Pre-U STPM Chemistry Term 1 CC039342a

342 Chemistry Term 1 STPM STPM Model Paper (962/1) 20 (a) Ammonia, carbon dioxide and hydrogen chloride are colourless gases at room conditions. The boiling points and solubility in water of the three gases are given in the table below. Ammonia, karbon dioksida dan hidrogen klorida adalah gas pada keadaan bilik. Takat didih dan keterlarutan dalam air ketiga-tiga gas diberikan dalam jadual di bawah. Gas Formula Boiling point/ °C Takat didih/ °C Solubility in water/mol dm–3 Keterlarutan dalam air/ mol dm–3 Ammonia Ammonia NH3 –32.8 18.0 Carbon dioxide Karbon dioksida CO2 –78.0 0.028 Hydrogen chloride Hidrogen klorida HCl –85.0 23.0 (i) Draw a dot-and cross diagram to show the arrangement of electrons in the ammonia, carbon dioxide and hydrogen chloride molecules. [3 marks] Lukis gambar rajah titik/silang untuk menunjukkan susunan elektron pada ammonia, karbon dioksida dan hidrogen klorida. [3 markah] (ii) Using the valence shell electron-pair repulsion theory, predict the shapes of ammonia and carbon dioxide. [3 marks] Dengan menggunakan teori penolakan pasangan-pasangan-elektron-valens, ramalkan bentuk ammonia dan karbon dioksida. [3 markah] (iii) Explain the variation of the boiling point and solubility in water of the three gases in terms of intermolecular forces. [5 marks] Terangkan perubahan takat-takat didih dan keterlarutan dalam air bagi ketiga-tiga gas dari segi daya antara molekul. [5 markah] (b) One of the most common sugar is glucose. A simplified structure of glucose is shown below. Salah satu gula yang paling biasa dijumpai adalah glukosa. Suatu struktur ringkas bagi glukosa ditunjukkan di bawah. H OH HO OH H OH H H H CH OH O 2 Honey can be assumed as containing water and glucose only. The composition by mass of water in honey is 28.5%, all of which is bonded to glucose by intermolecular forces. Madu boleh dianggap terdiri daripada air dan glukosa sahaja. Komposisi air mengikut jisim dalam madu ialah 28.5%, yang kesemuanya diikat kepada glukosa melalui daya antara molekul. (i) What type of intermolecular force binds the water molecule to glucose? Illustrate your answer by showing how one such bond is formed between water and glucose. [2 marks] Apakah jenis daya antara molekul yang mengikat molekul-molekul air kepada glukosa? Ilustrasikan jawapan anda dengan menunjukkan bagaimana satu ikatan tersebut dibentuk antara air dan glukosa. [2 markah] (ii) Calculate the number of water molecules bonded to each glucose molecule in honey. [2 marks] Hitung bilangan molekul air yang diikat kepada setiap molekul glukosa dalam madu. [2 markah]

343 Chemistry Term 1 STPM Group 1(I) 2(II) 3 4 5 6 7 8 9 10 11 12 13 (III) 14 (IV) 15(V) 16 (VI) 17 (VII) 18 (VIII) 6.9 Li 3 9.0 Be 4 23.0 Na 11 24.3 Mg 12 39.1 K19 40.1 Ca 20 85.5 Rb 37 87.6 Sr 38 133 Cs 55 137 Ba 56 [223] Fr 87 [226] Ra 88 a Xb 45.0 Sc 21 88.9 Y39 175.0 Lu 71 [262] Lr 103 47.9 Ti 22 50.9 V23 91.2 Zr 40 92.9 Nb 41 178 Hf 72 181 Ta 73 [261] Rf 104 [262] Db 105 52.0 Cr 24 95.9 Mo 42 184 W74 [263] Sg 106 54.9 Mn 25 55.8 Fe 26 [98] Tc 43 101 Ru 44 186 Re 75 190 Os 76 [264] Bh 107 [265] Hs 108 58.9 Co 27 103 Rh 45 192 Ir 77 [266] Mt 109 58.7 Ni 28 63.5 Cu 29 106 Pd 46 108 Ag 47 195 Pt 78 197 Au 79 [269] Uuu 110 [272] Uuu 111 65.4 Zn 30 10.8 B5 12.0 C6 14.0 N7 16.0 O8 19.0 F9 20.2 Ne 10 27.0 Al 13 28.1 Si 14 31.0 P15 32.1 S16 35.5 Cl 17 40.0 Ar 18 69.7 Ga 31 72.6 Ge 32 74.9 As 33 79.0 Se 34 79.9 Br 35 83.8 Kr 36 115 In 49 119 Sn 50 112 Sb 51 128 Te 52 127 I 53 131 Xe 54 204 Ti 81 207 Pb 82 209 Bi 83 [209] Po 84 [210] At 85 [222] Rn 86 4.0 He 2 112 Cd 48 201 Hg 80 139 La Lanthanides 57 Antinides140 Ce 58 141 Pr 59 144 Nd 60 [145] Pm61 150 Sm62 152 Eu 63 159 Tb 65 163 Dy 66 165 Ho 67 167 Er 68 169 Tm69 173 Yb 70 157 Gd 64 227 Ac 89 232 Th 90 231 Pa 91 238 U92 237 Np 93 [244] Pu 94 [243] Am95 [247] Bk 97 [251] Cf 98 [252] Es 99 [257] Fm 100 [258] Md 101 [259] No 102 [247] Cm 96 [277] Uub 112 1.0 H1 a = relative atomic mass X = atomic symbol b = proton (atomic) number Periodic Table

344 Chemistry Term 1 STPM Ionisation Energies (kJ mol–1) Z Name 1st 2nd 3rd 4th 5th 6th 7th 1 H 1310 2 He 2370 5250 3 Li 519 7300 11 800 4 Be 900 1760 14 800 21 000 5 B 799 2420 3660 25 000 32 828 6 C 1090 2350 4610 6220 37 830 42 270 7 N 1400 2860 4590 7480 9440 53 270 64 360 8 O 1310 3390 5320 7450 10 990 13 320 71 330 9 F 1680 3370 6040 8410 11 020 15 160 17 870 10 Ne 2080 3950 6150 9290 12 180 15 240 20 000 11 Na 494 4560 6940 9540 13 350 16 610 21 110 12 Mg 736 1450 7740 10 500 13 630 17 800 21 700 13 Al 577 1820 2740 11 600 14 830 18 380 23 290 14 Si 786 1580 3230 4360 16 090 15 P 1060 1900 2920 4960 3270 21 270 16 S 1000 2260 3390 4540 7010 8500 27 110 17 Cl 1260 2300 3850 5150 6540 9360 11 020 18 Ar 1520 2660 3950 5770 7240 8780 12 000 19 K 418 3070 4600 5860 20 Ca 590 1150 4940 6480 21 Sc 632 1240 2390 7110 8840 10 720 22 Ti 661 1310 2720 4170 9570 11 520 23 V 648 1370 2870 4600 6290 12 360 24 Cr 653 1590 2990 4770 6690 8740 25 Mn 716 1510 3250 5190 6990 9200 26 Fe 762 1560 2960 5400 7240 9600 27 Co 757 1640 3230 5100 7670 9840 28 Ni 736 1750 3390 5400 7280 10 400 29 Cu 745 1960 3550 5690 7710 9940 30 Zn 908 1730 3828 5980 7970 10 400 31 Ga 577 1980 2960 6190 32 Ge 762 1540 3300 4390 9020 35 Br 140 2080 3460 4850 5760 8550 9940 38 Sr 548 1060 4120 5440 50 Sn 707 1410 2940 3930 6980 53 I 1010 18 470 2040 4030 56 Ba 502 966 3390 82 Pb 716 1450 3080 4080 6640

345 Chemistry Term 1 STPM Electronegativity (Pauling’s Scale) Ionisation Constant of Weak Acids (mol dm–3) [HA + H2 O  H3 O+ + A– ] Acid Ka (298 K) pKa (298 K) [Al(H2 O)6 ]3+ 1.0 × 10–5 5.00 C6 H5 COOH 6.3 × 10–5 4.20 C6 H5 OH 1.3 × 10–10 9.89 CH3 CH2 COOH 1.3 × 10–5 5.11 CH3 COOH 1.8 × 10–5 4.74 [Fe(H2 O)6 ]2+ 1.8 × 10–7 6.74 [Fe(H2 O)6 ]3+ 6.7 × 10–3 2.17 H2 C2 O4 5.9 × 10–2 1.23 H2 CO3 4.3 × 10–7 6.37 H2 PO4 – 6.3 × 10–8 7.20 19 K 0.8 37 Rb 0.8 55 Cs 0.7 87 Fr 0.7 20 Ca 1.0 38 Sr 1.0 56 Ba 0.9 1 H 2.1 3 Li 1.0 11 Na 0.9 4 Be 1.5 12 Mg 1.2 21 Sc 1.3 39 Y 1.2 57 La 1.1 22 Ti 1.5 40 Zr 1.4 72 Hf 1.3 23 V 1.6 41 Nb 1.6 73 Ta 1.5 24 Cr 1.6 42 Mo 1.8 74 W 1.7 25 Mn 1.5 43 Tc 1.9 75 Re 1.9 26 Fe 1.8 44 Ru 2.2 76 Os 2.2 27 Co 1.9 45 Rh 2.2 77 Ir 2.2 28 Ni 1.9 46 Pd 2.2 78 Pt 2.2 29 Cu 1.9 47 Ag 1.9 79 Au 2.4 30 Zn 1.6 48 Cd 1.7 80 Hg 1.9 31 Ga 1.6 49 In 1.7 81 Tl 1.8 32 Ge 1.8 50 Sn 1.8 82 Pb 1.9 83 Bi 1.9 34 Se 2.4 52 Te 2.1 84 Po 2.0 35 Br 2.8 33 As 2.0 51 Sb 1.9 53 I 2.5 85 At 2.2 36 Kr 54 Xe 86 Rn 5 B 2.0 13 Al 1.5 6 C 2.5 14 Si 1.8 15 P 2.1 8 O 3.5 16 S 2.5 7 N 3.0 9 F 4.0 17 Cl 3.0 2 He 10 Ne 18 Ar 88 Ra 0.9

346 Chemistry Term 1 STPM Ionisation Constant of Weak Bases (mol dm–3) [B + H+  HB+] Acid Ka (298 K) pKa (298 K) H2 S 1.0 × 10–7 7.00 H2 SO3 1.2 × 10–8 7.92 H3 BO3 5.8 × 10–10 9.27 H3 PO4 7.5 × 10–3 2.12 HC2 O4 – 6.4 × 10–5 4.19 HClO 3.5 × 10–8 7.46 HCN 4.0 × 10–10 9.40 HCO3 – 7.0 × 10–11 10.2 HCOOH 1.8 × 10–4 3.74 HF 7.2 × 10–4 3.14 HIO3 1.7 × 10–1 0.77 HNO2 4.7 × 10–4 3.33 HPO4 2– 3.6 × 10–13 12.4 HS– 1.0 × 10–19 19.0 HSO3 – 6.2 × 10–8 7.21 HSO4 – 1.0 × 10–2 2.00 [Mg(H2 O)6 ]2+ 3.7 × 10–12 11.4 NH4 + 5.6 × 10–10 9.25 Base Kb (298 K) pKb (298 K) C6 H5 NH2 4.6 × 10–10 9.34 CH3 COO– 5.6 × 10–10 9.25 CH3 NH2 4.4 × 10–4 3.36 CN– 1.7 × 10–5 4.77 CO3 2– 2.1 × 10–4 3.68 F– 1.4 × 10–11 10.8 HCO3 – 2.3 × 10–8 7.64 HS– 1.0 × 10–7 7.00

347 Chemistry Term 1 STPM Base Kb (298 K) pKb (298 K) HSO3 – 5.9 × 10–13 12.2 N2 H4 1.0 × 10–6 6.00 NH3 1.8 × 10–5 4.74 SO3 2– 1.7 × 10–7 6.77 Solubility Product of Some Common Salts Compound Formula Ksp (Appropriate unit) Aluminium hydroxide Al(OH)3 1.9 × 10–33 Barium bromate Ba(BrO3 )2 2.4 × 10–4 Barium carbonate BaCO3 8.1 × 10–9 Barium chromate BaCrO4 2.4 × 10–10 Barium fluoride BaF2 1.7 × 10–6 Barium iodate Ba(IO3 )2 6.5 × 10–10 Barium oxalate BaC2 O4 1.2 × 10–7 Barium sulphate BaSO4 1.1 × 10–10 Beryllium hydroxide Be(OH)2 6.9 × 10–22 Cadmium carbonate CdCO3 1.0 × 10–12 Cadmium hydroxide Cd(OH)2 7.2 × 10–15 Cadmium oxalate CdC2 O4 1.5 × 10–8 Cadmium phosphate Cd3 (PO4 )2 2.5 × 10–33 Cadmium sulphide CdS 3.6 × 10–29 Calcium carbonate CaCO3 1.0 × 10–8 Calcium chromate CaCrO4 2.3 × 10–2 Calcium fluoride CaF2 3.4 × 10–11 Calcium hydroxide Ca(OH)2 8.0 × 10–6 Calcium oxalate CaC2 O4 1.8 × 10–9 Calcium sulphate CaSO4 6.1 × 10–5 Chromium(III) hydroxide Cr(OH)3 6.3 × 10–31 Copper(II) carbonate CuCO3 1.0 × 10–10

348 Chemistry Term 1 STPM Compound Formula Ksp (Appropriate unit) Copper(II) hydroxide Cu(OH)2 4.8 × 10–20 Copper(II) iodate Cu(IO3 )2 1.4 × 10–7 Copper(II) oxalate CuC2 O4 2.9 ×10–8 Copper(II) sulphide CuS 8.5 × 10–45 Iron(III) hydroxide Fe(OH)3 1.1 × 10–36 Iron(II) carbonate FeCO3 2 × 10–11 Iron(II) hydroxide Fe(OH)2 7.9 × 10–15 Iron(II) oxalate FeC2 O4 2.1 × 10–7 Iron(II) sulphide FeS 3.7 × 10–19 Lead bromide PbBr2 6.3 × 10–6 Lead carbonate PbCO3 3.3 × 10–14 Lead chromate PbCrO4 1.8 × 10–14 Lead chloride PbCl2 1.7 × 10–5 Lead fluoride PbF2 3.7 × 10–8 Lead hydroxide Pb(OH)2 1.0 × 10–16 Lead iodate Pb(IO3 )2 1.2 × 10–13 Lead iodide PbI2 7.5 × 10–9 Lead oxalate PbC2 O4 2.7 × 10–11 Lead sulphate PbSO4 1.6 × 10–8 Lead sulphide PbS 3.4 × 10–28 Lithium carbonate Li2 CO3 1.7 × 10–3 Lithium fluoride LiF 1.8 × 10–3 Magnesium carbonate MgCO3 2.6 × 10–5 Magnesium fluoride MgF2 7.1 × 10–9 Magnesium hydroxide Mg(OH)2 1.2 × 10–11 Magnesium oxalate MgC2 O4 8.5 × 10–5 Manganese carbonate MnCO3 9.0 × 10–11 Manganese hydroxide Mn(OH)2 4.0 × 10–14 Manganese sulphide MnS 4.3 × 10–22

349 Chemistry Term 1 STPM Compound Formula Ksp (Appropriate unit) Nickel hydroxide Ni(OH)2 5.5 ×10–16 Nickel sulphide NiS 1.4 × 10–24 Silver bromide AgBr 4.1 × 10–13 Silver carbonate Ag2 CO3 6.1 × 10–12 Silver chloride AgCl 1.5 × 10–10 Silver chromate Ag2 CrO4 1.2 × 10–12 Silver iodate AgIO3 0.9 × 10–8 Silver iodide AgI 1.5 × 10–16 Silver nitrite AgNO2 5.8 × 10–4 Silver oxalate Ag2 C2 O4 1.3 × 10–11 Silver sulphate Ag2 SO4 1.2 × 10–5 Silver sulphide Ag2 S 1.6 × 10–49 Strontium carbonate SrCO3 1.6 × 10–9 Strontium chromate SrCrO4 3.6 × 10–5 Strontium fluoride SrF2 2.8 × 10–9 Strontium oxalate SrC2 O4 5.6 × 10–8 Strontium sulphate SrSO4 2.7 × 10–7 Tin(II) sulphide SnS 1.0 × 10–28 Zinc hydroxide Zn(OH)2 4.5 × 10–17 Zinc carbonate ZnCO3 6.0 × 10–11 Zinc sulphide ZnS 1.2 × 10–23

350 Chemistry Term 1 STPM Chemical Constants Speed of light in vacuum c = 3.0 × 108 m s–1 Planck constant h = 6.63 × 10–34 J s = 3.99 × 10–13 kJ mol–1 s Molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Rydberg constant RH = 1.097 × 107 m–1 Avogadro constant L, NA = 6.02 × 1023 mol–1 Mass of proton mp = 1.67 × 10–27 kg Mass of neutron mn = 1.67 × 10–27 kg Mass of electron me = 9.11 × 10–31 kg Electronic charge e = –1.60 × 10–19 C Ionic product of water Kw = 1.0 × 10–14 mol2 dm–6 (at 298 K) Molar volume of gas Vm = 22.4 dm3 (at s.t.p.) = 24.0 dm3 (at room conditions) Specific heat capacity of water c = 4.18 J g–1 K–1 p = 3.142 ln x = 2.303 log10 x

351 Activation energy Tenaga pengaktifan The minimum amount of energy required for a chemical reaction to occur Allotropy Alotropi The phenomenon where an element can exist in more than one form. The different forms of the same element are called allotropes. For example, carbon (diamond, graphite); sulphur (rhombic and monoclinic); oxygen (oxygen and ozone, O3 ) Autocatalysis Autopemangkinan A reaction where one of the products can act as a catalyst for the reaction Avogadro’s law Hukum Avogadro At constant temperature and pressure, equal volume of all gases contains the same number of moles of gas. Avogadro’s number/constant Pemalar/nombor Avogadro The number of atoms in exactly 12.0000 g of carbon-12 Azeotrope Azeotrop A mixture of liquids with a constant boiling point where the composition of the vapour is the same as the composition of the liquid. The composition of the azeotrope remains unchanged on distillation. Boiling point Takat didih The temperature at which the vapour pressure of a liquid is equal to the atmospheric pressure. Bond energy Tenaga ikatan The amount of heat energy required to break a covalent bond in one mole of gaseous molecules. Boyle’s law Hukum Boyle The volume of a fixed mass of gas at constant temperature is inversely proportional to its pressure. Bronsted-Lowry acid Asid Bronsted-Lowry A proton donor Bronsted-Lowry base Bes Bronsted-Lowry A proton acceptor Buffer capacity Kapasiti penimbal The amount of acid or base that needs to be added to a buffer solution to change its pH by one unit. Buffer solution Larutan penimbal A solution whose pH does not change significantly when a small amount of acid or base is added to it. Catalyst Mangkin A substance that alters the rate of a chemical reaction without itself being consumed Charles’ law Hukum Charles The volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature. Common ion effect Kesan ion sepunya The shift in equilibrium caused by addition of a compound having an ion similar to one of the ions formed from the dissolved substance. Compound Sebatian A type of matter that is made up of more than one type of atoms

352 Chemistry Term 1 STPM Coordinate covalent bond Ikatan koordinat A covalent bond where both the shared electrons are contributed by one of the two bonded atoms. Coordination number (solid) Nombor koordinatan (pepejal) The number of atoms, molecules or ions surrounding a given particle in the solid lattice structure Covalent bond Ikatan kovalen The electrostatic attraction between two adjacent nuclei and the pair of electrons that are shared between them Critical point Titik genting The temperature above which a gas will not liquefy by increasing pressure only Critical pressure Tekanan genting The minimum pressure required to bring about liquefaction of a gas at the critical temperature Dalton’s law of partial pressure Hukum tekanan separa Dalton The total pressure of a mixture of gases that do not interact with one another is given by the sum of the partial pressures of all the gases present. Dative/coordinate bond Ikatan datif/koordinat A covalent bond where both the shared electrons are contributed by one of the two bonded atoms. Degenerate orbitals Orbital setenaga Orbitals having the same energy Degree of dissociation Darjah penceraian The fraction or percentage of the initial amount that has dissociated Dynamic equilibrium Keseimbangan dinamik The condition in which the rate of the forward reaction is equal to the rate of the reverse reaction and the amount of substances remains constant. Electron affinity Afiniti elektron The heat change when a gaseous atom receives an electron, per mole of the atom Electronegativity Keelektronegatifan The relative strength of an atom to attract electrons in a covalent bond to which it is bonded. Element Unsur A type of matter that is made up of one type of atom only Empirical formula Formula empirik It shows the simplest whole number ratio of atoms of each element present in a molecule of the compound. Equilibrium constant Pemalar keseimbangan The ratio of the equilibrium concentration of the products to the equilibrium concentration of the reactants each raise to the power of their respective stoichiometric coefficient. Fractional distillation Penyulingan berperingkat A process for separating liquid components in a liquid mixture based on their different boiling points Half-life Setengah hayat The time taken for the amount of a substance to decrease to half its initial value.

353 Chemistry Term 1 STPM Hybridisation Penghibridan/Pengacukan The mixing of two or more non-equivalent atomic orbitals to form a set of equivalent hybrid orbitals. Hydrogen bond Ikatan hidrogen The electrostatic attraction between a hydrogen atom bonded to an atom of a very electronegative element (N, O, F) and the lone pair electrons of another very electronegative atom (N, O, F). Ideal gas Gas unggul A gas that obeys the ideal equation perfectly under all conditions Ideal solution Larutan unggul A solution that obeys Raoult’s law Ionic bond Ikatan ion The electrostatic attraction between oppositely charged ions that are formed when electrons are transferred from one atom to the other. Ionisation energy Tenaga pengionan The minimum energy required to remove the most loosely held electron from an isolated atom (or ion) in its ground state, per mole of the atom or ion. Isotope Isotop Atoms of the same element i.e. with the same proton number but with different nucleon number. Le Chatelier’s principle Prinsip Le Chatelier When a system in dynamic equilibrium is subjected to a change, the position of equilibrium will shift in the direction to nullify the effect of the change so that equilibrium is reestablished. Lewis acid Asid Lewis A substance that accepts a lone pair of electrons from another substances. Lewis base Bes Lewis A substance that donates a lone pair of electrons to another substance. Ligand Ligan An atom, a molecule or an ion that is bonded to the metal atom or metal cation in a complex ion by coordinate bond. Matter Jirim Anything that has mass and occupies space. Melting point Takat lebur The temperature at which the solid state and liquid state are in equilibrium. Metallic bond Ikatan logam The electrostatic attraction between metal cations and the ‘sea’ of delocalised electrons in a solid lattice. Metalloid Metaloid An element with properties of metal and nonmetal Mole Mol The amount of substance that contains the same number of elementary particles (atoms, molecules, ions or electrons) as the number of atoms in exactly 12 g of carbon-12. Molecular formula Formula molekul It shows the actual number of the different types of atoms (indicated by their symbols) in one discrete molecule of the substance.

354 Chemistry Term 1 STPM Nucleon number Nombor nukleon The number of protons and neutrons in the nucleus of an atom Orbital Orbital A region in space where the probability of finding a particular electron is high (about 95% chances) Order of reaction Tertib tindak balas The order of reaction is the sum of the powers to which the concentration of the reactants is raised in an experimentally determined rate equation. Rate = k[A]x [B]y Order of reaction with respect to A is x. Order of reaction with respect to B is y. Overall order of reaction = x + y. Ostwald dilution law Hukum pencairan Ostwald The degree of dissociation of a weak electrolyte is inversely proportional to its concentration. At infinite dilution (where concentration approaches zero), all weak electrolytes are fully dissociated. Partial pressure Tekanan separa The pressure exerted by a gas, in a mixture of gases, if that gas alone occupies the same volume as the mixture at the same temperature. Pauli exclusion principle Prinsip kecualian Pauli No two electrons in an atom can have the same four quantum numbers. Two electrons occupying the same orbital must have opposite spins. Pi (p) bond Ikatan Pi A covalent bond formed by the sideways overlapping of atomic orbitals (usually the p orbitals). The electron density of the bond is concentrated above and below the plane containing the nuclei of the two bonded atoms. Proton number Nombor proton The number of protons in the nucleus of an atom Radioactivity Keradioaktifan The spontaneous disintegration of unstable nuclei with the emission of high energy radiation/particles Raoult’s law Hukum Raoult The partial vapour pressure of a liquid, A, in a mixture of immiscible liquids, is given by the product of the vapour pressure of pure A and the mole fraction of A in the mixture, at the same temperature. PA = XAP o A Rate-determining step Peringkat penentuan kadar The slowest step in a sequence of steps that leads to the formation of products. It is also the step with the highest activation energy. Relative atomic mass Jisim atom relatif The mass of one atom of the element relative to —–1 12 times the mass of atom of carbon-12 Relative isotopic mass Jisim isotop relatif The mass of one atom of the isotope relative to —–1 12 times the mass of atom of carbon-12 Relative molecular mass Jisim molekul relatif The mass of one molecule of the element or compound relative to —–1 12 times the mass of atom of carbon-12 Screening effect Kesan pengskrinan (penabiran) The repulsion between electronic shells that are filled with electrons. It also refers to the effect of shielding of the outermost electrons by inner electrons from the pull of the nucleus.

355 Chemistry Term 1 STPM Sigma (s) bond Ikatan sigma A covalent bond formed by the end-to-end overlapping of atomic orbitals. The electron density of the bond is concentrated in the region between the nuclei of the bonding atoms. Solubility product Hasil darab keterlarutan The product of the concentrations of the ions, in a saturated solution of a sparingly soluble salt, each raised to the power of its stoichiometric coefficient. sp hybridisation Penghibridan (pengacukan) sp The mixing of an s and a p orbital to form two equivalent hybrid orbitals sp2 hybridisation Penghibridan (pengacukan) sp2 The mixing of an s and two p orbitals to form three equivalent hybrid orbitals sp3 hybridisation Penghibridan (pengacukan) sp3 The mixing of an s and three p orbitals to form four equivalent hybrid orbitals Triple point Takat tripel (tigaan) The temperature and pressure at which the solid, liquid and vapour phases of a substance can exist in equilibrium Van der Waals forces Daya van der Waals The intermolecular forces that hold molecules of covalent compounds together in the solid or liquid state

357 Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry Quick Check 1.1 1 (a) 222 86Rn (b) 234 91Pa (c) 38 20Ca (d) 212 82Pb (e) 208 81 Tl; 208 82Pb 2 228 90Y Quick Check 1.2 1 39.985 2 (a) 1.083 times (b) 12C : 13C = 9.9 : 0.1 3 12.0038 Quick Check 1.3 1 110.32 2 (a) 3 (b) 206 82M; 207 82M; 208 82M (c) It is caused by 206M2+ ion 3 (a) 20 21 22 (b) 20.18 (c) 1.68 times Quick Check 1.4 1 158 160 162 Ratio: 158 : 160 : 162 = 1 : 2 : 1 2 (a) 3 : 2 (b) 12 3 [Not to scale] 14 18 28 32 4 14 16 18 30 32 46 48 50 14 16 18 30 32 46 48 50 N+ 16O+ 16O+ N16O+ N18O+ N16O16O+ N16O18O+ N18O2 O+ Quick Check 1.5 1 (a) 714.15 g (b) 28.88 g (c) 3369.8 g 2 (a) 23.46 g (b) 8.56 g (c) 1.15 g 3 (a) 5.56  104 (b) 3.34  1025 Quick Check 1.6 1 (a) CaCO3 + 2HCl → CaCl2 + H2 O + CO2 (b) 2.00 g (c) 80.0% 2 2 3 1.95 mol dm–3 STPM Practice 1 Objective Questions 1 D 2 C 3 C 4 B 5 B 6 A 7 C 8 B 9 C 10 C 11 D 12 B 13 D 14 D 15 C 16 A 17 B 18 D 19 D 20 B 21 B 22 B 23 C 24 B 25 D 26 B 27 D 28 D 29 C 30 B 31 D 32 A Structured and Essay Questions 1 (a) Number of protons in the nucleus of an atom. (b) Total number of protons and neutrons in the nucleus of an atom. (c) Atoms of the same element but with different nucleon number. 2 Refer to section isotopes. 3 (a) Mole ratio N : O = 30.4 14 : 69.9 16 = 2.1 : 4.4 = 1 : 2 Empirical formula is NO2 (b) The molecular peak = 92 (NO2 )n = 92 46n = 92 n = 2 Molecular formula is N2 O4 (c) m/e (46) = [NO2 ]+ or [N2 O4 ]2+ m/e (94) = [N16O18O]+ 4 (a) Refer to text (b) Refer to text (c) (i) 28 29 30 29.8 1.5 1.0 m/e (ii) Ar = (28  29.8) + (29  1.5) + (30  1.0) (29.8 + 1.5 + 1.0) = 28.11 (iii) 28.11 12 = 2.34 times ANSWERS

358 Chemistry Term 1 STPM 5 (a) Negative. [Positive particles are attracted towards the negative plate] (b) H+. It is lighter and hence deflected more. (c) (i) m e of 2 H+ = 2 ∴ Angle = 1 2  4° = 2° (ii) Zero. Neutron is neutral and is not deflected. 6 C2 H3 79Br79Br35Cl : 220 ≡ (1  1  3) = 3 C2 H3 79Br81Br35Cl : 222 ≡ 2(1  1  3) = 6 C2 H3 81Br81Br35Cl : 224 ≡ (1  1  3) = 3 C2 H3 79Br79Br37Cl : 222 ≡ (1  1  1) = 1 C2 H3 79Br81Br37Cl : 224 ≡ 2(1  1  1) = 2 C2 H3 81Br81Br37Cl : 226 ≡ (1  1  1) = 1 Relative abundance: 220 : 222 : 224 : 226 = (3 + 1) : (6 + 1) : (3 + 2) : 1 = 4 : 7 : 5 : 1 7 (a) Isotope with unstable nucleus and undergoes spontaneous disintegration to form nucleus of smaller isotopes. (b) (i) 227 88W and 227 89X (ii) A ≡ 4 4 2 He + 0 –1e (4 α-particles and 1 β-particle) B ≡ 4 2 He + 2 0 –1e (iii) They are isotopes. 8 (a) The amount of substance that contains the same number of particles as the number of atoms in 12 g of 12C. (b) (i) 180 g (ii) 4.8 × 10–3 180  (6.02  1023) = 1.61  1019 molecules (iii) Mass = 89.5 100  2.5 g = 2.24 g 9 (a) Atoms having the same number of protons and the same number of electrons, but different number of neutrons. (b) (i) 35 36 37 38 35CI+ 35CI1 H+ 37CI+ 37CI1 H+ (ii) 35Cl1 H2+ The energy required to form ion of +2 charge is much higher. 10 (a) 18 34 60 H2 16O CH3 18OH CH3 COOH (b) CH3 —C—18O— CH3 || O 11 (a), (b) Species m/e value P35Cl+ 66 P37Cl+ 68 P35Cl35Cl+ 101 P35Cl37Cl+ 103 P37Cl37Cl+ 105 P35Cl35Cl35Cl+ 136 P35Cl35Cl37Cl+ 138 P35Cl37Cl37Cl+ 140 P37Cl37Cl37Cl+ 142 (c) 66 : 68 = 3 : 1 103 : 105 : 107 = (3  3) : 2(3  1) : (1  1) = 9 : 6 : 1 138 : 140 : 142 : 144 = (3  3  3) : 3(3  3 1) : 3(3  1  1) : (1  1  1) = 27 : 27 : 9 : 1 12 (a) Due to the presence of isotopes. (b) Let the % abundance of 35X = a% % abundance of 37X = 100 – a% 35.5 = 35a + 37(100 – a) 100 a = 75% (c) 75% 35 37 25% 13 (a) Relative molecular mass of Y = 12 × 9.5 = 114 Let the molecular formula of Y be Cn H2n+2. 12n + 2n + 2 = 114 n = 8 Alkane Y is C8 H18. (b) C8 H18 + 25 2 O2 → 8CO2 + 9H2 O Volume of O2 required = 25 2 × 0.15 dm3 = 1.88 dm3 \ Volume of air needed = 100 20 × 1.88 = 9.40 dm3 Chapter 2 Electronic Structure of Atom Quick Check 2.1 1 (a) 602 kJ mol–1 (b) 1.51  1015 Hz (c) 1.99  10–7 m 2 (a) 3.0  10–16 m (b) 6.63  10–10 J (c) 3.99  1011 kJ mol–1 Quick Check 2.2 1 (a) 4.57  1014 Hz (b) Visible 2 (a) 5.08  1014 Hz (b) 202.9 kJ mol–1 3 x = 3.20  1015 Hz. y = 2.33  1014 Hz

359 Chemistry Term 1 STPM Quick Check 2.3 1 (a) Na4+(g) → Na5+(g) + e (b) Al11+(g) → Al12+(g) + e 2 (a) 13 (b) 3 (c) 2.8.3 3 Ionisation energy No. of electrons removed Quick Check 2.4 1 (a) 2.8.18.4 (b) 2.8.18.7 (c) 2.8.18.32.18.4 2 (a) [Ar] 3d6 4s 2 (b) Fe2+: [Ar] 3d6 Fe3+: [Ar] 3d5 (c) Fe3+. All the five 3d orbitals are singly occupied. STPM Practice 2 Objective Questions 1 A 2 C 3 D 4 C 5 C 6 A 7 C 8 D 9 B 10 A 11 C 12 C 13 B 14 A 15 A 16 A 17 C 18 A 19 B 20 A 21 B 22 A 23 C Structured and Essay Questions 1 E = hf 242 = (3.99  10–13)  f f = 6.07  1014 Hz Wavelength = c f = 3.0 × 108 6.07 × 1014 = 4.94  10–7 m or = 494 nm 2 (a) n = ∞ n = 4 n = 3 n = 2 n = 1 8.22 10.27 10.62 10.97 Wave number (fi 106 ) m–1 (b) (i) Ionisation energy of hydrogen is the minimum energy required to remove the lone electron in the hydrogen atom per mole of gaseous hydrogen atom under standard conditions (298 K and 101 kPa). H(g) → H+(g) + e (ii) E = hf = h c λ = (3.99  10–13)(3.0  108 )(10.97  106 ) = 1313.1 kJ mol–1 3 (a) n = ∞ n = 4 n = 3 n = 2 n = 1 X Frequency Y (b) Li2+ and He+. Both the ions are one electron species. 4 0 5 10 15 20 25 2.4 2.6 2.8 3.0 3.2 3.24 (× 10–2 ) 1 n2 f (× 105 )/s–1 (a) Using E = hf = (6.63  10–34)(3.24  1015) = 2.15  10–18 J (b) E = (2.15  10–18)(6.02  1023)  10–3 = 1294.3 kJ mol–1 5 (a) The minimum energy required to remove one electron from every atom in one mole of gaseous atoms. (b) This is because successive species have ever increasing proton electron ratio. This causes the remaining electrons to be more tightly held by the nucleus. (c) I.E. 496 4563 6913 9544 13 352 16 611 20 115 log I.E. 2.7 3.7 3.8 4.0 4.1 4.2 4.3 4 3 1 2 3 4 5 6 7 2 Order of electron removed Log(ionisation energy)

360 Chemistry Term 1 STPM D is in Group 1 of the Periodic Table. There is a very large increase between the 1st and 2nd I.E. 6 (a) 4s 3p 3s 2p 2s 1s Energy 3d (b) (i) 1s 2s 2p (ii) (iii) 7 (a) The third electron will have the same spin as one of the other two electrons and will experience repulsion. (b) A set of 3d orbitals that are fully half-filled has extra stability compared to a partially filled 3d orbitals. 8 (a) Hund’s rule states that in a set of degenerate orbitals electrons will occupy the orbitals singly first before pairing occurs. Pauli’s exclusion principle states that an orbital can accommodate a maximum of two electrons only with opposite spins. (b) The electronic configuration of N2– is 1s2 2s2 3p5 . 2p 2s 1s 9 Each orbital can accept two electrons only. The seven s electrons will have configuration of 1s 2 , 2s 2 , 3s 2 and 4s 1 . The twelve p electrons will have configuration of 2p6 and 3p6 . The ten d electrons will have configuration of 3d10. Electronic configuration of X is 1s 2 2s 2 2p6 3s 2 3p6 3d10 4s 1 . 10 (a) (i) n = ∞ n = 2 n = 1 1st Last 1st Last Balmer series Lyman series The lines in the Lyman series are in the ultraviolet region, while the lines in the Balmer series are in the visible region. (ii) Using the equation: ∆E = hf 987.3 × 103 = (6.63 × 10–34)(6.02 × 1023)(f) f = 2.47 × 1015 s–1 Or, Using the relationship: l = c — f = 3.00 × 108 —–———— 2.47 × 1015 = 1.21 × 10–7 m = 121 nm (b) (i) The electronic configuration is: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 4s 3d 3p 3s 2p 2s 1s Energy (ii) The electronic arrangement violates the Aufbau’s principle which states that electrons will fill available orbitals with lower energy first before orbitals with higher energies are filled. In an empty atom, the 4s orbital is of lower energy than the 3d. Thus, by right, two electrons must be filled in 4s before the 3d orbitals are filled to give it a configuration of 1s 2 2s 2 2p6 3s 2 3p6 3d4 4s 2 instead of 1s 2 2s 2 2p6 3s 2 3p6 3d5 4s 1 . This abnormality arises because a half-filled 3d subshell (3d5 ) is energetically more stable than a partially filled 3d sub-shell (3d4 ). Chapter 3 Chemical Bonding Quick Check 3.1 1 (a) (f) O S O O– O– x • x • x • x • x • x • Cl x x • x • Cl– Cl I Cl x • x • • • •     

361 Chemistry Term 1 STPM (b) (g) O S O– O– x • x • x • x • • • O O • O • x x x x x x •• x x x x x• x• (c) (h) I – I I x x • x x x x x x x •• •• •• x x x x x x C O x • xx x • • • x x (d) (i) P O O– O– x • x • x • x • x • O– Cl S x • x • x • S Cl x x x x x x x x (e) (j) Cl x • x • x • P+ Cl Cl Cl x • H C N x • • • x•x•x• 2 (a) (b) Cl Cl Al Cl x • x • x • Cl Al Cl x • x • x • x • • • • • x • x • Cl Cl Cl Cl Al 3 (a) Cl Be Cl x • x • (b) Presence of empty orbitals in beryllium and lone-pair electrons in the ammonia molecule. (c) H N Cl H N • x • H Be H Cl H H • • x • x • x x x x x x • x • 4 (a) (b) O N O x • x x x • • x x• x• x x x O • x • • O O N O • x • x • x • •• • N Quick Check 3.2 1 Pyramidal 8 Linear 2 Linear 9 Square planar 3 Tetrahedral 10 Bent 4 Trigon bipyramidal 11 Square-based pyramidal 5 Octahedral 12 Bent 6 Tetrahedral 13 Trigon planar 7 Pyramidal 14 Bent Quick Check 3.3 1 Nitrogen is more electronegative than phosphorous. The bonding electrons in NH3 are closer to one another and experience greater repulsion. 2 H2 S. Sulphur is more electronegative than Se. Quick Check 3.4 1 Polar 4 Non-polar 2 Polar 5 Non-polar 3 Non-polar 6 Non-polar Quick Check 3.5 Chromium can make use of the electrons from the 3d as well as 4s sub-shell to form metallic bonds. Calcium can only make use of electrons from the 4s sub-shell to form metallic bonds. Quick Check 3.6 1 H2  O2  CH4  Cl2 Number of electrons in the molecules increases. 2 Size of the molecules as well as the total number of electrons in the molecules increases from CH4 to SnH4 . As a result, the van der Waals forces get stronger. 3 (a) Compound Relative molecular mass Total number of electrons Boiling point/°C Silane 32.1 18 –112 Hydrogen sulphide 33.1 18 –61 (b) H2 S is polar but SiH4 is non-polar. 4 CH3 CN. CH3 CN is polar but C3 H8 is non-polar. Quick Check 3.7 (a) and (d) Quick Check 3.8 1 (a), (b), (d), (e) and (f) 2 Ethanol: Strong hydrogen bonding Dimethyl ether: Weak van der Waals forces 3 HCOOH can form more intermolecular hydrogen bonding compared to ethanol.

362 Chemistry Term 1 STPM 4 1-propanol. It is bigger and has more electrons than ethanol. Quick Check 3.9 1 CH3 COOH. The large non-polar C6 H5 group in C6 H5 COOH decreases the solubility of benzoic acid. 2 1,2,3-propantriol has three —OH group. As a result, it can form more hydrogen bonds with water molecule compared to ethanol. 3 C2 H5 OH. It can form hydrogen bonds with water molecules, but C2 H5 SH cannot. 4 (C2 H5 )3 N cannot form hydrogen bonds with water because there are no hydrogen atoms attached directly to the nitrogen atom. 5 NH3 can form hydrogen bonds with water molecules, but PH3 cannot. STPM Practice 3 Objective Questions 1 C 2 C 3 B 4 B 5 C 6 A 7 D 8 D 9 B 10 D 11 B 12 B 13 D 14 A 15 C 16 B 17 B 18 C 19 D 20 D 21 C 22 A 23 B 24 C 25 D 26 C 27 D 28 B 29 C 30 C 31 B 32 C 33 D 34 D 35 B 36 B 37 C 38 D 39 A 40 C 41 C 42 C Structured and Essay Questions 1 Ionic bond is the electrostatic attraction between a pair of opposite charged ions formed by the complete transfer of electrons from one atom to the other. An example is sodium chloride. Sodium transfers its lone valence electron to the chlorine atom. This results in the formation of Na+ and Cl– ions with octet configurations Na + Cl → Na+ + Cl– 2.8.1 2.8.7 2.8 2.8.8 Attraction between Na+ and Cl– results in the formation of the ionic bond. Covalent bond is the electrostatic attraction between adjacent nuclei and the electrons that are shared between them. An example is Cl2 . Each chlorine atom shares one of its unpaired valence electron with another chlorine atom so that both achieve octet configurations. Cl + Cl → Cl — Cl 2.8.7 2.8.7 2.8.8 2.8.8 The attraction between the two nuclei of chlorine atom and the shared pair of electrons hold the chlorine atoms together in Cl2 molecule. 2 (a) In the HF molecule, the hydrogen atom is bonded to a very small and highly electronegative fluorine atom. The covalent bond in H—F is greatly polarised. δ++H—Fδ– – The dipole-dipole attractions between HF molecules are stronger than the ‘ordinary’ van der Waals force. This dipole-dipole attraction is called hydrogen bonding. (b) NH3 and H2 O have intermolecular hydrogen bonding while weak van der Waals forces exist between CH4 molecules. This accounts for the higher boiling point of NH3 and H2 O. H2 O has two lone pair of electrons compared to only one for NH3 . Thus, H2 O can form more intermolecular hydrogen bonding with one another. On top of that, oxygen being more electronegative than nitrogen forms stronger hydrogen bonds. (c) HC—(CHOH)4 —CH2 —O—H H H O || O 3 (a) (i) HCl is polar but CO2 is non-polar. (ii) Intermolecular forces between NH3 molecules are the hydrogen bonds. Weak intermolecular van der Waals forces exist between HCl molecules (b) NH3 can form hydrogen bonds with water molecules. HCl reacts with water to form water-soluble ions: HCl(g) + H2 O(l) → H3 O+(aq) + Cl– (aq) CO2 is non-polar. 4 (a) Cl Al x • x • x • Cl Cl H N x • x • x H • H x x Be Cl x • Cl x • Trigonal planar Linear Pyramidal (b) The aluminium atom in AlCl3 and the beryllium atom in BeCl2 do not have eight electrons in their valence shells. (c) (i) AlCl3 •NH3 (ii) Cl Al x • x • x • Cl Cl H H N H • • • (d) (i) BeCl2 •2NH3 (ii) H N x • x • x H • H Cl Cl H Be • • • • H N H x x x • x

363 Chemistry Term 1 STPM (iii) The beryllium atom in BeCl2 is 4 electrons short of octet. Hence, it will combine with two NH3 molecules to achieve octet. 5 (a) O • H x • x x H x x x sp3 (b) N • H x • x x H x x • H sp3 (c) x• • x x x • H N • x π C x π sp sp 6 (a) CH3 O O N sp3 sp2 sp2 + – sp3 (b) Trigonal planar (c) – C N (d) The carbon-nitrogen bond in CN– is a triple bond while that in the CH3 NO2 is a single bond. 7 (a) Mixing of two or more non-equivalent atomic orbitals to form a set of equivalent hybrid orbitals. (b) (i) Nitrogen undergoes sp 3 hybridisation. N • H x x• x H x x • H (ii) Carbon undergoes sp 3 hybridisation. C •x Cl Cl Cl •x x• •x H 8 (a) In the reaction, NH3 accepts a proton from another NH3 to form NH4 +. Thus, one of the NH3 molecule acts as a Bronsted base (proton acceptor) and the other a Bronsted-Lowry acid (proton donor). (b) + H H H N H H H N H + H H H N H H H H N H + H H H N H – H H N H H N H NH3 : Trigonal pyramidal (3 bond-pairs + 1 lone-pair) NH4 + : Tetrahedral (4 bond-pairs) NH2 – : Bent (2 bond-pairs + 2 lone-pairs) (c) CI CI CI P CI CI N CI CI CI P CI CI CI N NCl3 : sp3 hybridisation PCl3 : sp3 hybridisation NCl3 has a larger bond angle. N is more electronegative than P, the bonding electrons will be closer to N in NCl3 compared to P in PCl3 . Thurs, the bond-pair repulsion in NCl3 is stronger and pushed the bonds further apart. 9 (a) (i) BeCl2 + 2NH3 → BeCl2 (NH3 )2 (ii) CI N CI H H H H H H N Be (iii) BeCl2 : sp hybridisation BeCl2 (NH3 )2 : sp3 hybridisation (b) (i) P H H H (ii) Ammonia. Nitrogen is more electronegative than phosphorus. As a result, the bonding electrons in the N—H bonds are nearer to the nitrogen atom than the phosphorus atom in PH3 . As a result, the three N—H bonds experience greater mutual repulsion which pushed them further apart than the P—H bonds in PH3 . 10 (a) H2 (g) + Cl2 (g) → 2HCl(g) (b) Boiling point HCI Mr of HX HBr HI HF

364 Chemistry Term 1 STPM (c) The intermolecular forces between the HCl molecules, the HBr molecules and the HI molecules are the weak van der Waals forces. The strength of the van der Waals force increases with the increase in the size of the hydrogen halide molecules as well as the number of electrons in the molecules from HCl to HI. This is reflected in the increase in the boiling point from HCl to HI. HF, despite its smallest size has the highest boiling point. This is due to the presence of hydrogen bonding between the HF molecules. 11 (a) Lead(II) chloride is an ionic compound with ionic bonds binding the Pb2+ and Cl– ions in the giant ionic lattice. Due to the large size of the Pb2+ ion, the ionic bond is rather weak. On the other hand, silicon dioxide exists in the form of giant covalent structure with strong covalent bonds holding the silicon and oxygen atoms together. A lot of energy is required to break the covalent bonds. This accounts for the higher melting point of silicon dioxide compared to lead(II) chloride. (b) (i) S O O O O- S OO O (ii) S O O Square planar (120°) Tetrahedron (109.5°) O S O- OO O 12 (a) (i) The melting point of ice decreases with increasing pressure shows that ice is less dense than water. The melting of ice is accompanied by a decrease in volume. H2 O(s)   H2 O(l) Volume decreases Increasing pressure would shift the equilibrium to the right-hand side and more ice will melt. This is because, in ice the water molecules are held rigidly in place by hydrogen bonds. This creates a lot of empty spaces in the ice structure. Hydrogen bond Empty space H H H H H H H H H H H H H H H H H H H H H H H H O O O O O O O O O O O O When ice melts, the rigid structure collapses and the mobile water molecules flow to fill up those empty spaces causing its volume to decrease. (ii) Hydrogen chloride is a covalent compound that exists as the simple HCl molecules with strong covalent bond holding the atoms together in the molecule. There are no ions present in liquid hydrogen chloride. As a result it does not conduct electricity. However, when hydrogen chloride dissolves in water it undergoes complete dissociation to produce free aqueous ions. HCl(g) + H2 O(l) → H3 O+(aq) + Cl– (aq) The production of the mobile aqueous ions is responsible for its electrical conductivity. (b) (i) Ion ClO2 – ClO3 – ClO4 – Oxidation state +3 +5 +7 (ii) O O– CI O O– O CI O O– O O CI Chapter 4 States of Matter Quick Check 4.1 1 0.179 dm3 2 143.2 kPa 3 27.8 kPa 4 Volume Pressure 400 K 300 K The higher the temperature, the higher is the pressure. 5 Pressure Condensation 1 V (Bent) (Trigonal pyramidal) (Tetrahedral)

365 Chemistry Term 1 STPM At high enough pressure, carbon dioxide undergoes condensation causing a very large decrease in the volume (or a large increase in 1 V ). After that, pressure has little effect on the volume because liquid carbon dioxide is difficult to compress. Quick Check 4.2 1 3.15 dm3 3 329 °C 2 323 °C 4 18.93 dm3 Quick Check 4.3 1 (a) 8.75 dm3 (b) 50.0 dm3 2 (a) 227 (b) 1481.8 dm3 3 C3 H6 4 C3 H8 Quick Check 4.4 1 5.00  10–3 m3 2 27.3 3 0.0173 m–3 4 31.08 dm3 Quick Check 4.5 1 (a) 9.01 kPa (b) 5.01 kPa 2 8.61 3 (a) Pressure/kPa Gas 120 72 48 CO N2 SO2 (b) (i) P(CO) = 120 kPa P(N2 ) = 72 kPa (ii) Total pressure = 192 kPa 4 (a) 498.6 kPa (b) 2493 kPa (c) 997.2 kPa The water vapour pressure is negligible. Quick Check 4.6 1 6.0 1023 mol–1 2 8 atoms 3 MS Quick Check 4.7 1 (a) Conversion of a substance from solid to gas/vapour without passing through the liquid phase (b) Reducing pressure to below 0.006 atm 2 (a) Vapour (b) Temperature Time Freezing A B Condensation (c) Increasing the pressure. Vapour → Liquid Quick Check 4.8 1 The molecules are more closely packed in solid CO2 compared to liquid CO2 . 2 Temperature Pressure/atm Gas Liquid Solid 3 (a) Temperature Pressure/atm (b) Melting of iron is accompanied by a decrease in volume. As a result, the melting point decreases with increasing pressure. 4 (a) & (b) Temperature/°C Pressure/atm 1 M.p. 132 Solid Liquid Gas B.p Critical point X (c) Yes. The critical temperature is higher than room temperature. (d) On the graph (e) (i) Nothing happens (ii) Solid → Liquid → Gas (f) Density of solid is higher than the liquid 5 CO2 , NH3 , C2 H5 OH, H2 O Quick Check 4.9 1 77.44 g 2 0.838 3 2.95 kPa 4 (a) x 2 (c) 3x 2 (b) 2x (d) 7.5x 2 Quick Check 4.10 1 100.17 °C 2 –0.084 °C 3 CaCl2 4 (a) Ethanoic acid undergoes association in benzene. 2CH3 COOH → (CH3 COOH)2 (b) Aluminium chloride undergoes association. 2AlCl3 → Al2 Cl6

366 Chemistry Term 1 STPM 5 (a) 60 (b) 119.9 (c) 88.3% STPM Practice 4 Objective Questions 1 B 2 C 3 C 4 D 5 B 6 D 7 D 8 B 9 D 10 B 11 B 12 D 13 D 14 C 15 C 16 C 17 B 18 D 19 C 20 A 21 A 22 A 23 B 24 D 25 B 26 B 27 B Structured and Essay Questions 1 (a) A gas that obeys the gas law: PV = nRT under all conditions (b) (i) When temperature increases, the kinetic energy of the particles increases. This leads to an increase in the rate of collision with the walls of the container. At the same time, the collisions are more energetic. (ii) Increasing pressure pushes the molecules closer to one another. This reduces the amount of empty space between the particles causing the volume to decrease. (iii) When temperature increases, the pressure exerted by the gas will increase. To maintain the original pressure, the volume occupied by the gas must increase so as to reduce the rate of collision. 2 (a) PV = nRT (b) Presence of intermolecular forces. The gas particles have finite volume. (c) Helium (d) At low pressures, intermolecular attraction operates between the molecules (In this case, it is the hydrogen bonding). Thus, the volume actually occupied by the gas is less than expected. This makes PV  nRT. P (a) (b) 1 PV nRT (e) High temperature and low pressure 3 Miscible: There are no intermolecular forces between the gas particles. Pressure: Collision of the gas particles with the walls of the containers. Compressible: No intermolecular forces and presence of a lot of empty space. 4 (a) No intermolecular forces between gas particles. The volume of the gas particles is negligible when compared to the volume of the container in which the gas is placed. (b) (i) Volume of one argon atom = 4 3 πr 3 = 4 3 π (0.192  10–9)3 m3 = 2.96  10–29 m3 Volume of one mole of argon atom = (2.96  10–29)(6.02  1023) = 1.78  10–5 m3 (ii) PV = nRT (101  103 )V = 1  8.31  273 V = 2.25  10–2 m3 (iii) % = 1.78 × 10–5 2.25 × 10–2  100% = 0.079% (iv) The volume of the gas molecules is indeed negligible when compared to the volume occupied by the gas. 5 (a) Increasing pressure pushes the gas particles closer to one another. The attractive force is then strong enough to hold the particles together causing it to condense. (b) Presence of intermolecular hydrogen bonding pulls the molecules closer. (c) There are no intermolecular forces between gas particles. They are free to move around and thus will distribute themselves throughout the container. (d) Going down the group, the size of the molecules gets bigger and the strength of the intermolecular forces between the molecules gets stronger causing the deviation to be more profound. (e) At high pressures, intermolecular repulsion exists between the molecules. This makes the gas more difficult to compress. At moderate pressures, intermolecular attraction exists between the molecules, thus making them easier to compress. 6 (a) Refer to text (b) P (× 105 )/Pa 4 8 15 20 PV (× 102 )/Pa m3 23.2 22.9 22.4 22.0 23 23.5 220 2 4 6 8 10 12 14 16 18 20 PV (× 102)/Pa m3 P(× 105 )/Pa

367 Chemistry Term 1 STPM PV at P = 0, is 23.5  102 Pa m3 Using PV = nRT 23.5  102 = n  8.31  298 ∴ n = 0.949 mol ∴ Mr = 32.5 0.949 = 34.2 7 (a) PV = nRT PV = —– m Mr RT P = —m V (—–1 Mr ) RT But —m V = d (density) \ P = —–d Mr RT or —d P = —– Mr RT (b) (i) When pressure increases, the volume occupied by the gas decreases. This causes the density  Mass Volume  to increase. (ii) Pressure (× 102 )/Pa 33.7 67.6 100.0 Density Pressure/kg m–3 Pa–1 2.85  10–5 2.86  10–5 2.93  10–5 2.7 × 10–5 2.8 × 10–5 2.9 × 10–5 3.0 × 10–5 0 10 20 30 40 50 60 70 80 90 100 P (× 102 )/Pa /kg m–3 Pa d –1 P d P at P = 0 is 2.80  10–5 Using d P = Mr RT ∴ Mr = (2.80  10–5)  (8.31  301.5) = 0.0702 kg mol–1 ∴ Relative molecular mass = 70.2 8 (a) Refer to text (b) (i) For argon: (2.0  105 )  2.60 = P  4.0 P(Ar) = 1.30  105 Pa For ethane: (1.2  10–5)  12.5 = P  4.0 P(Ethane) = 3.75  105 Pa For CO2 : (2.8  104 )  0.80 = P  4.0 P(CO2 ) = 5.60  103 Pa (ii) Total pressure = (1.30  105 ) + (3.75  105 ) + (5.60  103 ) Pa = 5.11  105 Pa (iii) P(Ar) = 1.30  105 Pa P(Ethane) = 3.75  105 Pa Total pressure = (1.30 + 3.75)  105 Pa = 5.05  105 Pa 9 Solid: • Strong attractive forces between particles. • Particles have no translational motion. • Particles vibrate and rotate in their mean position. Liquid: • Intermediate strength attractive forces between particles. • Particles can move freely in the body of the liquid but not out of it. Gas: • No intermolecular forces between particles. • Particles can move freely throughout the container where it is placed. 10 (a) The pressure exerted by water vapour which is in equilibrium with liquid water at a fixed temperature. The pressure is due to the collisions of the water vapour molecules with the walls of the container. (b) Pressure Temperature Increasing temperature increases the kinetic energy of the water molecules. More water will undergo evaporation. At the same time, the collisions between the vapour molecules and the walls of the container is more energetic. (c) Decreases. Fraction of the liquid surface for water to evaporate is decreased. 11 (a) Melting point: The temperature at which a solid is in equilibrium with its liquid at 101 kPa. Boiling point: The temperature at which the vapour pressure of a liquid is equal to the external pressure (101 kPa). (b) Temperature/°C Pressure/atm 0 100 Vapour Liquid Solid Sea 1.0

368 Chemistry Term 1 STPM (c) Presence of dissolved salts makes the orderly arrangement of the ice crystal structure more difficult to form. The volume of water in the sea is much larger. 12 (a) The temperature and pressure where the three phases of a substance can exist in equilibrium. (b) Refer to text (c) Decreases. Presence of another substance makes the solid crystal structure of CO2 more difficult to form. (d) Because it’s triple point pressure is higher than 1 atm. 13 (a) (i) Melting point (ii) Boiling point (b) Temperature Time a b c d (c) Water. Because the solid/liquid line has a negative slope. 14 (a) Refer to text (b) The triple point. It is the only conditions where solid, liquid and vapour can exist in equilibrium. (c) Freezing point refers to an external pressure of 1 atm. The triple point refers to when water is under its own vapour pressure (0.006 atm). (d) It is the temperature above which water vapour cannot be condensed by increasing pressure alone. 15 (a) Helium. It has the smallest size and the least number of electrons amongst all gases. As a result, the volume of the molecules and the intermolecular forces can be ignored. (b) –273 °C (c) and (d) Volume (d) p p’ X 0 Temperature / °C 16 (a) Temperature/°C Pressure/atm B A C T Solid Liquid Gas –57 31 5 73 (b) Increasing the pressure on the carbon dioxide gas pushes the molecules close together. The attractive force ‘binds’ the molecules in clusters causing the gas to liquefied. (c) When the liquefied carbon dioxide is released, it absorbs heat from the surroundings and vaporised. This causes the water vapour in the surrounding air to condense causing a white mist to form. Chapter 5 Reaction Kinetics Quick Check 5.1 1 57.0 s 2 (a) 0.0125 mol dm–3 min–1 (b) 0.25 mol dm–3 3 0.5 0.6 0.7 0.8 0.9 Rate = 0.021 Rate = 0.0085 Rate = 0.0040 1 0.4 0.3 0.2 0.1 10 Time/min [A]/mol dm–3 20 30 40 50 60 70 80 0 (a) 0.021 mol dm–3 min–1 (b) 0.0085 mol dm–3 min–1 (c) 0.0040 mol dm–3 min–1 Quick Check 5.2 1 (a) 4.5  10–3 mol dm–3 s–1 (b) 3.0  10–3 mol dm–3 s–1 2 (a) – 1 2 ∆[Cu2+] ∆t = – 1 6 ∆[CN– ] ∆t = 1 2 ∆[Cu(CN)2 – ] ∆t = ∆[(CN)2 – ] ∆t (b) (i) 0.168 mol dm–3 s–1 (ii) 0.028 mol dm–3 s–1 3 5.0  10–3 mol dm–3 s–1 Quick Check 5.3 1 (a) Rate = k[A][B]2 (b) mol–2 dm6 s–1 (c) (i) 0.5x (ii) 0.25x (iii) 18x

369 Chemistry Term 1 STPM Quick Check 5.4 1 –6 –4 –2 –8 –10 –12 –14 1.2 1.3 1.4 1.5 1.6 1.7 1.8 In k (× 10–3)/K–1 1 T Gradient = 9 0.42 × 10–3 178.1 kJ mol–1 Quick Check 5.5 1 Rate = k[NO]2 [O2 ] 2 Rate = k[A][B]2 3 (a) (i) +4 (ii) +5 (iii) +3 (b) (i) 2 (ii) 1 (c) 17.39 mol–2 dm6 s–1 (d) 0.34 mol dm–3 s–1 Quick Check 5.6 1 (a) 1.0 0.8 0.6 0.4 0.2 0 10 Time/min 20 30 40 50 [X]/mol dm–3 25 25 First order (b) 2.77  10–2 min–1 2 (a) 25 30 35 40 20 15 10 5 20 Time/min [Sucrose] (× 10–2)/mol dm–3 40 60 80 100120140 160 0 84 44 Second order (b) 9.97  10–4 mol–1 dm3 s–1 3 1.0 0.8 0.6 0.4 0.2 10 Time/min [A]/mol dm–3 0 5 15 20 25 30 35 40 Zero order 4 (a) 3 4 2 [N2O5]/mol dm–3 1 2 3 1 0 Rate/mol dm–3 s–1 Gradient = 2.6 2.35 = 1.11 s–1 First order (b) 1.11 s–1 5 (a) Rate [H2O2] (c) [H O2 2] Time (b) Rate [H2O2] 2 7 (a) Fraction of molecules having energy equal to or greater than the activation energy. (b) Ratio between 35 °C to 25 °C = 2.34 : 1 (c) The fraction of molecules having energy equal to or greater than the activation energy increases about 2 times for 10° rise in temperature. As a result, the rate will increase by about 2 times.

370 Chemistry Term 1 STPM STPM Practice 5 Objective Questions 1 C 2 D 3 C 4 C 5 B 6 C 7 C 8 C 9 C 10 B 11 B 12 D 13 C 14 C 15 B 16 C 17 B 18 C 19 A 20 A 21 C 22 B 23 C 24 A 25 D 26 A 27 A 28 B 29 B 30 A 31 D 32 C 33 B Structured and Essay Questions 1 (a) Catalyst (b) So that its concentration remains almost constant and will not affect the rate of reaction. (c) 10 8 6 4 2 0 10 20 30 40 50 60 70 80 [Ester] (× 10–1)/mol dm–3 Time/min t2 t1 (i) 20 minutes (ii) 19.5 minutes (d) First order 2 (a) Rate = k[NO]2 [O2 ] (b) 2NO + O2 9: Slow N2 O4 N2 O4 9: Slow 2NO2 (c) (i) 9.6 × 10–3 = k(0.80)2 (0.26) k = 5.7 × 10–2 mol–2 dm6 s–1 (ii) Let original rate = x x = k[NO]2 [O2 ] The new rate = k[ 1 — 4 NO]2 [2O2 ] = 1 — 8 k[NO]2 [O2 ] = 1 — 8 x (c) [H+ ] Time 3 (a) k = Rate constant A = Constant Ea = Activation energy R = Gas constant (b) (i) 1 0 –1 –2 3.1 3.2 3.3 In k (× 10–3)/K–1 1 T Gradient = = 1.8 0.22 × 10–3 –8.18 × 10–3 –8.18  103 = –Ea R ∴ Ea = 67 975.8 J = 67.98 kJ (ii) Adding a suitable catalyst (iii) First order. The rate constant has a unit of s–1. 4 (a) H2 O2 + 2H+ + 2I– → 2H2 O + I2 (b) As one of the reagent (c) Rate = k[IO– ][H+] However, the rate of formation of IO– is given by: Rate = k’[H2 O2 ][I– ] Substitute into the first rate equation: Rate = kk’[H2 O2 ][I– ][H+] 5 (a) H2 O2 + 2H+ + 2Cl– → 2H2 O + Cl2 (b) Energy Reaction pathway Uncatalysed reaction Catalysed reaction (c) Increasing the temperature 6 (a) Let the rate equation be: Rate = k[BrO3 – ]x [Br– ]y [H+]z Comparing experiment II and I:  0.60 0.40  x = 3.75 × 10–3 2.50 × 10–3 \ x = 1 Comparing experiment III and II:  0.56 0.28  y = 7.50 × 10–3 3.75 × 10–3 \ y = 1 Comparing experiment IV and I:  0.062 0.031 z = 1.0 × 10–2 2.50 × 10–3 z = 2 Rate equation is: Rate = k[BrO3 – ][Br– ][H+]2

371 Chemistry Term 1 STPM (b) Using experiment I: 2.50 × 10–3 = k[0.40][0.28][0.031]2 k = 23.2 mol–3 dm9 s–1 7 (a) Concentration of PCl5 /mol dm-3 1.0 0.6 Half-life/min 190* 190** [NOTE: * The time taken for the concentration to decrease from 1.0 to 0.5 mol dm-3. ** The time taken for the concentration to decrease from 0.6 to 0.3 mol dm-3] (b) Rate = k[PCl5 ] [Since the half-life is a constant it is first order with respect to PCl5 ] (c) The half-life is given by: T 1 — 2 = In 2 ——– k When temperature increases, the rate constant, k, also increases. This will cause the half-life to decrease. [Alternatively: When temperature increases, the rate of reaction will increase. Hence, it takes shorter time for the original concentration to decrease to half its original value.] Chapter 6 Chemical Equilibrium Quick Check 6.1 1 (a) (i) H2 O(g) + C(s)  CO(g) + H2 (g) (ii) Kp = P(H2 )P(CO) P(H2 O) Pa (b) (i) 180 kPa (ii) 281.7 kPa 2 (a) Catalyst (b) P(SO2 ) = 0.10 atm P(O2 ) = 0.050 atm (c) P/atm SO2 O2 SO3 2 1.9 1 0.1 0.05 (d) 7220 atm–1 3 (a) 7.79  106 Pa (b) 8.33  10–2 mol dm–3 (c) 2.73  106 Pa Quick Check 6.2 1 (a) Q = 4.78 atm–1  Kp System not in equilibrium (b) From the right to the left 2 (a) Q = 6.12 ≠ Kc (b) Net reaction: From right to left Quick Check 6.3 1 (b) and (e) only 2 (a) and (d) only 3 (a) Increases (b) Increases (c) No change 4 [H2 ] Kc (a) Increases No change (b) Decreases No change (c) No change No change (d) Decreases Decreases STPM Practice 6 Objective Questions 1 C 2 C 3 D 4 A 5 B 6 B 7 A 8 B 9 A 10 B 11 A 12 C 13 C 14 D 15 A 16 D 17 A 18 D 19 D 20 B 21 C 22 C 23 D 24 C 25 C 26 B Structured and Essay Questions 1 (a) Let the degree of dissociation = a. N2 O4  2NO2 Original/mol 1 0 Equilibrium 1 – a 2a Total particle present at equilibrium = (1 – a) + 2a = 1 + a ∴ 92 60 = 1 + a 1 a = 0.53 (or 53.0%) (b) Partial pressure of NO2 = 2a 1 + a P Partial pressure of N2 O4 = (1 – a) (1 + a) P Kp = P2 (NO2 ) P(N2 O4 ) =  2a (1 + a) P 2  (1 – a) (1 + a)  P = 4a2 P (1 + a)(1 – b) By substitution: Kp =  4(0.53)2 (1 + 0.53)(1 – 0.53) × 100 = 153 kPa

372 Chemistry Term 1 STPM (c) Let the degree of dissociation = b Kp = 4b2 (1 + b)(1 – b) × 1 000 156 = 4b2 (1 + b)(1 – b) × 1 000 ∴ b = 0.21 (or 21.0%) According to Le Chatelier’s Principle, increasing pressure would shift the equilibrium to the left-hand side as the number of moles of gaseous particles is less. This is would decrease the degree of dissociation of N2 O4 as shown by the calculation. 2 (a) When a reversible reaction has achieved dynamic equilibrium, the ratio of the molar concentrations of the products to the molar concentration of the reactants, both raised to their respective stoichiometric coefficient, is a constant at constant temperature. (b) Concentration, pressure and temperature. Only temperature will affect the equilibrium constant. (c) Kp = P2 (NH3 ) P(N2 )P3 (H2 ) N2 + 3H2  2NH3 Initial/mol 1 3 0 Final/mol 0.4 1.2 1.2 % NH3 = 1.2 2.8  100% = 42.9% % H2 = 1.2 2.8  100% = 42.9% % N2 = 100 – 2(42.9) = 14.2% Kp = (0.429 × P)2 (0.429 × P)3 (0.142P) = 16.42(P)–2 = 16.42  (3.0  104 )–2 = 1.82  10–8 kPa 3 (a) Low temperature and high pressure (b) It is not too low that the rate becomes too slow. It is not too high that the yield becomes too low. (c) (i) Vanadium(V) oxide (ii) No effect. It just shortens the time taken for equilibrium to established. (d) Energy Uncatalysed Catalysed (e) (i) Kp = P2 (SO3 ) P2 (SO2 )P(O2 ) = 0.802 (0.10)2 (0.70) = 91.4 atm–1 (ii) 2SO2 + O2  2SO3 Final/mol 0.10 0.70 0.80 Initial pressure of SO2 = 0.10 + 0.80 = 0.90 atm Initial pressure of O2 = 0.70 + 0.40 = 1.10 atm 4 (a) (i), (ii) & (iii) –288 kJ Ea Ea ’ Ea = Activation energy for the forward reaction Ea ' = Activation energy of the reverse reaction (b) Higher. Due to the smaller size of the chlorine atom, the Cl—Cl bond is stronger than the I—I bond. 5 (a) Kc = [PCl3 ][Cl2 ] [PCl5 ] Kp = P[PCl3 ]P[Cl2 ] P[PCl5 ] (b) At 200 °C: PCl5  PCl3 + Cl2 Original/mol 1 0 0 Final/mol 0.5 0.5 0.5 P(PCl3 ) = P(Cl2 ) = P(PCl5 ) = 0.5 1.5  200 = 66.67 kPa Kp = 66.67 kPa At 350 °C: PCl5  PCl3 + Cl2 Original/mol 1 0 0 Final/mol 0.15 0.85 0.85 P(PCl3 ) = P(Cl2 ) = 0.85 1.85  200 = 91.89 kPa P(PCl5 ) = 200 – 2(91.89) = 16.22 kPa Kp = (91.89)2 16.22 = 520.6 kPa (c) Endothermic. Kc increases with increasing temperature. (d) (i) Decreases (ii) Decreases (iii) No change

373 Chemistry Term 1 STPM 6 1 T (× 10–3)/K–1 2.86 1.97 1.82 ln Kc 1.22 7.60 8.70 6 5 7 8 9 4 3 2 1.8 2.0 2.2 2.4 2.6 2.8 In Kc 1 T (× 10–3)/K–1 1 Gradient = = 4.6 0.64 × 10–3 –7187.5 Gradient = – 7187.5 K – 7187.5 = –∆H R ∆H = +59 728 J or = +59.73 kJ 7 (a) 2SO2 + O2 2SO3 Initial/mol: 2 1 0 Final/mol: 0.2 0.1 1.8 Partial pressure of SO3 = 1.8 2.1 × 5 = 4.29 atm Partial pressure of SO2 = 0.2 2.1 × 5 = 0.48 atm Partial pressure of O2 = 5 – 4.29 – 0.48 = 0.23 Kp = 4.292 (0.48)2 (0.23) = 347.3 atm–1 (b) (i) Increasing temperature favours the reverse reaction which is endothermic. The yield of SO3 decreases. (ii) Increasing temperature increases both the forward and reverse reactions. As a result, the rate of attainment of equilibrium increases. 8 (a) The pressure of a gas in a mixture of gases is the pressure that will be exerted by that gas if it alone occupies the same volume at the same temperature. (b) Kp = p(Z) p(X)p2 (Y) (c) Kp = 1.2 (0.4)(0.9)2 = 3.7 (d) The volume occupied by the mixture will increase in order to maintain the original pressure. This amounts to decreasing the pressure on the original equilibrium mixture. As a result, the equilibrium will shift to the left-hand side to increase the number of moles of gas. 9 (a) (i) pV = nRT \ For an ideal gas: pV —––— nRT = 1 When a graph of pV —––— nRT is plotted against p, a straight line parallel to the p axis will be obtained and it cuts the y-axis at 1.0 (ii) Positive deviation occurs when carbon dioxide is under high pressure. The molecules are pushed so close to one another that the molecules repel one another and make the gas less compressible. [NOTE: In this case, the volume decrease less than expected and the product of pV becomes greater than nRT causing pV —––— nRT > 1] (b) (i) d p p Given: d = p1 M —––– RT 2 \ d —p = 1 M —––– RT 2 At constant temperature, 1 M —––– RT 2 = constant. Hence, a plot of d —p against p would give a straight line parallel to the p axis. (ii) Using the expression: d = p1 M —––– RT 2 0.57 = (101 × 10–3)1 M ————––––– 8.31 × 308 2 M = 1.44 × 10–2 kg mol–1 Or, = 14.4 g mol–1 The relative molecular mass of the gas = 14.4 10 (a) (i) The magnitude of the equilibrium constant decreases when the temperature increases. This shows that, the equilibrium shifts to the lefthand side when heat is supplied. Therefore, the reverse reaction is endothermic and the forward reaction is exothermic. (ii) Using the expression: ln K —––2 K1 = ∆H—––– R 1 1 —– T1 – 1 —– T2 2 ln 1.6 —–– 2.4 = ∆H—––– R 1 1 ——––– 1 000 – 1 ——––– 1 300 2 –0.41 = ∆H—––– 8.31 (2.31 × 10–4) ∆H = 1.48 × 104 J mol–1 Or, = –14.8 kJ mol–1

374 Chemistry Term 1 STPM (iii) A catalyst decreases the activation energy for the forward and the reverse reactions by the same amount. As a result, equilibrium is established in a shorter time. Ea Ea ' Time Energy Ea = Activation energy without catalyst Ea ' = Activation energy with catalyst However, catalyst does not affect the yield or equilibrium constant for the reaction. It also does not alter the enthalpy change of reaction. (b) The forward reaction is exothermic. Thus, it is favoured by low temperature. However, if the temperature is too low, the rate of reaction will be slow and it takes a longer time for equilibrium to establish. Thus, a compromise temperature of 450 o C is used where the yield is reasonable and the rate not too slow. In order to increase the rate further, finely divided iron is added as a catalyst. The forward reaction is accompanied by a decrease in the number of moles of gas. Hence, it is favoured by high pressure. However, if the pressure is too high, the cost of production will also be high as the pipes and storage tanks must be strong enough to withstand the high pressure. As a compromise, a pressure of 500 atm is used. Chapter 7 Ionic Equilibria and Solubility Equilibria Quick Check 7.1 (a) (b) (c) (d) (e) Cu2+ SO3 AlCl3 H+ H+ Quick Check 7.2 1 Conjugate base Conjugate acid Acid Base (a) NH4 + (b) H2 O (c) [Cu(H2 O)6 ] 2+ (d) HCOOH (e) H3 O+ (f) H2 O NH3 OH– [Cu(H2 O)5 OH]+ HCOO– H2 O OH– H2 O CN– H2 O HCOOH C2 H5 O– NH2 – H3 O+ HCN H3 O+ HCOOH2 + C2 H5 OH NH3 2 (a) (b) (c) (d) (e) (f) C2 H5 NH3 + CH3 CONH3 + NH2 OH2 + H3 O+ H2 CO3 H2 SO4 3 (a) (b) (c) (d) Cl– OH– C6 H5 COO– SO3 2– 4 HCO3 – + H2 O  H2 CO3 + OH– HCO3 – + H2 O  CO3 2– + H3 O+ Quick Check 7.3 1 Acid Ka /mol dm–3 pKa Iodoethanoic acid Nitrous acid Cynic acid Boric acid 3.16 3.14 3.44 9.23 6.9  10–4 7.2  10–4 3.6  10–4 5.8  10–10 2 (a) CH3 COOH + H2 O  CH3 COO– + H3 O+ (b) Ka = [CH3 COO– ][H+] [CH3 COOH] (c) 1.80  10–5 mol dm–3 (d) 4.74 3 (a) 1.70  10–6 mol dm–3 (b) N2 H4 + H2 O  N2 H5 + + OH– (c) 2.60  10–3 4 (a) HSO3 – + H2 O  SO3 2– + H3 O+ (b) Ka = [H+][SO3 2–] [HSO3 – ] (c) 7.21 Quick Check 7.4 1 (a) H2 CO3 + H2 O  H3 O+ + HCO3 – HCO3 – + H2 O  H3 O+ + CO3 2– (b) Ka, 1 = [H+][HCO3 – ] [H2 CO3 ] Ka, 2 = [H+][CO3 2–] [HCO3 – ] (c) 2.06  10–17 mol2 dm–6 2 (a) H3 PO4  H+ + H2 PO4 – H2 PO4 –  H+ + HPO4 2– HPO4 2–  H+ + PO4 3– (b) Ka, 1 = [H+][H2 PO4 – ] [H3 PO4 ] Ka, 2 = [H+][HPO4 2–] [H2 PO4 – ] Ka, 3 = [H+][PO4 3–] [HPO4 2–] (c) 1.99  10–22 mol3 dm–9

375 Chemistry Term 1 STPM Quick Check 7.5 1 (a) (b) (c) (d) (e) pH = 2.47 pH = 0.097 pH = –0.38 pOH = 1.25 pOH = –0.45 2 (a) H2 SO4  2H+ + SO4 2– (b) 0.50 mol dm–3 (c) 0.30 3 (a) 1.0 mol dm–3 (b) pOH = 0 4 (a) 1.34  10–3 mol dm–3 (b) 2.87 5 (a) NH3 + H2 O  NH4 + + OH– (b) [OH– ] = 5.1  10–3 mol dm–3 pOH = 2.29 Quick Check 7.6 1 2.93 2 11.1 3 (a) pH = 3.08; α = 4.12  10–3 (b) pH = 4.08; α = 4.12  10–2 (c) Dilution increases both the pH and the degree of dissociation. 4 (a) 7.87  10–6 mol dm–3 (b) 8.90 5 (a) (i) 2.87 (ii) 0.0134 (b) (i) 2.96 (ii) 0.0164 6 (a) 4.30  10–4 (b) 3.70  10–7 7 (a) 3.12  10–3 mol dm–3 (b) 3.21  10–12 mol dm–3 (c) 11.49 Quick Check 7.7 1 Acid Neutral Acid Base Acid Base Base (a) (b) (c) (d) (e) (f) (g) Quick Check 7.8 1 (a) NaOH + ClCH2 COOH → ClCH2 COONa + H2 O (b) 0.063 M (c) 1.56 2 (a) 0.143 M (b) NaOH + HXO4 → NaXO4 + H2 O (c) 0.200 mol dm–3 (d) 100.5 (e) Ar = 35.5; X ≡ Chlorine Quick Check 7.9 1 (d) and (f) only 2 (a) 4.91 (c) 4.22 (e) 4.86 (b) 4.91 (d) 9.67 (f) 4.47 Quick Check 7.10 1 (a) 4.84 (b) 4.87 (c) The change is negligible because the initial solution is a buffer solution. 2 (a) 4.69 (b) (i) 4.74 (ii) 4.65 Quick Check 7.11 1 Salt Solubility/g dm–3 BaSO4 2.44  10–3 Al(OH)3 2.26  10–33 Fe2 S3 2.04 10–16 Ca3 (PO4 ) 2 1.22 10–3 Salt Ksp Silver sulphate 5.00  10–7 M3 Barium carbonate 8.16  10–9 M2 Silver iodide 1.50  10–16 M2 Quick Check 7.12 1 3.55  10–3 g 2 1.00  10–10 mol2 dm–6 3 No Quick Check 7.13 1 The oxalate ions react with H+ to form undissociated H2 C2 O4 . C2 O4 2– + 2H+ → H2 C2 O4 The sulphate ion does not react with H+. 2 BaCO3 reacts with the dilute HCl in the stomach to form soluble BaCl2 . BaCO3 (s) + 2HCl(aq) → BaCl2 (aq) + H2 O(l) + CO2 (g) BaSO4 does not react with HCl. STPM Practice 7 Objective Questions 1 A 2 C 3 C 4 A 5 C 6 B 7 B 8 D 9 C 10 C 11 A 12 C 13 D 14 D 15 D 16 A 17 A 18 D 19 A 20 B 21 C 22 A 23 B 24 C 25 D 26 C 27 A 28 C 29 C 30 C 31 D 32 B 33 B 34 C 35 D Structured and Essay Questions 1 (a) The solubility product of a sparingly salt is defined as the product of the molar concentration of the constituent ions, in a saturated solution, each raised to the power of the stoichiometric coefficient. (b) (i) Ca(OH)2 + 2HCl → CaCl2 + 2H2 O M × 20 0.045 × 20.2 = 1 2 M = 0.023 mol dm–3 [OH– ] = 2  0.023 = 0.046 mol dm–3 pOH = –log(0.046) = 1.34 ∴ pH = 14 – 1.34 = 12.66

376 Chemistry Term 1 STPM (ii) Ca(OH)2 (s)  Ca2+(aq) + 2OH– (aq) Ksp = [Ca2+][OH– ]2 (iii) Ksp = (0.023)(0.046)2 = 4.87  10–5 mol3 dm–9 (iv) Ammonia is a weak base. The [OH– ] in aqueous ammonia is not sufficient to cause precipitation of calcium hydroxide. (v) No. of moles of Ca(OH)2 = 2.33 × 10–3 74.1 = 3.14  10–5 mol Solubility of Ca(OH)2 = 0.023 mol dm–3 Volume of water needed = 3.14 × 10–5 0.023 = 1.36  10–3 dm3 2 (a) (i) Ka = [RCOO– ][H3 O+] [RCOOH] (ii) Using: [H+] = Ka C = (7.5  10–4)(0.18) = 1.16  10–2 mol dm–3 pH = 1.94 (b) (i) A solution whose pH does not change significantly on the addition of a little acid or base. (ii) H+ (added) + RCOO– → RCOOH OH– (added) + RCOOH → RCOO– + H2 O (iii) Using the formula: pH = pKa – log [Acid] [Salt] = 3.12 – log 0.18 0.35 = 3.41 (c) On addition of sodium salt which provides the RCOO– ions, the following equilibrium will shift to the left: RCOOH  RCOO– + H+ The [H+] concentration decreases causing the pH to increase. 3 (a) Bronsted-Lowry defines an acid as a proton donor. When propanoic acid dissolves in water, the following equilibrium is set up. CH3 CH2 COOH(aq) + H2 O(l)  CH3 CH2 COO– (aq) + H3 O+(aq) In the above equilibrium, the CH3 CH2 COOH molecule donates a proton to H2 O molecule. As a result, CH3 CH2 COOH is a Bronsted-Lowry acid. (b) CH3 CH2 COOH(aq)  CH3 CH2 COO– (aq) + H+(aq) CH3 CH2 COONa(s) + aq 9: 100% CH3 CH2 COO– (aq) + Na+(aq) In a mixture of propanoic acid and sodium propanoate, there will be a large amount of undissociated CH3 CH2 COOH molecules and CH3 CH2 COO– ions. When a little acid is added to the mixture, the H+ added would react with the CH3 CH2 COO– ions to form undissociated CH3 CH2 COOH molecules. ABBBB ABBBBBBBBBBBB H+ (added) + CH3 CH2 COO– (aq) : CH3 CH2 COOH(aq) When a little base is added to the mixture, the added OH– would react with the undissociated CH3 CH2 COOH molecules to form CH3 COO– ions. OH– (added) + CH3 COOH(aq) : CH3 COO– (aq) + H2 O(l) As a result, the pH of the buffer solution remains constant. (c) Using the expression for acidic buffer solutions: pH = pKa – log [Propanoic acid] [Sodium propanoate] 5.9 = 4.87 – log 0.050 x x = 0.54 mol dm–3 ∴ Mass of sodium propanoate to be added to 500 cm3 propanoic acid = 1 2 (0.54 × 74)g = 20.0 g 4 (a) pKa = –log Ka (b) C6 H5 B(OH)2 + 2H2 O  C6 H5 B(OH)3 – + H3 O+ C6 H5 COOH + H2 O  C6 H5 COO– + H3 O+ (c) [H+] = ABBBBBBBBBBBBB (1.59 × 10–9)(0.12) = 1.38  10–5 mol dm–3 pH = 4.86 (d) Lower. Because benzoic acid is a stronger acid than phenylboronic acid. (e) (i) Volume of NaOH/cm3 pH 8.8 X 4.2 X 6 12 16.7 21.4 (ii) For benzoic acid: 25.0  M = 0.10  12 M = 4.80  10–2 mol dm–3 For phenylboronic acid: 25.0  M = 0.10  9.4 M = 3.76  10–2 mol dm–3 5 (a) The ammonium ion is a Bronsted-Lowry acid. NH4 +(aq) + H2 O(l)  NH3 (aq) + H3 O+(aq) (b) The O–H covalent bond in benzoic acid is strong. (c) Formation of a water-soluble complex with concentrated HCl. PbCl2 (s) + 2Cl– (aq)  [PbCl4 ]2–(aq) (d) Mg2+(aq) + 2NH3 (aq) + 2H2 O(l)  Mg(OH)2 (s) + 2NH4 +(aq) Addition of NH4 Cl shifts the equilibrium to the left causing the Mg(OH)2 to dissolve.

377 Chemistry Term 1 STPM 6 (a) pH = –log[H+] (b) pH = 3.10 –log[H+] = 3.10 [H+] = 7.94 × 10–4 mol dm–3 (c) (i) Phenolphthalein (ii) Let the concentration of HA = M mol dm–3 22.50 × 0.100 = M × 25.00 M = 0.090 mol dm–3 (d) HA is a weak acid that undergoes partial dissociation in aqueous solution. (e) Using: [H+] = Ka C (7.94 × 10–4)2 = 0.090Ka Ka = 7.00 × 10–6 mol dm–3 (f) pH 7 22.5 V Chapter 8 Phase Equilibria Quick Check 8.1 1 0.479 2 (a) 0.162 (b) 973.7 g 3 148.5 g Quick Check 8.2 1 37.68 kPa 2 (a) A solution that obeys Raoult’s law (b) (i) 38.56 kPa (ii) 34.90 kPa (iii) 27.67 kPa Quick Check 8.3 1 (a) 0.38 mole fraction of C6 H6 (b) 0.75 mole fraction of C6 H6 2 (a) 26.1 kPa (b) Mole fraction of hexane = 0.684 Mole fraction of heptane = 0.316 Quick Check 8.4 (a) (i) & (ii) 70 60 50 40 0 1 30 Pressure/kPa Mole fraction of P (b)(i) (a)(i) (a)(ii) (c) (b)(ii) ABBB (b) (i) 50 kPa (ii) 0.58 mole fraction of P (c) 0.56 mole fraction of P Quick Check 8.5 1 (a) (i) Mole fraction of X Pressure/kPa 0 100 46 35 V l (ii) Temperature/°C Boiling (b)(i) (b)(ii) point V l Composition of vapour 0 100 50 Mole fraction of X 2 (a) 90 80 70 60 0 0.2 0.4 0.6 0.8 1.0 Temperature/o C Mole fraction of X (b)(i) (b)(ii) V (b) (i) 71.0 °C (ii) 0.72 Quick Check 8.6 1 (a) P. P has a lower boiling point than Q. (b) (i) 77 °C (ii) Azeotrope (iii) Pure Q (c) The mixture with a constant boiling point. The composition of the vapour and the liquid mixture remains the same even on fractional distillation. 2 (a) Positive. The azeotrope has the lowest boiling point. (b) The intermolecular forces between water and ethanol molecules in the mixture are weaker than those between the molecules in their pure liquids. (c) (i) Vapour pressure 0 95 100 % by mass of ethanol

378 Chemistry Term 1 STPM (ii) Temperature/°C 0 0.11 100 % by mass of ethanol 78.5 78 100 95 (d) (i) Azeotrope (ii) Water 3 (a) 54.75 kPa (b) The mixture shows negative deviation. The bonds between the molecules in the mixture are stronger than those between molecules in the pure liquids. 4 (a) 0.11 (b) (i) Vapour pressure 0 0.11 1 Mole fraction of HCl (ii) Temperature/°C 0 0.11 1 Mole fraction of HCl – 84 108.6 100 (c) HCl reacts with water to form ions. HCl + H2 O → H3 O+ + Cl– The intermolecular forces in the mixture is stronger than those between HCl molecules and between H2 O molecules. (d) Hydrogen chloride STPM Practice 8 Objective Questions 1 B 2 A 3 C 4 C 5 D 6 A 7 D 8 D 9 B 10 D 11 C 12 A 13 C 14 C 15 D Structured and Essay Questions 1 (a) van der Waals forces (b) A solution that obeys Raoult’s law (c) The intermolecular forces between molecules of pure liquid (van der Waals forces) are of the same nature and strength as those between molecules in the mixture. (d) Temperature/°C Mole fraction of octane 1 126 98 0 (e) Fractional distillation 2 (a) A mixture with a constant boiling point. Its composition remains constant on distillation. (b) Temperature/°C 100 0 1 % HNO3 122 30 65 l l V V 87 (c) (i) Water (ii) The azeotrope (d) (i) Raoult’s law states that the partial vapour pressure of liquid A in a solution is equal to the product of the vapour pressure of pure A and the mole fraction of A in the mixture, at the same temperature. (ii) The intermolecular forces in the mixture (ion-dipole attraction) are stronger than those between molecules of the pure liquids (hydrogen bonds). 3 (a) Raoult’s law states that the partial vapour pressure of liquid A in a solution is equal to the product of the vapour pressure of pure A and the mole fraction of A in the mixture, at the same temperature. (b) (i) The intermolecular forces in the mixture (van der Waals forces) are of the same type and strength as those present between molecules of the pure liquids. (ii) (iii) Pressure 0 1 Mole fraction of benzene 0 1 Mole Fraction of 1-propanol Pressure Temperature Mole fraction of benzene 0 Liquid Vapour 1 (iv) (I): Benzene (because it has a lower boiling point) (II): Methylbenzene

379 Chemistry Term 1 STPM 4 (a) A mixture with a constant boiling point. Its composition remains constant on distillation. (b) Pressure 0 % HBr 100 A mixture of water and HBr shows negative deviation from Raoult’s law because the forces holding the H2 O and HBr molecules together in the mixture is the strong ion-dipole attraction, as compared to the weaker hydrogen bonds between water molecules and the van der Waals forces between HBr molecules. (c) Temperature (°C) 0 100 % HBr 47 60 126 –67 100 V l V l (d) (i) HBr (ii) The azeotrope 5 (a) 74 72 76 78 80 82 70 68 66 0 0.2 0.4 0.6 0.8 1.0 Mole fraction of X (d)(i) (d)(ii) l V l V Temperature/°C (b) Positive deviation. (Because the boiling point/ composition curve shows a minimum). (c) Boiling point = 67.0 °C Azeotrope = 0.72 mole fraction of X (d) (i) 71.4 °C (ii) 0.54 mole fraction of X (iii) The azeotrope (iv) Pure X 6 (a) (i) (ii) Pressure 0 1 Mole fraction of ethanol Temperature 0 1 78 82 Mole fraction of ethanol (b) The intermolecular forces between ethanol and cyclohexane molecules in the mixture are weaker than those between ethanol molecules, and between cyclohexane molecules in the pure liquids. Hence, more heat is absorbed to break the bonds and less heat is released when new bonds are form. The mixing process is endothermic. 7 (a) Positive (b) Azeotrope is a mixture with constant boiling point. The composition of the solution and the vapour are the same at all temperatures. (c) Composition: 0.6 mole fraction of P and 0.4 mole fraction of Q Boiling point: 60 °C (d) (i) 77 °C (ii) First distillate: The azeotrope (60% P and 40% Q) Last distillate: Pure Q (100% Q) STPM Model Paper (962/1) SECTION A 1 C 2 D 3 D 4 C 5 A 6 D 7 B 8 D 9 D 10 B 11 B 12 C 13 C 14 C 15 C SECTION B 16 (a) NH3 + H+ : NH4 + NH3 : H+ + NH2 – (b) + H H H N H – H H N + H H H N H – H H N (c) NH4 +. The N atom in NH4 + is surrounded by 4 bonding pairs while in the NH2 – , it surrounded by 2 lone-pairs and 2 bond pairs. The repulsion between lone pairs is stronger thus pushing the N9H bonds in NH2 – closer to one another. 17 (a) pH = –log[H+] (b) (i) pH = 3.1 ∴ [H+] = 7.9 × 10–4 mol dm–3 [HCl] = 7.9 × 10–4 mol dm–2

380 Chemistry Term 1 STPM (ii) Using the expression: [H+] = Kb C (7.9 × 10–4)2 = (1.8 × 10–5)C C = 3.5 × 10–2 mol dm–3 [CH3 COOH] = 3.5 × 10–2 mol dm–3 (c) CH3 COOH is a weak acid that dissociates partially in water. As a result, more of it must be present to provide the same H+ concentration. (d) (i) H9C≡C9H + NH2 – : H9C≡C– + NH3 (ii) In the above reaction, ethyne acts a proton donor (Bronsted-Lowry acid) while the NH2 – ion acts as a proton acceptor (Bronsted-Lowry base). SECTION C 18 (a) (i) Change in temperature: The Boltzmann distribution curve for a chemical reaction at two different temperatures is shown below: Energy Fraction of molecules Ea T1 T2 T1 >T2 Fraction of molecules with energy, E ≥ Ea at T2 Fraction of molecules with energy, E ≥ Ea at T1 For a reaction to occur, the reactant molecules must collide with one another with energy greater than the activation energy Ea . When the temperature increases (T2 > T1 ), the fraction of molecules having energy greater than the activation energy increases and also the collisions are now more energetic. This results in an increase in the rate of reaction. The reverse is true when the temperature is lowered, and the rate of reaction decreases. (ii) Presence of a catalyst: A catalyst provides an alternative pathway, with lower activation energy for the reaction to occur. Energy Fraction of molecules Activation energy with catalyst Activation energy without catalyst As can be seen from the Boltzmann distribution curve above, in the presence of a catalyst, more molecules have energy greater than the new (lower) activation energy. This would result in more ‘effective collision’ leading to an increase in the rate of reaction. (b) The order of reaction with respect to a given substance is the power to which the molar concentration of the substance is raised in an experimentally determined rate equation. (c) (i) Rate = k[NO2 Cl] (ii) Mechanism: NO2 Cl : NO2 + Cl Slow (Rate-determining step) NO2 Cl + Cl : NO2 + Cl2 Fast (iii) During the reaction, the colour of the mixture changes from yellow (NO2 Cl) to brown (NO2 ). 19 (a) A buffer solution is a solution that resists changes in pH on the addition of a little acid or base. (b) CH3 COOH(aq)  CH3 COO– (aq) + H+(aq) CH3 COONa(aq) : CH3 COO– (aq) + Na+(aq) Due to common ion effect, the mixture would contain a large amount/resoviour of undissociated CH3 COOH molecules and the CH3 COO– ions. On the addition of a little acid, the added H+ would react with the CH3 COO– ions to form undisoociated CH3 COOH molecules according to the equation: H+ (added) + CH3 COO– (aq) : CH3 COOH(aq) On the addition of a little base, the added OH– would react with CH3 COOH molecules to form CH3 COO– ions according to the equation: OH– (addad) + CH3 COOH(aq) : CH3 COO– (aq) + H2 O(l) In both cases, the added H+ or OH– are ‘destroyed’ and the pH of the buffer solution remains unchanged. (c) Using the expression for a buffer solution: pH = pKa – [CH3 COOH] [CH3 COO– ] Since the molarity of CH3 COOH and CH3 COONa are the same and the volume of the mixtures is constant (total = 50.0 cm3 ), the concentration of the species can be equated to their respective volumes in the mixture. Hence, pH = pKa – [V(CH3 COOH)] [V(CH3 COONa)] Rearranging: pH = –log [V(CH3 COOH)] [V(CH3 COONa)] + pKa Hence, a graph of pH against log [V(CH3 COOH)] [V(CH3 COONa)] would give a straight line with a negetive slope, and the intersect at the pH axis gives the value of pKa for ethanoic acid. pH 3.92 4.28 4.70 log [Methanoic acid] [Sodium methanoate] 0.95 0.6 0.18

381 Chemistry Term 1 STPM 6.0 5.0 4.5 4.0 3.5 3 0 0.2 0.4 0.6 0.8 1.0 pH 0.1 0.3 0.5 0.7 0.9 [CH3 COOH] [CH3 COOH– ] log When log [CH3 COOH] [CH3 COO– ] = 0 pH = 4.85 ∴ pKa of ethanoic acid = 4.85 –log Ka = 4.85 ∴ Ka = 1.42 × 10–5 mol dm–3 (d) (i) Solubility product of a sparingly soluble salt is the products of the molar concentration aqueous ion (raised to their respective stoichiometric coefficient) produced in a saturated solution of the salt at constant temperature. (ii) The common ion effect refers to the reduction in the solubility of a sparingly soluble when a soluble compound containing one of the ions of the salt is added to the saturated solution. Silver chloride is sparingly soluble in water. In a saturated solution of silver chloride, the following equilibrium is established. AgCl(s) + aq  Ag+(aq) + Cl– (aq) If a little Cl– ions (e.g. from dilute HCl) is added to the saturated solution, then according to Le Chatelier’s Principle, the equilibrium would shift to the left-hand side and as a result, less silver chloride will dissolved. The same effect is observed when a little Ag+(aq) ions (e.g. from AgNO3 ). Thus the addition of either Ag+ or Cl– (the common ions) would decrease the solubility of silver chloride in water. (iii) Mn2+(aq) + 2OH– (aq) 9: Mn(OH)2 (s) Ksp of Mn(OH)2 = [Mn2+(aq)][OH– (aq)]2 In the mixture: [MnSO4 ] = 1 2 (0.020) = 0.010 M [NH3 ] = 1 2 (0.050) = 0.025 M To calculate the OH– of the NH3 solution: [OH– ]= Kb C = (1.80 × 10–5)(0.025) = 6.71 × 10–4 M The ionic quotient of [Mn2+][OH– ]2 = (0.010)(6.71 × 10–4)2 = 4.50 × 10–9 mol3 dm–9 Since the ionic quotient is larger than the solubility product (2.00 × 10–13). Precipitation will occur. 20 (a) (i) O C O H H N H O C O H H H N H O C O H CI H H N H (ii) NH3 : 1 lone pair + 3 bond pair. Shape: trigonal pyramidal H N H H CO2 : 2 double bond pair. Shape: Linear O C O HCl: 1 bond pair + 3 lone pair Shape: Linear H CI (iii) All three molecules exist as simple covalent molecules with covalent bonds holding the atoms together in the molecules. The intermolecular forces between CO2 and HCl molecules are the weak van der Waals forces. However, the van der Waal’s force in CO2 is stronger because the molecule has more electrons (22) than HCl (18). This accounts for the higher boiling point of CO2 compared with HCl. Although NH3 has only 10 electrons, the intermolecular force is the stronger hydrogen bond that needs more energy to break. This accounts for NH3 having the highest boiling point. (b) (i) Hydrogen bond. O H H H OH HO OH H OH H H H O CH OH 2 (ii) Let the number of water molecules bonded to glucose in honey be x. Molecular formula: C6 H12O6 xH2 O % by weight of H2 O in honey = 18x 180 + 18x × 100% = 28.6 % ∴ x = 4 The number of H2 O molecules bonded to each glucose molecule in honey is 4.

382 Chemistry Term 1 STPM Index A Acid-base indicator 273 Activated complex 188 Activation energy 187 Allotropy 158 Arrhenius acid 254 Arrhenius base 254 Arrhenius collision theory 186 Atomic number 3 Atomic orbital, shape 49 Aufbau’s principle 51 Autocatalysis 199 Avogadro’s constant 24 Avogadro’s law 137 Azeotrope 324 B Base line, mass spectrum 21 Bohr’s model, Hydrogen atom 41 Bond energy 104 length 104 π-bond 80 σ-bond 80 Boyle’s law 132 Bronsted-Lowry acid 255 Bronsted-Lowry base 255 Buffer capacity 287 Buffer solution 280 C Catalyst 194 Charles’ law 135 Colligative property 169 Common ion effect 293 Compressibility factor, of gas 145 Conjugate acid-base pair 257 Convergence frequency, Line spectrum 44 Convergence limit, Line spectrum 44 Coordinate bond 105 Covalent bond 70 Critical point 164 D Dalton’s law of partial pressure 140 d-orbital 51 Dynamic equilibrium 223 E Electrolyte 253 Equivalent point 275 F Fractional distillation 316 H Half-life 203 Hund’s rule 52 Hybridisation 80 Hydrogen bond 116 I Ideal gas equation 139 Ideal solution 309 Ionic bonding 65 Isotope 4 Isotopic abundance 5 K Kinetic theory of gases 132 L Law of mass action 224

383 Chemistry Term 1 STPM Le Chatelier’s principle 233 Lewis acid 256 Lewis base 256 Line spectrum, Hydrogen 39 Liquid 147 M Mass number 3 Mass spectrometer 17 Maximum buffer capacity 288 Maxwell-Boltzmann distribution curve 143 Metallic bond 107 Methyl orange 274 Mole 24 Molecular ion 20 N Noble gas and equilibrium 236 Non-ideal solution 319 Nuclear fission 11 Nuclear fusion 12 Nuclear stability 10 Nucleon number 3 O Order of reaction 192, 201 Ostwald’s dilution law 253 P Particle α-particle 7 b-particle 7 Pauli’s exclusion principle 52 Phase diagram, water 161 Phenolphthalein 273 Planck’s equation 38 Proton number 3 R Radiation γ-radiation 8 Radioactive decay 6 Radioactive isotope 6 Rate of reaction 182 Rate law 191 Reaction mechanism 211 Reaction quotient 230 Reduced pressure distillation 330 Relative atomic mass 15 Relative formula mass 16 Relative isotopic mass 14 Relative molecular mass 16 S Salt hydrolysis 267 Saturated vapour pressure 148 Solid 152 Solubility product 292 Stratosphere ozone 232 Supercritical fluid 166 T Titration curve 276 Triple point 164 V Valence-shell electron-pair repulsion theory 91 van der Waals force 110 W Water softening 298 Z Zeolite 299

Notes