PREFACE Pre-U STPM Text Chemistry Term 2 is written based on the latest syllabus prepared by the Malaysian Examinations Council (MEC). The book is designed and well organised with the following features to help students understand the concepts taught. Analysis of STPM Papers (2015 – 2018) v Chapter 2015 2016 2017 2018 A B C A B C A B C A B C 7 Chemical Energetics 3 2 1 3 3 1 8 Electrochemistry 3 1 3 1 2 1 3 1 9 Periodic Table: Periodicity 2 3 1 1/3 2 1 2 10 Group 2 2 2 2/3 2 1 1 1 11 Group 14 1 1 1 2 2 3 12 Group 17 2 1 1 2 1 2 1 13 Transition Elements 2 1 2 1 2 1 1 1 Total 15 2 3 15 2 3 15 2 3 15 2 3 STPM Scheme of Assessment Term of Study Paper Code and Name Type of Test Mark (Weighting) Duration Administration First Term 962/1 Chemistry Paper 1 Written Test Section A 15 compulsory multiplechoice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 6. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 962/2 Chemistry Paper 2 Written Test Section A 15 compulsory multiplechoice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 7 to 13. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment vi CHAPTER CHEMICAL ENERGETICS 7 Learning earning Outcomes Students should be able to: Enthalpy changes of reaction, ΔH • explain that most chemical reactions are accompanied by enthalpy changes (exothermic or endothermic); • define enthalpy change of reaction, ΔH, and state the standard conditions; • define enthalpy change of formation, combustion, hydration, solution, neutralisation, atomisation, bond energy, ionisation energy and electron affinity; • calculate the heat energy change from experimental measurements using the relationship: heat change, q = mcΔT or q = mcθ; • calculate enthalpy changes from experimental results. Hess’ law • state Hess’ law, and its use to find enthalpy changes that cannot be determined directly, e.g. an enthalpy change of formation from enthalpy changes of combustion; • construct energy level diagrams relating the enthalpy to reaction path and activation energy; • calculate enthalpy changes from energy cycles. Born-Haber cycle • define lattice energy for simple ionic crystals in terms of the change from gaseous ions to solid lattice; • explain qualitatively the effects of ionic charge and ionic radius on the numerical magnitude of lattice energy values; • construct Born-Haber cycle for the formation of simple ionic crystals. Solubility of solids in liquids • construct energy cycles for the formation of aqueous solutions of ionic compounds; • explain qualitatively the influence on solubility of the relationship between enthalpy change of solution, lattice energy of solid and enthalpy change of hydration or other solvent-solute interaction. Chemical Energetics Standard Enthalpy Change of Neutralisation • Strong acids and strong bases • Hydrofluoric acid • Weak acids or weak bases Standard Ionisation Energy Thermochemical Equation Hess’ Law Enthalpy Change of Reaction, ∆H Standard Enthalpy Change of Atomisation Standard Enthalpy Change of Hydration Electron Affinity Standard Enthalpy Change of Combustion Standard Conditions in Enthalpy Measurement Standard Enthalpy Bond energy Change of Solution Lattice Energy • Factors • Ionic compounds • Covalent character Standard Enthalpy Change of Formation • Enthalpy change of reaction • Stability Concept Map STPM Scheme of Assessment Latest STPM scheme of Assessment starting 2012. Concept Map Provides an overall view of the concepts learnt in the chapter. Learning Outcomes A list of subtopics that students will learn in each chapter. ii Analysis of STPM Papers Shows past yearsʼ STPM questions according to the chapters.
Chemistry Term 2 STPM Chapter 10 Group 2 159 10 SUMMARY SUMMARY 10.3 Uses of Group 2 Compounds Compound Uses Magnesium oxide Refractory material in high temperature furnace Magnesium hydroxide (Milk of magnesia) As an antacid to relieve indigestion Magnesium sulphate As a laxative Calcium sulphate ‘Plaster of Paris’ Calcium hydroxide To neutralise acidic in soil Calcium carbonate Making writing chalks Barium carbonate Rat poison Barium sulphate As ‘barium meal’ in the X-ray of the digestive system Objective Questions 1 Which of the following statements regarding the Group 2 elements and their compounds is not true? A They are known as the alkaline metals. B Beryllium oxide is the only amphoteric oxide in Group 2. C The melting point of calcium is higher than that of magnesium. D Barium oxide is soluble in water. 2 Going down Group 2 of the Periodic Table, A the maximum oxidation state of the elements increases. B the polarising power of the cations decreases. C the electronegativity of the elements decreases. D the standard electrode potential becomes more negative. 3 Beryllium oxide is amphoteric. This is because A beryllium is a metalloid B Be2+ is unstable C Be2+ has a high charge density D Be2+ forms the [Be(OH)4]2- complex ion 1 The Group 2 elements are also known as the ‘alkaline earth metals’. 2 They have two electrons in their valence shell. 3 They exhibit a constant valence of +2 in all their compounds. 4 They are powerful reducing agents. The reducing strength increases down the group. 5 Their salts are white except when in combination with the oxo-anions of transition elements. 6 The carbonates, nitrates and hydroxides decompose when heated to form the corresponding oxides. 7 The thermal stability of the salts increases down the group. 8 The solubility of the sulphates decreases down the group. 9 Beryllium shows some properties that are not typical of the Group 2 elements: • Beryllium oxide and beryllium hydroxide are amphoteric. • Beryllium chloride is covalent. • Beryllium forms complexes. 10 STPM PRACTICE Chemistry Term 2 STPM Chapter 7 Chemical Energetics 22 7 3 Thus, when Na+ and Cl– ions come together to form solid sodium chloride, the same amount of energy must be released. Na+(g) + Cl– (g) ⎯→ Na+Cl– (s) ∆Ho = –770 kJ mol–1 This energy is called the lattice energy. 4 The lattice energy of an ionic compound is the heat energy released when one mole of the ionic solid is formed from its constituent gaseous ions under standard conditions. Example 7.11 The lattice energy of potassium chloride is –701 kJ mol–1. Write a thermochemical equation to represent this statement. Solution K+(g) + Cl– (g) ⎯→ KCl(s) ∆Ho = –701 kJ mol–1 Definition of lattice energy Quick Check 7.7 1 Write thermochemical equations to represent the following lattice energies. Compound ∆Ho lattice/kJ mol–1 (a) Sodium bromide –742 (b) Silver chloride –890 (c) Sodium oxide –2478 (d) Calcium fluoride –2600 (e) Aluminium oxide –15 920 Factors Affecting the Lattice Energy 1 The lattice energy is a measure of the strength of the ionic bonds in the ionic solid. 2 The stronger the ionic bonds, the more exothermic is the lattice energy. 3 The lattice energy depends on: (a) the charge on the ions, (b) the size of the ions. 4 The higher the charge on the ions, the stronger is the attraction between the ions. 5 The smaller the size of the ions, the shorter is the distance between the ions. This will increase the force of attraction between the ions. (charge on the cation) × (charge on the anion) Lattice energy ∝ ——————————————————— (cationic radius + anionic radius)2 6 The lattice energy of some ionic solids is given in the tables below: (a) Compound Charge on cation Charge on anion ∆Ho lattice /kJ mol–1 NaCl +1 –1 –770 Na2O +1 –2 –2478 MgO +2 –2 –3850 Al2O3 +3 –2 –15 920 2009/P1/Q19 2012/P1/Q19 2014/P2/Q3 VIDEO Lattice Energy QR Code Info/Video Scan QR code using mobile smartphone devices to access related information or videos Chemistry Term 2 STPM Chapter 10 Group 2 148 10 Reactions with Water 1 All Group 2 elements react with water to form their respective hydroxides with the liberation of hydrogen gas. M(s) + 2H2O ⎯→ M(OH)2 + H2(g) In the reaction, the Group 2 elements act as reducing agents and reduce water to hydrogen. 2H2O + 2e– ⎯→ H2 + 2OH– 2 Going down the group, the reactivity towards water increases as the reducing power of the elements increases. 3 Beryllium reacts slowly with steam: Be(s) + 2H2O(g) ⎯→ Be(OH)2(s) + H2(g) Δ 4 Magnesium reacts slowly with hot water, but rapidly with steam. Mg(s) + 2H2O(g) ⎯→ Mg(OH)2(s) + H2(g) Δ 5 Calcium reacts rapidly with hot water, but slowly with cold water. Ca(s) + 2H2O(l) ⎯→ Ca(OH)2(aq) + H2(g) Δ 6 Strontium and barium reacts vigorously with cold water. Sr(s) + 2H2O(l) ⎯→ Sr(OH)2(aq) + H2(g) Ba(s) + 2H2O(l) ⎯→ Ba(OH)2(aq) + H2(g) 7 All the hydroxides are bases except beryllium hydroxide, which is amphoteric. 8 For example, Mg(OH)2(s) + H2SO4(aq) ⎯→ MgSO4(aq) + 2H2O(l) Ba(OH)2(aq) + 2HCl(aq) ⎯→ BaCl2(aq) + 2H2O(l) 9 The solubility of the hydroxides increases down the group. Hydroxide Solubility Solubility/g per 1000 g H2O Be(OH)2 Insoluble Insoluble Mg(OH)2 Insoluble 0.012 Ca(OH)2 Sparingly soluble 1.2 Sr(OH)2 Soluble 10.0 Ba(OH)2 Soluble 47.0 10 An aqueous suspension of magnesium hydroxide (milk of magnesia) is used as antacid for the treatment of gastrointestinal discomfort by neutralising the acid in the stomach. 2018/P2/Q18(a) Reduction of water to hydrogen. Tips Exam Tips Exam Reactivity towards water increases as the reducing power of the elements increases. Beryllium hydroxide is amphoteric. Solubility Mr of M(OH)2 Quick Check 10.2 1 Explain why aqueous sodium hydroxide is not used as antacid? 2 An aqueous solution of barium hydroxide turns cloudy when exposed to air for sometimes. Explain with the aid of an equation what happens. 3 Explain why calcium hydroxide or calcium oxide are sometimes added to agricultural soils. Info Chem Barium hydroxide reacts with carbon dioxide to form insoluble barium carbonate. INFO Reactions of the Group 2 Elements with Water Past-year Questions tagging Enables students to know the topics frequently tested in the exam. Exam Tips Provides helpful tips for students in answering exam questions. Notes Provides explanation in a simple way to ease students’ understanding. Summary Summarises key concepts learnt in the chapter. STPM Practice A variety of examinationtype questions to check students’ understanding of the chapters learnt. Side Column Notes Provides additional information or pointers to improve students’ understanding. iii Quick Check Provides short questions for students to test their understanding of the concepts learnt in the subtopics. Info Chem Provides extra information that relates to the subtopics learnt. Example Gives step-by-step solutions to explain the example.
241 STPM Model Paper (962/2) Section A [15 marks] Bahagian A [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1 The enthalpy changes involved in the dissolution of an ionic solid is shown below: Perubahan entalpi yang terlibat apabila suatu pepejal ion larut dalam air ditunjukkan di bawah: Gaseous ions Ion bergas ∆H1 ∆H2 Ionic solid ∆H3 Aqueous solution Pepejal ion ⎯→ Larutan berair Which of the enthalpy changes is/are always negative? Yang manakah antara berikut sentiasa bernilai negatif? A ∆H1 C ∆H1 and/dan ∆H2 B ∆H2 D ∆H2 and/dan ∆H3 2 The standard enthalpy of combustion of aluminium is -835 kJ mol-1. Which of the following deduction is not correct? Perubahan entalpi pembakaran piawai aluminium adalah -835 kJ mol-1. Deduksi yang manakah tidak benar? A 150.3 kJ is liberated when 4.86 g of aluminium burns in excess oxygen. Apabila 4.86 g aluminium dibakar dalam oksigen berlebihan, 150.3 kJ dibebaskan. B -58.94 kJ of heat is involved in the formation of 3.60 g of aluminium oxide. -58.94 kJ terlibat dalam pembentukan 3.60 g aluminium oksida. C 242.5 g of aluminium has to be burned completely in excess oxygen to generate 3.00 × 104 kJ of heat. 242.5 g aluminium perlu dibakar dengan lengkap untuk menghasilkan 3.00 x 104 kJ haba. D The enthalpy change for the following reaction is -3340 kJ. 4Al(s) + 3O2 (g) → 2Al2 O3 (s) Perubahan entalpi bagi tindak balas berikut ialah -3340 kJ. 4Al(p) + 3O2 (g) → 2Al2 O3 (p) 3 The standard reduction potentials of two half-cells are given below: Keupayaan penurunan piawai bagi dua setengahsel diberikan berikut: Mn3+(aq) + e– Mn2+(aq) E° = + 1.51 V Ti3+(aq) + e– Ti2+(aq) E° = -0.37 V Which of the following is not correct? Yang manakah antara berikut tidak benar? A Under standard conditions, the most stable oxidation of Mn is +2 and for Ti is +3. Di bawah keadaan piawai, keadaan pengoksidaan yang paling stabil bagi Mn ialah +2 dan bagi Ti ialah +3. B In an electrochemical cell, the Mn3+/ Mn2+ is the cathode. Dalam suatu sel elektrokimia, Mn3+/Mn2+ merupakan katod. C The e.m.f. of the cell is +1.88 V. D.g.e. sel ialah +1.88 V. D The cell notation is: Notasi sel adalah: Pt(s) Ti3+(aq); Ti2+(aq) Mn3+(aq); Mn2+(aq) Pt(s) Pt(p) Ti3+(ak); Ti2+(ak) Mn3+(ak); Mn2+(ak) Pt(p) 4 Which of the following statements is not correct? Pernyataan yang manakah antara berikut adalah tidak benar? A The negative terminal of an electrochemical cell is called the anode. Kutub negatif suatu sel elektrokimia disebut anod. 256 Alkaline earth metal Logam alkali bumi The Group 2 elements Alkaline metal Logam alkali The Group 1 elements Anode Anod The electrode where oxidation occurs Anodisation Penganodan An electrolytic passivation process used to increase the thickness of the natural oxide layer on the surface of metal Catalyst Mangkin A substance that changes the rate of a chemical reaction but remains chemically unchanged at the end of it Cathode Katod The electrode where reduction occurs Charge density Ketumpatan cas The charge over size ratio of an ion Complex ion Ion kompleks A central metal ion bonded to a group of atoms, molecules or ions by coordinate (dative) bonds Corrosion Kakisan The gradual destruction of material, usually metal, by a chemical reaction with its environment Covalent radius Jejari kovalen Half the distance between the centres of two adjacent atoms that are held by a covalent bond Decomposition potential Keupayaan penguraian The minimum voltage that is required before electrolysis can occur Diagonal relationship Hubungan penjuru A relationship within the periodic table by which certain elements in the second period have a close chemical similarity to their diagonal neighbours in the next group of the third period. For example, Li and Mg; Be and Al. Effective nuclear charge Cas nukleus berkesan The net positive charge experienced by an electron in a multi-electron atom Electrochemical cell Sel elektrokimia A device used for generating an electromotive force (voltage) and current from spontaneous redox reactions Electrolysis Elektrolisis Decomposition of an electrolyte by electrical current Electrolytic cell Sel elektrolisis A device that uses electricity to cause a nonspontaneous redox reaction to occur Electron affinity Afiniti elektron The energy released when an electron is added to a gaseous atom per mole of the atom under standard conditions Electronegativity Keelektronegatifan The relative tendency of an atom to attract electron in a covalent bond towards itself Endothermic reaction Tindak balas endotermik A reaction where there is a net gain in energy Energy Tenaga The ability to do work 259 Chapter 7 Chemical Energetics Quick Check 7.1 1 (a) Ag+ (aq) + Cl– (aq) → AgCl(s) ∆H° = –66 kJ mol–1 (b) +1.32 kJ 2 (a) +354 kJ (b) –885 kJ 3 (a) –8.55 kJ (b) 29.12 °C 4 (a) –100 kJ mol–1 (b) 45.02 kJ 5 (a) –150.3 kJ (b) –58.94 kJ (c) 970.1 g 6 (a) 831.82 kJ (b) –2868.34 kJ Quick Check 7.2 1 (a) Ca(s) + C(s) + —3 2 O2(g) → CaCO3(s) (b) Cu(s) + S(s) + 5H2(g) + —9 2 O2(g) → CuSO4•5H2O(s) (c) 2C(s) + 2H2(g) + O2(g) → CH3COOH(l) (d) —1 2 N2(g) + O2(g) → NO2(g) (e) —1 2 H2(g) + —1 2 I2(s) → HI(g) (f) C(s) + 2Cl2(g) → CCl4(l) (g) Cd(s) + 2C(s) + N2(g) → Cd(CN)2(s) Quick Check 7.3 1 –876.0 kJ 2 +127 kJ mol–1 Quick Check 7.4 1 (a) Na(s) + —1 4 O2(g) → —1 2 Na2O(s) (b) H2C2O4(s) + —1 2 O2(g) → 2CO2(g) + H2O(l) (c) SO2(g) + —1 2 O2(g) → SO3(g) (d) NH3(g) + —3 4 O2(g) → —1 2 N2(g) + —3 2 H2O(l) 2 –720 kJ mol–1 3 91.65 °C 4 (a) 23.76 kJ (b) 570.24 kJ mol–1 5 (a) C2H5OH + 3O2 → 2CO2 + 3H2O (b) –1337.6 kJ mol–1 (c) 2.91 × 104 kJ Quick Check 7.5 1 29.1 °C 2 (a) CH3COOH + NaOH → CH3COONa + H2O (b) 1421.2 kJ (c) –56.8 kJ mol–1 Quick Check 7.6 1 –110 kJ mol–1 2 –110 kJ mol–1 3 (a) C(Diamond) C(Graphite) –394 kJ CO2 –396 kJ (b) –2 kJ mol–1 (c) Graphite 4 –848 kJ 5 +45.0 kJ mol–1 6 (a) 2C(s) + —1 2 O2(g) + 3H2(g) → C2H5OH(l) (b) –279 kJ mol–1 7 –1260 kJ mol–1 8 –334.6 kJ 9 (a) C2H5OH(l) + —1 2 O2(g) → CH3CHO(l) + H2O(l) (b) –200.2 kJ mol–1 10 –466 kJ mol–1 Quick Check 7.7 1 (a) Na+ (g) + Br– (g) → NaBr(s) (b) Ag+ (g) + Cl– (g) → AgCl(s) (c) 2Na+ (g) + O2–(g) → Na2O(s) (d) Ca2+(g) + 2F– (g) → CaF2(s) (e) 2Al3+(g) + 3O2–(g) → Al2O3(s) Quick Check 7.8 1 (a) MgO (b) MgSO4 (c) Al2O3 (d) LiF (e) Na2O Quick Check 7.9 1 –2050.6 kJ mol–1 2 –494 kJ mol–1 3 –325 kJ mol–1 Quick Check 7.10 1 (a) –18 kJ mol–1 (b) Soluble 2 –86.7 kJ mol–1 3 12.8 °C 4 (a) +81 kJ mol–1 (b) Insoluble STPM Practice 7 Objective Questions 1 B 2 B 3 B 4 A 5 A 6 D 7 B 8 C 9 B 10 C 11 D 12 A 13 B 14 C 15 C 16 C 17 C 18 A 19 B 20 D 21 B 22 B 23 B 24 A 25 C 26 C 27 D 28 A 29 B 30 C 31 C 32 A 33 B 34 A 35 D 36 B 37 C 38 A 39 B Structured and Essay Questions 1 (a) Ag+ (g) + Br– (g) → Ag+ Br– (s) (b) (i) (+285) + (+112) + (+731) + (–325) + ∆H = –100 ∆H = –903 kJ mol–1 (ii) Silver bromide has significant covalent character. (iii) Less exothermic. The size of I– is larger than Br– . (c) Less. The size of I– is larger than that of Br– . The electron is less strongly held by the nucleus. ANSWERS 279 Chemistry Term 1 STPM Index Index A Absorption 228 Active electrode 87 Adsorption 228 Alloy, tin 184 Aluminium nitride 132 Aluminium, recycle 94 Aluminium/air fuel cell 81 Amphiboles 180 Anode 57 Anodisation 95 Anticlockwise rule 62 Asbestos 180 Atomic radius 110 B Beryllium, anomalous behaviour of 156 Bidentate ligand 232 Black-and-white photography 201 Bond energy 13 Born-Haber cycle 21 C Catalyst, transition elements 226 Cathode 57 Cathodic protection 93 Cell diagram 57 Ceramics 183 Chain silicate 179 Clay 181 Complex 229 Complex, geometry of 233 Corrosion of iron 91 Covalent radius 110 D Daniell cell 39 d-d transition, complex 219 Diagonal relationship 157 Diaphragm cell 95 Disproportionation 44 Dynamic equilibrium 73 E Effective nuclear charge 111 Effluent, treatment of 96 Electric car 81 Electrochemical cell 57 Electrode potential 51 Electrolysis 82 Electron affinity 17 Electronegativity 122 Electroplating 96 Endothermic reaction 2 Enthalpy change of reaction 2 Enthalpy of dilution 10 Equilibrium constant 73 Exothermic reaction 2 Extraction of aluminium 93 F Faraday’s first law of electrolysis 89 Faraday’s second law of electrolysis 90 First ionisation energy 16 Framework silicate 181 Fuel cell 79 G Galvanisation 92 Giant structure silicate 181 Glass 182 Group (Periodic Table) 109 H Half-equation 39 Halogen 190 Hess’ law 19 Heterogeneous catalysis 228 Hexadentate ligand 233 Homogeneous catalysis 227 Hydrogen fuel cell 80 Hydroquinone 202 I Ionic radius 115 L Lattice energy 21 Ligand 231 M Metallic radius 110 Mica 181 Molar enthalpy of vaporisation Monodentate ligand 118 231 N Negative electrode potential 51 Nernst equation 69 Nuclear charge 110 O Overpotential 86 Oxidation 39 Oxidation number 41 Oxidation state 41 Oxidising agent 40 P Period (Periodic Table) 109 Photochromic lens 202 Positive electrode potential 52 Pyroxenes 179 R Redox reaction 39 Reducing agent 40 Reduction 39 Rusting of iron 91 S Sacrificial anode 93 Salt bridge 39 Screening effect 110 Second ionisation energy 16 Sheet silicate 180 Silicate 178 Silicone 177 Solubility product 74 Standard electrode potential 50 Standard electrode potential series 56 Standard enthalpy change of atomisation 12 Standard enthalpy change of formation 5 Standard enthalpy change of hydration 17 Standard enthalpy change of neutralisation 9 Standard enthalpy change of reaction 3 Standard enthalpy change of solution 18 Standard enthalpy of combustion 8 Standard hydrogen electrode 53 Standard ionisation energy 16 Standard reduction potential 52 T Talcum 181 Thermochemical equation 3 Transition element, definition 209 Z Zeolite 182 STPM Model Paper A model paper that follows the latest STPM exam format is provided for practice. Glossary Gives a list of important terms to ease the students’ understanding of their meanings. Answers Complete answers are provided. Index Provides a list of terms to enable easy and direct access to the subtopics. iv
Analysis of STPM Papers (2015 – 2018) v Chapter 2015 2016 2017 2018 A B C A B C A B C A B C 7 Chemical Energetics 3 2 1 3 3 1 8 Electrochemistry 3 1 3 1 2 1 3 1 9 Periodic Table: Periodicity 2 3 1 1/3 2 1 2 10 Group 2 2 2 2/3 2 1 1 1 11 Group 14 1 1 1 2 2 3 12 Group 17 2 1 1 2 1 2 1 13 Transition Elements 2 1 2 1 2 1 1 1 Total 15 2 3 15 2 3 15 2 3 15 2 3
STPM Scheme of Assessment Term of Study Paper Code and Name Type of Test Mark (Weighting) Duration Administration First Term 962/1 Chemistry Paper 1 Written Test Section A 15 compulsory multiplechoice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 6. 60 (26.67%) 15 15 30 1— 1 2 hours Central assessment Second Term 962/2 Chemistry Paper 2 Written Test Section A 15 compulsory multiplechoice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 7 to 13. 60 (26.67%) 15 15 30 1— 1 2 hours Central assessment vi
Term of Study Paper Code and Name Type of Test Mark (Weighting) Duration Administration Third Term 962/3 Chemistry Paper 3 Written Test Section A 15 compulsory multiplechoice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 14 to 21. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment 962/5 Chemistry Paper 5 Written Practical Test 3 compulsory structured questions to be answered. 45 (20%) 1— 1 2 hours Central assessment First, Second and Third Terms 962/4 Chemistry Paper 4 School-based Practical Assessment 13 compulsory experiments and one project to be carried out. 225 to be scaled to 45 (20%) Throughout the three terms School-based assessment vii
CONTENTS Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 7 Chemical Energetics 1 7.1 Enthalpy Changes of Reaction, ∆H 2 7.2 Hess’ Law 19 7.3 Born-Haber Cycle 21 7.4 The Solubility of Solids in Liquids 27 Summary 28 STPM Practice 7 29 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 8 Electrochemistry 38 8.1 Half-cell and Redox Equations 39 8.2 Standard Electrode Potential 50 8.3 Non-standard Cell Potentials 69 8.4 Fuel Cells 79 8.5 Electrolysis 82 8.6 Applications of Electrochemistry 91 Summary 97 STPM Practice 8 98 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 9 Periodic Table: Periodicity 108 9.1 Physical Properties of Elements of Period 2 and Period 3 109 9.2 Reactions of Period 3 Elements with Oxygen and Water 131 9.3 Acidic and Basic Properties of Oxides and Hydrolysis of Oxides 136 Summary 138 STPM Practice 9 139 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 10 Group 2 144 10.1 Selected Group 2 Elements and Their Compounds 145 10.2 Anomalous Behaviour of Beryllium 156 10.3 Uses of Group 2 Compounds 159 Summary 159 STPM Practice 10 159 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •11 Group 14 163 11.1 Physical Properties of Group 14 Elements 164 11.2 Tetrachlorides and Oxides of Group 14 Elements 168 11.3 Relative Stability of +2 and +4 Oxidation States of Group 14 Elements 174 11.4 Silicon, Silicone and Silicates 177 11.5 Tin Alloys 184 Summary 184 STPM Practice 11 185 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •12 Group 17 189 12.1 Physical Properties of Selected Group 17 Elements 190 12.2 Reactions of Selected Group 17 Elements 193 12.3 Reactions of Selected Halide Ions 199 12.4 Industrial Applications of Halogens and Their Compounds 201 Summary 202 STPM Practice 12 203 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •13 Transition Elements 207 13.1 Physical Properties of First Row Transition Elements 208 13.2 Chemical Properties of First Row Transition Elements 216 13.3 Nomenclature and Bonding of Complexes 229 13.4 Uses of First Row Transition Elements and Their Compounds 236 Summary 237 STPM Practice 13 237 • STPM Model Paper (962/2) 241 • Appendix 247 • Glossary 256 • Answers 259 • Index 279 viii
CHAPTER 7 CHEMICAL ENERGETICS Learning earning Outcomes Students should be able to: Enthalpy changes of reaction, ΔH • explain that most chemical reactions are accompanied by enthalpy changes (exothermic or endothermic); • define enthalpy change of reaction, ΔH, and state the standard conditions; • define enthalpy change of formation, combustion, hydration, solution, neutralisation, atomisation, bond energy, ionisation energy and electron affinity; • calculate the heat energy change from experimental measurements using the relationship: heat change, q = mcΔT or q = mcθ; • calculate enthalpy changes from experimental results. Hess’ law • state Hess’ law, and its use to find enthalpy changes that cannot be determined directly, e.g. an enthalpy change of formation from enthalpy changes of combustion; • construct energy level diagrams relating the enthalpy to reaction path and activation energy; • calculate enthalpy changes from energy cycles. Born-Haber cycle • define lattice energy for simple ionic crystals in terms of the change from gaseous ions to solid lattice; • explain qualitatively the effects of ionic charge and ionic radius on the numerical magnitude of lattice energy values; • construct Born-Haber cycle for the formation of simple ionic crystals. Solubility of solids in liquids • construct energy cycles for the formation of aqueous solutions of ionic compounds; • explain qualitatively the influence on solubility of the relationship between enthalpy change of solution, lattice energy of solid and enthalpy change of hydration or other solvent-solute interaction. Chemical Energetics Standard Enthalpy Change of Neutralisation • Strong acids and strong bases • Hydrofluoric acid • Weak acids or weak bases Standard Ionisation Energy Thermochemical Equation Hess’ Law Enthalpy Change of Reaction, ∆H Standard Enthalpy Change of Atomisation Standard Enthalpy Change of Hydration Electron Affinity Standard Enthalpy Change of Combustion Standard Conditions in Enthalpy Measurement Standard Enthalpy Change of Solution Bond energy Lattice Energy • Factors • Ionic compounds • Covalent character Standard Enthalpy Change of Formation • Enthalpy change of reaction • Stability Concept Map
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 2 7 7.1 Enthalpy Changes of Reaction, ∆H Endothermic and Exothermic Reactions 1 The absolute heat energy content or the enthalpy (symbol, H) of a substance cannot be measured. 2 For example, what is the enthalpy or heat content of 1 mole of sodium hydroxide or 1 mole of hydrochloric acid? 3 However, we can measure the heat change or enthalpy change when 1 mole of sodium hydroxide reacts with 1 mole of hydrochloric acid to form sodium chloride and water. NaOH(aq) + HCl(aq) ⎯→ NaCl(aq) + H2O(l) 4 This heat change is called enthalpy change of reaction, ∆H. Enthalpy change of reaction is the heat energy absorbed or released in a chemical reaction for the number of moles of reactants as indicated by the chemical equation. 5 Reactions where there is a net gain in heat energy are called endothermic reactions, and ∆H is given a positive value. The energy profile of an endothermic reaction is shown below: Energy Heat is absorbed = positive Reactants Products ∆H An endothermic reaction is accompanied by a drop in temperature as heat is absorbed from the surroundings. 6 Reactions where there is a net loss in heat energy are called exothermic reactions, and ∆H is given a negative value. The energy profile for an exothermic reaction is shown below: Energy Heat is given out = negative Products Reactants ∆H An exothermic reaction is accompanied by an increase in temperature as heat is released to the surroundings. Definition of enthalpy change of reaction Energy profile
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 3 7 Standard Conditions in Enthalpy Measurement 1 The enthalpy change of a reaction depends on (a) the temperature under which the experiment is carried out, (b) the external pressure, (c) the physical state of the substances involved, (d) the concentration of solutions, and (e) the type of allotrope of the reactants. 2 For example: 2H2(g) + O2(g) ⎯→ 2H2O(l) ∆H = –572 kJ But 2H2(g) + O2(g) ⎯→ 2H2O(g) ∆H = –483.6 kJ The heat energy released in the second reaction is less because heat has to be absorbed to vaporise the water formed. 3 Another example involves the combustion of carbon. Carbon has two allotropes, graphite and diamond. Graphite and diamond are made up of carbon atoms arranged differently in the solid state. C(graphite) + O2(g) ⎯→ CO2(g) ∆H = –394 kJ But C(diamond) + O2(g) ⎯→ CO2(g) ∆H = –396 kJ The enthalpy change is different in the two processes because different allotropic forms of carbon are involved. 4 To standardise thermochemical measurements, the following standard conditions are chosen. (a) Temperature is fixed at 25 °C or 298 K. (b) Pressure is fixed at 101 kPa or 1 atm. (c) The concentration of solutions involved is fixed at 1.0 mol dm–3. (d) For allotropic elements, the most stable allotrope at 298 K and 101 kPa is used. 5 Enthalpy change measured under these conditions is called standard enthalpy change of reaction, and is given a symbol of ∆H°. 6 The heat energy change during an experiment (q) can be calculated using the following relationship: q = mc∆T Thermochemical (Thermodynamic) Equation 1 When 12.0 g (1 mole) of graphite is burned in excess oxygen to form carbon dioxide, 394 kJ of heat is released. This can be represented by the following thermochemical equation: C(s) + O2(g) ⎯→ CO2(g) ∆H = –394 kJ 2 The value of ∆H in a thermochemical equation depends on: (a) the direction of the reaction, (b) the stoichiometry of the equation. 3 For example, ∆H for the following reaction: CO2(g) ⎯→ C(s) + O2(g) is –(–394) = +394 kJ 4 The ∆H for the following reaction: 2C(s) + 2O2(g) ⎯→ 2CO2(g) is 2(–394) = –788 kJ 2H2O(l) → 2H2O(g) ∆H = +88.4 kJ Info Chem A substance under standard conditions (25 °C; 1 atm) is said to be in its standard state. For example, the standard state of nitrogen is diatomic gas whereas the standard state of water is liquid. Graphite Diamond –394 –396 CO2 2018/P2/Q1
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 4 7 Example 7.1 Write a thermochemical equation for each of the following reactions: (a) When 1 mole of methane is burned in oxygen to form carbon dioxide and water, 890 kJ of heat is released. (b) 52.0 kJ of heat is absorbed when a mole of hydrogen reacts with 1 mole of solid iodine to form gaseous hydrogen iodide. (c) 48.6 kJ of heat is released when 1 mole of copper reacts with 1 mole of sulphur to form copper(II) sulphide. Solution (a) CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(l) ∆H = –890 kJ (b) H2(g) + I2(s) ⎯→ 2HI(g) ∆H = +52.0 kJ (c) Cu(s) + S(s) ⎯→ CuS(s) ∆H = –48.6 kJ Example 7.2 Given the following thermodynamic equation: 2C2H2(g) + 5O2(g) ⎯→ 4CO2(g) + 2H2O(l) ∆H = –2600 kJ Calculate (a) the enthalpy change when 1 mol of C2H2 is burned completely in oxygen, (b) the enthalpy change when 1 g of C2H2 is burned completely in oxygen, (c) the mass of C2H2 that needs to be burned to generate 10 000 kJ of heat energy. Solution (a) Enthalpy change per mole C2H2 = ——– –2600 2 = –1300 kJ (b) 1 mol (26 g) of C2H2 = –1300 kJ ∴ 1 g of C2H2 = ——– –1300 26 = –50 kJ (c) 1 g of C2H2 liberates 50 kJ when burned. ∴ Mass of C2H2 needed to generate 10 000 kJ = 10 000 ——– 50 × 1 g = 200 g Exam Tips Exam Tips Always indicate whether ∆H is positive or negative. Quick Check 7.1 1 When 1 mole of Ag+(aq) ions react with 1 mole of Cl– (aq) ions to form silver chloride solid, 66 kJ of heat is released. (a) Write a thermochemical equation to represent this reaction. (b) What is the heat change when 0.02 mole of silver chloride dissolves completely in water? 2 Given the following thermochemical equation: NH3(g) + HCl(g) ⎯→ NH4Cl(s) ∆H = –177 kJ (a) Calculate the heat change for the following reaction: 2NH4Cl(s) ⎯→ 2NH3(g) + 2HCl(g) (b) Calculate the heat change when 85 g of ammonia reacts with 328.5 g of hydrogen chloride gas.
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 5 7 3 Given: NaOH(aq) + HCl(aq) ⎯→ NaCl(aq) + H2O(l) ∆H = –57.0 kJ (a) Calculate the heat change when 2 moles of NaOH react with 0.15 mole of HCl. (b) 50 cm3 of 0.10 M NaOH is added to 50 cm3 of 0.10 M of HCl in a calorimeter. Both solutions were initially at 28.6 °C. Calculate the final temperature of the mixture. [Density of solution = 1.2 g cm–3. Specific heat of solution = 4.6 J g–1 K–1] 4 Given: 2Ag(s) + Br2(l) ⎯→ 2AgBr(s) ∆H = –200 kJ (a) Calculate the heat change when 1 mole of silver bromide is formed from the reaction between silver and bromine. (b) How much heat is liberated when 84.6 g of AgBr is formed from silver and bromine. 5 Given: 4Al(s) + 3O2(g) ⎯→ 2Al2O3(s) ∆H = –3340 kJ (a) Calculate the heat change when 4.86 g of aluminium is burned completely in excess oxygen. (b) Calculate the heat change when 3.6 g of aluminium oxide is formed from the above experiment. (c) How much aluminium has to be burned in excess oxygen to generate heat of 30 000 kJ? 6 When 16.24 g of ethene (C2H4) is burned in excess oxygen, the heat released raises the temperature of 5 kg of water by 39.8 °C. (a) Calculate the heat released in the experiment. (b) Calculate the enthalpy change (in kJ) of the following reaction: 2C2H4(g) + 6O2(g) ⎯→ 4CO2(g) + 4H2O(l) [Specific heat of water = 4.18 J g–1 K–1] Standard Enthalpy Change of Formation, ∆Ho f 1 The standard enthalpy change of formation of a substance is the heat change when one mole of the substance is formed from its constituent elements under standard conditions. 2 For example, the standard enthalpy change of formation of carbon dioxide is –394 kJ mol–1. This means that when one mole of carbon dioxide gas is formed from the combination of its elements (carbon and oxygen), 394 kJ of heat energy is liberated under standard conditions (298 K and 101 kPa): C(s) + O2(g) ⎯→ CO2(g) ∆Ho f = –394 kJ mol–1 3 The standard enthalpy change of formation of aluminium oxide (–1670 kJ mol–1) refers to the process: 2Al(s) + 3 ––2 O2(g) ⎯→ Al2O3(s) ∆Ho f = –1670 kJ mol–1 4 The standard enthalpy change of formation can have either positive or negative values. For example, H2(g) + 1 ––2 O2(g) ⎯→ H2O(l) ∆Ho = –286 kJ mol–1 2C(s) + 2H2(g) ⎯→ C2H4(g) ∆Ho = +53 kJ mol–1 5 The standard enthalpy change of formation of elements is zero. H2(g) ⎯→ H2(g) ∆Ho = 0 Na(s) ⎯→ Na(s) ∆Ho = 0 Definition of standard enthalpy change of formation ∆H o f Elements ⎯⎯→ compounds The carbon in this case refers to graphite. ∆Ho f of elements = 0 2013/P2/Q18 2017/P2/Q2 2011/P1/Q19 2016/P2/Q16
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 6 7 6 However, for elements that exist in more than one allotropic form, only the most stable allotrope under standard conditions is given ‘zero’ standard enthalpy change of formation. 7 For example, the two common allotropes of carbon are graphite and diamond. Under standard conditions, graphite is the stable allotrope. Its standard enthalpy change of formation is zero, but the standard enthalpy change of formation of diamond is +2 kJ mol–1. Allotrope Graphite Diamond ∆Ho f /kJ mol–1 0 +2.0 The standard enthalpy of formation of diamond refers to the following process: C(graphite) ⎯→ C(diamond) ∆Ho = +2 kJ mol–1 8 The standard enthalpy change of formation of some substances cannot be determined directly from experiments. Such standard enthalpy change of formation is calculated from other known enthalpy changes using Hess’ law. (Refer to Section 7.2) Example 7.3 Write thermodynamic equations to represent the standard enthalpy change of formation for the following substances. (a) NH3 (–46.1 kJ mol–1) (b) NO (+90.2 kJ mol–1) (c) Al2O3 (–1700 kJ mol–1) Solution (a) 1 ––2 N2(g) + 3 ––2 H2(g) ⎯→ NH3(g) (b) 1 ––2 N2(g) + 1 ––2 O2(g) ⎯→ NO(g) (c) 2Al(s) + 3 ––2 O2(g) ⎯→ Al2O3(s) Info Chem The ∆H o f of Al3+(aq) ion is –531.0 kJ mol–1. This refers to the following process: Al(s) + aq ⎯→ Al3+(aq) + 3e– ∆H o = –531.0 kJ mol–1 Graphite is the stable allotrope. Quick Check 7.2 1 Write thermodynamic equations to represent the standard enthalpy change of formation for the following substances. (a) CaCO3 (–1207 kJ mol–1) (e) HI (+26 kJ mol–1) (b) CuSO4•5H2O (–2278 kJ mol–1) (f) CCl4 (–107 kJ mol–1) (c) CH3COOH (–484.5 kJ mol–1) (g) Cd(CN)2 (+162.3 kJ mol–1) (d) NO2 (+34 kJ mol–1) Standard Enthalpy Change of Formation and Enthalpy Change of Reaction 1 We can use the standard enthalpy change of formation to calculate the enthalpy change of a reaction using the following equation: ∆Ho reaction = [sum of ∆Ho f of the products] – [sum of ∆Ho f of the reactants] 2013/P2/Q1 2008/P1/Q18
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 7 7 Quick Check 7.3 1 Calculate the standard enthalpy change for the reaction: CH3COOH(l) + 2O2(g) → 2CO2(g) + 2H2O Given: ∆Ho f of CO2 = –394 kJ mol–1 ∆Ho f of H2O = –286 kJ mol–1 ∆Ho f of CH3COOH = –484 kJ mol–1 2 The ∆Ho f of Ag+(g) and Cl– (g) are +105 kJ mol–1 and –167 kJ mol–1 respectively. Calculate the ∆Ho f of AgCl from the equation: Ag+(g) + Cl– (g) ⎯→ AgCl(s) ∆Ho = –65 kJ Standard Enthalpy Change of Formation and Stability 1 Generally, substances with large negative standard enthalpy change of formation are more stable than those with large positive standard enthalpy change of formation. 2 The standard enthalpy change of formation of water is –286 kJ mol–1: H2(g) + 1 ––2 O2(g) ⎯→ H2O(l) ∆Ho f = –286 kJ mol–1 This indicates that water is energetically more stable than its elements hydrogen and oxygen. 3 On the other hand, the standard enthalpy change of formation of nitrogen trichloride, NCl3, is +230 kJ mol–1. 1 ––2 N2(g) + 3 ––2 Cl2(g) ⎯→ NCl3(l) ∆Ho f = +230 kJ mol–1 Nitrogen trichloride is energetically less stable than its elements nitrogen and chlorine. 4 The table below lists the standard enthalpy change of formation of the hydrides of chlorine, bromine and iodine. Compound HCl HBr HI ∆Ho f /kJ mol–1 –92.0 –36.0 +26.5 5 From the standard enthalpy change of formation above, we can predict that HI is the least stable of the three hydrides followed by HBr and HCl. This indeed is the case if we look at the percentage dissociation of the hydrides into their constituent elements at different temperatures. Temperature/°C Percentage dissociation HCl HBr Hl 200 – – 13 1000 0.01 0.4 25 2000 0.4 4.2 30 The more negative the ∆Ho f, the more stable the substance. 2 For example, we can calculate the standard enthalpy change of the following reaction: Fe2O3(s) + 2Al(s) ⎯→ Al2O3(s) + 2Fe(s) if we are given the standard enthalpy change of formation of Al2O3 (–1670 kJ mol–1) and Fe2O3 (–822 kJ mol–1): Fe2O3(s) + 2Al(s) ⎯→ Al2O3(s) + 2Fe(s) ∆Ho f : –822 0 –1670 0 Hence, ∆Ho reaction = (–1670 + 0) – (–822 + 0) = –848 kJ Info Chem Under standard conditions, water will not decompose spontaneously to hydrogen and oxygen. 2Al + 2Fe + —3 2 O2 0 –822 –848 –1670 Fe2O3 + 2Al Al2O3 + 2Fe
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 8 7 Standard Enthalpy Change of Combustion, ∆Ho c 1 The standard enthalpy change of combustion is the heat released when one mole of a substance is burned in excess oxygen under standard conditions. 2 For example, the standard enthalpy change of combustion of ethanol is –1367 kJ mol–1. This means that when 1 mole of ethanol is burned completely in excess oxygen (or air), 1367 kJ of heat is released: C2H5OH(l) + 3O2(g) ⎯→ 2CO2(g) + 3H2O(l) ∆Ho = –1367 kJ mol–1 3 The standard enthalpy change of combustion of some substances is given in the table below. Substance ∆Ho c /kJ mol–1 Carbon (graphite) –394 Hydrogen –286 Benzene –3267 Hexane –4163 Aluminium –835 Magnesium –602 Glucose –2820 Note that the the standard enthalpy change of combustion is always exothermic. 2011/P1/Q45 Definition of standard enthalpy change of combustion Exam Tips Exam Tips The standard enthalpy of combustion of graphite is also the standard enthalpy of formation of carbon dioxide. C(s) + O2(g) → CO2(g) ∆H = –394 kJ mol–1 Quick Check 7.4 1 Write balanced equations to represent the standard enthalpy change of combustion for the following substances: (a) Sodium (c) Sulphur dioxide (b) Oxalic acid, H2C2O4 (d) Ammonia 2 When 1.00 g of methanol, CH3OH, is burned completely in oxygen, 22.5 kJ of heat is released. Calculate the standard enthalpy change of combustion of methanol. 3 The standard enthalpy change of combustion of ethanol is –1367 kJ mol–1. In an experiment, 1.25 g of ethanol is burned and the heat released is used to heat up 100 g of water initially at 25 °C. If the process is 75% efficient, calculate the final temperature of the water. [Specific heat capacity of water = 4.18 J g–1 K–1] 4 When 0.14 g of magnesium is burned in a calorimeter containing 300 g of water, the temperature of water rose from 25 °C to 26.1 °C. If the heat capacity of the calorimeter is 1.77 kJ K–1, calculate (a) the heat released in kJ per gram of magnesium. (b) the heat released in kJ mol–1 of magnesium. [Specific heat capacity of water = 4.18 J g–1 K–1] 5 When 11.5 g of ethanol is completely burned in excess oxygen, the heat released raises the temperature of 1.6 kg of water from 28 °C to 78 °C. [Specific heat of water = 4.18 J mol–1 K–1] (a) Write a thermochemical equation to represent the enthalpy of combustion of ethanol. (b) Calculate the standard enthalpy of combustion of ethanol. (c) How much heat would be released when 1 kg of ethanol is completely burned in excess oxygen?
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 9 7 Standard Enthalpy Change of Neutralisation 1 Neutralisation is a reaction between an acid and a base to form salt and water. For example, NaOH(aq) + HCl(aq) ⎯→ NaCl(aq) + H2O(l) CH3COOH(aq) + KOH(aq) ⎯→ CH3COOK(aq) + H2O(l) 2 During neutralisation, heat is released. 3 The standard enthalpy change of neutralisation is defined as the heat energy released when one mole of water is formed from the reaction between an acid and a base under standard conditions. 4 The standard enthalpy change of neutralisation between sodium hydroxide and hydrochloric acid is –57.3 kJ mol–1. This heat change refers to the reaction: NaOH(aq) + HCl(aq) ⎯→ NaCl(aq) + H2O(l) ∆Ho = –57.3 kJ mol–1 5 The table below lists the standard enthalpy of neutralisation between different pairs of acids and bases. Acid HCl HCl HNO3 HNO3 CH3COOH* HF* HCl H3SO4 Base NaOH KOH NaOH KOH NaOH NaOH NH3* NaOH ∆Ho/kJ mol–1 –57.3 –57.3 –57.3 –57.3 –56.1 –68.6 –53.4 –66.8 [* Either weak acids or weak bases] Example 7.4 100 cm3 of 1.00 mol dm–3 NaOH is added to 100 cm3 of 1.00 mol dm–3 CH3COOH in a calorimeter. The temperature increases by 6.7 °C. (a) Calculate the standard enthalpy of neutralisation between NaOH and CH3COOH. (b) Calculate the temperature change when 250 cm3 of 1.00 mol dm–3 NaOH is added to 100 cm3 of 1.50 M CH3COOH. [Density of solution = 1.00 g cm–3. Specific heat of solution = 4.2 J g–1 °C– ] Solution (a) Heat change in the experiment = mc∆T = (100 + 100) × 4.2 × 6.7 = 5628 J No. of moles of NaOH= No. of moles of CH3COOH = ——100 1000 × 1 = 0.10 mol No. of moles of H2O formed = 0.10 mol ∴ Heat change per mole of water = ——5628 0.10 × 10–3 kJ = 56.28 kJ ∴ Standard enthalpy of neutralisation = –56.28 kJ mol–1 (b) No. of moles of NaOH = ——250 1000 × 1.0 = 0.25 mol No. of moles of CH3COOH = ——250 1000 × 1.5 = 0.15 mol The number of moles of water formed = 0.15 mol Hence, heat released = 0.15 × (56.28 × 103 ) J = 8442 J Using: q = mc∆T 8442 = (100 + 150) × 4.18 × T ∴ T = 8.1 °C Definition of standard enthalpy change of neutralisation Calculate the number of moles of water formed. 2014/P2/Q2 Info Chem Enthalpy of neutralisation is always negative
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 10 7 Quick Check 7.5 1 125 cm3 of 0.10 M NaOH is added to 100 cm3 of 0.10 M HNO3 initially at 28.5 °C. Calculate the final temperature of the mixture. [∆Ho neutralisation (NaOH + HCl) = –57.3 kJ mol–1. Specific heat capacity of the solution = 4.2 J g–1 K–1] 2 50.0 cm3 of 0.50 M sodium hydroxide is added to 50.0 cm3 of 0.50 M ethanoic acid in a plastic cup initially at 28.4 °C. The highest temperature recorded is 31.8 °C. (a) Write a balanced equation for the reaction between sodium hydroxide and ethanoic acid. (b) Calculate the amount of heat change during the experiment. (c) Calculate the standard enthalpy of neutralisation between sodium hydroxide and ethanoic acid. [Density of the solution = 1.00 g cm–3; specific heat capacity of the solution = 4.18 J g–1 K–1] Standard Enthalpy of Neutralisation between Strong Acids and Strong Bases 1 The standard enthalpy of neutralisation between strong acids and strong bases is a constant (–57.3 kJ mol–1). 2 This is because all strong acids and strong bases dissociate completely in water to form aqueous ions: NaOH(aq) ⎯→ Na+(aq) + OH– (aq) KOH(aq) ⎯→ K+(aq) + OH– (aq) HCl(aq) ⎯→ H+(aq) + Cl– (aq) HNO3(aq) ⎯→ H+(aq) + NO3 – (aq) 3 As a result, neutralisation between strong acids and strong bases involve the same reaction, which is the combination of H+(aq) and OH– (aq) to form H2O. The other ions present are just spectator ions (in bold face) that do not take part in the reactions. Na+(aq) + OH– (aq) + H+(aq) + Cl– (aq) →Na+(aq) + Cl– (aq) + H2O(l) K+(aq) + OH– (aq) + H+(aq) + NO3 – (aq) →K+(aq) + NO3 – (aq) + H2O(l) The overall reaction is: H+(aq) + OH– (aq) ⎯→ H2O(l) Hence, the heat released is the same. 4 However, the standard enthalpy of neutralisation between sulphuric acid and sodium hydroxide is more exothermic than expected. 1 ––2 SO4 2–(aq) + H+(aq) + OH– (aq) + Na+(aq) → Na+(aq) + 1 ––2 SO4 2–(aq) + H2O(l) ∆Ho = –66.8 kJ mol–1 5 The heat released when 1 mole of water is formed from the above reaction should be 57.3 kJ mol–1. So, where does the extra 9.5 kJ mol–1 comes from? 6 This extra heat released is the enthalpy change of dilution of sulphuric acid. 7 When aqueous sodium hydroxide is added to sulphuric acid, the acid is diluted in the process. Spectator ions Na+ and SO4 2– are spectator ions. Heat of dilution of sulphuric acid is significant
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 11 7 8 The heat released due to dilution for sulphuric acid is significant. This makes the standard enthalpy change of dilution more exothermic. [The ∆Ho for the reaction = heat of neutralisation + heat of dilution] 9 Heat of dilution of other acids is insignificant and is ignored. Standard Enthalpy of Neutralisation Involving Weak Acids or Weak Bases 1 The standard enthalpy of neutralisation involving weak acids or weak bases is less than 57.3 kJ mol–1. For example: CH3COOH(aq) + NaOH(aq) ⎯→ CH3COONa(aq) + H2O(l) ∆Ho = –56.1 kJ mol–1 NH3•H2O (aq) + HCl(aq) ⎯→ NH4Cl(aq) + H2O(l) ∆Ho = –53.4 kJ mol–1 NaOH(aq) + HCN(aq) ⎯→ NaCN(aq) + H2O(l) ∆Ho = –11.7 kJ mol–1 2 This is because weak acids are only partially dissociated in water. For example, CH3COOH(aq) CH3COO– (aq) + H+(aq) On addition of a strong base such as NaOH, the OH– from NaOH will react with the H+ from the dissociation of CH3COOH. H+(aq) + OH– (aq) ⎯→ H2O(l) ∆Ho = negative 3 Removal of H+ causes the above equilibrium to shift to the right and more CH3COOH molecules dissociate. However, the dissociation process requires absorption of heat energy. CH3COOH(aq) ⎯→ CH3COO– (aq) + H+(aq) ∆Ho = positive As a result, the total heat released is less than expected. 4 Similarly, for the case of ammonia, a weak base: NH3(aq) + H2O(l) NH4 +(aq) + OH– (aq) On addition of a strong acid such as HCl, the H+ from HCl will react with OH– from ammonia to form water. OH– (aq) + H+(aq) ⎯→ H2O(l) ∆Ho = negative 5 Removal of OH– causes the above equilibrium to shift to the right and more ammonia molecules dissociate. However, this dissociation requires absorption of energy. NH3(aq) + H2O(l) ⎯→ NH4 +(aq) + OH– (aq) ∆Ho = positive As a result, the total heat released is less than expected. Standard Enthalpy of Neutralisation Involving Hydrofluoric Acid, HF 1 Hydrofluoric acid, HF, is a weak acid that dissociates partially in water. HF(aq) H+(aq) + F– (aq) 2 Hence, we would expect the standard enthalpy of neutralisation between HF and a strong base such as NaOH to be less than 57.3 kJ mol–1. 3 However, the experimental standard enthalpy of neutralisation between NaOH and HF is more negative than expected. HF(aq) + NaOH(aq) ⎯→ NaF(aq) + H2O(l) ∆Ho = –68.6 kJ mol–1 This is in contradiction to our discussions above. Part of the heat released is reabsorbed to cause dissociation of the weak acids. 2008/P1/Q19 The degree of dissociation of a 0.1 M CH3COOH is about 1.6%.
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 12 7 4 This is because the dissociation of hydrogen fluoride molecule is an exothermic process. 5 Dissociation of HF involves the breaking of the H—F bond and energy needs to be absorbed. However, the F– ions, due to its small size, has very high hydration energy. F– (g) + aq ⎯→ F– (aq) ∆Ho = –506 kJ mol–1 This makes the whole process of dissociation exothermic. HF(aq) ⎯→ H+(aq) + F– (aq) ∆Ho = negative As a result, the standard enthalpy change of neutralisation is more negative than expected. Standard Enthalpy Change of Atomisation, ∆Ho atm 1 The standard enthalpy change of atomisation of an element is the enthalpy change when one mole of free gaseous atom is formed from the element under standard conditions. 2 The standard enthalpy change of atomisation of sodium is +107 kJ mol–1. This refers to the following process: Na(s) ⎯→ Na(g) ∆Ho = +107 kJ mol–1 3 The standard enthalpy change of atomisation of chlorine is +121 kJ mol–1. This refers to: 1 ––2 Cl2(g) ⎯→ Cl(g) ∆Ho = +121 kJ mol–1 4 The table below lists the standard enthalpy change of atomisation of some elements. Process ∆Ho atm /kJ mol–1 —1 2 H2(g) → H(g) +218.0 —1 2 l2(s) → l(g) +106.0 —1 2 O2(g) → O(g) +247.0 —1 2 N2(g) → N(g) +472.8 C(s, graphite) → C(g) +716.7 C(s, diamond) → C(g) +714.8 Na(s) → Na(g) +107.0 Mg(s) → Mg(g) +147.7 Al(s) → Al(g) +326.4 Hg(l) → Hg(g) +61.3 5 The standard enthalpy change of atomisation of the Noble gases (He, Ne, Ar, Kr, Xe and Rn) is zero, because all of them exist as monatomic gases at standard conditions. 6 The standard enthalpy change of atomisation of diatomic gases (e.g. O2, Cl2, N2) is equal to half the value of their bond energies. Cl—Cl(g) ⎯→ 2Cl(g) ∆Ho = +242 kJ mol–1 N N(g) ⎯→ 2N(g) ∆Ho = +945.6 kJ mol–1 O = O(g) ⎯→ 2O(g) ∆Ho =+494.0 kJ mol–1 The hydration energy of F – is highly exothermic. Info Chem The enthalpy change refers to one mole of chlorine atom (the product) and not one mole of chlorine molecule (the reactant). Standard enthalpy change of atomisation for monatomic gases is zero. 2008/P2/Q5(b) 2014/P2/Q18 2015/P2/Q1 Info Chem The standard enthalpy change of atomisation of a molecular compound is the enthalpy change when one mole of the compond is converted into its free gasesous atoms under standard conditions E.g. ∆Ho atm of silicon tetrachloride: SiCl4(l) → Si(g) + 4Cl(g)
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 13 7 7 The processes that are involved in the atomisation of a solid element comprise the following steps (sodium is used as an example): (a) Heat is absorbed by one mole of the solid to increase its temperature from 298 K to its melting point (371 K). (b) The enthalpy of fusion is absorbed (+2.6 kJ mol–1) to change one mole of the solid to liquid at its melting point. (c) Heat is absorbed to increase the temperature of one mole of the liquid sodium to its boiling point (1156 K). (d) The enthalpy of vaporisation is absorbed (+96.8 kJ mol–1) to change one mole of liquid sodium to vapour at its boiling point. (e) This is summarised in the diagram below: Na(g) Na(s) at 298 K Na(s) at melting point (371 K) Na(l) at 371 K Na(l) at boiling point (1156 K) Enthalpy of vaporisation (+96.8 kJ mol–1) Enthalpy change of atomisation Enthalpy of fusion (+2.6 kJ mol–1) Heat is absorbed Heat is absorbed Bond Energy 1 Bond energy is defined as the enthalpy change when one mole of a covalent bond is broken in the gaseous state. 2 For example, the bond energy of the H—H bond refers to the process: H—H(g) ⎯→ 2H(g) ∆Ho = +436 kJ mol–1 3 The table lists the bond energy of the covalent bonds in some diatomic molecules. Bond Bond energy/kJ mol–1 H—H +436 F—F +158 Cl—Cl +242 Br—Br +192 I—I +151 H—Cl +431 H—Br +366 H—I +299 O=O +496 NN +945 4 However, the same type of bond in different molecules can have different bond energies depending on the chemical environment of the bond. 2018/P2/Q2
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 14 7 5 As an example, consider water, H2O, which has two O—H bond. The energy required to break the two O—H bonds is given below: H—O—H(g) ⎯→ H(g) + O—H(g) ∆Ho = +499 kJ mol–1 O—H(g) ⎯→ O(g) + H(g) ∆Ho = +428 kJ mol–1 6 Similarly, the energy required to break the O—H bond in ethanol and ethanoic acid will not be the same as those in the H2O molecule. C2H5O—H(g) ⎯→ C2H5O(g) + H(g) CH3COO—H(g) ⎯→ CH3COO(g) + H(g) 7 For such cases, scientists use the average value for the same type of bond taken over a large number of different molecules. Examples are the C—C bond, N—H bond, C=O bond and many others. Bond Bond energy/kJ mol–1 CC +838 C=O +743 C=C +612 O—H +463 C—H +413 N—H +388 C—O +358 C—C +348 C—Cl +338 C—I +238 C—Br +176 N—N +163 8 Generally, (a) the bigger the size of the atoms, the weaker the bond, (b) a triple bond is stronger than a double bond, which in turn is stronger than a single bond. 9 We can use the bond energy to calculate the enthalpy change of a reaction and also to predict which bond(s) will break first in a chemical reaction. Example 7.5 The enthalpy of atomisation of methane, CH4, is +1663.5 kJ mol–1. (a) Write an equation to represent the heat of atomisation of methane. (b) Calculate the average C—H bond energy. Solution (a) CH4(g) ⎯→ C(g) + 4H(g) ∆Ho = +1663.5 kJ mol–1 (b) In the atomisation of methane, four C—H bonds are broken. ∴ Average C—H bond energy = 1663.5 ——– 4 = +415.9 kJ mol–1
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 15 7 Example 7.6 Use the bond energies in the table given in the text to calculate the enthalpy change for the following reaction. N2(g) + 3H2(g) ⎯→ 2NH3(g) Solution Bond breaking: (1 × NN) + (3 × H—H) = +945 + 3(436) = +2253 kJ Bond forming: (6 × N—H)= –6 × (388) = –2328 kJ [Note: Heat is released when a covalent bond is formed.] Net heat change = +2253 + (–2328) = –75 kJ Example 7.7 Methane reacts with chlorine in the presence of ultraviolet light to produce chloromethane. CH4(g) + Cl2(g) ⎯→ CH3Cl(g) + HCl(g) The two initiation steps proposed are: CH4(g) ⎯→ CH3(g) + H(g) (i) Cl2(g) ⎯→ 2Cl(g) (ii) Determine the most probable initiation step. Solution Step (i) involves the breaking of a C—H bond. CH4(g) ⎯→ CH3(g) + H(g) ∆Ho = +413 kJ mol–1 Step (ii) involves the breaking of a Cl—Cl bond. Cl—Cl(g) ⎯→ 2Cl(g) ∆Ho = +242 kJ mol–1 Since step (ii) requires less energy than step (i), the most probable initiation step is step (ii). Example 7.8 Calculate the enthalpy change for the following process: C(s) + 2H2(g) + 1 ––2 O2(g) ⎯→ CH3OH(g) Given: Element Carbon Hydrogen Oxygen ∆Hatomisation/kJ mol–1 +716 +217 +247 Bond C—H C—O O—H Bond energy/kJ mol–1 +413 +358 +463
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 16 7 Solution C(g) + 4H(g) + O(g) 1 C(s) + 2H2(g) + —O2(g) 2 716 + 4(217) + 247 3(C—H) + (C—O) + (O—H) ⎯→ CH3OH(g) X Based on Hess’ law: (Refer to Section 7.2) [716 + 4(217) + 247] – [3(413) + 358 + 463] = X X = –229 kJ mol–1 [Note: –229 kJ mol–1 is not the standard enthalpy of formation of methanol. This is because under standard conditions, methanol is a liquid and not a gas/vapour.] Standard Ionisation Energy 1 If sufficient energy is supplied to an electron in an atom or ion, it is possible to remove that electron completely from the attraction of the nucleus. We say that the electron is being ionised. The energy required for the process is called the ionisation energy. 2 The first ionisation energy (1st I.E.) is the minimum energy required to remove the most loosely held electron from a free gaseous atom to form a unipositive ion, per mole of atoms, under standard conditions. M(g) ⎯→ M+(g) + e– ∆Ho = 1st I.E. For example, Na(g) ⎯→ Na+(g) + e– ∆Ho = +496 kJ mol–1 Mg(g) ⎯→ Mg+(g) + e– ∆Ho = +738 kJ mol–1 Cl(g) ⎯→ Cl+(g) + e– ∆Ho = +1260 kJ mol–1 3 The second ionisation energy (2nd I.E.) is the minimum energy required to remove the most loosely held electron from a gaseous unipositive ion to form a dipositive ion, per mole of the atom, under standard conditions. M+(g) ⎯→ M2+(g) + e– ∆Ho = 2nd I.E. For example, Na+(g) ⎯→ Na2+(g) + e– ∆Ho = +4563 kJ mol–1 Mg+(g) ⎯→ Mg2+(g) + e– ∆Ho = +1451 kJ mol–1 Cl+(g) ⎯→ Cl2+(g) + e– ∆Ho = +2300 kJ mol–1 4 Higher ionisation energies are defined in the same manner. For example, the 4th ionisation energy of magnesium refers to the process: Mg3+(g) ⎯→ Mg4+(g) + e– ∆Ho = +10 500 kJ mol–1 5 Note that the 4th ionisation energy of magnesium is not: Mg(s) ⎯→ Mg4+(g) + 4e– The above process actually involves the following steps: [Heat of atomisation] + [1st + 2nd + 3rd + 4th ionisation energy] Ionisation energy is the minimum energy required to remove the most loosely held electron from a gaseous atom or a gaseous ion, per mole of the atom or ion, under standard conditions.
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 17 7 Example 7.9 Write equations to represent the following processes: (a) The 3rd ionisation energy of chlorine (b) The 14th ionisation energy of bromine (c) The sum of the 4th and 5th ionisation energy of calcium Solution (a) Cl2+(g) ⎯→ Cl3+(g) + e– (b) Br13+(g) ⎯→ Br14+(g) + e– (c) Ca3+(g) ⎯→ Ca5+(g) + 2e– Electron Affinity 1 When an electron is added to an atom, a negative charged ion (called anion) is formed. The enthalpy change for this process is called the electron affinity (EA). 2 The first electron affinity is the enthalpy change when every atom in one mole of free gaseous atom accepts an electron to form a uninegative ion under standard conditions. M(g) + e– ⎯→ M – (g) ∆Ho = 1st EA For example, H(g) + e– ⎯→ H– (g) ∆Ho = –72.8 kJ mol–1 Cl(g) + e– ⎯→ Cl– (g) ∆Ho = –364.0 kJ mol–1 O(g) + e– ⎯→ O– (g) ∆Ho = –141.1 kJ mol–1 S(g) + e– ⎯→ S– (g) ∆Ho = –200 kJ mol–1 3 The second electron affinity is defined in the same way: M– (g) + e– ⎯→ M2–(g) ∆Ho = 2nd EA Second and higher electron affinities are all positive. For example, O– (g) + e– ⎯→ O2–(g) ∆Ho = +844 kJ mol–1 S– (g) + e– ⎯→ S2–(g) ∆Ho = +532 kJ mol–1 4 This is because energy is required to overcome the repulsion between the negative charge on the ion and the approaching electron. 5 Electron affinity should not be confused with electronegativity. The latter refers to the tendency of an atom to attract the bonding electrons in a covalent bond to which the atom is bonded. Standard Enthalpy Change of Hydration 1 The standard enthalpy change of hydration is the enthalpy change when one mole of free gaseous ions dissolves in water to form hydrated ions of infinite dilution under standard conditions. 2 Infinite dilution means there is no more enthalpy change when more water is added to the solution. Ions + water ⎯→ Ion(aq) ∆Ho = hydration energy 3 Water is a polar molecule. The positive end of the water molecule will be attracted to the anions, while the negative end of the water molecule will be attracted to the cations, to form ion-dipole attractive force. Definition of electron affinity Exam Tips Exam Tips The first electron affinity is always exothermic. e Repulsion S– 2007/P1/Q19 2018/P2/Q19(a) Definition of standard enthalpy change of hydration Exam Tips Exam Tips Second electron affinity and higher are all endothermic.
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 18 7 4 The dipole in the water molecule can be represented as shown in the diagram on the left. 5 The ion-dipole attraction can be visualised as in the diagrams below: δ+ + δ– δ– δ– δ– δ– δ– δ– δ– δ+ – δ+ δ+ δ+ δ+ δ+ δ+ δ+ δ+ Formation of these bonds is accompanied by release of energy. 6 For example, Na+(g) + aqueous ⎯→ Na+(aq) ∆Ho = –406 kJ mol–1 Cl– (g) + aqueous ⎯→ Cl– (aq) ∆Ho = –364 kJ mol–1 7 The magnitude of the hydration energy is a measure of the strength of the ion-dipole attraction. The stronger the attraction, the more exothermic is the hydration energy. 8 This in turn depends on the charge and size of the ions or the charge density ( Charge ————— Ionic radius) of the ions. 9 For example, Ion H+ K+ Na+ Li+ Ca2+ Mg2+ Al3+ Charge +1 +1 +1 +1 +2 +2 +3 Ionic radius/nm – 0.133 0.095 0.060 0.099 0.065 0.050 Charge density – 7.5 10.5 16.7 20.2 30.8 60.0 ∆Ho/kJ mol–1 –1090 –322 –406 –519 –1650 –1920 –4690 Ion F – Cl – Br – l– Ionic radius/nm 0.136 0.181 0.195 0.216 Charge density 7.4 5.5 5.1 4.6 ∆Ho/kJ mol–1 –506 –364 –335 –293 Standard Enthalpy Change of Solution, ∆Ho solution 1 A solution is a homogeneous mixture of two or more substances. The substance in larger quantity is the solvent, while those in lesser quantity is the solute. 2 For example, when sodium chloride dissolves in water to form aqueous sodium chloride, sodium chloride is the solute and water is the solvent. 3 The solubility of a solute in a solvent depends on: (a) the nature of the solute (ionic or covalent), (b) the nature of the solvent (polar or non-polar), (c) temperature, (d) pressure (if it involves gases). 4 For example, iodine (a non-polar covalent element) is sparingly soluble in water (a polar solvent) but dissolves completely in tetrachloromethane (a non-polar solvent). δ+ δ– Factors affecting the solubility of a substance
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 19 7 5 When a substance dissolves in water, heat energy is either absorbed or released. The heat energy change is called the standard enthalpy change of solution. 6 The standard enthalpy change of solution is the enthalpy change when one mole of a substance dissolves in water to form aqueous solution of infinite dilution under standard conditions. 7 The table below lists the standard enthalpy change of solution of some common substances. Substance ∆Ho solution/kJ mol–1 NaCl +6.0 NaBr –0.58 Nal –7.5 NaOH –44.5 CaCl2 –81.3 NH4Cl +15.2 CuSO4 –62.7 NH4NO3 +26.2 7.2 Hess’ Law 1 The total enthalpy change of a chemical reaction depends only on the difference between the enthalpy of the products and that of the reactants. It does not depend on how the reaction is completed. ∆H = H(products) – H(reactants) 2 This concept is given in Hess’ law which states that the overall enthalpy change of a reaction is independent of its pathway. 3 Consider the thermochemical cycle on the right. The cycle shows three different ways to convert substance W to Z. By Hess’ law: ∆H1 + ∆H3 = ∆H2 + ∆H4 = ∆H5 4 Hess’ law is especially useful to calculate enthalpy changes that cannot be measured directly from experiments. For example, lattice energy or the enthalpy of formation of certain substances. Example 7.10 Calculate the enthalpy change for the reaction: 2C(s) + H2(g) ⎯→ C2H2(g) from the following data. C(s) + O2(g) ⎯→ CO2(g) ∆Ho = –394 kJ H2(g) + 1 ––2 O2(g) ⎯→ H2O(l) ∆Ho = –286 kJ C2H2(g) + 5 ––2 O2(g) ⎯→ 2CO2(g) + H2O(l) ∆Ho = –1300 kJ Definition of standard enthalpy change of solution Info Chem Cold packs are used to reduce fever and to treat certain muscular discomfort. It contains ammonium nitrate and water. When the two substances are mixed, heat is absorbed from the affected region thus lowering the temperature. 2007/P2/Q6(b) 2010/P1/Q19 2013/P2/Q18 Hess’ law is derived from the law of conservation of energy which states that energy can neither be created nor destroyed. W X Y Z ∆H2 ∆H5 ∆H3 ∆H1 ∆H4
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 20 7 Solution The problem can be solved by two methods: Thermochemical cycle (using Hess’ law) or the algebraic method. (a) Thermochemical cycle method: x 2C(s) + H2(g) + 1 ––2 O2(g) ⎯→ C2H2(g) + 1 ––2 O2(g) 2(–394) + (–286) –1300 2CO2(g) + H2O(l) By Hess’ law: x + (–1300) = 2(–394) + (–286) x = +226 kJ (b) In the algebraic method, we arrange the thermodynamic equations given (reversing some if necessary) so that when they are added together yields the equation that we wanted. In the following example, the physical states are omitted for clarity. 2C + 2O2 ⎯→ 2CO2 –394 × 2 H2 + 1 ––2 O2 ⎯→ H2O –286 2CO2 + H2O ⎯→ C2H2 + 2 1 ––2 O2 +1300 —————————————————————— 2C(s) + H2(g) ⎯→ C2H2(g) ∆Ho = (–394 × 2) + (–286) + (1300) = +226 kJ Quick Check 7.6 1 Calculate the standard enthalpy change of the following reaction: C(s) + 1 ––2 O2(g) ⎯→ CO(g) Given: C(s) + O2(g) ⎯→ CO2(g) –394 kJ CO(g) + 1 ––2 O2(g) ⎯→ CO2(g) –284 kJ 2 Calculate the standard enthalpy change for the following reaction: 5C(s) + 5H2(g) ⎯→ C5H10(l) Given: C(s) + O2(g) ⎯→ CO2(g) –394 kJ H2(g) + 1 ––2 O2(g) ⎯→ H2O(l) –286 kJ C5H10(l) + —– 15 2 O2(g) ⎯→ 5CO2(g) + 5H2O(l) –3290 kJ 3 Given: C(graphite) + O2(g) ⎯→ CO2(g) –394 kJ C(diamond) + O2(g) ⎯→ CO2(g) –396 kJ (a) Draw an energy profile for both the reactions. (b) Calculate the standard enthalpy change for the reaction: C(diamond) ⎯→ C(graphite) (c) Which allotrope of carbon is more stable? INFO Hess’ Law
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 21 7 4 Calculate the ∆Ho for the reaction: 2Al(s) + Fe2O3(s) ⎯→ Al2O3(s) + 2Fe(s) Given: 2Fe(s) + 3 ––2 O2(g) ⎯→ Fe2O3(s) ∆Ho = –822 kJ 2Al(s) + 3 ––2 O2(g) ⎯→ Al2O3(s) ∆Ho = –1670 kJ 5 Calculate the standard enthalpy change of formation of benzene, C6H6, from the following data: ∆Ho f of CO2 = –394 kJ mol–1 ∆Ho f of H2O = –286 kJ mol–1 C6H6(l) + —– 15 2 O2(g) ⎯→ 6CO2(g) + 3H2O(l) ∆Ho = –3267 kJ 6 Ethanol has the molecular formula of C2H5OH. (a) Write a thermochemical equation to represent the formation of ethanol from its elements. (b) Calculate the standard enthalpy change of formation of ethanol using the following data: C(s) + O2(g) ⎯→ CO2(g) –394 kJ H2(g) + 1 ––2 O2(g) ⎯→ H2O(l) –286 kJ C2H5OH(l) + 3O2(g) ⎯→ 2CO2(g) + 3H2O(l) –1367 kJ 7 Glucose is a monosaccharide with the molecular formula of C6H12O6. Calculate the ∆Ho f of glucose from the following information. C(s) + O2(g) ⎯→ CO2(g) –394 kJ H2(g) + 1 ––2 O2(g) ⎯→ H2O(l) –286 kJ C6H12O6(s) + 6O2(g) ⎯→ 6CO2(g) + 6H2O(l) –2820 kJ 8 Hydrogen sulphide gas reduces sulphur dioxide gas according to the equation: 2H2S(g) + SO2(g) ⎯→ 3S(s) + 2H2O(l) Use the following data to calculate the enthalpy change of the above reaction. Compound H2S(g) SO2(g) H2O(l) ∆Ho f /kJ mol–1 –20.2 –197.0 –286.0 9 Under controlled conditions, ethanol (C2H5OH) can be oxidised by atmospheric oxygen to ethanal (CH3CHO). (a) Write a balanced equation for the atmospheric oxidation of ethanol. (b) Given the standard enthalpy of combustion of ethanol and ethanal are –1367.3 kJ mol–1 and –1167.1 kJ mol–1, calculate the enthalpy change for the oxidation process. 10 The standard enthalpies change of combustion of carbon, hydrogen and ethane-1,2-diol (CH2OHCH2OH) are –394 kJ mol–1, –286 kJ mol–1 and –1180 kJ mol–1. Calculate the standard enthalpy change of formation of ethane-1,2-diol. 7.3 Born-Haber Cycle Lattice Energy, ∆Ho lattice 1 In solid sodium chloride, strong ionic bonds hold the Na+ and Cl– ions together in a fixed orderly arrangement. 2 Energy is required to break the ionic bonds and separate the ions. Na+Cl– (s) ⎯→ Na+(g) + Cl– (g) ∆Ho = +770 kJ mol–1 + – – + + – + – – + – – + – – + – – + + – + + – + + – 2014/P2/Q1 2015/P2/Q2 2016/P2/Q2 2017/P2/Q1
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 22 7 3 Thus, when Na+ and Cl– ions come together to form solid sodium chloride, the same amount of energy must be released. Na+(g) + Cl– (g) ⎯→ Na+Cl– (s) ∆Ho = –770 kJ mol–1 This energy is called the lattice energy. 4 The lattice energy of an ionic compound is the heat energy released when one mole of the ionic solid is formed from its constituent gaseous ions under standard conditions. Example 7.11 The lattice energy of potassium chloride is –701 kJ mol–1. Write a thermochemical equation to represent this statement. Solution K+(g) + Cl– (g) ⎯→ KCl(s) ∆Ho = –701 kJ mol–1 Definition of lattice energy Quick Check 7.7 1 Write thermochemical equations to represent the following lattice energies. Compound ∆Ho lattice/kJ mol–1 (a) Sodium bromide –742 (b) Silver chloride –890 (c) Sodium oxide –2478 (d) Calcium fluoride –2600 (e) Aluminium oxide –15 920 Factors Affecting the Lattice Energy 1 The lattice energy is a measure of the strength of the ionic bonds in the ionic solid. 2 The stronger the ionic bonds, the more exothermic is the lattice energy. 3 The lattice energy depends on: (a) the charge on the ions, (b) the size of the ions. 4 The higher the charge on the ions, the stronger is the attraction between the ions. 5 The smaller the size of the ions, the shorter is the distance between the ions. This will increase the force of attraction between the ions. (charge on the cation) × (charge on the anion) Lattice energy ∝ ——————————————————— (cationic radius + anionic radius)2 6 The lattice energy of some ionic solids is given in the tables below: (a) Compound Charge on cation Charge on anion ∆Ho lattice /kJ mol–1 NaCl +1 –1 –770 Na2O +1 –2 –2478 MgO +2 –2 –3850 Al2O3 +3 –2 –15 920 2009/P1/Q19 2012/P1/Q19 2014/P2/Q3 VIDEO Lattice Energy
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 23 7 Quick Check 7.8 1 For each pair of the following ionic solids, suggest which will have a more exothermic lattice energy. (a) MgO and MgCl2 (c) Al2O3 and Al2S3 (e) NaCl and Na2O (b) MgSO4 and BaSO4 (d) LiF and KCl The Lattice Energy of Ionic Compounds 1 The lattice energy of an ionic compound (e.g. NaCl) cannot be determined directly from experiment because it would involve bringing separately one mole of Na+ gaseous ions and one mole of Cl– gaseous ions together and combining them to form NaCl solid. 2 Instead, the lattice energy is calculated from other known enthalpy changes (that can be determined from experiments) and by using Hess’ Law. 3 The thermochemical cycle used to determine the lattice energy is called the Born-Haber cycle. 4 Let us use sodium chloride as an example. The standard enthalpy change of formation of sodium chloride = –411 kJ mol–1. This refers to the following process: Na(s) + 1 ––2 Cl2(g) ⎯→ NaCl(s) ∆Ho = –411 kJ mol–1 This reaction can be assumed as comprising the following hypothetical steps. The actual reaction does not follow these individual steps. However, it makes our calculations easier to follow. (a) Atomisation of sodium: Na(s) ⎯→ Na(g) ∆Ho = +108 kJ mol–1 (b) Ionisation of sodium: Na(g) ⎯→ Na+(g) + e– ∆Ho = +494 kJ mol–1 (c) Atomisation of chlorine: 1 ––2 Cl2(g) ⎯→ Cl(g) ∆Ho = +121 kJ mol–1 (d) Formation of Cl– ion: first electron affinity Cl(g) + e– ⎯→ Cl– (g) ∆Ho = –364 kJ mol–1 (e) Combination of Na+(g) and Cl– (g) to form NaCl(s) Na+(g) + Cl– (g) ⎯→ NaCl(s) ∆Ho lattice The higher the charge on the ions, the more exothermic is the lattice energy. (b) Compound Cationic radius/nm Anionic radius/nm ∆Ho lattice /kJ mol–1 NaCl 95 181 –770 NaBr 95 196 –731 Nal 95 216 –684 The smaller the interionic distance, the more exothermic is the lattice energy. 2009/P2/Q5(b) 2014/P2/Q1 2017/P2/Q3
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 24 7 5 The Born-Haber cycle is constructed as follows: Na+ (g) + Cl(g) Na(s) + Cl 2 (g) Na(g) + Cl 2 (g) Na+ (g) + Cl 2 (g) +121 1 2 1 2 +494 +108 1 2 –411 Na+ Cl– (s) –364 ∆H(lattice) Na+ (g) + Cl – (g) Based on Hess’ law: (+108) + (+494) + (+121) + (–364) + ∆Ho lattice = –411 ∴ ∆Ho lattice = –770 kJ mol–1 6 Sometimes, the Born-Haber cycle can be drawn as shown below: Cl Na+ (g) Na(g) +494 1 2 +108 –411 ∆Ho lattice Na+ – Na(s) (s) Cl– (g) –364 +121 Cl(g) Cl(g) Simplified Born-Haber cycle: Cation + anion Several steps Lattice energy ∆Hformation Elements Ionic solid Example 7.12 Construct a Born-Haber cycle for the formation of calcium oxide. Name all the enthalpy changes in your diagram. Solution Ca2+(g) + O(g) Ca(g) + O2 (g) Ca+ (g) + O2 (g) Ca2+(g) + O2 (g) D 1 2 1 2 C B 1 2 Ca2+(g) + O2–(g) Ca(s) + O2 (g) A 1 2 H Ca2+O2–(s) E Ca2+(g) + O– (g) F G
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 25 7 Or: Ca2+(g) + O2–(g) C F Ca+(g) + O– (g) B E Ca(g) + O(g) CaO(s) A D 1 Ca(s) + —O2(g) 2 H G Process Enthalpy change A Atomisation of Ca B First ionisation energy of Ca C Second ionisation energy of Ca D Atomisation of oxygen E First electron affinity of oxygen F Second affinity electron of oxygen G Lattice energy H Enthalpy of formation Quick Check 7.9 1 Construct the Born-Haber cycle for barium chloride and use it to determine the lattice energy of barium chloride. Given: Process ∆Ho/kJ mol–1 Atomisation of Ba +190 First ionisation energy of Ba +502 Second ionisation energy of Ba +966 Atomisation of chlorine +121 First electron affinity of chlorine –354 Formation of barium chloride –858.6 2 Use the following data to calculate the lattice energy of potassium hydride, KH. Process ∆Ho/kJ mol–1 Atomisation of K +90 First ionisation energy of K +420 Atomisation of hydrogen +218 First electron affinity of hydrogen –78 Formation of potassium hydride +156
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 26 7 3 Use the following data to calculate the first electron affinity of fluorine. Process ∆Ho/kJ mol–1 Atomisation of sodium +108 First ionisation energy of sodium +494 Atomisation of fluorine +158 Lattice energy of NaF –930 Enthalpy of formation of NaF –495 Lattice Energy and Covalent Character 1 The lattice energy of an ionic compound can be calculated theoretically by assuming that the bonds in the solid are pure ionic, and the ions exist as perfect spheres and taking into consideration of the electrostatic forces of attraction and repulsion. 2 The table below lists the theoretical and the experimental lattice energies of some sodium halides and silver halides. (The experimental values are determined from Born-Haber cycles). Compound NaCl NaBr AgCl AgBr Experimental value/kJ mol–1 –775 –740 –910 –900 Theoretical value/kJ mol–1 –774 –738 –770 –755 3 There is very close agreement between the experimental and theoretical values for the sodium halides. 4 However, there is significant difference between the experimental and theoretical values for the silver halides. 5 This shows that the assumption that sodium halides are pure ionic is correct, but not so for the silver halides. 6 Looking at the electronegativities of sodium and the halogens: Element Na Cl Br Electronegativity 0.9 3.0 2.8 We find that the difference in electronegativity between sodium and the halogen is large. Compound NaCl NaBr Difference in electronegativity 2.1 1.9 Hence, the bonds in the sodium halides are predominantly ionic. 7 However, if we look at the electronegativities of silver and the halogens: Element Ag Cl Br Electronegativity 1.9 3.0 2.8 The difference between silver and the halogen is small. Compound AgCl AgBr Difference in electronegativity 1.1 0.9 2018/P2/Q19(b)(ii) Ions as perfect spheres with even distribution of electron density Cation Anion Theoretical lattice energy assumed that the compound is 100% ionic. Info Chem Ionic bonds are formed between elements with a large difference in their electronegativities. Covalent bonds are formed between elements with a small difference in their electronegativities.
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 27 7 Hence, the bonds in the silver halides have significant covalent character. This causes the discrepancies in the theoretical and experimental values. Ag+ Cl– 7.4 The Solubility of Solids in Liquids 1 The dissolution of an ionic solid in water can be visualised as occurring in two hypothetical steps. Take the example of sodium chloride. NaCl(s) + aq ⎯→ Na+(aq) + Cl– (aq) ∆Ho = +6.0 kJ mol–1 (a) The first step involves the breaking of the ionic bonds that hold the ions together in the solid. The magnitude of the energy required is numerically the same as the lattice energy. –Lattice energy Na+Cl– (s) ⎯⎯⎯⎯⎯→ Na+(g) + Cl– (g) ∆Ho = +776 kJ mol–1 (b) The second step involves the hydration of the ions to form hydrated ions. The energy involved is the hydration energy of the cations and that of the anions. Na+(g) + aq ⎯→ Na+(aq) ∆Ho = –406 kJ mol–1 Cl– (g) + aq ⎯→Cl– (aq) ∆Ho = –364 kJ mol–1 (c) The overall heat change is: NaCl(s) → Na+(aq) + Cl– (aq) ∆Ho = (+776) + (–406 – 364) = +6.0 kJ mol–1 (d) The process can be summarised by the following thermochemical cycle: Na+(g) + Cl– (g) –Lattice energy Hydration energy ∆H (solution) Na+Cl– (s) Na+(aq) + Cl– (aq) Based on Hess’ Law: (–Lattice energy) + (Hydration energy) = enthalpy of solution –(–776) + (–406 – 364) = ∆Ho solution ∴ Ho solution = +6 kJ mol–1 2 Generally, salts with large positive enthalpy change of solution are less soluble than those with large negative enthalpy change of solution. Significant amount of covalent characters in the silver halides Info Chem The energy involved is sometimes called the ‘lattice dissociation energy’. 2013/P2/Q3 2015/P2/Q1 2016/P2/Q1 2018/P2/Q19(b)(i)
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 28 7 Quick Check 7.10 1 Given: Lattice energy of NaI = –695 kJ mol–1 Hydration energy of Na+ = –406 kJ mol–1 Hydration energy of I– = –307 kJ mol–1 (a) Calculate the standard enthalpy change of a solution of sodium iodide. (b) Comment on the solubility of sodium iodide in water. 2 When 3.05 g of anhydrous calcium chloride is dissolved in 150 cm3 of water at 298 K, the temperature of the solution rises to 301.8 K. Calculate the standard enthalpy change of solution of calcium chloride. [Density of water = 1.00 g cm–3. Specific heat capacity of water = 4.18 J g–1 K–1] 3 The standard enthalpy change of solution of ammonium chloride is +15.2 kJ mol–1. 45 g of ammonium chloride is added to 250 g of water initially at 25.0 °C. Calculate the final temperature of water. 4 Given the following data: Lattice energy of AgCl = –910 kJ mol–1 Hydration energy of Ag+ = –465 kJ mol–1 Hydration energy of Cl– = –364 kJ mol–1 (a) Calculate the standard enthalpy change of solution of silver chloride. (b) Comment on the solubility of silver chloride based on your answer to (a). SUMMARY SUMMARY 1 Thermochemistry deals with heat energy changes during chemical reactions. 2 Reactions accompanied by an increase in heat energy are called endothermic reactions. 3 Reactions accompanied by a decrease in heat energy are called exothermic reaction. 4 Endothermic reactions are accompanied by a drop in temperature. 5 Exothermic reactions are accompanied by a rise in temperature. 6 Hess’ Law states that the overall enthalpy change of a reaction is independent of its pathway. 7 In calorimetry, the heat change of a reaction is calculated using the formula: q = mc∆T 8 The standard conditions for thermochemical measurements are: (a) temperature 298 K (b) pressure 1 atm or 101 kPa 9 The standard enthalpy of formation of a substance is the heat change when one mole of the substance is formed from its constituent elements under standard conditions. 10 The standard enthalpy change of combustion is the heat released when one mole of a substance is burned in excess oxygen under standard conditions. 11 The standard enthalpy change of neutralisation is defined as the heat energy released when one mole of water is formed from the reaction between an acid and a base. 12 The standard enthalpy change of atomisation of an element is the enthalpy change when one mole of free gaseous atoms is formed from the element under standard conditions. 13 The first ionisation energy (1st I.E) is the minimum energy required to remove the most loosely held electron from a free gaseous atom to form a unipositive ion, per mole of atoms, under standard conditions. 14 The first electron affinity is the enthalpy change when every atom in one mole of
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 29 7 free gaseous atom accepts an electron to form a uninegative ion under standard conditions. 15 The lattice energy of an ionic compound is the heat energy released when one mole of the ionic solid is formed from its constituent gaseous ions under standard conditions. 16 The standard enthalpy change of hydration is the enthalpy change when one mole of free gaseous ions dissolves in water to form hydrated ions of infinite dilution under standard conditions. 17 The standard enthalpy change of solution is the enthalpy change when one mole of a substance dissolves in water to form aqueous solution of infinite dilution under standard conditions. 18 The standard enthalpy change of a reaction can be calculated from the standard enthalpy change of formation using the following formula: ∆Ho = S∆Ho f of the products – S∆Ho f of the reactants 19 The factors that affect the lattice energy of ionic compounds are: (a) the charge on the ions, (b) the distance between ions. 20 The standard enthalpy change of hydration depends on: (a) the charge on the ion, (b) the size of the ion. 21 The standard enthalpy change of solution is given by: ∆Ho solution = – lattice energy + hydration energy STPM PRACTICE 7 Objective Questions 3 The standard enthalpy of formation of diamond is +2 kJ mol–1. What is the standard enthalpy of formation of graphite? A –2 kJ mol–1 C +2 kJ mol–1 B 0 D Insufficient data 4 Which of the following is expected to have the highest hydration energy? A Al3+ C Li+ B Be2+ D H 5 Which bond has the highest standard bond energy? A CO C CN B NN D CC 6 Which of the following statements regarding an exothermic reaction is not necessarily true? A The activation energy of the reverse reaction is higher than that of the forward reaction. B The temperature of the surroundings increases. C Heat is given out during the reaction. D It will happen spontaneously. 1 Given the following ∆Ho : S(s) + O2(g) → SO2(g) ∆Ho = –297 kJ 2S(s) + 3O2(g) → 2SO3(g) ∆Ho = –790 kJ What is the enthalpy change for the conversion of sulphur dioxide to sulphur trioxide? A –196 kJ mol–1 C +98 kJ mol–1 B –98 kJ mol–1 D +196 kJ mol–1 2 The interionic distances in sodium chloride and magnesium oxide are almost the same. However, the lattice energy of magnesium oxide is almost five times that of sodium chloride. This is because A the ionisation energy of magnesium is higher than that of sodium. B magnesium oxide consists of doubly charged ions whereas sodium chloride consists of singly charged ions. C sodium chloride is soluble in water but magnesium oxide is not. D magnesium oxide is more stable than sodium chloride.
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 30 7 7 Methane and octane are two important fuels. Which of the following statements is true when 1 gram each of methane and octane are burned separately in air? A The heat energy produced in both cases is the same. B The mass of water produced by the octane is less than that of methane. C The mass of oxygen consumed by methane is less than that of octane. D The mass of carbon dioxide produced by octane is less than that of methane. 8 Given the following: Enthalpy of formation of methane = –75.0 kJ mol–1 Enthalpy of atomisation of carbon = +715.0 kJ mol–1 Enthalpy of atomisation of hydrogen = +436.0 kJ mol–1 What is the mean C—H bond energy (in kJ mol–1) in the methane molecule? A –1662.0 C +415.5 B –415.5 D +1662.0 9 Calculate the standard enthalpy change for the following reaction: 3C2H2(g) → C6H6(l) Given that the standard enthalpies of combustion of C2H2 and C6H6 are –1300 and –3270 kJ mol–1. A –1970 C +630 B –630 D +1970 10 The enthalpies of solution of sodium chloride and silver chloride are +4.0 kJ mol-1 and +82 kJ mol-1. What could be deduced from this information? A NaCl is more stable towards heat than AgCl B NaCl is ionic but AgCl is covalent C NaCl is more soluble in water than AgCl D Ag+ has higher hydration energy than Na+ 11 Given the standard enthalpies of formation of iron(II) chloride and iron(III) chloride are –341 and –405 kJ mol–1 respectively. What is the standard enthalpy change for the following reaction? 2FeCl3(s) → 2FeCl2(s) + Cl2(g) A –128 C +64 B –64 D +128 12 Consider the following thermochemical equations: CH4 + 2O2 → CO2 + 2H2O ∆H = –890 kJ 2CH3OH + 3O2 → 2CO2 + 4H2O ∆H = –1450 kJ Which of the following statements is correct when 1 mole of methane and 1 mole of methanol were completely burnt in separate experiments? A Methane produces more heat energy. B Methanol requires more oxygen for combustion. C Methanol produces more water. D Methane produces more carbon dioxide. 13 Consider the following thermochemical equations: NH4NO3(s) + aq → NH4 +(aq) + NO3 – (aq) ∆H = + 25.0 kJ mol–1 CaSO4(s) + aq → Ca2+(aq) + SO4 2–(aq) ∆H = – 66.0 kJ mol–1 When a mixture of the two solids is added to 250 cm3 of water, the temperature of the water remains unchanged. If the mass of ammonium nitrate in the mixture is 0.514 g, what is the mass of calcium sulphate in the mixture? A 0.135 g C 0.514 g B 0.331 g D 1.331 g 14 The enthalpy change of the solution of NaOH is -44.5 kJ mol-1. Which enthalpy changes are involved in the solubility of NaOH? I Enthalpy of neutralisation II Lattice energy III Hydration energy IV Electron affinity V Enthalpy of formation A I and III C II and III B II, III and IV D I only 15 Which of the following processes is numerically the same as lattice energy but algebraically the opposite? A Na(s) + —1 2 Cl2(g) → NaCl(s) B KCl(s) → K+(aq) + Cl– (aq) C AgCl(s) → Ag+(g) + Cl– (g) D Cl2(g) → 2Cl– (g) 16 Calculate the standard enthalpy change of formation of iodine trichloride, ICl3, from the following data: I2(s) + Cl2(g) → 2ICl(s) ∆H = +14 kJ mol–1 ICl(s) + Cl2(g) → ICl3(s) ∆H = –88 kJ mol–1 A –60 kJ mol–1 C –81 kJ mol–1 B –74 kJ mol–1 D –162 kJ mol–1
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 31 7 17 The table below shows the standard enthalpy change of neutralisation for various pairs of acids and bases. Acid Base ∆Ho neutralisation/kJ mol–1 HCl KaOH –57.3 X NaOH –68.6 HNO3 Y –57.3 HCl Z –54.2 What are the identities of X, Y and Z ? X Y Z A H2SO4 NH3 NaOH B CH3COOH KOH NH3 C HF KOH NH3 D H2SO4 NaOH NH3 18 The Born-Haber cycle for the formation of calcium oxide is shown below: + Ca 2+ + O2- (g) C B A F E D G H Ca + + O- (g) Ca + O(g) Ca + 1 2 O2 (g) CaO(s) (s) A C F H A Enthalpy of atomisation Ionisation energy Electron affinity Enthalpy of formation B Enthalpy of atomisation Electron affinity Electron affinity Lattice energy C Enthalpy of vaporisation Ionisation energy Enthalpy of atomisation Enthalpy of combustion D Ionisation energy Electron affinity Electronegativity Enthalpy of formation 19 The standard enthalpy change of combustion of carbon (graphite) is –286 kJ mol–1. What is the enthalpy change of formation of carbon dioxide? A –572 kJ mol–1 C +286 kJ mol–1 B –286 kJ mol–1 D +572 kJ mol–1 20 For which of the following reactions is the enthalpy change equal to the heat of formation of the product? A 2C(s) + O2(g) → 2CO(g) B N2(g) + 3H2(g) → 2NH3(g) C H(g) + Cl(g) → HCl(g) D Si(s) + O2(g) → SiO2(s) 21 Given the following thermodynamic equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) ∆H = -2600 kJ Which deduction is not correct about the reaction? A The standard enthalpy change of combustion of C2H2 is -1300 kJ mol-1 B Presence of strong CC covalent bond in C2H2 C C2H2 is thermally less stable than CO2 and H2O D The enthalpy change for the following reaction is +2600 kJ 4CO2(g) + 2H2O(l) → 2C2H2(g) + 5O2(g) 22 Consider the reaction: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) Which of the following species has zero standard enthalpy change of formation? A NH3(g) C NO(g) B O2(g) D H2O(l) 23 The standard enthalpy change of formation of diamond is +2 kJ mol–1. What is the standard enthalpy change of formation of graphite? A –2 kJ mol–1 B 0 C +2 kJ mol–1 D Insufficient data 24 What is the enthalpy change for the following reaction? 3Fe2O3(s) + CO(g) → CO2(g) + 2Fe3O4 Given: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ∆H = –28.0 kJ 3Fe(s) + 4CO2(g) → 4CO(g) + Fe3O4(s) ∆H = +12.5 kJ A –59.0 kJ C +40.5 kJ B –15.5 kJ D +109 kJ 25 Given that the standard enthalpy of solution of ammonium chloride is +14.8 kJ mol-1. Calculate the change in temperature when 7.91 g of ammonium chloride is dissolved in 100 g of water. [Specific heat capacity of the solution is 4.2 J g-1 K-1] A +5.2 K C -5.2 °C B +5.2 °C D -5.2 K
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 32 7 26 The standard enthalpy of formation of CCl4 is –140 kJ mol-1 and the standard enthalpies of atomisation of C and CCl4 are +715 kJ mol-1 and +1304 kJ mol-1 respectively. What is the bond energy of Cl2 in kJ mol-1? A +449 C +224.5 B -449 D -224.5 27 The enthalpy of solution of sodium hydroxide is –44.4 kJ mol–1. When a sample of sodium hydroxide dissolves in 250 cm3 of water, the temperature of water changes by 14 K. What is the mass of sodium hydroxide used in the experiment? A 0.85 g C 10.8 g B 2.75 g D 13.9 g 28 Which of the following corresponds to the standard enthalpy change of formation of a substance? A C(graphite) → C(diamond) B H2(g) + O2(g) → H2O2(g) C Na(s) + Cl(g) → NaCl(s) D 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) 29 Which of the following equations corresponds to the lattice energy of lead(II) chloride? A Pb(s) + Cl2(g) → PbCl2(s) B Pb2+(g) + 2Cl- (g) → PbCl2(s) C Pb(s) + 2Cl(g) → PbCl2(s) D PbCl2(s) → Pb2+(g) + 2Cl- (g) 30 Calculate the enthalpy change of the following reaction: HCOOH(l) → CO2(g) + H2(g) Given the enthalpy of formation (in kJ mol–1) of HCOOH and CO2 are –409 kJ mol–1 and –394 kJ mol–1 respectively. A –803 C +15 B –15 D +803 31 The standard enthalpies of formation of graphite and diamond are 0 kJ mol–1 and +2 kJ mol–1. Which statement explains the difference in these values? A Diamond is hard while graphite is soft and slippery. B Diamond is more stable than graphite. C Graphite is a stable allotrope at room conditions. D The carbon atoms in graphite undergo sp2 hybridisation whereas the carbon atoms in diamond undergo sp3 hybridisation. 32 Which of the following compounds has the most negative lattice energy? A LiF C LiI B NaCl D KF 33 The standard enthalpies of reactions of several processes are given below: Process ∆H/kJ mol–1 Combustion of hydrogen –286 Combustion of ethyne, CHCH –1299 Combustion of ethane, C2H6 –1560 Hydrogenation of ethene, CH2=CH2 –157 What is the standard enthalpy change (in kJ mol–1) for the hydrogenation of ethyne to ethene? CH CH(g) + H2(g) → CH2 = CH2(g) A –132 C –468 B –154 D –3302 34 Consider the energy profile below: kJ mol–1 H2(g) + —1 2 O2(g) 0 H2O(g) x H2O(l) y H2O(s) z Which of the following statement(s) is/are correct? I The standard enthalpy of vaporisation of water = x – y. II The standard enthalpy of formation of water = –y. III The standard enthalpy of combustion of hydrogen = z. A I only C I and III B II only D II and III 35 The standard enthalpy of formation of CuSO4•5H2O is –x kJ mol–1. Which of the following equations corresponds to the formation of CuSO4•5H2O under standard conditions?
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 33 7 A CuSO4(s) + 5H2O(l) → CuSO4•5H2O(s) B CuSO4(s) + 5H2O(l) → CuSO4•5H2O(aq) C Cu(s) + S(s) + 9O(g) + 10H(g) → CuSO4•5H2O(s) D Cu(s) + S(s) + 9 ––2O2(g) + 5H2(g) → CuSO4•5H2O(s) 36 The standard enthalpy of combustion of carbon is –396 kJ mol–1. It can be concluded that I the standard enthalpy of atomisation of carbon is +396 kJ mol–1 II the standard enthalpy of formation of carbon dioxide is –396 kJ mol–1 III the carbon-oxygen bond energy is +——396 2 kJ mol–1 A I only B II only C III only D II and III 37 Which of the following enthalpy changes is always positive? A First electron affinity B Enthalpy of neutralisation C Enthalpy of atomisation D Lattice energy 38 The standard enthalpy of neutralisation between sodium hydroxide and hydrofluoric acid is –68.6 kJ mol–1. NaOH(aq) + HF(aq) → NaF(aq) + H2O(l) Whatis the estimated enthalpy of neutralisation (in kJ mol–1) between sodium hydroxide and ethanoic acid? A –56.1 C –57.3 B –68.6 D –116.6 39 Which compound has the highest lattice energy? A Al2(SO4)3 C NaCl B MgO D SiO2 Structured and Essay Questions 1 (a) Write a balanced equation to represent the lattice energy of silver bromide. (b) Given: Process ∆Ho/kJ mol–1 Atomisation of silver +285 Bond energy of bromine +224 First ionisation energy of silver +731 First electron affinity of bromine –325 Enthalpy change of formation of silver bromide –100 (i) Calculate the lattice energy of silver bromide. (ii) The theoretical lattice energy of silver bromide is less than that calculated in (b)(i) . Suggest why is it so. (iii) How would you compare the lattice energy of silver iodide to that of silver bromide? Explain your answer. (c) In black-and-white photography, a bromide ion in the silver bromide emulsion absorbs light energy and releases an electron which reduces a silver ion to silver atom. Br– → Br + e– Predict whether it would require more or less energy if a silver iodide emulsion is used instead of silver bromide. Explain your answer. 2 The thermochemical cycle for the formation of magnesium oxide is shown below: Mg2+(g) Mg(g) ∆H ∆H ∆H ∆H ∆H ∆H 2 1 2 1 6 5 Mg(s) MgO(s) O2–(g) 4 3 O2(g) O(g) (a) What is the name given to such a cycle?
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 34 7 (b) Name the enthalpy changes represented as ∆H1: ∆H4: ∆H2: ∆H5: ∆H3: ∆H6: (c) Using appropriate values from the Data Booklet and the following information, calculate ∆H5 for magnesium oxide. ∆H1 = +150 kJ mol–1 ∆H4 = +702 kJ mol–1 ∆H3 = +248 kJ mol–1 ∆H6 = –603 kJ mol–1 (d) Would you expect the value for ∆H5 for calcium oxide to be more or less compared to magnesium oxide. Explain your answer. 3 (a) Define standard enthalpy change of formation of a compound. (b) The standard enthalpy change of formation of nitrogen monoxide is +90 kJ mol–1. (i) Write a thermochemical equation to represent the standard enthalpy change of formation of nitrogen monoxide. (ii) Based on the equation in (b)(i), explain why the standard enthalpy change of formation of nitrogen monoxide has a positive value. (c) Calculate the standard enthalpy change of formation of anhydrous aluminium chloride, Al2Cl6(s) from the following data: 2Al(s) + 6HCl(aq) → Al2Cl6(aq) + 3H2(g) –1007.0 kJ mol–1 H2(g) + Cl2(g) → 2HCl(g) –185.0 kJ mol–1 HCl(g) + aqueous → HCl(aq) –73.0 kJ mol–1 Al2Cl6(s) + aqueous → Al2Cl6(aq) –646.0 kJ mol–1 4 (a) Define standard enthalpy change of combustion of a compound. (b) The standard heat of combustion of ethane is –1540 kJ mol–1. (i) Write a thermochemical equation to illustrate the combustion of ethane. (ii) Assuming that the process is 80% efficient, what is the volume of ethane gas measured at room conditions required to raise the temperature of 40.5 kg of water from 25 °C to 85 °C? [Specific heat of water = 4.2 J g–1 K–1. 1 mole of gas occupies 24 dm3 at room conditions] (c) The standard enthalpies of combustion of methane and hydrogen are –890 kJ mol–1 and –286 kJ mol–1 respectively. (i) Write a balanced equation to represent the enthalpy of combustion of both methane and hydrogen. (ii) Calculate the heat released per gram of each substance. (iii) Use your result in (c)(ii), comment the efficiency of methane and hydrogen as fuel. 5 (a) Hydrazine, N2H4, burns in excess oxygen to produce nitrogen and steam. The standard enthalpy change of combustion of hydrazine is –540 kJ mol–1. (i) What is the oxidation state of nitrogen in hydrazine? (ii) Write a balanced equation for the complete combustion of hydrazine. (iii) 3.5 g of liquid hydrazine is burned in a bomb calorimeter. Calculate the change in temperature if the total heat capacity of the bomb calorimeter and its content is 6.25 kJ K–1. (b) (i) Write a balanced equation for the formation of hydrazine from its elements under standard conditions. (ii) Calculate the standard enthalpy change of formation of hydrazine using the following data: ∆Hf of H2O –286 kJ mol–1 Heat of vaporisation of H2O +44 kJ mol–1 6 The theoretical and experimental lattice energies of sodium chloride and silver chloride are given in the table below: Compound NaCl AgCl Theoretical lattice energy / kJ mol-1 -774 -770 Experimental lattice energy / kJ mol-1 -775 -910
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 35 7 (a) Define lattice energy. (b) Explain the differences between the theoretical and experimental lattice energies of the two salts. [Electronegativity of Na, Ag and Cl are 0.9, 1.9 and 3.0 respectively] 7 (a) Define standard enthalpy change of combustion. (b) Octane, C8H18, is used as a fuel. (i) Write a balanced equation for the combustion of octane in excess oxygen. (ii) Calculate the standard enthalpy change of combustion using the following information. ∆Hf of CO2 –394 kJ mol–1 ∆Hf of H2O –286 kJ mol–1 ∆Hf of C8H18 –250 kJ mol–1 8 (a) When 0.020 mole of sodium carbonate, Na2CO3, was added to 30.0 cm3 of approximately 1 M sulphuric acid (in excess), the temperature rose by 5.2 K. [The density of all solution = 1.00 g cm–3. The specific heat of all solutions is 4.18 J g–1 °C–1] (i) Write a balanced equation for the reaction. (ii) Calculate the enthalpy change per mole of sodium carbonate. (b) When 0.020 mole of sodium hydrogencarbonate, NaHCO3, was added to 30.0 cm3 of the same sulphuric acid, the temperature fell by 3.9 K. (i) Write an equation for the reaction. (ii) Calculate the enthalpy change per mole of sodium hydrogencarbonate. (c) When sodium hydrogencarbonate is heated, it decomposes into sodium carbonate, water and carbon dioxide. (i) Write a balanced equation for the decomposition of sodium hydrogencarbonate. (ii) Calculate the enthalpy of decomposition of sodium hydrogencarbonate. 9 (a) Using ethane as an example, explain what is meant by the standard enthalpy change of formation of a compound. (b) Given the following enthalpies of combustion: ∆Hc of hydrogen –286 kJ mol–1 ∆Hc of carbon –394 kJ mol–1 ∆Hcof ethane –1560 kJ mol–1 Calculate the standard enthalpy change of formation of ethane. 10 (a) When 2.00 g of ethanol is burned, it raises the temperature of 200 g of water from 25 °C to 75 °C. The process is known to be only 68% efficient. (i) Write a balanced equation for the complete combustion of ethanol. (ii) Calculate the standard enthalpy of combustion of ethanol. (b) Using the value calculated in (a)(ii) and the following data, calculate the standard enthalpy change of formation of ethanol. ∆Hf of CO2 –394 kJ mol–1 ∆Hf of H2O –286 kJ mol–1 11 Chlorine dioxide, ClO2, is formed when silver chlorate(V) reacts with chlorine according to the equation: 2AgClO3(s) + Cl2(g) → 2AgCl(s) + 2ClO2(g) + O2(g) ∆Ho = 0 kJ mol–1 (a) Calculate the standard enthalpy change of formation of chlorine dioxide using the following data: ∆Hf of AgClO3(s) –25 kJ mol–1 ∆Hf of AgCl(s) –125.8 kJ mol–1 (b) Comment on the answer in (a). 12 Dinitrogen tetroxide dissociates to nitrogen dioxide according to the equation: N2O4(g) 2NO2(g) ∆H = + 57.0 kJ mol–1 Given the standard enthalpy change of formation of dinitrogen tetroxide is +9.7 kJ mol–1, calculate the standard enthalpy change of formation of nitrogen dioxide.
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 36 7 13 (a) State Hess’ Law. (b) Show how would you use the following sets of data to verify Hess’ Law: Set Process ∆Ho/kJ I Zn(s) + S(s) → ZnS(s) –167 ZnS(s) + 2O2(g) → ZnSO4(s) –795 II Zn(s) + —1 2 O2(g) → ZnO(s) –355 S(s) + O2(g) → SO2(g) –293 SO2(g) + —1 2 O2(g) → SO3(g) –92 ZnO(s) + SO3(g) → ZnSO4(s) –222 14 (a) Urea, CO(NH2)2, can be hydrolysed by water to carbon dioxide and ammonia according to the equation: CO(NH2)2(aq) + H2O(l) → CO2(aq) + 2NH3(aq) Calculate the standard enthalpy change for the hydrolysis of urea using the following data. Compound H2O(l) CO2(aq) NH3(aq) CO(NH2)2(aq) ∆Hf /kJ mol–1 –286 –414 –81 –320 (b) When heated, methylhydrazine, CH3NHNH2, and nitrogen tetroxide, N2O4 react according to the equation: 4CH3NHNH2(l) + N2O4(l) → 4CO2(g) + 9N2(g) + 12H2O(l) (i) Calculate the enthalpy change for the reaction. Given: Compound CH3NHNH2(l) N2O4(l) CO2(g) H2O(l) ∆Ho f /kJ mol–1 +56.2 –20.0 –394.0 –286.0 (ii) Explain why a mixture of methylhydrazine and dinitrogen tetroxide is sometimes used as an explosive. 15 The standard enthalpy of hydration of copper(II) sulphate refers to the following process: CuSO4(s) + 5H2O(l) → CuSO4•5H2O(s) However, this enthalpy change cannot be measured directly. When 4.0 g of anhydrous copper(II) sulphate is added to 50.0 g of water, the temperature changes from 27.0 °C to 35.0 °C. When 4.0 g of the hydrated solid is added to 50.0 g of water, the temperature changes from 27.0 °C to 25.7 °C. (a) Calculate the heat change when 4.0 g of anhydrous copper(II) sulphate dissolves in 50.0 g of water. (b) Calculate the enthalpy of solution, in kJ mol–1, of anhydrous copper(II) sulphate. (c) Calculate the heat change when 4.0 g of hydrated copper(II) sulphate dissolves in 50.0 g of water. (d) Calculate the enthalpy change of solution, in kJ mol–1, of hydrated copper(II) sulphate. (e) Calculate the enthalpy of hydration of anhydrous copper(II) sulphate. 16 The standard enthalpies of formation (∆Hf°) of carbon monoxide and carbon dioxide are given below: ∆Hf° CO = -99 kJ mol-1 ∆Hf° CO2 = -394 kJ mol-1 (a) Write the equations to show the standard enthalpy of formation of (i) carbon monoxide, (ii) carbon dioxide. (b) (i) Draw the Lewis structure for carbon monoxide and carbon dioxide, and state their general shapes. (ii) Explain in terms of bonding, why the standard enthalpy of formation of carbon dioxide is more exothermic than that of carbon monoxide. (c) Draw an energy cycle to determine the value of the enthalpy change of the reaction for the following equation: 2CO(g) + O2(g) → 2CO2(g) Hence, calculate the enthalpy change of this reaction.
Chemistry Term 2 STPM Chapter 7 Chemical Energetics 37 7 17 (a) Define the lattice energy of potassium hydride, KH. (b) Draw a Born-Haber cycle for potassium hydride. (c) Use the following information, calculate the lattice energy of potassium hydride. Process ∆Ho/kJ mol–1 Atomisation of potassium +90 First ionisation energy of potassium +418 Bond energy of hydrogen +436 First electron affinity of hydrogen –78 Formation of potassium hydride +78 (d) How would you expect the lattice energy of sodium hydride compared to that of potassium hydride. Explain your answer. 18 Strontium reacts with chlorine when heated to produce strontium chloride, SrCl2. (a) The standard enthalpy change of formation of strontium chloride is –829 kJ mol–1. Write a thermochemical equation to represent the process. (b) Draw a Born-Haber cycle for the formation of strontium chloride and use it to calculate the lattice energy of strontium chloride. Given: Process ∆Ho/kJ mol–1 Atomisation of Sr +165 Bond energy of Cl2 +242 1st ionisation energy of Sr +550 2nd ionisation energy of Sr +1064 1st electron affinity of Cl –349 (c) Theoretical enthalpies of formation of SrCl(s) and SrCl3(s) are: ∆Hf SrCl(s) = –200 kJ mol–1 ∆Hf SrCl3(s) = +570 kJ mol–1 (i) Comment on the relative stability of these two compounds with respect to (I) their constituent elements (II)SrCl2(s) (ii) How would you expect the lattice energy of the hypothetical compound SrCl3 compared to that of SrCl2. Explain your answer. 19 (a) Construct a labelled Born-Haber cycle for sodium chloride in the form of an energy diagram. (b) Predict the stability of sodium chloride under standard conditions. 20 (a) Born-Haber cycle, which is a type of Hess’ Law cycle, is used to calculate the lattice energy of an ionic solid. (b) Construct a Born-Haber cycle for silver iodide. 21 (a) The standard enthalpy of hydration of some of the Group 1 cations is given in the table below. Ion Li+ Na+ K+ ∆Hhydration/kJ mol–1 –524.0 –400.0 –318.0 Comment on the trend. (b) The standard enthalpy of atomisation of hydrogen fluoride is +562 kJ mol–1. Calculate the energy required to break the covalent bond in a hydrogen fluoride molecule. [Avogadro constant, L = 6.02 × 1023 mol–1]
CHAPTER 8 ELECTROCHEMISTRY Concept Map Learning earning Outcomes Students should be able to: Half-cell and redox equations • explain the redox processes and cell diagram (cell notation) of the Daniell cell; construct redox equations. Standard electrode potential • describe the standard hydrogen electrode; • use the standard hydrogen electrode to determine standard electrode potential (standard reduction potential), E°; • calculate the standard cell potential using the E° values, and write the redox equations; • predict the stability of aqueous ions from E° values; • predict the power of oxidising and reducing agents from E ° values; • predict the feasibility of a reaction from E° cell value and from the combination of various electrode potentials: spontaneous and non-spontaneous electrode reactions. Non-standard cell potentials • calculate the non-standard cell potential, Ecell, of a cell using the Nernst equation. Fuel cells • describe the importance of the development of more efficient batteries for electric cars in terms of smaller size, lower mass and higher voltage, as exemplified by hydrogen-oxygen fuel cell. Electrolysis • compare the principles of electrolytic cell to electrochemical cell; • predict the products formed during electrolysis; • state the Faraday’s first and second laws of electrolysis; • state the relationship between the Faraday constant, the Avogadro constant and the electronic charge; • calculate the quantity of electricity used, the mass of material and/or gas volume liberated during electrolysis. Applications of electrochemistry • explain the principles of electrochemistry in the process and prevention of corrosion (rusting of iron); • describe the extraction of aluminium by electrolysis, and state the advantages of recycling aluminium; • describe the process of anodisation of aluminium to resist corrosion; • describe the diaphragm cell in the manufacture of chlorine from brine; • describe the treatment of industrial effluent by electrolysis to remove Ni2+, Cr3+ and Cd2+; • describe the electroplating of coated plastics. Electrochemistry Corrosion • Rusting of iron • Preventing rusting Oxidation Number • Redox reactions Standard Electrode Potential Series Daniell Disproportionation Cell Electrochemical Cells • Construction • Fuel cells • Batteries for electric cars Constructing Redox Equations • Methods • Oxidation number • Ion-electron • Balancing redox equations in basic solutions Uses of The Standard Electrode Potential Series • Compare strength of oxidising agents and reducing agents • Predicting feasibility of redox reactions • Predict stability of aqueous ions • Calculating e.m.f. of electrochemical cell Electrolysis • Cells • Molten lead(II) bromide electrolysis • Factors affecting electrolysis products • Quantitative electrolysis Electrode Potential • Standard hydrogen electrode • Measuring standard electrode potential Factors Affecting Standard Electrode Potential • Concentration • Nernst equation • Pressure • Nernst equation and – Electrochemical cell – Equilibrium constant – Solubility product • pH • Complex formation Applications of Electrolysis in Industries • Extraction of aluminium • Anodisation • Manufacturing chlorine • Effluent treatment • Electroplating
Chemistry Term 2 STPM Chapter 8 Electrochemistry 39 8 8.1 Half-cells and Redox Equations 1 Electrochemistry deals with reactions involving the transfer of electrons from one chemical species to another, and the electrical energy that is produced or used during these reactions. 2 Such reactions are called redox reactions, which is a short-form for reduction and oxidation. 3 Oxidation is defined as a process of electron loss. For example, Fe ⎯→ Fe2+ + 2e– 2I– ⎯→ I2 + 2e– 4 Reduction is defined as a process of electron gain. For example, Fe3+ + e– ⎯→ Fe2+ Cl2 + 2e– ⎯→ 2Cl– 5 When electrons are made to move through a solid conductor, an electrical current is generated and flows in the opposite direction. Daniell Cell 1 The conversion of chemical energy into electrical energy is demonstrated by the Daniell cell. A simplified diagram of the Daniell cell is shown below. Salt bridge ZnSO4 (aq) V Zn Cu CuSO4 (aq) + 2 The cell consists of two half-cells. The zinc half-cell consists of a zinc rod partially immersed in aqueous zinc sulphate, and the copper half-cell consists of a copper electrode partially immersed in aqueous copper sulphate. 3 The two half-cells are joined using a salt bridge which is a glass tubing filled with saturated potassium chloride. 4 When the circuit is closed, an electric current flow from the copper electrode to the zinc electrode as indicated by the voltmeter. As a result, there is a corresponding flow of electrons from the zinc electrode to the copper electrode. 5 As time passes, the zinc electrode decreases in size, while the copper electrode increases in size, and the blue colour of the copper sulphate solution slowly fades. 6 As current is drawn from the cell, the zinc electrode in the zinc halfcell dissolves to produce aqueous Zn2+ ions. Zn(s) → Zn2+(aq) + 2e– This is called a half-equation. + – Current Electron The Daniell cell converts chemical energy into electrical energy. Sometimes, saturated potassium nitrate is used as the salt bridge. Electricity flows from copper to zinc. Electrons flow from zinc to copper. Blue colour of CuSO4 fades.
Chemistry Term 2 STPM Chapter 8 Electrochemistry 40 8 7 The electrons travel through the external circuit to the copper half-cell, where the Cu2+ aqueous ions accept the electrons and get converted to copper which adhere to the copper electrode. Cu2+(aq) + 2e– → Cu(s) 8 As more and more Cu2+ aqueous ions are removed from the electrolyte, the blue colour fades. 9 Thus, there is a flow of electrons from the zinc electrode to the copper electrode, and correspondingly a current flow from the copper electrode to the zinc electrode through the connecting wires. 10 The copper electrode forms the positive terminal and the zinc electrode forms the negative terminal of the Daniell cell. [By convention, electric current flows from the positive terminal to the negative terminal of a cell]. Electrons Negative terminal Zinc Copper Positive terminal Current 11 The overall cells reaction that produces the electricity is: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) In the reaction: (a) Zinc undergoes oxidation by losing electrons. (b) Copper(II) ions get reduced by gaining electrons. (c) Zinc supplies electrons to copper(II) ions. Zinc is called a reducing agent. (d) Copper(II) ions remove (or accept) electrons from zinc. Copper(II) ion is an oxidising agent. 12 Reducing agents are electron donors. Oxidising agents are electron acceptors. 13 Note that zinc, as a reducing agent, undergoes oxidation, while copper(II), as an oxidising agent, undergoes reduction. 14 By definition, anode is the electrode where oxidation occurs, and cathode is the electrode where reduction occurs. 15 Thus, zinc is the anode of the cell, and copper is the cathode of the Daniell cell. Electrons Anode/negative Zinc Copper Cathode/positive Current 16 Note that oxidation and reduction must occur simultaneously. We cannot have one without the other. The electrons supplied by the reducing agent must be received by the oxidising agent. 17 The function of the salt bridge is to maintain electrical neutrality of the system. As zinc dissolves to form Zn2+(aq) ions and Cu2+(aq) ions are reduced to copper, there is surplus of positive charge in the zinc half-cell while there is a surplus of negative charge in the copper half-cell. Info Chem The blue colour of aqueous copper(II) sulphate is due to the presence of Cu2+ aqueous ions or more correctly the Cu(H2O)6 2+ aqueous ions. Electricity flows from the positive terminal to the negative terminal. Reducing agent gets oxidised. Oxidising agent gets reduced. Exam Tips Exam Tips The cell notation for the Daniell cell is: Zn(s) Zn2+(aq) Cu2+(aq) Cu(s)
Chemistry Term 2 STPM Chapter 8 Electrochemistry 41 8 18 The salt bridge allows the migration of Zn2+(aq) ions into the copper half-cell and migration of SO4 2–(aq) ions from the copper half-cell into the zinc half-cell at the same time. ZnSO4 (aq) CuSO4 (aq) Zn2+ SO4 2– Zn SO4 2+ 2– 19 The Daniell cell can be represented by the following cell diagram: Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) 20 Summary: Oxidation is the process of electron loss. Reduction is the process of electron gain. Oxidising agents are electron acceptors. Reducing agents are electron donors. Example 8.1 Identify the oxidising agents and reducing agents in the following reactions by writing the appropriate half-equations. (a) Cl2 + 2Fe2+ → 2Cl– + 2Fe3+ (b) Cu + 2Ag+ → Cu2+ + 2Ag Solution (a) Cl2 + 2e– → 2Cl– Fe2+ → Fe3+ + e– Cl2 is the oxidising agent (electron acceptor). Fe2+ is the reducing agent (electron donor). (b) Cu → Cu2+ + 2e– Ag+ + e– → Ag Cu is the reducing agent (electron donor). Ag+ is the oxidising agent (electron acceptor). The salt bridge completes the circuit and helps to maintain electrical neutrality and also prevents the electrolytes from the two half-cells from mixing. Quick Check 8.1 Identify the oxidising agents and reducing agents in the following reactions. (a) Mg + Fe2+ → Mg2+ + Fe (c) Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+ (b) Cl2 + Sn2+ → 2Cl– + Sn4+ (d) S2O8 2– + 2I– → 2SO4 2– + I2 Oxidation Number (O.N.) or Oxidation State 1 All atoms, either in elements, compounds or ions, can be assigned an oxidation number or oxidation state. 2 All atoms in their respective elemental state are assigned oxidation numbers of ‘zero’. 2015/P2/Q19(b) All elements have ‘zero’ oxidation number. Another name for the cell diagram is cell notation.
Chemistry Term 2 STPM Chapter 8 Electrochemistry 42 8 For example: Element H2 Na Al Fe C Cl2 S8 N2 Oxidation number/state 0 0 0 0 0 0 0 0 3 In simple ions, the oxidation number of the atom/element concerned is the same as the charge on the ion. For example: Element Cu+ Mg2+ Al3+ Sn4+ Cl– O2– N3– Oxidation number +1 +2 +3 +4 –1 –2 –3 4 For a polyatomic ion (or molecular ion), the sum of all the oxidation numbers of the elements in the ion is equal to the net charge on the ion. For example: SO4 2– : (oxidation number of S) + 4(oxidation number of O) = –2 Cr2O7 2– : 2(oxidation number of chromium) + 7(oxidation number of O) = –2 MnO4 – : (oxidation number of Mn) + 4(oxidation number of O) = –1 5 For a neutral covalent molecule, the more electronegative element would be assigned a negative oxidation number, while the less electronegative element will have a positive oxidation number. The sum of the oxidation numbers of all the atoms in the molecule is zero. For example: CO2 : (oxidation number of C) + 2(oxidation number of oxygen) = 0 NaAl(OH)4 : (O.N. of Na) + (O.N. of Al) + 4(O.N. of O) + 4(O.N. of H) = 0 6 Certain elements have fixed oxidation number in all their compounds. For example, all elements in Group 1, Group 2 and Group 13 have fixed oxidation number of +1, +2 and +3 respectively. Compound Na2O MgCl2 Al2(SO4)3 O. N. of the metallic element +1 +2 +3 7 Fluorine (F), the most electronegative element, has a fixed oxidation state of –1 in all its compounds. Compound AlF3 Na3AlF6 SbF5 O. N. of F –1 –1 –1 8 Hydrogen (H) is assigned an oxidation number of +1 for all its compounds, except when it combines with elements from Group 1, 2 and 13 (to form metal hydrides), where it has an oxidation number of –1. For example: Compound CH4 SiH4 NH3 H2O HCl HNO3 Oxidation number of hydrogen +1 +1 +1 +1 +1 +1 Compound Lithium hydride, LiH Magnesium hydride, MgH2 Aluminium hydride, AlH3 Oxidation number of hydrogen –1 –1 –1 Oxidation number of simple ions. Sum of all the oxidation numbers must give the charge of the ion. Fluorine exhibits a constant oxidation number of –1 in all its compounds. Some elements have variable oxidation states