(CH3)3C] Step 3: rate3 k3[(CH3)3COH2] (c) (CH3)3C and (CH3)3COH2 (d) The rate-determining step is step 1. The rate law for this step agrees with the rate law observed with k k1. 14.85 4.61 104 J/mol 14.89(a) Rate increases 2.5 times. (b) Rate is halved. (c) Rate decreases by a factor of 0.01. (d) Rate does not change. 14.90 second order 14.93 57 yr 14.96(a) 0.21 h1; 3.3 h (b) 6.6 h (c) If the concentration of sucrose is relatively low, the concentration of water remains nearly constant even with small changes in the amount of water. This gives an apparent zero-order reaction with respect to water. Thus, the reaction is first order overall because the rate does not change with changes in the amount of water. 14.99(a) 0.68 mol/L (b) 0.57 14.102 71 kPa 14.106 7.3 103 J/mol 14.108(a) 2.4 1015mol/L (b) 2.4 1011 mol/Ls 14.111(a) 2.8 days (b) 7.4 days (c) 4.5 mol/m3 14.114(a) rate1 1.7 105 mol/Ls; rate2 3.4 105 mol/Ls; rate3 3.4 105 mol/Ls (b) zero order with respect to S2O82; first order with respect to I (c) 4.3 104 s1 (d) rate (4.3 104 s1)[I] 14.117(a) 7 104 cells/L (b) 2.0 101 min 14.120(a) Use the Monod equation. 1.60⫻10⫺4
Reaction progress Cl
Cl
N
1.40⫻10⫺4
O
14.65 The rate of an overall reaction depends on the rate of the slowest step. The rate of the overall reaction will be lower than the average of the individual rates because the average includes higher rates as well. 14.69 The probability of three particles colliding with one another with the proper energy and orientation is much less than the probability for two particles. 14.70 No, the overall rate law must contain only reactants (no intermediates), and the overall rate is determined by the slow step. 14.72(a) A(g) B(g) C(g) ¡ D(g) (b) X and Y are intermediates. Molecularity Rate Law (c) Step A(g) B(g) ¡ X(g) bimolecular rate1 k1[A][B] X(g) C(g) ¡ Y(g) bimolecular rate2 k2[X][C] Y(g) ¡ D(g) unimolecular rate3 k3[Y] (d) yes (e) yes 14.74 The proposed mechanism is valid because the individual steps are chemically reasonable, they add to give the overall equation, and the rate law for the mechanism matches the observed rate law. 14.77 No. A catalyst changes the mechanism of a reaction to one with lower activation energy. Lower activation energy means a faster reaction. An increase in temperature does not influence the
Growth rate (s⫺1)
(b) 3 kJ (c)
1.20⫻10⫺4 1.00⫻10⫺4 8.00⫻10⫺5 6.00⫻10⫺5 4.00⫻10⫺5 2.00⫻10⫺5 0 0
0.25 0.5 0.75 1 Substrate concentration (kg/m3)
(b) 8.2 103 cells/m3 (c) 8.4 103 cells/m3 14.125(a) 8.0 101 s (b) 59 s (c) 2.6 102 s
Chapter 15 Answers to Boxed Reading Problem: B15.1 (a) Enzyme 3 (d) If F inhibited enzyme 6, then the second branch would not take place when enough F was made. • 15.1 If the change is one of concentrations, it results temporarily in more products and less reactants. After equilibrium is re-established, Kc remains unchanged because the ratio of
Appendix G • Brief Answers to Selected Problems
products and reactants remains the same. 15.7 The equilibrium constant expression is K po2. If the temperature remains constant, K remains constant. If the initial amount of Li2O2 present is sufficient to reach equilibrium, the amount of O2 obtained will be constant. p2HI 15.8(a) Q pH2 pI2
Concentration
[H2]
[HI]
Time
The value of Q increases as a function of time until it reaches the value of K. (b) no 15.11 Yes. If Q1 is for the formation of 1 mol NH3 from H2 and N2, and Q2 is for the formation of NH3 from H2 and 1 mol of N2, then Q2 Q12. 15.12(a) 4NO(g) O2(g) Δ 2N2O3(g) p2N2O3 Q 4 pNOpO2 (b) SF6(g) 2SO3(g) Δ 3SO2F2(g) p3SO2F2 Q pSF6 p2SO3 (c) 2SC1F5(g) H2(g) Δ S2F10(g) 2HCl(g) pS2F10 p2HCl Q 2 pSClF5 pH2 15.14(a) 2NO2Cl(g) Δ 2NO2(g) Cl2(g) p2NO2 pCl2 Q 2 pNO2Cl (b) 2POCl3(g) Δ 2PCl3(g) O2(g) p2PCl3 pO2 Q 2 pPOCl3 (c) 4NH3(g) 3O2(g) Δ 2N2(g) 6H2O(g) p2N2 p6H2O Q 4 pNH3 p3O2 15.16(a) 7.9 (b) 3.2 105 15.18(a) 2Na2O2(s) 2CO2(g) Δ 2Na2CO3(s) O2(g) pO2 Q 2 pCO2 (b) H2O(l) Δ H2O(g) Q pH2O (c) NH4Cl(s) Δ NH3(g) HCl(g) Q pNH3 pHCl 15.20(a) 2NaHCO3(s) Δ Na2CO3(s) CO2(g) H2O(g) Q pCO2 pH2O (b) SnO2(s) 2H2(g) Δ Sn(s) 2H2O(g) p2H2O Q 2 pH2 (c) H2SO4(l) SO3(g) Δ H2S2O7(l) 1 Q pSO3
15.23(a) (1) C12(g) F2(g) (2) 2C1F(g) 2F2(g) Overall: Cl2(g) 3F2(g) p2C1F3 p2C1F Q2 2 2 Q1 pC12 pF2 pC1F pF2 (b) Qoverall
A-35
Δ 2C1F(g) Δ 2C1F3(g) Δ 2ClF3(g) p2ClF3 Qoverall pCl2 p3F2
p2ClF3
pCl2 p3F2 15.25 Kc and K are equal when ngas 0. 15.26(a) smaller (b) Assuming that RT 1 (T 12.2 K), K Kc if the amount (mol) of gaseous products exceeds the amount of reactant at equilibrium, and K Kc if there are more moles of gaseous reactants than gaseous products. 15.27(a) 3 (b)-1 (c) 3 15.29(a) 3.2 (b) 28.5 15.31(a) 0.15 (b) 3.5 107 15.33 The reaction quotient (Q) and equilibrium constant (K) are determined by the ratio [products]/[reactants]. When Q K, the reaction proceeds to the right to form more products. 15.35 no; to the left 15.38 At equilibrium, equal concentrations of CFCl3 and HCl exist, regardless of starting reactant concentrations, because the product coefficients are equal. 15.40(a) The approximation applies when the change in concentration from initial concentration to equilibrium concentration is so small that it is insignificant; this occurs when K is small and initial concentration is large. (b) This approximation should not be used when the change in concentration is greater than 5%. This can occur when [reactant]initial is very small or when change in [reactant] is relatively large due to a large K. 15.41 50.8 PCl5(g) Δ PCl3(g) ⴙ Cl2(g) 15.43 Concentration (M) Initial 0.075 0 0 Change x x x Equilibrium 0.075 x x x 15.45 28 bar 15.47 0.33 bar 15.49 3.5103mol/L 15.51 [HI] 0.0152 mol/L; [I2] 0.00328 mol/L 15.53 [I2]eq [Cl2]eq 0.0200 mol/L; [ICl]eq 0.060 mol/L 15.55 6.01 106 15.58 Equilibrium position refers to the specific concentrations or pressures of reactants and products that exist at equilibrium, whereas equilibrium constant is the overall ratio of equilibrium concentrations or pressures. Equilibrium position changes as a result of a change in reactant and product concentrations. 15.59(a) B, because the amount of product increases with temperature (b) A, because the lowest temperature will give the least product 15.63(a) shifts toward products (b) shifts toward products (c) does not shift (d) shifts toward reactants 15.65(a) more F and less F2 (b) more C2H2 and H2 and less CH4 15.67(a) no effect (b) less H2 and O2 and more H2O 15.69(a) no change (b) increase volume 15.71(a) amount decreases (b) amounts increase (c) amounts increase (d) amount decreases 15.73 2.0 15.76(a) lower temperature, higher pressure (b) Q decreases; no change in K (c) Reaction rates are lower at lower temperatures, so a catalyst is used to speed up the reaction. 15.78(a) pN2 31 bar; pH2 93 bar; ptotal 174 bar (b) pN2 18 bar; pH2 111 bar; ptotal 179 bar; not a valid argument 15.81 0.206 bar 15.84(a) 3 103 bar (b) high pressure; low temperature (c) 2 105 (d) No, because water condenses at a higher temperature. 15.87(a) 0.016 bar (b) Kc 5.6 102; po2 0.17 bar 15.89 12.5 g CaCO3 15.93 Both concentrations increased by a factor of 2.2. 15.95(a) 3.0 1014 bar (b) 0.013 pg CO/L
A-36
Appendix G • Brief Answers to Selected Problems
15.97(a) 98.0% (b) 99.0% (c) 2.60 105 J/mol 15.99(a) 2CH4(g) O2(g) 2H2O(g) ÷ 2CO2(g) 6H2(g) (b) 1.76 1029 (c) 3.02 1023 (d) 48 bar 15.100(a) 4.0 1021 bar (b) 5.5 108 bar (c) 29 N atoms/L; 4.0 1014 H atoms/L (d) The more reasonable step is N2(g) H(g) ¡ NH(g) N(g). With only 29 N atoms in 1.0 L, the first reaction would produce virtually no NH(g) molecules. There are orders of magnitude more N2 molecules than N atoms, so the second reaction is the more reasonable step. 15.103(a) pN2 0.780 bar; pO2 0.210 bar; pNO 2.67 1016 bar (b) 0.990 bar (c) Kc K 4.35 1031 15.105(a) 1.26 103 (b) 794 (c) 10.6 kJ/mol (d) 1.2 104 J/mol 15.109(a) 1.52 (b) 0.975 bar (c) 0.2000 mol CO (d) 0.01093 mol/L.
Chapter 16 16.2(a) All Arrhenius acids have H in their formula and produce hydronium ion (H3O) in aqueous solution. All Arrhenius bases have OH in their formula and produce hydroxide ion (OH) in aqueous solution. (b) Neutralization occurs when each H3O ion combines with an OH ion to form two molecules of H2O. Chemists found the reaction of any strong base with any strong acid always had a H of 56 kJ/mol H2O produced. 16.4(a) The Brønsted-Lowry theory defines acids as proton donors and bases as proton acceptors, while the Arrhenius definition looks at acids as containing ionizable hydrogen atoms and at bases as containing hydroxide ions. In both definitions, an acid produces H (H3O) ions and a base produces OH ions when added to water. (b) Ammonia and carbonate ion are two Brønsted-Lowry bases that are not Arrhenius bases because they do not contain OH ions. BrønstedLowry acids must contain an ionizable hydrogen atom in order to be proton donors, so a Brønsted-Lowry acid is also an Arrhenius acid. 16.7 An amphiprotic species is one that can lose a proton to act as an acid or gain a proton to act as a base. The dihydrogen phosphate ion, H2PO4, is an example H2PO4 (aq) OH (aq) ¡ H2O(aq) HPO2 4 (aq) In the presence of a strong acid (HCl), the dihydrogen phosphate ion acts like a base by accepting hydrogen: H2PO4 (aq) HCl(aq) ¡ H3PO4(aq) Cl (aq) 16.8(a) Strong acids and bases dissociate completely into ions when dissolved in water. Weak acids and bases dissociate only partially. (b) The characteristic property of all weak acids and bases is that a great majority of the molecules are undissociated. 16.9 a, c, and d 16.11 b and d. [H3O ][CN ] 16.13(a) Ka [HCN] [H3O ][CO2 3 ] (b) Ka [HCO3 ] [H3O ][HCOO ] (c) Ka [HCOOH] [H3O ][NO2 ] 16.15(a) Ka [HNO2] [H3O ][CH3COO ] (b) Ka [CH3COOH] [H3O ][BrO2 ] (c) Ka [HBrO2]
16.17(a) H3PO4(aq) H2O(l) Δ H2PO4 (aq) H3O (aq) [H3O ][H2PO4 ] Ka [H3PO4] (b) C6H5COOH(aq) H2O(l) Δ C6H5COO (aq) H3O (aq) [H3O ][C6H5COO ] Ka [C6H5COOH] (c) HSO4 (aq) H2O(l) Δ SO2 4 (aq) H3O (aq) 2 [H3O ][SO4 ] Ka [HSO4 ] 16.19(a) Cl (b) HCO3 (c) OH 16.21(a) NH4 (b) NH3 (c) C10H14N2H 16.23(a) HCl H2O Δ Cl H3O acid base conjugate base conjugate acid Conjugate acid-base pairs: HCl/Cl– and H3O/H2O H3SO4 (b) HClO4 H2SO4 Δ ClO4 acid base conjugate base conjugate acid Conjugate acid-base pairs: HClO4/ClO4– and H3SO4/H2SO4 (c) HPO2 4 H2SO4 Δ H2PO4 HSO4 base acid conjugate acid conjugate base Conjugate acid-base pairs: H2SO4/HSO4– and H2PO4–/HPO42– 16.25(a) NH3 H3PO4 Δ NH4 H2PO4 base acid conjugate acid conjugate base Conjugate acid-base pairs: H3PO4/H2PO4–; NH4/NH3 (b) CH3O NH3 Δ CH3OH NH2 base acid conjugate acid conjugate base Conjugate acid-base pairs: NH3/NH2–; CH3OH/CH3O– 2 (c) HPO2 4 HSO4 Δ H2PO4 SO4 base acid conjugate acid conjugate base Conjugate acid-base pairs: HSO4–/SO42–; H2PO4–/HPO42– 16.27(a) OH (aq) H2PO4 (aq) Δ H2O(l) HPO2 4 (aq) Conjugate acid-base pairs: H2PO4–/ HPO42– and H2O/OH– 2 (b) HSO4 (aq) CO2 3 (aq) Δ SO4 (aq) HCO3 (aq) − – 2– 2− Conjugate acid-base pairs: HSO4 /SO4 ; HCO3 /CO3 16.29 K 1; HS HCl Δ H2S Cl K 1; H2S Cl Δ HS HCl 16.31 K 1 for both a and b 16.33 57 Kc 1 for both a and b 16.35 CH3COOH HF HIO3 HI 16.37(a) weak acid (b) strong base (c) weak acid (d) strong acid 16.39(a) strong base (b) strong acid (c) weak acid (d) weak acid 16.44(a) The acid with the smaller Ka (4 105) has the higher pH, because less dissociation yields fewer hydronium ions. (b) The acid with the larger pKa (3.5) has the higher pH, because it has a smaller Ka and, thus, lower [H3O]. (c) Lower concentration (0.01 mol/L) contains fewer hydrogen (hydronium) ions. (d) A 0.1 mol/Lweak acid solution contains fewer hydronium ions. (e) The 0.01 mol/Lbase solution contains more hydroxide ions, so fewer hydronium ions. (f) The solution that has pOH 6.0 has the higher pH: pH 14.0 6.0 8.0. 16.45(a) 12.05; basic (b) 11.13; acidic 16.47(a) pH 2.212; acidic (b) pOH 0.708; basic 16.49(a) [H] 1.4 1010 mol/L, [OH–] 7.1 105 mol/L, pOH 4.15 (b) [H] 2.7 105 mol/L, [OH] 3.7 1010 mol/L, pH 4.57
Appendix G • Brief Answers to Selected Problems
16.51(a) [H] 1.7 105 mol/L, [OH] 5.9 1010 mol/L, pOH 9.23 (b) [H] 4.5 109 mol/L, [OH] 2.2 106 mol/L; pH 8.35 16.53 4.8 104 mol OH/L 16.55 1.4 104 mol OH 16.58(a) Rising temperature increases the value of Kw. (b) Kw 2.5 1014; pH pOH 6.80; [H] [OH] 1.6 107mol/ 16.70(a) A strong acid is 100% dissociated, so the acid concentration will be very different after dissociation. (b) A weak acid dissociates to a very small extent, so the acid concentration before and after dissociation is nearly the same. (c) same as b, but with the extent of dissociation greater (d) same as a 16.71 No. HCl is a strong acid and dissociates to a greater extent than the weak acid CH3COOH. The Ka of the acid, not the concentration of H3O, determines the strength of the acid. 16.74 1.5 105 16.76 11.79 16.78 11.45 16.80 [H] [NO2] 2.1 102 mol/L; [OH] 4.8 1013 mol/L 16.82 [H] [ClCH2COO] 0.041 mol/L; [ClCH2COOH] 1.21 mol/L; pH 1.39 16.84(a) 11.19 (b) 5.56 16.86(a) 8.78 (b) 4.66 16.88(a) [H] 6.0 103 mol/L; pH 2.22; [OH] 1.7 1012 mol/L; pOH 11.78 (b) 1.9 104 16.90 2.2 107 16.92(a) 2.37 (b) 11.53 16.94(a) 2.290 (b) 12.699 16.96 1.1% 16.98 [H] [HS] 9 105 mol/L; pH 4.0; [OH] 1 1010 mol/L; pOH 10.0; [H2S] 0.10 mol/L; [S2] 1 1017 mol/L 16.101 1.5% 16.102 [OH] 5.5 104 mol/L; pH 10.74 16.104 As a nonmetal becomes more electronegative, the acidity of its binary hydride increases. The electronegative nonmetal attracts the electrons more strongly in the polar bond, shifting the electron density away from H, thus making H more easily transferred to a water molecule to form H3O. 16.107 Chlorine is more electronegative than iodine, and HClO4 has more oxygen atoms than HIO. 16.108(a) H2SeO4 (b) H3PO4 (c) H2Te 16.110(a) H2Se (b) B(OH)3 (c) HBrO2 16.112(a) 0.5 mol/L AlBr3 (b) 0.3 mol/L SnCl2 16.114(a) 0.2 mol/L Ni(NO3)2 (b) 0.35 mol/L Al(NO3)3 16.117 NaF contains the anion of the weak acid HF, so F acts as a base. NaCl contains the anion of the strong acid HCl. H2O
16.119(a) KBr(s) ¡ K (aq) Br (aq); neutral H2O (b) NH4I(s) ¡ NH4 (aq) I (aq); acidic H2O (c) KCN(s) ¡ K (aq) CN (aq); basic H2O
16.121(a) Na2CO3(s) ¡ 2Na (aq) CO2 3 (aq); basic H2O (b) CaCl2(s) ¡ Ca2 (aq) 2C1(aq); neutral H2O (c) Cu(NO3)2(s) ¡ Cu2 (aq) 2NO3 (aq); acidic 16.123(a) A solution of strontium bromide is neutral because Sr2 is the conjugate acid of a strong base, Sr(OH)2, and Br– is the conjugate base of a strong acid, HBr, so neither change the pH of the solution. (b) A solution of barium acetate is basic because CH3COO– is the conjugate base of a weak acid and therefore forms OH– in solution whereas Ba2 is the conjugate acid of a strong base, Ba(OH)2, and does not influence solution pH. The base-dissociation reaction of acetate ion is CH3COO (aq) H2O(l) Δ CH3COOH(aq) OH (aq). (c) A solution of dimethylammonium bromide is acidic because (CH3)2NH2 is the conjugate acid of a weak base and therefore forms H3O in solution whereas Br– is the conjugate base of a strong acid and does not influence the pH of the solution. The
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acid-dissociation reaction for methylammonium ion is (CH3)2NH2 (aq) H2O(l) Δ (CH3)2NH(aq) H3O (aq). 16.125(a) NH4 (aq) H2O(l) Δ NH3(aq) H3O (aq) 2 PO3 4 (aq) H2O(l) Δ HPO4 (aq) OH (aq) Kb Ka; basic (b) SO2 4 (aq) H2O(l) Δ HSO4 (aq) OH (aq) Na gives no reaction; basic (c) ClO (aq) H2O(l) Δ HClO(aq) OH (aq) Ligives no reaction; basic 16.127(a) Fe(NO3)2 KNO3 K2SO3 K2S (b) NaHSO4 NH4NO3 NaHCO3 Na2CO3 16.129 Since both bases produce OHions in water, both bases appear equally strong. CH3O(aq) H2O(l) ¡ OH(aq) CH3OH(aq) and NH2(aq) H2O(l) ¡ OH(aq) NH3(aq). 16.131 Ammonia, NH3, is a more basic solvent than H2O. In a more basic solvent, weak acids such as HF act like strong acids and are 100% dissociated. 16.133 A Lewis acid is an electron-pair acceptor while a Brønsted-Lowry acid is a proton donor. The proton of a Brønsted-Lowry acid is a Lewis acid because it accepts an electron pair when it bonds with a base. All Lewis acids are not Brønsted-Lowry acids. A Lewis base is an electron-pair donor and a Brønsted-Lowry base is a proton acceptor. All Brønsted-Lowry bases are Lewis bases, and vice versa. 16.134(a) No, for example : Ni2 (aq) 4H2O(l) ÷ Ni(H2O)42(aq); Water is a very weak BrønstedLowry base, but forms the Zn complex fairly well and is a reasonably strong Lewis base. (b) cyanide ion and water (c) cyanide ion 16.137(a) Lewis acid (b) Lewis base (c) Lewis acid (d) Lewis base 16.139(a) Lewis acid (b) Lewis base (c) Lewis base (d) Lewis acid 16.141(a) Lewis acid: Na; Lewis base: H2O (b) Lewis acid: CO2; Lewis base: H2O (c) Lewis acid: BF3; Lewis base: F 16.143(a) Lewis (b) Brønsted-Lowry and Lewis (c) none (d) Lewis 16.146 3.5 108 to 4.5 108 mol/L H; 5.2 107 to 6.6 107 mol/L OH 16.147(a) Acids vary in the extent of dissociation depending on the acid-base character of the solvent. (b) Methanol is a weaker base than water since phenol dissociates less in methanol than in water. (c) C6H5OH (solvated) CH3OH(l) Δ CH3OH2 (solvated) C6H5O (solvated) (d) CH3OH(l) CH3OH(l) Δ CH3O (solvated) CH3OH2 (solvated) 16.150(a) SnCl4 is the Lewis acid; (CH3)3N is the Lewis base (b) 5d 16.151 pH 5.00, 6.00, 6.79, 6.98, 7.00 16.157 3 1018 16.160 10.48 16.162 2.41 16.164 amylase, 2 107 mol/L; pepsin, 1 102 mol/L; trypsin, 3 1010 mol/L 16.168 1.47 103 16.170(a) Ca2 does not react with water; CH3CH2COO (aq) H2O (l) Δ CH3CH2COOH(aq) OH (aq); basic (b) 9.08 16.176 4.5 105 16.179(a) The concentration of oxygen is higher in the lungs, so the equilibrium shifts to the right. (b) In an oxygendeficient environment, the equilibrium shifts to the left to release oxygen. (c) A decrease in [H3O] shifts the equilibrium to the right. More oxygen is absorbed, but it will be more difficult to remove the O2. (d) An increase in [H3O] shifts the equilibrium to the left. Less oxygen is bound to Hb, but it will be easier to remove it. 16.181(a) 1.012 mol/L (b) 0.004 mol/L (c) 0.4% 16.182(a) 10.0 (b) The pKb for the 3 amine group is much smaller than that for the aromatic ring, thus the
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Appendix G • Brief Answers to Selected Problems
Kb is significantly larger (yielding a much greater amount of OH). (c) 4.7 (d) 5.1
Chapter 17 Answers to Boxed Reading Problem: B17.2 (a) 64 mol (b) 6.28 (c) 3.9103 g HCO3 • 17.2 The acid component neutralizes added base, and the base component neutralizes added acid, so the pH of the buffer solution remains relatively constant. The components of a buffer do not neutralize one another because they are a conjugate acid-base pair. 17.7 The buffer range, the pH over which the buffer acts effectively, is greatest when the buffer-component concentration ratio is 1; the range decreases as this ratio deviates from 1. 17.9(a) Ratio and pH increase; added OHreacts with HA. (b) Ratio and pH decrease; added Hreacts with A. (c) Ratio and pH increase; added Aincreases [A]. (d) Ratio and pH decrease; added HA increases [HA]. 17.11 [H3O] 5.6 106 mol/L; pH 5.25 17.13 [H3O] 5.2 104 mol/L; pH 3.28 17.15 3.89 17.17 10.03 17.19 9.47 17.21(a) Ka2 (b) 10.55 17.23 3.6 17.25 0.20 17.27 3.37 17.29 8.82 17.31(a) 4.81 (b) 0.66 g KOH 17.33(a) HOOC(CH2)4COOH/HOOC(CH2)4COO or C6H5NH3/C6H5NH2 (b) H2PO4/HPO42 or H2AsO4/ HAsO42 17.35(a) HOCH2CH(OH)COOH/HOCH2CH(OH) COO or CH3COOC6H4COOH/CH3COOC6H4COO (b) C5H5NH/C5H5N 17.38 1.6 17.42 To see a distinct colour in a mixture of two colours, you need one to have about 10 times the intensity of the other. For this to be the case, the concentration ratio [HIn]/[In] has to be greater than 10/1 or less than 1/10. This occurs when pH pKa 1 or pH pKa 1, respectively, giving a pH range of about 2 units. 17.44 The equivalence point in a titration is the point at which the amount (mol) of base is stoichiometrically equivalent to the amount (mol) of acid. The endpoint is the point at which the added indicator changes colour. If an appropriate indicator is selected, the endpoint is close to the equivalence point, but they are not usually the same. The pH at the endpoint, or colour change, may precede or follow the pH at the equivalence point, depending on the indicator chosen. 17.46(a) initial pH: strong acid–strong base weak acid–strong base weak base–strong acid (b) pH at equivalence point: weak base– strong acid strong acid–strong base weak acid–strong base 17.48 At the centre of the buffer region, the concentrations of weak acid and conjugate base are equal, so the pH pKa of the acid. 17.52 pH range from 7.5 to 9.5 17.54(a) bromthymol blue (b) thymol blue or phenolphthalein 17.56(a) methyl red (b) bromthymol blue 17.58(a) 1.0000 (b) 1.6368 (c) 2.898 (d) 3.903 (e) 7.00 (f) 10.10 (g) 12.05 17.60(a) 2.91 (b) 4.81 (c) 5.29 (d) 6.09 (e) 7.41 (f) 8.76 (g) 10.10 (h) 12.05 17.62(a) 59.0 mL and 8.54 (b) 66.0 mL and 7.13, total 132.1 mL and 9.69 17.64(a) 123 mL and 5.17 (b) 194 mL and 5.80 17.72 Fluoride ion is the conjugate base of a weak acid and reacts with H2O: F (aq) H2O(l) Δ HF(aq) OH (aq). As the pH increases, the equilibrium shifts to the left and [F] increases. As the pH decreases, the equilibrium shifts to the right and [F] decreases. The changes in [F] influence the solubility of BaF2. Chloride ion is the
conjugate base of a strong acid, so it does not react with water and its concentration is not influenced by pH. 17.74 The compound precipitates. 17.75(a) Ksp [Ag]2[CO32] (b) Ksp [Ba2][F]2 (c) Ksp [Cu2][HS][OH] 17.77(a) Ksp [Ca2][CrO42] (b) Ksp [Ag][CN] (c) Ksp [Ni2][HS][OH] 17.79 1.3 104 17.81 2.8 1011 17.83(a) 2.3 105 mol/L (b) 4.2 109 mol/L 17.85(a) 1.7 103 mol/L (b) 2.0 104 mol/L 17.87(a) Mg(OH)2 (b) PbS (c) Ag2SO4 17.89(a) CaSO4 (b) Mg3(PO4)2 (c) PbSO4 17.91(a). AgCl(s) Δ Ag (aq) Cl (aq); The chloride ion is the anion of a strong acid, so it does not react with H3O. No change with pH. (b) SrCO3(s) Δ Sr2 (aq) CO2 3 . The strontium ion is the cation of a strong base, so pH will not affect its solubility. The carbonate ion acts as a base: CO2 3 (aq) H2O(l) Δ HCO3 (aq) OH (aq); also 2 CO2(g) forms and escapes: CO3 (aq) 2H3O(aq) ¡ CO2(g) 3H2O(l). Therefore, the solubility of SrCO3 will increase with addition of H3O(decreasing pH). 17.93(a) Fe(OH)2(s) Δ Fe2 (aq) 2OH (aq) The OH ion reacts with added H3O: OH(aq) H3O(aq) ¡ 2H2O(l). The added H3O consumes the OH, driving the equilibrium toward the right to dissolve more Fe(OH)2. Solubility increases with addition of H3O (decreasing pH). (b) CuS(s) H2O(l) Δ Cu2 (aq) HS (aq) OH (aq). Both HS and OH are anions of weak acids, so both ions react with added H3O. Solubility increases with addition of H3O (decreasing pH). 17.95 yes 17.97 yes 17.100(a) Fe(OH)3 (b) The two metal ions are separated by adding just enough NaOH to precipitate iron(III) hydroxide. (c) 2.0 107 mol/L 17.103 No, because it indicates that a complex ion forms between the lead ion and hydroxide ions: (aq) Pb2 (aq) nOH (aq) Δ Pb(OH)2n n 17.104 Hg(H2O)2 4 (aq) 4CN (aq) Δ Hg(CN)2 4 (aq) 4H2O(l) 2 17.106 Ag(H2O)2 (aq) 2S2O3 (aq) Δ Ag(S2O3)3 2 (aq) 2H2O(l) 15 17.108 8.1 10 mol/l 17.110 0.05 mol/L 17.112 1.0 1016 mol/L Zn2; 0.025 mol/L Zn(CN)42; 0.049 mol/L CN 17.114 0.035 L of 2.00 mol/L NaOH and 0.465 L of 0.200 mol/L HCOOH 17.116(a) 0.99 (b) assuming the volumes are additive: 0.468 L of 1.0mol/L HCOOH and 0.232 L of 1.0 mol/L NaOH (c) 0.34 mol/L 17.119 1.3 104 mol/L 17.122(a) 14 (b) 1 g from the second beaker 17.125(a) 0.088 (b) 0.14 17.126 0.260 mol/L TRIS; pH 8.53 17.128 Lower the pH below 6.6 17.132 8 105 17.134(a) V (mL)
pH
pH/ V
Vavg (mL)
0.00 10.00 20.00 30.00 35.00 39.00 39.50 39.75 39.90 39.95 39.99
1.00 1.22 1.48 1.85 2.18 2.89 3.20 3.50 3.90 4.20 4.90
0.022 0.026 0.037 0.066 0.18 0.62 1.2 2.67 6 18
5.00 15.00 25.00 32.50 37.00 39.25 39.63 39.83 39.93 39.97
Appendix G • Brief Answers to Selected Problems
pH
pH/ V
Vavg (mL)
7.00 9.40 9.80 10.40 10.50 10.79 11.09 11.76 12.05 12.30 12.43 12.52
200 200 10 10 0.67 1.2 0.60 0.17 0.058 0.025 0.013 0.009
40.00 40.01 40.03 40.08 40.18 40.38 40.75 43.00 47.50 55.00 65.00 75.00
Change in pH per unit volume
V (mL)
40.00 40.01 40.05 40.10 40.25 40.50 41.00 45.00 50.00 60.00 70.00 80.00 (b) 210 190 170 150 130 110 90 70 50 30 10 ⫺10 30
35 40 45 50 Average volume (mL) Maximum slope (equivalence point) is at Vavg 40.00 mL. 17.136 4.05 17.142 H2CO3/HCO3 and H2PO4/HPO42; [HPO42]/[H2PO4] 5.8 17.143 3.8 17.145(a) 58.2 mL (b) 7.85 mL (c) 6.30 17.147 170 mL 17.151 5.68 17.153 3.9 109 g Pb2/100 mL blood 17.155 No NaCl will precipitate. 17.156(a) A and D (b) pHA 4.35; pHB 8.67; pHC 2.67; pHD 4.57 (c) C, A, D, B (d) B
Chapter 18 Answer to Boxed Reading Problem: B18.2 –12.6 kJ/mol • 18.2 (a) A spontaneous process occurs by itself, whereas a nonspontaneous process requires a continuous input of energy to make it happen. (b) It is possible to cause a nonspontaneous process to occur, but the process stops once the energy source is removed. A reaction that is nonspontaneous under one set of conditions may be spontaneous under a different set of conditions. 18.5 The transition from liquid to gas involves a greater increase in dispersal of energy and freedom of motion than does the transition from solid to liquid. 18.6(a) (i)In an exothermic reaction, surrS 0. (ii) In an endothermic reaction,
surrS 0. (b) A chemical cold pack for injuries is an example of an application using a spontaneous endothermic process. 18.8 a, b, and c 18.10 a and b 18.12(a) positive (b) negative (c) negative 18.14(a) positive (b) positive (c) positive 18.16(a) negative (b) negative (c) positive 18.18(a) positive (b) negative (c) positive 18.20(a) positive (b) negative (c) positive 18.22(a) Butane. The double bond in 2-butene restricts freedom of rotation. (b) Xe(g). It has the greater molar mass. (c) CH4(g). Gases have greater entropy than liquids. 18.24(a) C2H5OH(l). It is a more complex molecule. (b) KClO3(aq). Ions in solution have their energy more dispersed than those in a solid. (c) K(s). It has a greater molar mass. 18.26(a) Diamond graphite charcoal. Freedom of motion is least in the network solid; more freedom between
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graphite sheets; most freedom in amorphous solid. (b) Ice liquid water water vapour. Entropy increases as a substance changes from solid to liquid to gas. (c) O atoms O2 O3. Entropy increases with molecular complexity. 18.28(a) ClO4(aq) ClO3(aq) ClO2(aq); decreasing molecular complexity (b) NO2(g) NO(g) N2(g). N2 has lower standard molar entropy because it consists of two of the same atoms; the other species have two different types of atoms. NO2 is more complex than NO. (c) Fe3O4(s) Fe2O3(s) Al2O3(s). Fe3O4 is more complex and more massive. Fe2O3 is more massive than Al2O3. 18.31 For a system at equilibrium, univS sysS surrS 0. For a system moving to equilibrium, univS 0. 18.32 S of Cl2O 2(S of HClO) S of H2O rS 18.33(a) negative; S 172.4 J/molK (b) positive; S 141.6 J/molK (c) negative; S 837 J/molK 18.35 S 93.1 J/molK; yes, the positive sign of S is expected because there is a net increase in the number of gas molecules. 18.37 S 311 J/molK; yes, the negative entropy change matches the decrease in moles of gas. 18.39 75.6 J/molK 18.41 242 J/molK 18.44 97.2 J/molK 18.46 A spontaneous process has univS 0. Since the absolute temperature is always positive, sysG must be negative ( sysG 0) for a spontaneous process. 18.48 H is positive and S is positive. Melting is an example. 18.49 The entropy changes little within a phase. As long as the substance does not change phase, the value of S is relatively unaffected by temperature. 18.50(a) 1138.0 kJ/mol (b) 1379.4 kJ/mol (c) 224 kJ/mol 18.52(a) 1138 kJ/mol (b) 1379 kJ/mol (c) 226 kJ/mol 18.54(a) Entropy decreases ( S is negative) because the amount (mol) of gas decreases. The combustion of CO releases energy ( H is negative). (b) 257.2 kJ/mol or 257.3 kJ/mol, depending on the method 18.56(a) 0.409 kJ/molK (b) 197 kJ/mol 18.58(a) rH 90.7 kJ/mol; rS 221 J/molK (b) At 28C,
G 24.3 kJ/mol; at 128C, G 2.2 kJ/mol; at 228C,
G 19.9 kJ/mol (c) For the substances in their standard states, the reaction is nonspontaneous at 28C, near equilibrium at 128C, and spontaneous at 228C. 18.60 rH 30.91 kJ/mol,
rS 93.15 J/molK, T 331.8 K 18.62(a) rH 241.826 kJ/mol, rS 44.4 J/molK, rG 228.60 kJ/mol (b) Yes. The reaction will become nonspontaneous at higher temperatures. (c) The reaction is spontaneous below 5.45 103 K. 18.64(a) G is a relatively large positive value. (b) K 1. Q depends on initial conditions, not equilibrium conditions. 18.67 The standard free energy change,
G, applies when all components of the system are in their standard states; G G when all concentrations equal 1 mol/L and all partial pressures equal 1 bar. 18.68(a) 1.7 106 (b) 3.89 1034 (c) 1.26 1048 18.70(a) 6.57 10173 (b) 4.46 1015 (c) 3.46 104 18.72 4.89 1051 18.74 3.36 105 18.76 2.7 104 J/mol; no 18.78(a) 2.9 104 J/mol (b) The reverse direction, formation of reactants, is spontaneous, so the reaction proceeds to the left. (c) 7.0 103 J/mol; the reaction proceeds to the left to reach equilibrium. 180.80(a) no T (b) 163 kJ/mol (c) 1 102 kJ/mol 18.83(a) spontaneous (b) (c) (d) (e) , not spontaneous (f) 18.87(a) 2.3 102 (b) The treatment is to administer oxygen-rich air; increasing the concentration of oxygen shifts the equilibrium to the left, in the direction of HbO2.
A-40
Appendix G • Brief Answers to Selected Problems
18.89 370. kJ/mol 18.91(a) 2N2O5(g) 6F2(g) ¡ 4NF3(g) 5O2(g) (b) rG 569 kJ/mol (c) rG 5.60 102 kJ/mol 18.93 rH 137.14 kJ/mol; rS 120.3 J/molK; rG 101.25 kJ/mol 18.101(a) 1.67 103 J/mol (b) 7.37 103 J/mol (c) 4.04 103 J/mol (d) 0.19 18.105(a) 465 K (b) 6.59 104 (c) The reaction rate is higher at the higher temperature. The shorter time required (kinetics) overshadows the lower yield (thermodynamics).
Chapter 19 Answers to Boxed Reading Problem: B19.1(a) reduction: Fe3 e ¡ Fe2; oxidation: Cu ¡ Cu2 e (b) Overall: Fe3 Cu ¡ Fe2 Cu2 • 19.1 Oxidation is the loss of electrons and results in a higher oxidation number; reduction is the gain of electrons and results in a lower oxidation number. 19.3 No, one half-reaction cannot take place independently because there is a transfer of electrons from one substance to another. If one substance loses electrons, another substance must gain them. 19.6 To remove H ions from an equation, add an equal number of OH ions to both sides to neutralize the H ions and produce water. 19.8(a) Spontaneous reactions, for which sysG 0, take place in voltaic cells (also called galvanic cells). (b) Nonspontaneous reactions, for which sysG 0, take place in electrolytic cells. 19.10(a) Cl (b) MnO4 (c) MnO4 (d) Cl (e) from Clto MnO4 (f) 8H2SO4(aq) 2KMnO4(aq) 10KCl(aq) ¡ 2MnSO4(aq) 5Cl2(g) 8H2O(l) 6K2SO4(aq) 19.12(a) ClO3(aq) 6H(aq) 6I(aq) ¡ Cl(aq) 3H2O(l) 3I2(s) Oxidizing agent is ClO3 and reducing agent is I. (b) 2MnO4(aq) H2O(l) 3SO32(aq) ¡ 2MnO2(s) 3SO42(aq) 2OH(aq) Oxidizing agent is MnO4and reducing agent is SO32. (c) 2MnO4(aq) 6H(aq) 5H2O2(aq) ¡ 2Mn2(aq) 8H2O(l) 5O2(g) Oxidizing agent is MnO4 and reducing agent is H2O2. 19.14(a) Cr2O72(aq) 14H(aq) 3Zn(s) ¡ 2Cr3(aq) 7H2O(l) 3Zn2(aq) Oxidizing agent is Cr2O72 and reducing agent is Zn. (b) MnO4(aq) 3Fe(OH)2(s) 2H2O(l) ¡ MnO2(s) 3Fe(OH)3(s) OH(aq) Oxidizing agent is MnO4and reducing agent is Fe(OH)2. (c) 2NO3(aq) 12H(aq) 5Zn(s) ¡ N2(g) 6H2O(l) 5Zn2(aq) Oxidizing agent is NO3 and reducing agent is Zn. 19.16(a) 4NO3(aq) 4H(aq) 4Sb(s) ¡ 4NO(g) 2H2O(l) Sb4O6(s) Oxidizing agent is NO3 and reducing agent is Sb. (b) 5BiO3(aq) 14H(aq) 2Mn2(aq) ¡ 5Bi3(aq) 7H2O(l) 2MnO4(aq) Oxidizing agent is BiO3 and reducing agent is Mn2. (c) Pb(OH)3(aq) 2Fe(OH)2(s) ¡ Pb(s) 2Fe(OH)3(s) OH(aq) Oxidizing agent is Pb(OH)3 and reducing agent is Fe(OH)2. 19.18(a) 5As4O6(s) 8MnO4(aq) 18H2O(l) ¡ 20AsO43(aq) 8Mn2(aq) 36H(aq) Oxidizing agent is MnO4 and reducing agent is As4O6.
(b) P4(s) 6H2O(l) ¡ 2HPO32(aq) 2PH3(g) 4H(aq) P4 is both the oxidizing agent and the reducing agent. (c) 2MnO4(aq) 3CN(aq) H2O(l) ¡ 2MnO2(s) 3CNO(aq) 2OH(aq) Oxidizing agent is MnO4 and reducing agent is CN. 19.21(a) Au(s) 3NO3(aq) 4Cl(aq) 6H(aq) ¡ AuCl4(aq) 3NO2(g) 3H2O(l) (b) Oxidizing agent is NO3 and reducing agent is Au. (c) HCl provides chloride ions that combine with the gold(III) ion to form the stable AuCl4 ion. 19.22(a) A (b) E (c) C (d) A (e) E (f) E 19.25(a) An active electrode is a reactant or product in the cell reaction. (b) An inactive electrode does not take part in the reaction and is present only to conduct a current. (c) Platinum and graphite are commonly used as inactive electrodes. 19.26(a) A (b) B (c) A (d) Hydrogen bubbles will form when metal A is placed in acid. Metal A is a better reducing agent than metal B, so if metal B reduces H in acid, then metal A will also. 19.27(a) Oxidation: Zn(s) ¡ Zn2(aq) 2e Reduction: Sn2(aq) 2e ¡ Sn(s) Overall: Zn(s) Sn2(aq) ¡ Zn2(aq) Sn(s) (b) Voltmeter e⫺
e⫺ Salt bridge
Zn (⫺)
Sn (⫹)
1 mol/L Zn2⫹
mol/L Anion flow
Cation flow
19.29(a) left to right (b) left (c) right (d) Ni (e) Fe (f) Fe (g) 1 mol/L NiSO4 (h) K and NO3 (i) neither (j) from right to left (k) Oxidation: Fe(s) ¡ Fe2(aq) 2e Reduction: Ni2(aq) 2e ¡ Ni(s) Overall: Fe(s) Ni2(aq) ¡ Fe2(aq) Ni(s) 19.31(a) Reduction: Fe2(aq) 2e ¡ Fe(s) Oxidation: Mn(s) ¡ Mn2(aq) 2e Overall: Fe2(aq) Mn(s) ¡ Fe(s) Mn2(aq) (b) Voltmeter e⫺
e⫺ Salt bridge
Mn (⫺) 1 mol/L Mn2⫹
Fe (⫹)
Anion flow
Cation flow
1 mol/L Fe2⫹
19.33(a) Al(s) 兩 Al3(aq) 储 Cr3(aq) 兩 Cr(s) (b) Pt 兩 SO2(g) 兩 SO42–(aq), H(aq) 储 Cu2(aq) 兩 Cu(s) 19.36(a) A negative Ecell indicates that the redox reaction is not spontaneous at the standard state, that is, G 0. (b) The reverse reaction is spontaneous with Ecell 0. 19.37(a) Similar to other state functions, E changes sign when a reaction is reversed. (b) Unlike G, H, and S, E (the ratio of energy
Appendix G • Brief Answers to Selected Problems
to charge) is an intensive property. When the coefficients in a reaction are multiplied by a factor, the values of G, H, and S are multiplied by that factor. However, E does not change because both the energy and charge are multiplied by the factor and thus their ratio remains unchanged. 19.38(a) Oxidation: Se2(aq) ¡ Se(s) 2e Reduction: 2SO32(aq) 3H2O(l) 4e ¡ S2O32(aq) 6OH(aq) (b) Eanode Ecathode Ecell 0.57 V 0.35 V 0.92 V 19.40(a) Br2 Fe3 Cu2 (b) Ca2 Ag Cr2O72 19.42(a) Co(s) 2H(aq) ¡ Co2(aq) H2(g) Ecell 0.28 V; spontaneous (b) 2Mn2(aq) 5Br2(l) 8H2O(l) ¡ 2MnO4(aq) 10Br(aq) 16H(aq) Ecell 0.44 V; not spontaneous (c) Hg22(aq) ¡ Hg2(aq) Hg(l) Ecell 0.07 V; not spontaneous 19.44(a) 2Ag(s) Cu2(aq) ¡ 2Ag(aq) Cu(s) Ecell 0.46 V; not spontaneous (b) Cr2O72(aq) 3Cd(s) 14H(aq) ¡ 2Cr3(aq) 3Cd2(aq) 7H2O(l) Ecell 1.73 V; spontaneous (c) Pb(s) Ni2(aq) ¡ Pb2(aq) Ni(s) Ecell 0.12 V; not spontaneous 19.46 3N2O4(g) 2Al(s) ¡ 6NO2(aq) 2Al3(aq) Ecell 0.867 V (1.66 V) 2.53 V 2Al(s) 3SO42(aq) 3H2O(l) ¡ 2Al3(aq) 3SO32(aq) 6OH(aq) Ecell 2.59 V SO42(aq) 2NO2(aq) H2O(l) ¡ SO32(aq) N2O4(g) 2OH(aq) Ecell 0.06 V Oxidizing agents: Al3 N2O4 SO42 Reducing agents: SO32 NO2 Al 19.48 2HClO(aq) Pt(s) 2H(aq) ¡ Cl2(g) Pt2(aq) 2H2O(l) Ecell 0.43 V 2HClO(aq) Pb(s) SO42(aq) 2H(aq) ¡ Cl2(g) PbSO4(s) 2H2O(l) Ecell 1.94 V Pt2(aq) Pb(s) SO42(aq) ¡ Pt(s) PbSO4(s) Ecell 1.51 V Oxidizing agents: PbSO4 Pt2 HClO Reducing agents: Cl2 Pt Pb 19.50 yes; C A B 19.53 A(s) B(aq) ¡ A(aq) B(s) with Q [A]/[B]. (a) [A] increases and [B] decreases. (b) Ecell decreases. (c) Ecell Ecell (RT/nF) ln ([A]/[B]); Ecell Ecell when (RT/nF) ln ([A]/[B]) 0. This occurs when ln ([A]/[B]) 0, that is, [A] equals [B]. (d) Yes, when [A] [B]. 19.55 In a concentration cell, the overall reaction decreases the concentration of the more concentrated electrolyte because it is being reduced. Reduction occurs in the cathode compartment. 19.56(a) 3 1035 (b) 4 1031 19.58(a) 1 1067 (b) 6 109 19.60(a) 2.03 105 J/mol (b) 1.7 105 J/mol 19.62(a) 3.82 105 J/mol (b) 5.6 104 J/mol 19.64 E 0.28 V; G 2.7 104 J/mol 19.66 Ecell 0.054 V; G 1.0 104 J/mol 19.68 8.7 105 mol/L 19.70(a) 0.05 V
A-41
(b) 0.50 mol/L (c) [Co2] 0.91 mol/L; [Ni2] 0.09 mol/L 19.72 A; 0.083 V 19.74 Electrons flow from the anode, where oxidation occurs, to the cathode, where reduction occurs. The electrons always flow from the anode to the cathode no matter what type of battery. 19.76 A D-sized alkaline battery is larger than an AAA-sized one, so it contains greater amounts of the cell components. (a) The cell potential is an intensive property and does not depend on the amounts of the cell components. (b) The total charge, however, does depend on the amount of cell components, so the D-sized battery produces more charge. 19.78 The Teflon spacers keep the two metals separated so that the copper cannot conduct electrons that would promote the corrosion (rusting) of the iron skeleton. 19.81 Sacrificial anodes are made of metals with E less than that of iron, 0.44 V, so they are more easily oxidized than iron. Only b, f, and g will work for iron: a will form an oxide coating that prevents further oxidation; c will react with groundwater quickly; d and e are less easily oxidized than iron. 19.83 To reverse the reaction requires 0.34 V with the cell in its standard state. A 1.5 V cell supplies more than enough potential, so the Cd metal is oxidized to Cd2 and Cr metal plates out. 19.85 The oxidation number of N in NO3 is 5, the maximum O.N. for N. In the nitrite ion, NO2, the O.N. of N is 3, so nitrogen can be further oxidized. 19.87(a) Br2 (b) Na 19.89 I2 gas forms at the anode; magnesium (liquid) forms at the cathode. 19.91 Bromine gas forms at the anode; calcium metal forms at the cathode. 19.93 copper and bromine 19.95 iodine, zinc, and silver 19.97(a) Anode: 2H2O(l) ¡ O2(g) 4H(aq) 4e Cathode: 2H2O(l) 2e ¡ H2(g) 2OH(aq) (b) Anode: 2H2O(l) ¡ O2(g) 4H(aq) 4e Cathode: Sn2(aq) 2e ¡ Sn(s) 19.99(a) Anode: 2H2O(l) ¡ O2(g) 4H(aq) 4e Cathode: NO3(aq) 4H(aq) 3e ¡ NO(g) 2H2O(l) (b) Anode: 2Cl(aq) ¡ Cl2(g) 2e Cathode: 2H2O(l) 2e ¡ H2(g) 2OH(aq) 19.101(a) 3.75 mol e (b) 3.62 105 C (c) 28.7 A 19.103 0.275 g Ra 19.105 9.20 103 s 19.107(a) The sodium and sulfate ions conduct a current, facilitating electrolysis. Pure water, which contains very low (107mol/L) concentrations of H and OH, conducts electricity very poorly. (b) The reduction of H2O has a more positive half-potential than does the reduction of Na; the oxidation of H2O is the only reaction possible because SO42 cannot be oxidized. Thus, it is easier to reduce H2O than Na and easier to oxidize H2O than SO42. 19.109 62.6 g Zn 19.111(a) 3.3 1011 C (b) 4.7 1011 J (c) 1.2 104 kg 19.114 64.3 mass % Cu 19.115(a) 8 days (b) 32 days (c) $1300 19.118 (a) 2.4 104 days (b) 2.1 g (c) $CAD 7.3 105 19.121(a) Pb/Pb2: Ecell 0.13 V; Cu/Cu2: Ecell 0.34 V (b) The anode (negative electrode) is Pb. The anode in the other cell is platinum in the standard hydrogen electrode. (c) The precipitation of PbS decreases [Pb2], which increases the potential. (d) 0.13 V 19.124 The three steps equivalent to the overall reaction M(aq) e ¡ M(s) are
H is hydrationH (1) M(aq) ¡ M(g) (2) M(g) e ¡ M(g) H is IE (3) M(g) ¡ M(s)
H is atomizationH
A-42
Appendix G • Brief Answers to Selected Problems
The energy for step 3 is similar for all three elements, so the difference in energy for the overall reaction depends on the values for hydrationH and IE. The Li ion has a much greater hydration energy than Na and K because it is smaller, with a larger charge density that holds the water molecules more tightly. The energy required to remove the waters surrounding Li offsets the lower ionization energy, making the overall energy for the reduction of lithium larger than expected. 19.125 The very high and very low standard electrode potentials involve extremely reactive substances, such as F2 (a powerful oxidizer) and Li (a powerful reducer). These substances react directly with water because any aqueous cell with a voltage of more than 1.23 V has the ability to electrolyze water into hydrogen and oxygen. 19.127(a) 1 105 s (b) 1.5 104 kWh (c) $1.3 19.129 If metal E and a salt of metal F are mixed, the salt is reduced, producing metal F because E has the greatest reducing strength of the three metals; F D E. 19.131(a) Cell I: 4 mol electrons; G 4.75105 J/mol Cell II: 2 mol electrons; G 3.94 105 J/mol Cell III: 2 mol electrons; G 4.53 105 J/mol (b) Cell I: 13.2 kJ/g Cell II: 0.613 kJ/g Cell III: 2.63 kJ/g Cell I has the highest ratio (most energy released per gram) because the reactants have very low mass, while Cell II has the lowest ratio because the reactants have large masses. 19.135(a) 9.7 g Cu (b) 0.56 mol/L Cu2 19.137 Sn2(aq) 2e ¡ Sn(s) Cr3(aq) e ¡ Cr2(aq) Fe2(aq) 2e ¡ Fe(s) U4(aq) e ¡ U3(aq) 19.141(a) 3.5 109mol/L (b) 0.3 mol/L 19.143(a) Nonstandard cell: (8.314 J/mol ⴢ K)(298 K) ln [Ag]waste Ewaste Ecell (1)(96485 C/mol) Standard cell: (8.314 J/mol ⴢ K)(298 K) ln [Ag]standard Estandard Ecell (1)(96485 C/mol) (b) [Ag ]waste c e
a
Estandard Ewaste 0.0256783
b
d ([Ag]standard)
(c) If both silver ion concentrations are in the same units, in this case ng/L, the “conversions” cancel and the equation derived in part (b) applies if the standard concentration is in ng/L. Conc.(Ag )waste c e
a
Estandard Ewaste 0.0256783
b
d ([Ag]standard)
where C is concentration in ng/L (d) 900 ng/L (e) [Ag ]waste ea
(Estandard Ewaste)(zF/R)Tstandard ln[Ag ]standard Twaste
water: Li, Ba, Na, Al, and Mn. Metals with potentials lower than that of hydrogen (0.00 V) can displace H2 from acid: Li, Ba, Na, Al, Mn, Zn, Cr, Fe, Ni, Sn, and Pb. Metals with potentials greater than that of hydrogen (0.00 V) cannot displace H2: Cu, Ag, Hg, and Au. 19.150(a) 5.4 1011 (b) 0.20 V (c) 0.43 V (d) 8.3 104 mol/L NaOH 19.153(a) 1.25 105 kJ (b) 1.28 104 L (c) 9.97 104 s (d) 166 kWh (e) $23.2 19.154 2.94
Chapter 20 • 20.2(a) Carbon’s electronegativity is midway between the most metallic and most nonmetallic elements of Period 2. To attain a filled outer shell, carbon forms covalent bonds to other atoms in molecules, network covalent solids, and polyatomic ions. (b) Since carbon has four valence electrons, it forms four covalent bonds to attain an octet. (c) To reach the He electron configuration, a carbon atom must lose four electrons, requiring too much energy to form the C4 cation. To reach the Ne electron configuration, the carbon atom must gain four electrons, also requiring too much energy to form the C4 anion. (d) Carbon is able to bond to itself extensively because its small size allows for close approach and great orbital overlap. The extensive orbital overlap results in a strong, stable bond. (e) The COC bond is short enough to allow sideways overlap of unhybridized p orbitals of neighbouring C atoms. The sideways overlap of p orbitals results in the bonds that are part of double and triple bonds. 20.3(a) H, O, N, P, S, and halogens (F, Cl, Br, I) (b) Heteroatoms are atoms of any element other than carbon and hydrogen. (c) More electronegative than C: N, O, F, Cl, and Br; less electronegative than C: H and P. Sulfur and iodine have the same electronegativity as carbon. (d) Since carbon can bond to a wide variety of heteroatoms and to carbon atoms, it can form many different compounds. 20.6 The COH and COC bonds are unreactive because electronegativities are close and the bonds are short. The COI bond is reactive because it is long and weak. The C“O bond is reactive because oxygen is more electronegative than carbon and the electron-rich bond makes it attract electron-poor atoms. The COLi bond is also reactive because the bond polarity results in an electron-rich region around carbon and an electron-poor region around lithium. 20.7(a) An alkane is an organic compound consisting of carbon and hydrogen in which there are no multiple bonds between carbon atoms, only single bonds. A cycloalkane is an alkane in which the carbon chain is arranged in a ring. (b) The general formula for an alkane is CnH2n2. The general formula for a cycloalkane is CnH2n. (elimination of two hydrogen atoms is required to form the additional bond between carbon atoms in the ring) CH3 20.9(a) CH3
CH
CH
b
19.145 (a) 1.08 103 C (b) 0.629 g Cd, 1.03 g NiO(OH), 0.202 g H2O; total mass of reactants 1.86 g (c) 10.1% 19.147 Li Ba Na Al Mn Zn Cr Fe Ni Sn Pb Cu Ag Hg Au. Metals with potentials lower than that of water (0.83 V) can displace H2 from
CH2
CH2
CH2
CH2
CH3
CH3
(b)
CH2
CH3
1 6
2 3
5 4
CH3
(c) 3,4-dimethylheptane (d) 2,2-dimethylbutane
Appendix G • Brief Answers to Selected Problems
20.11(a) 4-methylhexane means a 6 C chain with a methyl group on the 4th carbon: CH3 6
5
CH3
4
CH2
CH2
CH 3
2
CH2
CH3 1
Numbering from the end carbon to give the lowest value for the methyl group gives the correct name of 3-methylhexane. (b) 2-ethylpentane means a five-carbon chain with an ethyl group on the second carbon. Numbering the longest chain gives the correct name, 3-methylhexane. 1 CH 3 2 CH 2
CH
CH3
3
CH2
CH2 5
4
CH3 6
20.25(a) Constitutional isomers are those with different sequences of bonded atoms. They are not stereoisomers. (b) Geometric isomers are a type of stereoisomers where there is a different orientation of groups around a double bond or a cyclic structure. (c) Optical isomers are a type of stereoisomers where a molecule and its mirror image cannot be superimposed on each other. They rotate the plane of polarized light in the opposite direction. 20.28(a) trans, labelled Z (b) trans, labelled E. 20.32 H, D, O, P, Br, I 20.33(a) asymmetric (b) symmetric (c) asymmetric (d) symmetric (e) symmetric (f) asymmetric 20.36 The compound 2-methylhex-2-ene does not have cis-trans isomers because the #2 carbon atom is attached to two identical methyl (–CH3) groups. S 20.38(a) 4 1 3
(b)
Cl 1
CH3 CH2
CH2
CH2
Me
Me 1
2S 3R
OH
4
H HO
CH2
R and E
2
(c)
CH3
CH
4
Me
HO
3
Correct name is methylcyclohexane. (d) 3,3-methyl-4-ethyloctane means an 8 C chain with two methyl groups attached to the third carbon and one ethyl group to the fourth carbon. C
2
F
CH3
CH2
Br
HO
(c) 2-methylcyclohexane means a 6 C ring with a methyl group on carbon #2:
CH3
A-43
2
2
3
3
CH3
O
H 4
OH 1
CH3
20.41
Correct name is 4-ethyl-3,3-dimethyloctane. 20.13
Me Me
1
4 3
Cl
20.15
3
Me
1
OH para
3
OH meta
OH ortho
20.44(a) MeO
H
F
Br
(b)
H
H
C H C chiral carbon Cl
SMe
C H
CH3
CH3 CH2 C C chiral carbon CH3 H
CH3 chiral carbon
CH3
20.46(a) 3-Bromohexane is optically active. chiral carbon
20.18 B is the most stable as it minimizes the interaction between the largest substituents (methyl groups). 20.20 B is the least stable. A and C are equally stable. Br
H H
Me
Me
H
H
H
A
H
Me
Me
H
Br
Br
B
H
H Cl
H
H H t-Bu
H
H H A
H H
H
H
t-Bu
H
H H
H H B
CH2
CH3
Br
Br
C
Me
CH2
CH3
H
C C
CH3
H
H
Me
C
Cl
CH2
CH3
Cl
(c) 1,2-Dibromo-2-methylbutane is optically active. CH3
20.48(a)
CH2
C
Br
Br H
chiral carbon CH2 CH3
H C
H H
CH2
H
Cl Cl 20.22 Carbon Br CH3 Fischer projection: skeleton: 20.23 Structure B is more stable as the bulky groups (t-butyl and methyl) are in the equatorial position, which reduces the 1,3 diaxial interactions. H
CH
CH2
(b) 3-Chloro-3-methylpentane is not optically active.
Me
Me
CH3
CH3CH2
C
H
CH3 C
CH3
cis-pent-2-ene
CH3CH2
C H
trans-pent-2-ene
A-44
Appendix G • Brief Answers to Selected Problems H
(b)
H
H
C
C
(c)
CH3 C
C
CH3
C
cis-cyclohexylprop-1-ene
CH2
C
C H
H
cis-hex-3-ene
CH2
C
CH2
CH3 CH CH2
CH3
C
CH3
CH3 CH CH
CH3
CH
CH
C
C H
H
cis-1,2-dichloroethene
CH3
CH3 CH2
C
Cl
trans-1,2-dichloroethene
C C
C
C
C
C
C
C
C
C
C
C
C
C
C
C C
C
C
C
C
C
C
C C
C
C
C
C
C
C C
C
C C
C
C
C
C
C C
C C
C
C
C
C
C
C
C
C
C
C
C C
C C
CH2
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
CH3
CH2
CH
CH2
CH3
CH
C
CH3 CH CH2
CH2
CH2
CH3
CH
CH3
CH3
CH3
CH2
CH3
CH3
CH
CH
CH2
CH3
CH3
CH2
CH2
CH2
CH3
CH2
CH2
C
CH
CH3 CH2 CH3
CH
CH3
CH3 CH2 CH3
CH
CH2
CH3
CH2
C
CH2
CH3
CH
CH3
CH3
CH3 CH2
CH
C
CH2
CH2
CH3
CH
CH
CH
CH3
CH3 CH3 CH3
CH2
CH
CH2
C
CH3
CH2
CH
CH
CH3
CH2
CH2
CH3
CH3
CH3
C
C
CH2
CH3
CH3
CH3 CH3
C
CH
CH3
C
C
C
CH3
C
C
C
C
C
C
CH3
CH
C
CH
CH3
CH3
CH
CH
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
CH
C C
C
CH3
CH3
C
CH2
C C
CH3
CH3
CH3
CH3 CH3 CH2
CH2 CH2
CH3 CH2 CH2 CH
CH CH CH3 CH2 CH CH2 CH2
CH2
CH2 CH2
CH2 CH2
CH2 CH
CH3
CH3
C
C
C
C
(c)
CH2 C
CH3
CH3
CH2 C
CH
C
20.57(a) (b)
H2C HC
CH C
CH2 CH
CH3
CH3
CH2
C C
CH2
CH
CH3 CH3
C
C
CH
CH3 CH3
CH3
C C
CH
CH2
C
C
C
CH2
(b)
C
C
C
CH3
CH3
C C
C
C
CH2
CH
C
(b)
C
C
C
C
C C
C
CH2
CH3 CH2 CH2
CH
C
C
C
C
CH3
C
C
C
C
C
CH3
20.52(a)An alkene is a hydrocarbon with at least one double bond between two carbon atoms. An alkyne is a hydrocarbon with at least one triple bond between two carbon atoms. (b) For an alkene, assuming only one double bond, the general formula is CnH2n. For an alkyne, assuming only one triple bond, the general formula is CnH2n–2. 20.53(a) C C C C C C C C C C C C
C
CH2
H
Cl
C
H
CH3
CH2
trans-hex-3-ene
(c) no geometric isomers Cl Cl (d) C
C
H
C
C
C C
20.55(a)
H
C
C
C
C
C
trans-cyclohexylprop-1-ene
CH3
C
C
C
C
C
C
H
(c) no geometric isomers 20.50(a) no geometric isomers (b) CH3 CH2 CH2 CH3
C
C
C
CH3 C
(c) Structure is correct 20.59 Due to resonance structures, all the bonds in benzene are equivalent and the same, having partial double bond and partial single bond characteristic. 20.62 Methyl: ortho; hydroxy: para;
Appendix G • Brief Answers to Selected Problems
bromo: meta; chloro: meta; fluoro: ortho. 20.64(a) 2-bromo-4methylphenol; (b) 5-bromo-2-chloro-3-methylaniline; (c) 1-bromo-2,6-dichloro-3-fluoro-5-methyl-4-ethenylbenzene 20.65 A is not aromatic, B is not aromatic, C is aromatic, D is not aromatic. 20.68(a) 4,4-dimethylpent-2-yne (b) (2Z,5E)hepta-2,5-diene (if counted from right to left) or (2E,5Z)-hepta2,5-diene (if counted from left to right) (c) (E)-5-bromopent3-en-1-yne Cl 20.69 Cl Cl
20.81
CH2
NH2
CH3
CH3
CH
CH2
NH2
CH3
CH3
CH2
CH
NH2
CH3
C
NH2
CH3
CH2
CH2
NH
CH3
CH3
CH
NH
CH3
N
CH3
CH3 CH2
NH
CH2
CH3
CH3
CH2
CH3
20.84 DU 6 20.86 DU 4 20.89 DU 5 O
Cl
1,4-dichlorobenzene (p-dichlorobenzene)
1,3-dichlorobenzene (m-dichlorobenzene)
1,2-dichlorobenzene (o-dichlorobenzene)
CH2
CH3
CH3 Cl
HO
CH2
CH3
Cl
20.71
CH3
A-45
OH
C(CH3)3 C7H6O2
(H3C)3C
20.93 CH3
alkene
alcohol
(b)
(E)-pent-2-ene
2-methylbut-1-ene
2-methylbut-2-ene
OH
OH
O
£
±£
£
haloalkane
HO
OH
C
CH2
(Z)-pent-2-ene
20.95 DU 2
O Cl
3-methylbut-1-ene
pent-1-ene
£
£
20.72 C B D A 20.75 (a) haloalkane (b) nitrile (c) carboxylic acid (d) aldehyde 20.77(a) CH3 CH CH CH2 OH
carboxylic acid
aldehyde
alkyne
aromatic ring O
(c)
C
O
CH3
N£
£ ±
aromatic ring
C
£
CH2
nitrile
C
O
C
CH3
£ketone CH2 £ CH3 (trans)-1-tert-butyl3-methylcyclohexane
O CH2
CH2
C
Most Stable
O H
CH3
CH2
CH
C
CH3
H
CH3
O CH3
CH
CH2
C
CH3
CH3
ester
20.80 Aldehydes: CH2
ketone
O
(e)
CH3
OH
20.96
O
N
O
amide H
alkene
(d)
OH
CH3
CH3 O H
CH3
CH3
C
H
C
Most Stable
CH3
(cis)-1-tert-butyl4-methylcyclohexane
Ketones: C
CH2
CH2
O C
CH
O
CH3
CH3
CH3
CH2
C
CH2
CH3
CH3
CH3
CH3 H3C
O
CH3
CH3
CH3
(trans)-1, 3-dimethylcyclohexane
Equally Stable
A-46
Appendix G • Brief Answers to Selected Problems
21.37(a)
Chapter 21 • 21.4(a) elimination (b) addition H⫹ £ 21.6(a) CH3CH2CH CHCH2CH3 ⫹ H2O ±
Et
Me OH
(b)
OH
(b)
±£ CH3CH CH2 ⫹ CH3CH2OH ⫹ KBr hυ £ CHCl2CH3 ⫹ 2HCl (c) CH3CH3 ⫹ 2Cl2 ± 21.8(a) oxidation (b) reduction (c) reduction 20.10(a) oxidized (b) oxidized 21.12 Step 1 substitution Step 2 addition 21.13 C 21.14(a) stepwise (b) exothermic (c) step A 21.16(a) heterolytic (b) hemolytic (c) heterolytic 21.18(b) is the most nucleophilic 21.24(b) It forms a more stable carbocation. 21.27 As the reaction goes by an SN1 mechanism, there is a planar carbocation intermediate which can be attacked from either side giving both enantiomers and resulting in a racemic mixture. 21.29(b). The azide (N 3 ) is negatively charged (making it highly nucleophilic). 21.32 OMe
and
Br
Me
CH3CH2CH2CHCH2CH3
I
CH3CHBrCH3 ⫹ CH3CH2OK
1. NaH 2. MeI
and
(c)
OH
21.38 (b) 21.41 (b) 21.43 The most basic is (b), the least basic is (d). The more stable the anion, the less basic it is. 21.47(a). In the reaction with the unhindered base (MeO) the SN2 reaction will dominate. Me 21.49 Me
Et
H
Leaving group
Me
±±±£
⫹
HCI
NaN3 ⫹
Me
R
CH
CH2
H2O, H⫹ H⫹
±±±£
Br
Br
HBr CH
CH3
OH⫺ Br
HBr
±±±£
O
Me
OH
R
21.56 B C A D. 21.57(a)
Leaving group O
O
Me
Me
(b) Nucleophile
t-BuONa ± ±±£
21.53
O Cl
Anti-periplanar conformation
Me
I
Br
⫹
Me
H
Me
OH
H
Et
21.51 The sterically hindered base will abstract the more accessible proton.
Me
21.33(a) Nucleophile
±±±£
I
OMe 1. TsCl 2. NaOMe
Me
Et
Me
Me
I
Me
and
Me
OH
NaN3
N3
Me
⫹
(b)
NaBr
Br Br2
±±±£
(c)
Br Leaving group
Nucleophile
O O
(c)
H N Me ⫹
Br HBr
⫹ HO(C
N
±±£
O
±±±£
O)Me
(d)
OH
O
HCl, H O
2 ±±± £
(d) Nucleophile
21.60
OH
Br
O
±±±£
⫹
Me
Leaving group
HOTs
Me
OTs
Me
21.61
(e) Leaving group O
Me
Me
HO Nucleophile OH
Me
Me
Me 1. BH3, THF 2. H2O2
±±±±±£
±±±£ O
OH
R
CH
CH3
Appendix G • Brief Answers to Selected Problems
21.63
(b)
Me Me
O
Me
O H3C
O
Me Me
21.66
21.84(a)
H
O
O
O
C
OH
OH
CH
C
H
N
H
NH3
CH2
CH3
CH3
O
21.68 In general, aromatic rings are much less reactive (more stable) than compounds with isolated C“C double bonds because of their delocalized electrons. The benzene ring can undergo electrophilic substitution, nucleophilic substitution, elimination, and addition reactions, while the alkenes mainly undergo addition reactions. 21.73 The aniline is a good ortho/ para director and under weak acidic conditions (right hand arrow) dominates the course of the reaction. Under strongly acidic conditions, the aniline is protonated, making it a weak meta director, and the methyl group (an ortho/para directing group) and the protonated aniline give the alternative product (left hand arrow). 21.74 The C“C bond is nonpolar while the C“O bond is polar, since oxygen is more electronegative than carbon. Therefore, in the case of addition to a C“O bond, the electron-rich group will bond to the carbon and the electron-poor group will bond to the oxygen, resulting in one product, while the addition to the alkene depends on the structure of the alkene.
(c)
O H
C
OH
H
N H
21.85(a) CH3
O
CH2
OH
Br ±±£ CH3
CH2
OH
CH2
CH
C
OH
±±3±±±± ±±±£ H
O CH3
CH2
C
CH2
O
Br
(b) CH3
CH2
C CH3
CH
CN
±±£
CH3
CH2
CH3 N
CH
CH3
OH C
O
H3O, H2O
±±±±£ CH3 CH2 CH 21.91(a) 116.2 g/mol (b) 102.2 g/mol
21.92(a) (b)
±±±±£
⫹
CH2 CH3 O
CH3
Me
Me
A-47
±£
CH3CHO C6H5
MgBr
CH3
CH3 CH3
CH2
C
CH3
C6H5CH(OH)CH3 CH3
CH
O
H O
2 ±± £
MgBr OH CH3
±±£
⫹
O
21.78 Alcohols undergo substitution at a saturated carbon while acids undergo substitution at the carboxyl carbon. 21.81(a) O C
H H
N
CH3
±£
CH3
C
±£
H
N
CH3
H2O
(b) CH3
O CH2
CH2
C
O
±£
CH3
H H
O
H2O eliminated
O CH3
CH2
CH2
C
±£
CH CH3 O
CH3 CH H2O CH3
(c)
CH3
O
H
C
O
£
H
H H
O
H2O eliminated O
CH3
C
CH
O
CH2
CH2
CH
CH3
CH3
CH2
C
CH
CH3
CH3 CH3
±£
MgBr
21.93(a)The functional group in ibuprofen is the carboxylic acid group COOH. The chiral centre is RC*H(CH3)COOH (b) React the aldehyde with methyl Grignard reagent, CH3MgBr to get the alcohol. React the alcohol with HBr to get the brominated product. React the bromide with cyanide ion to produce the nitrile. Then hydrolyze the nitrile with aqueous HCl to get the carboxylic acid. 21.94(a) Perform an acid-catalyzed dehydration of the alcohol (elimination), followed by bromination of the double bond (addition of Br2) (b) The product is an ester, so a carboxylic acid is needed to prepare the ester. First, oxidize one mole of ethanol to ethanoic acid (acetic acid). Then, react one mole of ethanoic acid with a second mole of ethanol to form the ester: OH⫺ 21.97(a) ⫹ OH⫺
CH3 H2O
CH3
MgBr
O
H
H2O eliminated
CH3
OR
O O
C
(c) CH3MgBr and C6H5CHO (d) HCHO (methanal, also known as formaldehyde) (e) CH3 C CH3 CH3 CH2
⫹
CH3
CH2
Cl
±£
H H
H Cl
±£
A-48
Appendix G • Brief Answers to Selected Problems
(b) Carbon 1 is sp2 Carbon 2 is sp3 hybridized. hybridized. Carbon 4 is sp2 Carbon 3 is sp3 hybridized. hybridized. Carbon 5 is sp3 Carbon 6 and 7 are sp2 hybridized. hybridized. (c) Carbon atoms 2, 3, and 5 are chiral centres as they are each bonded to four different groups.
(b) Br
±£
Br
Br
⫹
⫺
Br ⫹ Br
±£ Br
(c)
C
O
O
H
O
O
OH ±±±O±± £
±£
Chapter 22
O
Answers to Boxed Reading Problems: B22.1 ddATP: four complementary chain pieces; ddCTP: three complementary chain pieces
21.99(a) O
O H⫹
±£
OH ⫹ HO
O
esterification
(b) Br ⫹ NaOH
±£
OH substitution (SN2)
(c)
±£
⫹ (CH3)3CONa
elimination (E2)
Cl
21.101 O
O
⫹
NH2
±±±£
N H
Cl reduction
±±±±£
N H
21.104 (a) Br
• 22.1 1.8 105g/mol 22.5 1.7 106 pm 22.8 Rg 8292 pm 22.11 The glass transition temperature is the midpoint of the range of temperatures when a semicrystalline substance, like a polymer, changes from a molten or rubber-like substance into a hard brittle solid. The glass transition temperature for the polymer given is 6C. 22.13 Both branching and crosslinking are bifurcations in a linear polymer, where instead of extending in only one linear direction, now two or more chains exist. In cross-linked polymers, these branches connect two chains together, whereas in a normal branched polymer they do not connect chains together. 22.15 a larger polymer made up from blocks of polymerized monomer units. 22.21 Nylon: an amine and a carboxylic acid; polyester: a carboxylic acid and an alcohol (b) H H 22.23(a) H H
H⫹, H2O
CH3Li
±±±±£
±±±±£
R
R
R OH
p-Tosyl chloride
±±±±±±±£
H
Cl
CH3, OH
n
C
C
H
CH3
CH3 HO
Si
R
OH
±±£
O
1,2 hydride shift
(b) The reaction rate would be doubled. (c) No effect.
CH
OH
O Mel
NaH ± ±£
ONa ±±£
A
OMe
C4H7NaO2
OH
O
Me
MeMgBr
±±±£ A
C6H12O
±£
CH
CH3
C11
3
CH
C
O
5
CH3
CH
CH2 C 7 ketone
£
ester
C
6
C
O
O
ester
n
H2O
n
CH3
CH2
22.32 Isotatic has a regular structure making its solid form more regular and semicrystalline. As this makes the solid more stable it will have a higher melting point than the atatic variant. 22.34 Polyamide polymers. They can form hydrogen bonds that stabilize the solid form and raise the glass transition temperature. 22.36(a) condensation (b) addition (c) condensation (d) condensation 22.42(a) O O H O H H2N
O
C
£
CH3
4
£
alkene
alkene
2
CH 1
O
21.110(a) CH2
C5H10O2
± £
21.108
⫹
O
CH3
21.106 O
Si
22.28(a) C6H5CHCH2 can be used to produce polystyrene, (CH(C6H5) CH2)n. (b) CH2 CH
OTs
O
n
CH3
CH3
⌬
±±±±£
R
Pyridine
C
22.26 n
O
C
CH
C
N
OH
C
N
CH2
CH2 C
CH
O
CH2
N NH
CH
HN
C
OH
Appendix G • Brief Answers to Selected Problems
(b)
H
O ⴙ
H3N
CH
C
O C
CH
N
CH2
H
H N
22.90 Kevlar is formed from 1,4-phenylenediamine with terephthaloyl chloride.
O CH
ⴚ
O
C
CH2
H2N
NH2
SH
OH
22.44(a) AATCGG (b) TCTGTA 22.48(a) Water is eliminated when the peptide is formed (b) glycine: 4, alanine: 1, valine: 3, proline: 6, serine: 7, arginine: 49 (c) 10,700 g/mol 22.52 Nucleoside sugar nucleobase; nucleotide sugar nucleobase phosphate O
HO
N
O
NH N H
O HO
NH
O H O
O
O P O
N
O H
H
O
OH
HO
Nucleoside
OH
C
N
⫹
N
Me
Me
Me
O
O
O Cl ⫹ H2N
Cl
Me Me
22.81 O-H bonds are readily exchangeable in CDCl3. The protons attached to the hydroxyl groups are clearly seen. When the sample is shaken with D2O, the deuterium in the D2O can exchange with OH groups of the sample, converting them to OD groups. As deuterium atoms cannot be seen in 1H NMR, the OD groups can no longer been seen. 22.83 two 22.86 two; one high medium shifted peak (approx. 2.5 ppm), multiplicity is a quartet (the CH2); one high low shifted peak (approx. 1.0 ppm), multiplicity is a triplet (the CH3)
NH2
O
Nucleotide
22.71 As chlorine is a highly electronegative atom, it increases the positive charge on the carbon of the carbonyl compared to propan-2-one (acetone). This in turn makes the carbonyl more polarized and increases the bond strength. As stronger bonds vibrate at larger wavenumbers, it increases the frequency of the absorption. 22.77 four 22.80 O
Cl
22.95(a)The complementary sequence to GATCGACTA would be: CTAGCTGAT. The RNA sequence would be different since RNA contains U instead of T, therefore the sequence would be: CUAGCUGAU. (b) It requires 3 bases in order to code for one amino acid, therefore this sequence could code for 3 amino acids. (c) GC would not be a good technique to separate this nucleotide sequence from a mixture of other nucleotides since the molecular weight of the sequence and polarity mean that the nucleotide is not volatile enough. Liquid chromatography or better yet gel electrophoresis would be better separation techniques. 22.97(a) A condensation polymerization.
H
22.55 The nitrogen-containing bases form hydrogen bonds to their complementary bases. The flat, N-containing bases stack above each other, which allows extensive dispersion forces. The exterior negatively charged sugar-phosphate chains form ion-dipole and hydrogen bonds with water molecules in the aqueous surroundings, which also stabilizes the structure. 22.58 Dispersion forces are present between the nonpolar tails of the lipid molecules within the bilayer. The polar heads interact with the aqueous surroundings through hydrogen bonds and ion-dipole forces. 22.61 Secondary 22.64 (a) Both R groups are from cysteine, which can form a disulfide bond (covalent bond). (b) Lysine and aspartic acid give a salt link. (c) Asparagine and serine will hydrogen bond. (d) Valine and phenylalanine interact through dispersion forces. ⫺ 22.66 O O H H C
Cl
H HO
Nucleobase
NH
OH
H
A-49
R
±£
⫹NH
H 2
Cl
R
±£
O
O C
NH ⫺ R Cl O
O⫺
HO
⫹
±£
(CH2)8
N H
(CH2)6
H N
H ⫹ nHCL
n
(b) Rubber gloves need to be flexible at room temperature, not hard and brittle, like glass, which is how this polymer becomes when it is cooled below its Tg (50 C). Therefore, it will not be suitable for making rubber gloves. (c) Since this is a polymer, Size Exclusion Chromatography (SEC) would likely be the best technique to analyze it, though it might also be possible to use liquid chromatography.
Chapter 23 Answers to Boxed Reading Problems: B23.2(a) (1) NO O3 ¡ XO O2 [slow], (2) XO O ¡ X O2 [fast] (b) The rate-determining step is the slow step. (b) rate 3 107 molecule/cm3s • 23.2 Fe from Fe2O3; Ca from CaCO3; Na from NaCl; Zn from ZnS 23.3(a) Differentiation refers to the processes involved in the formation of Earth into regions (core, mantle, and crust) of differing composition. Substances separated according to their densities, with the more dense material in the core and the less dense in the crust. (b) O, Si, Al, and Fe (c) O 23.7 Plants produced O2, slowly increasing the oxygen concentration in the atmosphere and creating an environment for oxidizing metals. The oxygen-free decay of plant and animal material created large fossil fuel deposits. 23.9 Fixation refers to the process of converting a substance in the atmosphere into a form more readily usable by organisms. Carbon and nitrogen; fixation of carbon dioxide gas by plants and fixation of nitrogen gas by nitrogen-fixing bacteria. 23.12 Atmospheric
A-50
Appendix G • Brief Answers to Selected Problems
nitrogen is fixed by three pathways: atmospheric, industrial, and biological. Atmospheric fixation requires high-temperature reactions (e.g., initiated by lightning) to convert N2 into NO and other oxidized species. Industrial fixation involves mainly the formation of ammonia, NH3, from N2 and H2. Biological fixation occurs in nitrogen-fixing bacteria that live on the roots of legumes. Human activity is an example of industrial fixation. It contributes about 17% of the fixed nitrogen. 23.14(a) the atmosphere (b) Plants excrete acid from their roots to convert PO43 ions into more soluble H2PO4 ions, which the plant can absorb. Through excretion and decay, organisms return soluble phosphate compounds to the cycle. 23.17(a) 1.1 103 L (b) 4.2 102 m3 23.18(a) The iron(II) ions form an insoluble salt, Fe3(PO4)2, that decreases the yield of phosphorus. (b) 8.8 t 23.20(a) Roasting consists of heating the mineral in air at high temperatures to convert the mineral to the oxide. (b) Smelting is the reduction of the metal oxide to the free metal using heat and a reducing agent such as coke. (c) Flotation is a separation process in which the ore is removed from the gangue by exploiting the difference in density in the presence of detergent. The gangue sinks to the bottom and the lighter ore-detergent mix is skimmed off the top. (d) Refining is the final step in the purification process to yield the pure metal. 23.25(a) Slag is a byproduct of steelmaking and contains the impurity SiO2. (b) Pig iron is the impure product of iron metallurgy (containing 3–4% C and other impurities). (c) Steel refers to iron alloyed with other elements to attain desirable properties. (d) The basic-oxygen process is used to purify pig iron and obtain carbon steel. 23.27 Iron and nickel are more easily oxidized and less easily reduced than copper. They are separated from copper in the roasting step and converted to slag. In the electrorefining process, all three metals are in solution, but only Cu2 ions are reduced at the cathode to form Cu(s). 23.30 According to Le Châtelier’s principle, the system shifts toward formation of K when the potassium gas is removed as it is produced. 23.31(a) Ehalf-cell 3.05 V, 2.93 V, and 2.71 V for Li, K, and Na, respectively. In all of these cases, it is energetically more favorable to reduce H2O to H2 than to reduce M to M. (b) 2RbX Ca ¡ CaX2 2Rb, where H IE1(Ca) IE2(Ca) 2IE1(Rb) 929 kJ/mol. Based on the IEs and positive H for the forward reaction, it seems more reasonable that Rb metal will reduce Ca2 than the reverse. (c) If the reaction is carried out at a temperature greater than the boiling point of Rb, the product mixture will contain gaseous Rb, which can be removed from the reaction vessel; this would cause a shift in the equilibrium to form more Rb as product. (d) 2CsX Ca ¡ CaX2 2Cs, where H IE1(Ca) IE2(Ca) 2IE1(Cs) 983 kJ/mol. This reaction is more unfavorable than for Rb, but Cs has a lower boiling point. Ca can be used to separate gaseous Cs from molten CsX if the reaction is carried out at a temperature between the boiling point of Cs and Ca. 23.32(a) 4.5 104 L (b) 1.30 108 C (c) 1.69 106 s 23.35(a) Mg2 is more difficult to reduce than H2O, so H2(g) would be produced instead of Mg metal. Cl2(g) forms at the anode due to overvoltage. (b) The fH of MgCl2(s) is 641.6 kJ/mol. High temperature favors the reverse (endothermic) reaction, the formation of magnesium metal and chlorine gas. 23.37(a) Sulfur dioxide is the reducing agent and is oxi-
dized to the 6 state (SO42). (b) HSO4(aq) (c) H2SeO3(aq) 2SO2(g) H2O(l) ¡ Se(s) 2HSO4(aq) 2H(aq) 23.42(a) O.N. for Cu: in Cu2S, 1; in Cu2O, 1; in Cu, 0 (b) Cu2S is the reducing agent, and Cu2O is the oxidizing agent. 23.44(a) 1.3 106 C (b) 1.2 103 A 23.47 2ZnS(s) C(graphite) ¡ 2Zn(s) CS2(g); rG 463 kJ/mol. Since
rG is positive, this reaction is not spontaneous at standardstate conditions. 2ZnO(s) C(s) ¡ 2Zn(s) CO2(g); rG 242.0 kJ/mol. This reaction is also not spontaneous, but is less unfavorable. 23.48 The formation of sulfur trioxide is very slow at ordinary temperatures. Increasing the temperature can speed up the reaction, but because the reaction is exothermic, increasing the temperature decreases the yield. Adding a catalyst increases the rate of the reaction, allowing a lower temperature to be used to enhance the yield. 23.51(a) Cl2, H2, and NaOH (b) The mercury-cell method yields higher purity NaOH, but releases some Hg, which is released into the environment. 23.52(a) G 142 kJ/mol; yes (b) The rate of the reaction is very low at 25C. (c) 500CG 53 kJ/mol, so the reaction is spontaneous. (d) K25C 7.8 1024. K500C 3.8 103 (e) 1.05 103 K 23.53 3 102kg Cl2 23.56(a) P4O10(s) 6H2O(l) ¡ 4H3PO4(l) (b) 1.52 23.58(a) 9.007 109 g CO2 (b) The 4.3 1010 g CO2 produced by automobiles is much greater than that from the blast furnace. 23.60(a) If [OH] 1.1 104 mol/L (i.e., if pH 10.04), Mg(OH)2 will precipitate. (b) 1 (To the correct number of significant figures, all the magnesium precipitates.) 23.61(a) K25C(step 1) 1 10168; K25C(side rxn) 7 10228 (b) K900C(step 1) 4.5 1049; K900C(side rxn) 1.4 1063 (c) $5.8 107 (d) $4.2 107 23.64(1) 2H2O(l) 2FeS2(s) 7O2(g) ¡ 2Fe2(aq) 4SO42(aq) 4H(aq) increases acidity (2) 4H(aq) 4Fe2(aq) O2(g) ¡ 4Fe3(aq) 2H2O(l) 3 (3) Fe (aq) 3H2O(l) ¡ Fe(OH)3(s) 3H(aq) increases acidity (4) 8H2O(l) FeS2(s) 14Fe3(aq) ¡ 15Fe2(aq) 2SO42(aq) 16H(aq) increases acidity 23.65 Density of ferrite: 7.86 g/cm3; density of austenite: 7.55 g/cm3 23.67(a) Cathode: Naⴙ(l) eⴚ¡ Na(l) Anode: 4OHⴚ(l) ¡ O2(g) 2H2O(g) 4eⴚ (b) 50% 23.70(a) nCO2(g) nH2O(l) ¡ (CH2O)n(s) nO2(g) (b) 27 L (c) 7.6 104 L 23.71 73 mg/L 23.72 892 kg Na3AlF6 23.73(a) 23.2 min (b) 13 effusion steps 23.75(a) 1.890 t Al2O3 (b) 0.3339 t C (c) 100% (d) 74% (e) 2.813 103 m3 23.80 Acid rain increases the leaching of phosphate into the groundwater, due to the protonation of PO43ⴚ to form HPO42ⴚ and H2PO4ⴚ. (a) 6.4 10ⴚ7mol/L (b) 1.1 10ⴚ2 mol/L 23.81(a) 1.00 mol % (b) 238.9 g/mol 23.83 density of silver 10.51 g/cm3; density of sterling silver 10.2 g/cm3
Chapter 24 • 24.2(a) 1s22s22p63s23p64s23d104p65s24dx (b) 1s22s22p6 3s23p64s23d104p65s24d105p66s24f 145dx 24.4(a) Five;
Appendix G • Brief Answers to Selected Problems
(b) Examples are Mn, [Ar]4s23d5, and Fe3, [Ar]3d5. 24.6(a) The elements should increase in size as they increase in mass from Period 5 to Period 6. Because there are 14 inner transition elements in Period 6, the effective nuclear charge increases significantly; so the atomic size decreases, or “contracts.” This effect is significant enough that Zr4 and Hf 4 are almost the same size but differ greatly in atomic mass. (b) The atomic size increases from Period 4 to Period 5, but stays fairly constant from Period 5 to Period 6. (c) Atomic mass increases significantly from Period 5 to Period 6, but atomic radius (and thus volume) increases slightly, so Period 6 elements are very dense. 24.9(a) A paramagnetic substance is attracted to a magnetic field, while a diamagnetic substance is slightly repelled by one. (b) Ions of transition elements often have partially filled d orbitals whose unpaired electrons make the ions paramagnetic. Ions of main-group elements usually have a noble-gas configuration with no partially filled levels. (c) Some d orbitals in the transition element ions are empty, which allows an electron from one d orbital to move to a slightly higher energy one. The energy required for this transition is small and falls in the visible wavelength range. All orbitals are filled in ions of main-group elements, so enough energy would have to be added to move an electron to the next principal energy level, not just another orbital within the same energy level. This amount of energy is very large and much greater than the visible range of wavelengths. 24.10(a) 1s22s22p63s23p64s23d3 (b) 1s22s22p63s23p64s2 3d104p65s24d1 (c) [Xe]4f145d106s2 24.12(a) [Xe]4f 145d 66s2 (b) [Ar]3d74s2 (c) [Kr]4d105s1 24.14(a) [Ar], no unpaired electrons (b) [Ar]3d 9, one unpaired electron (c) [Ar]3d5, five unpaired electrons (d) [Kr]4d 2, two unpaired electrons 24.16(a) 5 (b) 4 (c) 7 24.18 Cr, Mo, and W 24.20 in CrF2, because the chromium is in a lower oxidation state 24.22 Atomic size increases slightly down a group of transition elements, but nuclear charge increases much more, so the first ionization energy generally increases. The reduction potential for Mo is lower, so it is more difficult to oxidize Mo than Cr. In addition, the ionization energy of Mo is higher than that of Cr, so it is more difficult to remove electrons from Mo. 24.24 CrO3, with Cr in a higher oxidation state, yields a more acidic aqueous solution. 24.28(a) seven (b) This corresponds to a half-filled f subshell. 24.30(a) [Xe]5d16s2 (b) [Xe]4f 1 (c) [Rn]5f 117s2 (d) [Rn]5f 2 24.32(a) Eu2: [Xe]4f 7; Eu3: [Xe]4f 6; Eu4: [Xe]4f 5. The stability of the half-filled f subshell makes Eu2 most stable. (b) Tb2: [Xe]4f9; Tb3: [Xe]4f8; Tb4: [Xe]4f7. Tb should show a 4 oxidation state because that gives a half-filled subshell. 24.34 Gd has the electron configuration [Xe]4f75d16s2 with eight unpaired electrons. Gd3 has seven unpaired electrons: [Xe]4f7. 24.37 The coordination number indicates the number of ligand atoms bonded to the metal ion. The oxidation number represents the number of electrons lost to form the ion. The coordination number is unrelated to the oxidation number. 24.39 2, linear; 4, tetrahedral or square planar; 6, octahedral 24.42 The complex ion has a negative charge. 24.45(a) hexaaquanickel(II) chloride (b) tris(ethylenediamine)chromium(III) perchlorate (c) potassium hexacyanomanganate(II) 24.47(a) 2, 6 (b) 3, 6 (c) 2, 6 24.49(a) potassium dicyanoargentate(I) (b) sodium tetrachlorocadmate(II) (c) tetraammineaquabromocobalt(III)
A-51
bromide 24.51(a)1, 2 (b) 2, 4 (c) 3, 6 24.53(a) [Zn(NH3)4]SO4 (b) [Cr(NH3)5Cl]Cl2 (c) Na3[Ag(S2O3)2] 24.55(a) 4, two ions (b) 6, three ions (c) 2, four ions 24.57(a) [Cr(H2O)6]2(SO4)3 (b) Ba[FeBr4]2 (c) [Pt(en)2]CO3 24.59(a) 6, five ions (b) 4, three ions (c) 4, two ions 24.61(a) The nitrite ion forms linkage isomers because it can bind to the metal ion through the lone pair on the N atom or any lone pair on either O atom. ⫺ O N O (b) Sulfur dioxide molecules form linkage isomers because the lone pair on the S atom or any lone pair on either O atom can bind the central metal ion. O S O (c) Nitrate ions have an N atom with no lone pair and three O atoms, all with lone pairs that can bind to the metal ion. But all of the O atoms are equivalent, so these ions do not form linkage isomers. O
⫺
O
N O
24.63(a) geometric isomerism H
H CH3
H
H
N
CH3
Br
N
Br
Pt CH3
Pt
H
Br
Br
N
N
H H
H
CH3
(b) geometric isomerism H H
H
H
N
H
F
H N
Pt H H
Cl
N
F Pt
Cl
N H
H
H H
(c) geometric isomerism H H
H
H
N
F
H
H N
Pt H
H H
Cl
H N
Pt Cl
O
H
H
F Pt
F
O
Cl
H
H
24.65(a) geometric isomerism Cl
2⫺
Cl
Cl
Pt
2⫺
Br Pt
Br
Br
Br
Cl
(b) linkage isomerism H3N
NH3 NH3
2⫺
H3N
NH3 NH3
2⫺
Cr
Cr H3N
NO2
H3N
NH3
ONO NH3
(c) geometric isomerism NH3 H3N
2⫹
NH3
I
H3N
I
I
Pt H3N
2⫹
I
Pt
NH3
O
NH3 NH3
24.67 The compound with the traditional formula is CrCl3. 4NH3; the actual formula is [Cr(NH3)4Cl2]Cl. 24.69(a) K[Pd(NH3)Cl3] (b) [PdCl2(NH3)2] (c) K2[PdCl6]
H
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Appendix G • Brief Answers to Selected Problems
(d) [Pd(NH3)4Cl2]Cl2 24.71(a) dsp2 (b) sp3 24.74 absorption of orange or yellow light 24.75(a) The crystal field splitting energy ( ) is the energy difference between the two sets of d orbitals that result from electrostatic effects of ligands on a central transition metal atom. (b) In an octahedral field of ligands, the ligands approach along the x, y, and z axes. The dx2 y2 and dz2 orbitals are located along the x, y, and z axes, so ligand interaction is higher in energy. The other orbital-ligand interactions are lower in energy because the dxy, dyz, and dxz orbitals are located between the x, y, and z axes. (c) In a tetrahedral field of ligands, the ligands do not approach along the x, y, and z axes. The ligand interaction is greater for the dxy, dyz, and dxz orbitals and lesser for the dx2 y2 and dz2 orbitals. Therefore, the crystal field splitting is reversed, and the dxy, dyz, and dxz orbitals are higher in energy than the dx2 y2 and dz2 orbitals. 24.78 If is greater than Epairing, electrons will pair their spins in the lower energy set of d orbitals before entering the higher energy set of d orbitals as unpaired electrons. If
is less than Epairing, electrons will occupy the higher energy set of d orbitals as unpaired electrons before pairing in the lower energy set of d orbitals. 24.80(a) no d electrons (b) eight d electrons (c) six d electrons 24.82(a) five (b) ten (c) seven 24.84 y
because it absorbs yellow light, which has higher energy (shorter ) than red light. 24.101 Hg is [Xe]4f 145d106s1 and Cu is [Ar]3d10. The mercury(I) ion has one electron in the 6s orbital that can form a covalent bond with the electron in the 6s orbital of another mercury(I) ion. In the copper(I) ion there are no electrons in the s orbital, so these ions cannot bond with one another. 24.102(a) 6 (b) 3 (c) two (d) 1 mol 24.109 geometric (cis-trans) and linkage isomerism NCS
NH3 Pt
NCS
NH3
cis-diamminedithiocyanatoplatinum(II) SCN
NH3 Pt
SCN
NH3
cis-diamminediisothiocyanatoplatinum(II) SCN
NH3 Pt NH3
NCS
cis-diamminethiocyanatoisothiocyanatoplatinum(II) NCS
NH3 Pt
H3N
y
SCN
trans-diamminedithiocyanatoplatinum(II) SCN
x
NH3 Pt
x H3N
NCS
trans-diamminediisothiocyanatoplatinum(II) SCN
d x2–y2
In an octahedral field of ligands, the ligands approach along the x, y, and z axes. The dx2 y2 orbital is located along the x and y axes, so ligand interaction is greater. The dxy orbital is offset from the x and y axes by 90, so ligand interaction is less. The greater interaction of the dx2 y2 orbital results in its higher energy. 24.86 a and d cannot form high- and low-spin complexes
H3N
(b)
SCN
trans-diamminethiocyanatoisothiocyanatoplatinum(II)
24.110 (a) [Co(NH3)4(H2O)Cl] 2+tetraammineaquachlorocobalt(III) ion 2 geometric isomers 2⫹
Cl
24.88(a)
NH3 Pt
d xy
H3N
(c)
NH3
Co
NH3
H3N
H2O
24.90(a)
(c)
24.92 [Cr(H2O)6]3 [Cr(NH3)6]3 [Cr(NO2)6]3 24.94 A violet complex absorbs yellow-green light. The light absorbed by a complex with a weaker field ligand would be at a lower energy and higher wavelength. Light of lower energy than yellow-green light is yellow, orange, or red. The colour observed would be blue or green. 24.97 The H2O ligand is weaker than the NH3 ligand. The weaker field ligand results in a lower splitting energy and absorbs visible light of lower energy. The hexaaqua complex appears green because it absorbs red light. The hexaammine complex appears violet
NH3 NH3
trans Cl and H2O
(b)
OH2
H3N
Co H3N
2⫹
Cl
cis Cl and NH3
(b) [Cr(H2O)3Br2Cl] triaquadibromochlorochromium(III) 3 geometric isomers Cl
Cl Br
H2O
H2O Cr
Cr Br
OH2 H2O
Br’s trans
Cl Br
H2O
OH2
H2O Cr
Br H2O Br’s cis H2O’s facial
H2O
Br Br Br’s cis H2O’s meridional
Appendix G • Brief Answers to Selected Problems (c) [Cr(NH3)2(H2O)2Br2]⫹diamminediaquadibromochromium(III) ion 6 isomers (5 geometric) ⫹ NH3 NH3 Br H2O Br H2O Cr
⫹
Cr OH2
Br
Br
H2O
NH3
NH3
All pairs are trans NH3
Only NH3’s are trans ⫹
NH3
NH3
H2O
H2O
Cr
⫹
Br Cr
OH2
Br
Br
Br
NH3 OH2
Only H2O’s are trans NH3
Only Br’s are trans NH3
⫹
H3N
NH3
H2O Cr
⫹
OH2 Cr
Br
H2O
OH2
Br
Br
Br
All pairs are cis. These are optical isomers of each other.
24.115(a) no optical isomers (b) no optical isomers (c) no optical isomers (d) no optical isomers (e) optical isomers 24.116 Pt[P(C2H5)3]2Cl2 C2H5 C2H5 C2H5
P
P
C2H5
Pt Cl
Cl
cis-dichlorobis(triethylphosphine)platinum(II) C2H5 C2H5
P
Cl Pt
Cl
P
C2H5
C2H5
trans-dichlorobis(triethylphosphine)platinum(II)
24.118(a) The first reaction shows no change in the number of particles. In the second reaction, the number of reactant particles is greater than the number of product particles. A decrease in the number of particles means a decrease in entropy. Based on entropy change only, the first reaction is favored. (b) The ethylenediamine complex will be more stable with respect to ligand exchange in water because the entropy change for that exchange is unfavourable (negative).
Chapter 25 Answers to Boxed Reading Problems: B25.1 In the s-process, a nucleus captures a neutron sometime over a long period of time. Then the nucleus emits a beta particle to form another element. The stable isotopes of most heavy elements up to 209Bi form by the s-process. The r-process very quickly forms less stable isotopes and those with A greater than 230 by multiple neutron captures, followed by multiple beta decays. B25.3 The simultaneous fusion of three nuclei is a termolecular process. Termolecular processes have a very low
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probability of occurring. The bimolecular fusion of 8Be with 4 210 0 He is more likely. B25.4 210 83Bi ¡ 84Po 1 (nuclide A); 210 206 4 206 1 84 Po ¡ 82Pb 2 (nuclide B); 82 Pb 30n ¡ 209 209 210 0 82Pb (nuclide C); 82 Pb ¡ 83Bi 1 (nuclide D) • 25.1(a) Chemical reactions are accompanied by relatively small changes in energy; nuclear reactions are accompanied by relatively large changes in energy. (b) Increasing temperature increases the rate of a chemical reaction but has no effect on a nuclear reaction. (c) Both chemical and nuclear reaction rates increase with higher reactant concentrations. (d) If the reactant is limiting in a chemical reaction, then more reactant produces more product and the yield increases. The presence of more radioactive reactant results in more decay product, so a higher reactant concentration increases the yield. 25.2(a) 95.02% (b) The atomic mass is larger than the isotopic mass of 32S. Sulfur-32 is the lightest isotope. 25.4(a) Z down by 2, N down by 2 (b) Z up by 1, N down by 1 (c) no change in Z or N (d) Z down by 1, N up by 1 (e) Z down by 1, N up by 1. A different element is produced in all cases except c . 25.6 A neutron-rich nuclide decays by decay. A neutron-poor nuclide undergoes positron decay or electron capture. 4 230 232 232 0 25.8(a) 234 92U ¡ 2 90Th (b) 93Np 1e ¡ 92U (c) 12 0 12 27 27 23 0 7N ¡ 1 6C 25.10(a) 12Mg ¡ 1 13Al (b) 12Mg 0 23 103 103 48 0 ¡ 1 11Na (c) 46Pd 1 e ¡ 45Rh 25.12(a) 23V 0 107 107 210 0 ¡ 48 22Ti 1 (b) 48Cd 1 e ¡ 47Ag (c) 86Rn ¡ 206 4 186 186 225 0 84Po 2 25.14(a) 78Pt 1 e ¡ 77Ir (b) 89Ac ¡ 221 4 129 129 0 87Fr 2 (c) 52Te ¡ 53I 1 25.16(a) Appears stable because its N and Z values are both magic numbers, but its N/Z ratio ( 1.50) is too high; it is unstable. (b) Appears unstable because its Z value is an odd number, but its N/Z ratio ( 1.19) is in the band of stability, so it is stable. (c) Unstable because its N/Z ratio is too high. 25.18(a) The N/Z ratio for 127 I is 1.4; it is stable. (b) The N/Z ratio for 106Sn is 1.1; it is unstable because this ratio is too low. (c) The N/Z ratio is 1.1 for 68As. The ratio is within the range of stability, but the nuclide is most likely unstable because there are odd numbers of both protons and neutrons. 25.20(a) alpha decay (b) positron decay or electron capture (c) positron decay or electron capture 25.22(a) decay (b) positron decay or electron capture (c) alpha decay 25.24 Stability results from a favorable N/Z ratio, even numbered N and/or Z, and the occurrence of magic numbers. The N/Z ratio of 52Cr is 1.17, which is within the band of stability. The fact that Z is even does not account for the variation in stability because all isotopes of chromium have the same Z. However, 52Cr has 28 neutrons, so N is both an even number and a magic number for this isotope only. 25.28 seven emissions and four emissions 25.31 No, it is not valid to conclude that t1/2 equals 1 min because the number of nuclei is so small. Decay rate is an average rate and is only meaningful when the sample is macroscopic and contains a large number of nuclei. For the sample containing 6 1012 nuclei, the conclusion is valid. 25.33 2.56 102 Ci/g 25.35 1.4 108 Bq/g 25.37 1 1012 day1 25.39 2.31 107 yr1 25.41 1.49 mg 25.43 2.2 109 yr 25.45 27 dpm 25.47 1.0 106 yr 25.50 Neither radiation nor neutron beams have charge, so neither is deflected by a magnetic or electric field. Neutron beams differ from radiation in
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Appendix G • Brief Answers to Selected Problems
that a neutron has a mass approximately equal to that of a proton. It was observed that a neutron beam induces the emission of protons from a substance; radiation does not cause such emission. 25.52 Protons are repelled from the target nuclei due to interaction with like (positive) charges. Higher energy is required to overcome the repulsion. 25.53(a) 105B 42 ¡ 10n 2 1 29 242 4 137N (b) 28 14Si 1H ¡ 0n 15P (c) 96Cm 2 ¡ 1 244 2 0n 98Cf 25.58 Ionizing radiation is more dangerous to children because their rapidly dividing cells are more susceptible to radiation than adults’ slowly dividing cells. 25.60(a) 5.4 107 rad (b) 5.4 109 Gy 25.62(a) 7.5 1010 Gy (b) 7.5 105 mrem (c) 7.5 1010 Sv 25.65 1.86 103 rad 25.67 NAA does not destroy the sample, while chemical analyses do. Neutrons bombard a nonradioactive sample, inducing some atoms within the sample to be radioactive. The radioisotopes decay by emitting radiation characteristic of each isotope. 25.73 Energy is released when a nucleus forms from nucleons. The nuclear binding energy is the quantity of energy holding 1 mol of nuclei together. This energy must be absorbed to break up the nucleus into nucleons and is released when nucleons come together. 25.75(a) 1.861 104 eV (b) 2.981 1015 J 25.77 7.6 1011 J 25.79(a) 7.976 MeV/nucleon (b) 127.6 MeV/atom (c) 1.23134 1010 kJ/mol 25.81(a) 8.768 MeV/nucleon (b) 517.3 MeV/ atom (c) 4.99128 1010 kJ/mol 25.85 Radioactive decay is a spontaneous process in which unstable nuclei emit radioactive particles and energy. Fission occurs as the result of highenergy bombardment of nuclei with small particles that cause the nuclei to break into smaller nuclides, radioactive particles, and energy. All fission events are not the same. The nuclei
split in a number of ways to produce several different products. 25.88 The water serves to slow the neutrons so that they are better able to cause a fission reaction. Heavy water is a better moderator because it does not absorb neutrons as well as light water does; thus, more neutrons are available to initiate the fission process. However, D2O does not occur naturally in great abundance, so its production adds to the cost of a heavywater reactor. 25.93(a) 1.1 1029 kg (b) 9.9 1013 J (c) 5.9 108 kJ/mol; this is approximately 1 million times larger than a typical enthalpy of reaction. 25.95 8.0 103 yr 25.98 1.35 105mol/L 25.100 6.2 102 25.102(a) 5.99 h (b) 21% 25.104(a) 0.999 (b) 0.298 (c) 5.58 106 (d) Radiocarbon dating is more reliable for the fraction in part (ii) because a significant amount of 14C has decayed and a significant amount remains. Therefore, a change in the amount of 14C will be noticeable. For the fraction in part (i), very little 14C has decayed, and for (iii) very little 14C remains. In either case, it will be more difficult to measure the change, so the error will be relatively large. 25.106 6.579 h 25.110 4.969 109 L/h 25.113 81 yr 25.115(a) 0.15 Bq/L (b) 0.27 Bq/L (c) 3.3 d; a total of 12.8 d 25.118 7.4 s 25.121 1926 25.124 6.27 105 eV, 6.05 107 kJ/mol 25.127(a) 2.07 1017 J/atom (b) 1.45 107 H atoms (c) 1.4960 105 J (d) 1.4959 105 J (e) No, the captain should continue using the current technology. 25.130(a) 0.043 MeV, 2.9 1011 m (b) 4.713 MeV 25.134(a) 3.26 103 d (b) 3.2 103 s (c) 2.78 1011 yr 25.136(a) 1.80 1017 J (b) 6.15 1016 kJ (c) The procedure in part (b) produces more energy per kilogram of antihydrogen. 25.137 9.316 102 MeV 25.141 7.81 d
