Book 3 ans

HKDSE CHEMISTRY – A Modern View (Chemistry) Coursebook 3 Suggested answers Chapter 25 Simple molecular substances with non-octet structures and shapes of simple molecules  Class Practice  Chapter Exercise Chapter 26 Bond polarity  Class Practice  Chapter Exercise Chapter 27 Intermolecular forces  Class Practice  Chapter Exercise Chapter 28 Structures and properties of molecular crystals  Class Practice  Chapter Exercise  Page Number 1 2 4 5 7 9 11 12 14 Part Exercise Chapter 29 Chemical cells in daily life  Class Practice  Chapter Exercise Chapter 30 Simple chemical cells  Class Practice  Chapter Exercise Chapter 31 Redox reactions © Aristo Educational Press Ltd. 2010 17 18 21 22  Class Practice  Chapter Exercise Chapter 32 Redox reactions in chemical cells  Class Practice  Chapter Exercise Chapter 33 Electrolysis  Class Practice  Chapter Exercise 24 26 29 30 32 33 Chapter 34 Importance of redox reactions in modern ways of living  Class Practice  Chapter Exercise  36 37 39 Part Exercise Chapter 35 Energy changes in chemical reactions  Class Practice  Chapter Exercise 42 43 Chapter 36 Standard enthalpy change of combustion, neutralization, solution and formation  Class Practice  Chapter Exercise Chapter 37 Hess’s Law  Class Practice  Chapter Exercise  Part Exercise 48 51 53 44 46 © Aristo Educational Press Ltd. 2010 HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 25 Simple molecular substances with non-octet structures and shapes of simple molecules Class Practice A25.1 (a) BCl3 NCl3 IF3 (b) BCl3. The central boron atom has only six outermost shell electrons. IF3. The central iodine atom has 10 outermost shell electrons. A25.2 Molecule No. of Spatial electron arrangement pairs of electron pairs 4 Tetrahedral 4 Tetrahedral 4 Tetrahedral No. of lone No. of Shape pairs bond pairs CH4 NH3 H2O 0 1 2 4 3 2 Tetrahedral Trigonal pyramidal V-shaped © Aristo Educational Press Ltd. 2010 1 B: (b) A: WZ3 B: YZ3 (c) A. 20. 2010 2 .5° trigonal bipyramidal. 22. 2. eight eight six 10 12 non-octet linear. 109. 90° 2. 11. 13.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 25 Simple molecular substances with non-octet structures and shapes of simple molecules Chapter Exercise 1. 90°. 120° octahedral. 17. 12. 18. The central atom (W) has only six electrons in its outermost shell. 10. 3. 2 1. 4. (a) (b) © Aristo Educational Press Ltd. 19. 9. 7. 120° tetrahedral. (a) A: . (a) X: Trigonal pyramidal Y: Trigonal bipyramidal (b) 107° (c) 90°(for Cl(axial) − Cl(equatorial) bond angles) and 120° (for Cl− Cl bond angles P− P− within the plane of triangle) 23. 16. 14. 180° trigonal planar. 6. 15. 5. 3 shapes D B C A D A 21. 8. 2010 3 . © Aristo Educational Press Ltd.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 24. all the O=S=O bond angles are about 120°. (a) SO2: SO3: (b) SO2: V-shaped SO3: Trigonal planar (c) Since there are three groups of electrons around the central atoms (sulphur) in both SO2 and SO3. The situations are shown below: In a non-uniform electrostatic field. As the dipole Cl moments of the three polar bonds cancel out each other.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 26 Bond polarity Class Practice A26. 2010 4 . the molecule is polar. © Aristo Educational Press Ltd. the molecule is non-polar. the resultant dipole moments of NBr3 and NF3 are pointing to different directions. (b) As the order of electronegativity is F > N > Br. As there is a Cl resultant dipole moment arising from the three polar bonds. the nitrogen end of NBr 3 will point to the positive pole while the nitrogen end of NF3 will point to the negative pole.2 (a) NCl3 has three polar N− bonds and is trigonal pyramidal in shape.1 (a) (b) (c) A26. BCl3 has three polar B− bonds and is trigonal planar in shape. 12. (b) (c) (d) 17. smaller polar. symmetrically deflection C B C B C A 14.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 26 Bond polarity Chapter Exercise 1. ‘covalent bond involving fluorine is polar in nature’ 2. (c) It decreases. 3. X. 4.5 (e) 0–2. 6. Z. 9. 13. 8. (a) The electronegativity values of elements do not follow the trend of changing masses. The covalent bond in a fluorine molecule (F− is non-polar. 7. 11. (a) Y. less. 2010 5 . (a) (b) (c) 16. ‘all compounds of fluorine must be polar’ (b) 1. (b) It increases.5–3. 10. 2. cancel out polar. 5. (d) 2. (a) Mistakes: 1.8 (f) C and F (g) Cl and Br 15. Polar X− bonds may cancel out their individual dipole moments to give F © Aristo Educational Press Ltd. attract polar non-polar larger. F) 2. etc. SF6. 2010 6 .HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 a non-polar molecule e. CF4.g. © Aristo Educational Press Ltd. So CH2Cl2 has a higher boiling point. Cl2 has a higher boiling point. On the other hand. since the molecular size of Cl2 is larger than that of F2. (a) CH3F is polar in nature and its molecules are attracted by both dipole-dipole forces and dispersion forces. (a) Non-polar molecules (b) Dispersion forces (c) (i) The statement is correct as butane has a higher boiling point than that of propane. CH3F has a higher boiling point than that of C2H6. Thus. Since the molecular size of CH2Cl2 is larger than that of ClF. The strength of dispersion forces increases with increasing molecular size as there is a greater chance of uneven distribution of electrons in a larger molecule.1 (a) When the electron distribution around the two iodine nuclei is uneven. ClF and CH2Cl2 are polar in nature and their molecules are attracted by both dipole-dipole forces and dispersion forces. instantaneous dipole is formed in which the side with more electrons will carry a partial negative charge.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 27 Intermolecular forces Class Practice A27. Butane (straight-chain hydrocarbon) has a long. Similarly. C2H6 is non-polar in nature and its molecules are attracted by dispersion forces only. 2. Cl2 has a higher boiling point. (b) A27. As a result. ClF and CH2Cl2 have higher boiling points than those of F2 and Cl2.2 1. thin shape. (b) Although Cl2 is non-polar and its molecules are attracted by dispersion forces only. F2 and Cl2 are nonpolar in nature and their molecules are attracted by dispersion forces only. 3. As a result. the dispersion forces between CH2Cl2 molecules are larger. (ii) The statement is incorrect as butane has a higher boiling point than that of 2-methylpropane. resulting in larger dispersion forces. 2010 7 . On the other hand. A27.3 (a) CH3OH © Aristo Educational Press Ltd. the larger dispersion forces in Cl2 outweigh the dipole-dipole forces in HCl. This contributes to a larger contact surface area between butane molecules. 2010 8 . 1. A27.4-trichlorobenzene molecules are attracted by dipole-dipole forces.2. Therefore.4-trichlorobenzene. (b) Glucose molecules are attracted to each other by extensive hydrogen bonds. On the other hand. In ice.4-trichlorobenzene cannot form any hydrogen bond with water. As a result.2.4 (a) In glucose. As the intermolecular forces in hydrogen fluoride are stronger than those in hydrogen chloride. Less energy is needed to melt 1.5 (a) In diamond. On the other hand.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (b) (c) A27. more energy is needed to separate water molecules during boiling. glucose is very soluble in water. © Aristo Educational Press Ltd. water molecules are held together by weak intermolecular forces (hydrogen bonds and van der Waals’ forces). the hydrogen atoms and oxygen atoms of the five hydroxyl groups (− OH) on the molecule can form hydrogen bond with the oxygen atoms and hydrogen atoms of water molecules respectively.2. carbon atoms are held together by strong covalent bonds and much energy is needed for separating the atoms during melting. it is more difficult for hydrogen fluoride molecules to escape into the atmosphere. 1. A considerable amount of energy is needed for separating glucose molecules in melting. (c) Hydrogen chloride molecules are held together by van der Waals’ forces but hydrogen fluoride molecules are held together by hydrogen bonds. so less energy is needed for separating molecules during melting. (b) Water molecules are held together by hydrogen bonds while oxygen molecules are held together by dispersion forces only. induced molecular. molecules. 3. 13. Ethanol and ethane-1. 20.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 27 Intermolecular forces Chapter Exercise 1. (a) Dipole-dipole forces and dispersion forces (b) Dispersion forces (c) Since the molecular sizes of HCl and F2 do not differ much. 8. 2. This is because much 9 © Aristo Educational Press Ltd. iscosity hydrogen bonds high proteins. − The melting point and boiling point of water are much higher than those of simple molecular substances without hydrogen bonds. 7. Therefore. 15. This gives rise to a more prominent instantaneous dipole and larger dispersion forces. 2010 . 11. 5. 16. Among them. less energy is needed to separate the molecules in boiling.2-diol have higher boiling points as their molecules are held together by hydrogen bonds. 4.2-diol molecule can form two hydrogen bonds per molecule while each ethanol molecule can form one hydrogen bond per molecule. 22. Chloroethane has a lower boiling point than ethanol because there are dipole-dipole forces but not hydrogen bonds between chloroethane molecules. Ethane is non-polar. 17. (d) The boiling point increases down the group. olarity hydrogen. the dispersion forces among their molecules are similar in strength. deoxyribonucleic acids D A A A B A 18. surface. electrostatic an der Waals’ forces. 21. the presence of dipole-dipole forces in HCl strengthens the attractions among molecules of HCl and so HCl has a higher boiling point. 9. 19. ydrogen bonding Dipole-dipole forces Dispersion forces. 12.2-diol has a higher boiling point than ethanol. electrons energy surface. the higher is the chance of uneven distribution of electrons. Their molecules are held together by hydrogen bonds. 14. ethane-1. (a) A and B. It has the lowest boiling point. However. 6. (e) The larger the atom. 10. each ethane-1. As there is no hydrogen bond between molecules of C. (b) Lower. (a) GroupVIII/0 (b) Dispersion forces (c) The intermolecular forces arise from the uneven distribution of electrons within the atoms. 2010 10 . Regarding to its viscosity. That is. This is because the strong hydrogen bonds hold water molecules together and do not allow them to move past one another easily. water is more viscous than most molecular liquids. The existence of extensive hydrogen bonds makes the surface tension of water exceptionally high. water molecules are held strongly together to form an unstretchable surface. © Aristo Educational Press Ltd.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 − − energy is needed to overcome the strong hydrogen bonds between water molecules and separate them. less energy is needed during melting. (a) Carbon nanotubes are electrically conductive because of the presence of delocalized electrons in their structures. In melting diamond and graphite. (b) Graphite has delocalized electrons in its structure but diamond and buckminsterfullerene do not. In a carbon nanotube. 2010 11 . (d) During boiling. (c) As buckminsterfullerene molecules are held by dispersion forces. 2. (b) Buckminsterfullerene is spherical in shape but carbon nanotube is cylindrical.3 (a) Both of them have molecules held together by dispersion forces. © Aristo Educational Press Ltd. (b) Carbon nanotubes have very high tensile strength. all hydrogen bonds have to be broken before the water molecules can escape as steam. In buckminsterfullerene.4 1. (a) In diamond. (c) To overcome a certain amount of hydrogen bonds and separate the water molecules. A28.2 (a) Diamond and graphite were discovered before fullerene. A28. strong covalent bonds have to be overcome. pentagonal patterns of atoms are found between the hexagonal patterns.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 28 Structures and properties of molecular crystals Class Practice A28.1 (a) Liquid water (b) The regular open network structure of ice allows the formation of maximum number of hydrogen bonds (four for each water molecule). pentagonal patterns of atoms are found only near the two ends of molecules. (b) Buckminsterfullerene (c) Covalent bond (d) Dispersion forces (or van der Waals’ forces) A28. carbon atoms are held strongly by covalent bonds but the buckminsterfullerene molecules are held by weak dispersion forces. 4. 2. (a) (b) (c) (d) It is an element as it is composed of carbon atoms only. (b) Covalent bond between O atom and H atom (c) Hydrogen bond between O atom of a water molecule and H atom of another water molecule 16. Furthermore. 8. Hence. 13. close. tensile. 3. (a) Ice has a lower density than water. (a) a b a b a b * Label all other bonds accordingly. Z −Buckminsterfullerene (b) Carbon nanotubes are good electrical conductors due to the movement of delocalized electrons. (b) Ice floats on water. 17. C60 has larger dispersion forces between its molecules and thus a higher boiling point than that of C20. 10. benzene. solvents. hydrogen oxygen open hydrogen carbon. conductors D B D B A C 15. there are hexagons over the surface of C60. 5. 14. insulator cylindrical. strong. hard. (a) X − Carbon nanotube. order four. 18. 12. carbon. Because of the strong © Aristo Educational Press Ltd. (c) Ice possesses an open structure in which H2O molecules are separated farther and there are more spaces between molecules. hollow spherical. 6. 2010 12 .HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 28 Structures and properties of molecular crystals Chapter Exercise 1. 7. each carbon atom in a carbon nanotube is covalently-bonded to three neighbouring carbon atoms. 9. C60 has a larger molecular size than C20. Y − Diamond. 11. melting. 12 pentagons and 0 hexagon Other than pentagons. This prevents heat loss from water and helps maintain the water temperature. carbon nanotubes can withstand high tension. 2010 13 .HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 covalent bonds holding carbon atoms along the tube axis. © Aristo Educational Press Ltd. potassium) atoms (d) No. There are no strong covalent bonds holding the carbon atoms along a specific direction.g. (c) Some metal (e. 4. © Aristo Educational Press Ltd. 12. 9. The molecule should adopt a linear shape to minimize the repulsion between the two groups of electrons. 2010 14 . 3. C A C A C B A C C A 11. (a) WY2 (b) The electron diagram of WY2 is: There are two double bonds surrounding the central atom W. SF6 −90° (c) SF2 is polar while SF6 is non-polar. the molecule adopts a trigonal pyramidal shape with a bond angle of about 107°. (c) X and Z. (a) (b) SF2 −105°. (a) (b) C–B–B–C (c) 13. 8. 5. 10. 6. 2. Due to the repulsion between the lone pair of electrons and the three bond pairs of electrons. 7.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Part VI Microscopic world II Part Exercise 1. HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (d) (e) It is non-polar as the effects of all polar S− bonds cancel each other. F 14. (a) (i) Non-polar (ii) Non-polar (iii) Polar (b) Iodine is non-polar. It dissolves well in tetrachloromethane because tetrachloromethane is also non-polar. However, iodine dissolves slightly in water which is polar. (The amount of energy released from the formation of intermolecular forces (dispersion forces) between iodine and tetrachloromethane molecules is large enough to compensate for that required to break the intermolecular forces (dispersion forces) between tetrachloromethane molecules. On the other hand, the amount of energy released from the formation of intermolecular forces (dispersion forces) between iodine and water molecules is not large enough to compensate for that required to break the hydrogen bonds between water molecules.) (c) Tetrachloromethane cannot dissolve in water. (d) Two separate layers will be seen. The yellow colour of the aqueous layer fades while the tetrachloromethane layer turns violet. 15. (a) In ice, water molecules are arranged orderly in a regular open network structure because of the extensive hydrogen bonding. In this case, water molecules are farther apart than in liquid water and thus ice takes up a larger volume. (b) Each H2O molecule can form two hydrogen bonds but each HF molecule can form one hydrogen bond only. Thus, more energy is needed to separate water molecules in the boiling process. (c) Although both compounds can form hydrogen bonds, CH3CH2CH2CH2OH has a larger size than CH3CH2OH and it has larger dispersion forces between molecules. Thus, more energy is needed to separate CH3CH2CH2CH2OH molecules in the boiling process. (d) Hexane is non-polar. It cannot dissolve in water which is polar. (Hexane cannot form strong hydrogen bonds with water and thus it is not soluble in water.) 16. (a) H2O and CH3CH2CH2CH2CH3 (b) H2O NH3 Cl2 CH3CH2CH2CH2CH3 Hydrogen bond Hydrogen bond Dispersion force Dispersion force (c) Each H2O molecule can form two hydrogen bonds but each NH3 molecule can form one hydrogen bond only. Thus, more energy is needed to separate water molecules in the boiling process. (d) CH3CH2CH2CH2CH3 should have a higher boiling point as it has a larger size than Cl2 and thus larger dispersion forces between molecules. © Aristo Educational Press Ltd. 2010 15 HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 17. (a) (b) (c) (d) (e) The roots of the plants can absorb the fertilizer dissolved in water from the soil. 18. (a) C60 has a higher boiling point than that of C28 since C60 has a larger size and so larger dispersion forces between molecules. More energy is needed to separate the molecules in boiling. (b) Ammonia has a higher boiling point than nitrogen. Ammonia molecules are attracted by hydrogen bonds while nitrogen molecules are attracted by weaker dispersion forces. More energy is needed in boiling ammonia. 19. (a) Either by using very high pressures to compress the hydrogen gas or very low temperatures to turn the hydrogen gas into a liquid. (b) Dispersion forces (c) Hydrogen in molecular form can be released easily and used as fuel. © Aristo Educational Press Ltd. 2010 16 HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 29 Chemical cells in daily life Class Practice A29.1 A little water should be present so that the ions present in the electrolyte would become mobile to allow flow of electricity. A29.2 1. Its electrolyte is potassium hydroxide, which is alkaline. Its positive electrode consists of manganese(IV) oxide (mixed with a little powdered graphite). 2. Hearing aids are small and they need very small cells. Silver oxide cells (shaped like a button) would serve the purpose. 3. Silver and its compounds (e.g. silver oxide) are expensive. 4. HK$100 000. Explanation: − − − Each silver oxide cell produces 2.79 × 10 4 kWh (=180 × 10 3 × 1.55 × 10 3 kWh) electricity. To produce one ‘unit’ of electricity (1 kWh), about 3585 (= 1 ) cells are needed. Therefore, the approximate cost is HK$30 × 3585 2.79 × 10 − 4 = HK$107 550 (about HK$100 000). A29.3 (a) A flat discharge curve. (b) Alkaline manganese cell/silver oxide cell/nickel-cadmium rechargeable cell/lithium-ion rechargeable cell/nickel-metal hydride rechargeable cell (any ONE) © Aristo Educational Press Ltd. 2010 17 10. 2. (− zinc ) Electrolyte: ammonium chloride 27. and can work even when the voltage and current are not steady. 18. 22. mercury. 4. 6. nickel-metal hydride rechargeable. 23. 21. silver oxide cells. lithium-ion rechargeable cells. rechargeable. 16. 3. Charging old zinc-carbon cells may cause cells to overheat or even explode. (d) Electrodes: (+) graphite. ithium primary nickel-cadmium rechargeable. lithium primary cells (any TWO) Secondary cells: nickel-cadmium rechargeable cells. 15. 5. lkaline manganese. 9. (b) Primary cells: zinc-carbon cells. 8. electricity electrolyte. 2010 18 . lead-acid accumulator zinc-carbon. 20. Leadacid accumulator. 7. paste inc-carbon. 25. 24. ilver oxide. (a) B New cells may have a higher voltage than the old cells. 17. 12 V voltage alkaline manganese (a) size of cell (b) shape of cell (c) type of cell nvironment. alkaline manganese. nickel-metal hydride rechargeable cells (any TWO) (c) Zinc-carbon cell is more commonly used because its price is low and a portable radio does not draw a large current. recycling D A B B A D C D B C A B C B B C 26. (b) A The acidic/alkaline electrolyte of the cells may corrode the metal case © Aristo Educational Press Ltd. (a) Primary cells are not rechargeable whereas secondary cells can be recharged and used again.7 V. 13. 12. alkaline manganese cells. 19. thus charging the old cells during operation. lithium-ion rechargeable. 11.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 29 Chemical cells in daily life Chapter Exercise 1. nickel-metal hydride rechargeable. 14. 3. The electrolyte leaks out when holes appear in the metal case. (e) D Small children may mistake small-sized button cells as sweets. Most other cell types are normally larger in size and too heavy for use in watches. and has no memory effect. (d) C Alkaline manganese cells cannot be recharged. 2010 19 ./Nickel-metal hydride rechargeable cell is more environmentally friendly. It is lightweight and small so is suitable for use in watches. Watches./Alkaline manganese cell is more expensive but more durable. heat and hydrogen may be produced and hence lead to explosion. however. and are lightweight. (b) Lithium-ion rechargeable cell It has a high charge capacity but lightweight. (Any ONE) 29. Measures must be taken to prevent the watch to stop when there is no light. (Any ONE) (d) Zinc-carbon cell/alkaline manganese cell/nickel-metal hydride rechargeable cell (any ONE) Zinc-carbon cell is cheap. while a lithium button cell can power a watch for about 5 years. (a) Lithium-ion cells are rechargeable. These are advantageous when they are used in © Aristo Educational Press Ltd. 30. They can provide a large and steady current. Since a watch consumes little energy. (d) The solar cell is not a chemical cell. so watches powered by solar cells are rare. (b) Calculators are always used when there is light. (a) Silver oxide cell/lithium button cell These cells are small enough to fit into a watch. (c) Zinc-carbon cell/alkaline manganese cell/nickel-metal hydride rechargeable cell (any ONE) Zinc-carbon cell is cheap. high-voltage of mobile phones./Nickel-metal hydride rechargeable cell is more environmentally friendly. (c) C Heat of fire may cause chemicals in cells to expand. so the expansion of chemicals under pressure may lead to explosion./Alkaline manganese cell is more expensive but more durable. The solar cell produces electricity when there is light. and has a high voltage.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (negative electrode) of the cells. and meets the demands for steady high-current. The secondary cell stores the excess electrical energy and powers the watch when there is no light. The electrical energy is not stored in the form of chemical energy. 28. The energy change involved in a solar cell is a physical change instead of a chemical change. Most cells are tightly sealed to prevent leakage. During recharging. The electrolyte can also cause skin burns. must continue to run even in the dark. It can be recharged for about 1200 cycles and a life span of about 3 years. Chemicals inside cells may be flammable (hydrogen produced at electrodes during discharging). The acidic/alkaline electrolyte can damage metal contacts/parts of the appliance. a silver oxide cell can power a watch for about 1–2 years. or oxidizing (depolarizer added) and can hence cause fire to burn more fiercely. (a) A silver oxide cell is normally made into a button-sized cell. It is most suitable for daily recharge irrespective of whether it has been fully discharged. (c) A solar cell powered watch must have another secondary chemical cell inside. These may include mercury in low-priced zinc-carbon cells. − It is a wet cell.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 portable electronic devices like mobile phones. − Other remaining chemicals in spent dry cells. Advantages of a lead-acid accumulator: − It provides a large current and a steady voltage. the following measures can be taken: − Use mains electricity instead of cells whenever possible. − It is rechargeable. Besides. since starting the car needs a large and steady current. difficult to carry. are wasted and cause burden to landfill sites. − Do not use mercury-containing zinc-carbon cells or alkaline manganese cells. For use in lowend electrical appliances like portable radios. Disadvantages of a lead-acid accumulator: − It is heavy and large in size. 31. 32. (b) Lithium-ion cells are expensive (each costs at least HK$200). − Use nickel-metal hydride rechargeable cells or lithium-ion rechargeable cells instead of nickel-cadmium rechargeable cells. − Dry cells contain toxic chemicals which may pollute the environment. and cadmium in nickel-cadmium rechargeable cells. so can be used for many years without the need of replacement. − Use rechargeable cells instead of disposable cells. hand torches (flashlights) and motorized toys. The acid electrolyte does not leak out unless the car overturns. frequent replacement of battery is inconvenient particularly when the car is travelling a long journey between cities. To minimize pollution. standard-sized and low-priced zinc-carbon cells or alkaline manganese cells are more affordable and more readily available. (Any FOUR) © Aristo Educational Press Ltd. turning the cell upside down may cause leakage of acid. 2010 20 . when disposed of. Suitability of using it as a car battery: − It is very suitable for use as a car battery. The car is powerful enough to carry the battery along. − Return expired rechargeable cells to the manufacturer or the Environmental Protection Department for recycling of the remaining chemicals. notebook computers and digital cameras. − The disadvantages do not affect its suitability for use as a car battery. Copper is more reactive than silver. Zinc is more reactive than copper. It is because copper is higher than silver in the E.1 (a) Copper to silver. A grey deposit forms on the copper rod and no voltage can be recorded in the multimeter. It is because electrons flow from copper rod to silver rod in the external circuit.S. (b) From left to right. © Aristo Educational Press Ltd. A30. the silver ions can come into contact with the copper electrode and a direct displacement reaction occurs on the copper surface.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 30 Simple chemical cells Class Practice A30. 2010 21 .C. Copper would lose electrons to silver when they are connected together.2 Mg/Ag > Zn/Ag > Fe/Ag > Pb/Ag > Cu/Ag A30. thus it loses electrons more readily. Electrons flow from zinc to copper.3 (a) Negative electrode. thus it loses electrons more readily. (c) Copper half-cell: − Cu(s) → Cu2+(aq) + 2e Silver half-cell: − Ag+(aq) + e →Ag(s) (d) Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) (e) If the porous pot is not used to separate the electrolytes. (b) Copper foil (c) Yes. HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 30 Simple chemical cells Chapter Exercise 1. dilute acids. cations. (ii) The iron electrode slowly dissolves and the green iron(II) solution becomes darker. 14. igital. 8.m. higher voltmeter. metals. 6. metals. Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) From iron electrode to copper electrode (or from left to right).f. 2010 22 . (a) Dilute sulphuric acid (or any other dilute acid. 4. complete. 10. or an acidic salt solution like ammonium chloride.). (d) (i) Bubbles appear at the copper electrode. electrolyte. (d) (e) © Aristo Educational Press Ltd. 13. (a) (b) (c) To complete the circuit by allowing ions to move from one half-cell to another half-cell. 3. 5. (ii) The zinc electrode dissolves gradually. 17. aqueous solution (c) reactivity.f. 11. 19.) (b) Positive electrode: copper electrode Negative electrode: zinc electrode (c) From zinc electrode to copper electrode (or from left to right). losing salt bridge. air. positive electromotive force (or e. ultimeter voltage (or e. charges porous pot D C C C B A C C A 18. 7. and to balance the charges in the solutions of the two halfcells. (e) (i) The copper electrode is quickly covered with a brown deposit. Positive electrode: copper electrode Negative electrode: iron electrode (i) A brown solid deposit forms on the copper electrode and the blue colour of the copper(II) solution becomes paler.). 2. 12. 15. 16. (ii) Bubbles appear at the zinc electrode and the zinc electrode dissolves gradually.m. Electrochemical Series (a) similar (b) lose. wire negative. 9. magnesium and zinc one by one in turn to the circuit as shown below: 22. (d) The voltmeter would give a positive reading. − clip metal strip under test silver sheet filter paper moistened with sodium chloride solution − − − − The metal strip under test and the silver sheet is separated by a piece of filter paper moistened with sodium chloride solution. Connect the metals aluminium. (b) The filter paper separates the two metal strips. Besides.HKDSE CHEMISTRY A Modern View (Chemistry) 2+ − Coursebook 3 20. (a) A metal couple is a combination of two different metals connected to an external circuit. Measure and record the voltage read out from the digital voltmeter for each metal tested. The zinc strip loses electrons and these electrons pass through the external circuit to the copper container. it allows ions to pass through to complete the circuit. (a) Zn(s) → Zn (aq) + 2e (b) Copper container − Cu2+(aq) + 2e → Cu(s) (c) The porous pot is used to separate the two electrolyte solutions and prevent them from direct mixing. (e) Removing the filter paper would cause a short circuit. copper. The higher the voltage reading. Once the zinc rod is coated with copper. The zinc strip is therefore the negative pole and the copper container is the positive pole. the higher position the metal is in the Electrochemical Series. Besides. preventing a short circuit. 21. thus the voltmeter reading drops to zero. Since the same reference electrode (silver) is used. (d) There would be a displacement reaction occurring at the surface of the zinc electrode immediately: Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq). 23 © Aristo Educational Press Ltd. This is because magnesium is in a higher position than copper in the Electrochemical Series. iron. so the voltage of the cell drops to zero. (f) The voltage would be higher. (e) Yes. lead. the salt solution allows movement of ions to complete the circuit. (c) From copper to silver. the Electrochemical Series of these metals can be obtained by arranging the metals in the order of voltage reading. both electrodes become the same (copper electrodes) and the electrolyte is also the same. 2010 . N.N. of N) + (− × 3 = 0 1) ∴ O.N.2 (a) O. Defined in terms of oxygen electron oxidation number 2. of S = 0 (b) (+2) × 3 + (O. of S) + (− × 3 = 0 2) ∴ O.N. of N = +3 A31. (iii) & (iv): not applicable (c) (i) Redox (ii) Cl2 (iii) FeSO4 (or Fe2+) (iv) Iron © Aristo Educational Press Ltd. of S) + (− × 4 = 0 2) ∴ O. of S = +4 (e) (+1) × 2 + (O. of N) + (+1) × 4 = +1 ∴ O. of Cr = +6 (i) (O.N.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 31 Redox reactions Class Practice A31.N. A31. of Cr = +6 (f) For NH4+. of C = +4 (h) (O. of N) × 2 = 0 ∴ O.1 (a) It involves a transfer of electrons (from Fe(s) to Cu2+(aq)). (b) Fe(s) is being oxidized.N. 2010 24 . of Cr) × 2 + (− × 7 = − 2) 2 ∴ O.3 1. of N = − 3 (c) (+1) × 2 + (O. of C) + (− × 3 = − 2) 1 ∴ O.N. (c) Cu2+(aq) is the oxidizing agent. Fe(s) loses electrons to others.N. of N = − 3 (g) (+1) + (O. of S = +6 (d) (+1) × 2 + (O. of Cr) + (− × 4 = 0 2) ∴ O.N.N.N.N.N. Cu2+(aq) gains electrons (or Cu2+(aq) is reduced). Oxidation +O − −e increases Reduction −O − +e decreases (a) (i) Redox (ii) O2 (iii) CH4 (iv) Carbon (b) (i) Not redox (ii).N.N. (O.N. 6 (a) Ag(s) + 2HNO3(aq) →AgNO3(aq) + NO2(g) + H2O(l) − (b) Ag(s) + 2H+(aq) + NO3 (aq) →Ag+(aq) + NO2(g) + H2O(l) A31. (a) Zn to Cu (b) Mg to Ag (c) Zn to Pb © Aristo Educational Press Ltd.5 (a) K+(aq) − (b) F (aq) (c) Fe2+(aq) A31. Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) 2. However. 2010 25 . (b) C(s) + H2SO4(l) → CO2(g) + 2SO2(g) + 2H2O(l) A31.8 1.7 (a) No reaction. the second statement is not a correct explanation of the first. Both statements are correct. A correct explanation would be: ‘The oxidation number of nitrogen in ammonia can be increased when ammonia reacts with a strong oxidizing agent.’ Cl2(g) + 2e → 2Cl (aq) − − 2Br (aq) → Br2(aq) + 2e − − Cl2(g) + 2Br (aq) → 2Cl (aq) + Br2(aq) − − MnO4 (aq) + 8H+(aq) + 5e (aq) → Mn2+(aq) + 4H2O(l) − Fe2+(aq) → Fe3+(aq) + e − MnO4 (aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) − − A31.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 3.4 (a) (i) (ii) (iii) (b) (i) (ii) (iii) A31. 3. 27. 23. 7. donating (a) 0 (b) ionic charge (c) 0 (d) ionic charge (a) increases (b) decreases Electrochemical (a) weak (b) increases (c) strong (d) oxidizing (e) reducing (f) strong (g) decreases (h) weak oxidizing. 6. 4. 22. 21. oxidizing. 20. 19. 14. 28. 25. 2. SO2(g) B C D D C C C C D B B B D D B C B C C A 5. NO2(g). 16. 15. 9. transfer loses. 17. 13. 8. 2010 26 . © Aristo Educational Press Ltd. accepting (b) reduces. gains (a) oxidizes. 26. 24. 12.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 31 Redox reactions Chapter Exercise 1. 11. 10. 18. brown fumes appear at the mouth of the test tube. and a yellow solution is formed. 32. zinc is oxidized. and the solution turns blue. (e) Redox reaction. (Chlorine is both oxidized and reduced in this reaction. (d) Redox reaction. (c) Not a redox reaction. reducing agent: NO. reducing agent: C. reducing agent: Cl2.) 31. Oxidizing agent: O2. (ii) The orange dichromate solution becomes green. (d) If the plug of cotton wool is removed. and bromine is formed which is yellow. (a) (i) MnO4 (aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(l) − + 5 × (Fe2+(aq) → Fe3+(aq) + e ) ____________________________________________________________ − MnO4 (aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) − − (ii) Cr2O72 (aq) + 14H+(aq) + 6e → 2Cr3+(aq) + 7H2O(l) − − + 3 × (2Br (aq) → Br2(aq) + 2e ) ____________________________________________________________ − − Cr2O72 (aq) + 14H+(aq) + 6Br (aq) → 2Cr3+(aq) + 3Br2(aq) + 7H2O(l) (b) (i) The purple permanganate solution is decolorized.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 29. − − © Aristo Educational Press Ltd. Copper is the reducing agent. so the nitrogen monoxide formed will not be converted to nitrogen dioxide. (a) (b) (c) (d) (e) (f) (g) (h) (i) 0 +2 +3 − 3 +4 +6 +7 +5 +1 30. (a) Colourless bubbles are evolved from the copper turnings. chlorine is oxidized. nitrogen monoxide is oxidized. (b) Redox reaction. 2010 27 . carbon is oxidized. Oxidizing agent: CuSO4. (a) Redox reaction. It is reduced as the oxidation number of nitrogen changes from +5 to +2. (b) 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(aq) +4H2O(l) (c) Dilute nitric acid is the oxidizing agent. Oxidizing agent: Cl2. reducing agent: Zn. The resultant solution looks yellowish-green. It is oxidized as the oxidation number of copper changes from 0 to +2. Oxidizing agent: HNO3. (e) 2NO(g) + O2(g) → 2NO2(g) (f) The cotton wool plug prevents oxygen in air from entering the test tube. (e) The water in the moistened broken porcelain chips can absorb unreacted hydrochloric acid vapour (hydrogen chloride gas). 2010 28 . the chlorine gas would leave the flask through the funnel rather than through the side arm to react with the potassium iodide. (a) Bubbles form immediately. It is reduced as the oxidation number of chlorine changes from 0 to –1.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 33. Otherwise. − − (b) 2MnO4 (aq) + 16H+(aq) + 10Cl (aq) → 2Mn2+(aq) + 5Cl2(aq) + 8H2O(l) (c) Permanganate ion is the oxidizing agent. and green fumes are evolved. It is reduced as the oxidation number of manganese changes from +7 to +2. It is oxidized as the oxidation number of iodine changes from –1 to 0. © Aristo Educational Press Ltd. − − (g) Cl2 + 2I → 2Cl + I2 (h) Chlorine is the oxidizing agent. (i) The experiment should be carried out in the fume cupboard as chlorine is a poisonous gas. It is oxidized as the oxidation number of chlorine changes from –1 to 0. (f) The potassium iodide crystals turn black/dark brown. Potassium iodide is the reducing agent. Hydrochloric acid is the reducing agent. which is highly soluble in water. (d) The stopcock should be closed. 1 (a) Anode: zinc electrode Cathode: silver electrode (b) Silver electrode (cathode) (c) Zinc electrode (anode) (d) Ionic half equation for the oxidation: − Zn(s) → Zn2+(aq) + 2e Ionic half equation for the reduction: − Ag+(aq) + e →Ag(s) A32. U is the negative electrode. (c) No.3 1. Br (aq) loses electrons at electrode U and is oxidized to Br2(aq). Hence. carbon does not participate in the electrode reaction. (e) Ionic half equation for reaction at electrode U: − − 2Br (aq) → Br2(aq) + 2e Ionic half equation for reaction at electrode V: − − MnO4 (aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(l) − − (f) 2MnO4 (aq) + 10Br (aq) + 16H+(aq) → 2Mn2+(aq) + 5Br2(aq) + 8H2O(l) A32. A metal-metal ion system is set up. 2010 29 . As shown by the half equation. Furthermore. (Any ONE) (c) Methanol is flammable. It is because − − the oxidizing agent MnO4 (aq) will gain electrons while the reducing agent Br (aq) will lose electrons. (a) A paste of ammonium chloride is used so that ions are mobile and can conduct electricity. © Aristo Educational Press Ltd. − (c) V is the positive electrode and U is the negative electrode. Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) 2. U is the anode and V is the cathode.4 − (a) At anode: CH3OH + H2O → 6H+ + CO2 + 6e − At cathode: 3O2 + 12H+ + 12e → 6H2O (b) Methanol is a liquid which is easier to handle than gaseous hydrogen during refilling. methanol is a colourless liquid like water. (b) If the cell is dry. As shown by the half equation. the ions inside are no longer mobile. (e) At the zinc cup (the anode).HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 32 Redox reactions in chemical cells Class Practice A32. (b) Oxidation takes place at electrode U. A32. it may catch fire./Methanol poses a lower risk of explosion than hydrogen. zinc dissolves to form zinc ions. Electrons thus move in the external circuit from U to V. Therefore. zinc participates in the electrode reaction. Electricity cannot pass through and so the cell will not produce electrical energy. if carelessly handled. If it is not stored or labelled properly.2 (a) Electrons flow from electrode U to electrode V in the external circuit. there is a danger of accidental poisoning. (d) At electrode U: solution around electrode turns yellow/brown. At electrode V: solution around electrode becomes pale purple (partially decolorized). yet it is highly poisonous. (d) Yes. 6. oxidation. cathode − Zn(s) → Zn2+(aq) + 2e − 2MnO2(s) + 2H+(aq) + 2e → Mn2O3(s) + H2O(l) secondary − − Pb (s) + SO42 (aq) ⇌ PbSO4(s) + 2e − − PbO(s) + 4H+(aq) + SO42 (aq) + 2e ⇌ PbSO4(s) + H2O(l) primary hydrogen. − (d) Fe2+(aq) → Fe3+(aq) + e (e) The colour of the iron(II) sulphate solution slowly changes from pale green to yellow.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 32 Redox reactions in chemical cells Chapter Exercise 1. 8. 2010 30 . − 19. (e) During storage. 12. the reaction at the cathode cannot take place. 4. 13. 2. oxygen oxidant. Therefore. (f) Electrons flow from the right electrode to the left electrode in the external circuit. © Aristo Educational Press Ltd. 10. 3. The cell will stop functioning. (a) At the zinc electrode (anode): − Zn(s) → Zn2+(aq) + 2e At the copper electrode (cathode): − Cu2+(aq) + 2e → Cu(s) (b) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (c) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (d) Yes. catalyst B A D D D D C C B 5. all the NH4+(aq) ions will be used up. 14. 7. zinc can be used up even when the cell is not being used. (a) Dilute sulphuric acid − − (b) Cr2O72 (aq) + 14H+(aq) + 6e → 2Cr3+(aq) + 7H2O(l) (c) The colour of the acidified potassium dichromate solution slowly changes from orange to green. (a) (b) (c) (d) 2MnO2(s) + 2H+(aq) + 2e → Mn2O3(s) + H2O(l) − Zn(s) → Zn2+(aq) + 2e Ammonium chloride After a zinc-carbon cell has been used for a certain time. 15. 17. 9. Therefore. 18. cathode negative. No more H+(aq) ions are available. the zinc casing of the cell will react with H+(aq) ions present in the electrolyte. 11. 16. they are the same. − − H2(g) + 2OH (aq) → 2H2O(l) + 2e Oxidation. (2) In the Daniell cell. Conventional cells either run out (primary cells) and need to be discarded. They can be used basically forever (by refilling the fuel). the reaction generates energy mainly in the form of electricity. In the beaker. There is no direct mixing of the chemicals. the Cu2+ (aq) and Zn(s) are in direct contact and solid copper deposits directly on the surface of the zinc rod. (a) (b) (c) (d) (e) (f) (g) Hot. concentrated potassium hydroxide solution. In the beaker. © Aristo Educational Press Ltd.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (e) (1) In the Daniell cell. Solid copper deposits on the surface of the copper container. − − O2(g) + 2H2O(l) + 4e → 4OH (aq) 2H2(g) + O2(g) → 2H2O(l) The electrons flow from the left hand side to the right hand side (clockwise) in the external circuit. Cu2+(aq) and Zn(s) are separated by a porous partition. or need recharging (secondary cells) which is timeconsuming. It is because the oxidation number of H changes from 0 to +1. (h) The fuel of fuel cells can be refilled easily. 2010 31 . As a catalyst. the reaction generates energy mainly in the form of heat. 20. © Aristo Educational Press Ltd. 2010 32 .5 (a) At copper cathode: hydrogen At platinum anode: oxygen (b) At graphite cathode: hydrogen At graphite anode: oxygen (c) At graphite cathode: copper At graphite anode: chlorine (d) Electrolysis products cannot be predicted because the concentration of solution and material of electrodes used are not specified. at anode: chlorine (ii) No electrolysis takes place. A33. Acidic effluents usually contain sulphuric acid. OH (aq). OH (aq). The blue colour of the solution becomes paler. H+(aq). (b) (a) (b) (c) (d) (e) A33. metal. H+(aq). 2. − At cathode: Mg2+(l) + 2e → Mg(l) − − At anode: 2Cl (l) → Cl2(g) + 2e A33. concentration A33.3 (a) Cations: Na+(aq).HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 33 Electrolysis Class Practice A33. if it is economical to do so. (b) Molten magnesium chloride contains mobile ions that conduct electricity while solid magnesium chloride does not conduct electricity. (c) For (a)(i).6 1. Besides. Remove the metal ions from effluents by chemical treatment. oxygen.1 (a) (i) At cathode: magnesium. hydrogen. cation. anions: OH (aq) (b) (i) Hydrogen (ii) Oxygen (c) Volume ratio of hydrogen : oxygen = 2 : 1 A33. so it would prevent the neutralization reaction from going on.2 − − H+(aq). (a) (i) At cathode: Cu2+(aq) + 2e → Cu(s) − At anode: Cu(s) → Cu2+(aq) + 2e (ii) Overall equation: not applicable (iii) The intensity of blue colour of the solution remains unchanged. chlorine sodium amalgam oxygen ionizes (dissolves) Cu2+(aq) − − 2. halide ions. it is difficult to remove the insoluble calcium sulphate. The calcium sulphate formed is only slightly soluble in water.4 1. SO42 . 5. OH . (a) Ions present: H+. decomposition chemical. anode. product at anode: O2 − − (b) Ions present: H+.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 33 Electrolysis Chapter Exercise 1. 20. 19. At the anode: Some brown gas bubbles appear. Product at cathode: H2. electroplated. health Environmental Protection C A B B A D A B 21. OH . heavy metal. anode. 3. product at anode: O2 − − (c) Ions present: H+. electricity cathode.. 2.S. Cu2+. 2010 33 . OH . 6. anode. − (d) At the cathode: Pb2+(l) + 2e → Pb(l) − − At the anode: 2Br (l) → Br2(g) + 2e 22. 14. concentration. product at anode: Cl2 − − (d) Ions present: H+. 10. 16. cathode E. OH . product at anode: O2 − − © Aristo Educational Press Ltd. Product at cathode: Cu. corrosion cathode anode thicken lkalis. electrodes electroplating. product at anode: O2 − − (e) Ions present: H+. 4. gain. anode positive ions. Product at cathode: H2. 17. 12. Na+. 8. 18. 7. OH . cathode. 13. Cl . Na+. (a) Carbon (b) Lead(II) ions and bromide ions (c) At the cathode: A silver colour appears. cathode. lose. 9. 15. Cl . Product at cathode: H2. rganic. Cl . Cu2+. 11.C. SO42 . Product at cathode: Cu. when 2 moles of electrons pass through the electrolytic cell. 2010 34 . (3) The electrolyte should be an aqueous solution of nickel(II) sulphate instead of dilute sulphuric acid. (3) Increase the concentration of the electrolyte. At the cathode. OH (aq) is preferentially − discharged as it is a stronger reducing agent than Cl (aq). (b) Two methods to shorten the time needed for electrolysis: (1) Use a larger current (higher voltage). when hydrogen ions are discharged. source is the anode. the actual volume of chlorine collected is less than the volume of hydrogen collected. (c) The relative volume of gases shown in the diagram is reasonable. (a) At electrode D: − − 2Cl (aq) → Cl2(g) + 2e At electrode E: − 2H+(aq) + 2e → H2(g) (b) Judging from the equations in (a). (c) In electrolytic cell A: − − 4OH (aq) → O2(g) + 2H2O(l) + 4e In electrolytic cell B: − Cu(s) → Cu2+(aq) + 2e (d) This is due to the difference in material used to make the electrodes. the solution near the anode turns red and then colourless. (a) The three mistakes are: (1) The door handle should be connected to the negative terminal (cathode). the electrode connected to the positive pole of the d.c. chlorine gas is produced. Cu(s) is preferentially discharged. Carbon is used as electrodes for electrolysis of halides as there is no direct reaction between carbon and the halogens.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 23. 1 mole of chlorine is liberated at the anode and at the same time 1 mole of hydrogen is liberated at the cathode. (a) Graphite (carbon) electrodes should be used. During electrolysis. When an − inert electrode (graphite or platinum) is used. During electrolysis of halides. (d) At the anode. (2) Increase the surface area of the piece of nickel metal connected to the anode. The solution near the cathode turns blue. (2) The piece of nickel metal should be connected to the positive terminal (anode). (b) The left electrode. The theoretical ratio of the volumes of gases collected over the electrodes should be 1:1. if platinum electrodes are used. Since hydrogen is insoluble in water whereas chlorine is quite soluble in water. (f) There would be colourless bubbles (of oxygen) appearing on the surface of the graphite anode. Since chlorine water is acidic and bleaching. When copper metal is used as the anode. (e) The copper anode becomes smaller. when the chloride ions are discharged. 25. 24. (Any TWO) © Aristo Educational Press Ltd. an excess of hydroxide ions are present. Cu(s) is a stronger − − reducing agent than OH (aq) and Cl (aq). they can be easily corroded by the respective halogens produced at the anode. of aluminium.g. nickel A thin layer of protective aluminium oxide coating on iron) is coated on the item. − Both processes involve electrolysis with the purpose of beautifying and/or strengthening the surface of metal items. The item to be electroplated is connected Aluminium item to be anodized is to the cathode during electroplating. © Aristo Educational Press Ltd. Similarities: − Both electroplating and anodizing are industrial processes. connected to the anode during anodizing. The oxide layer can be dyed to a to another colour. A thin layer of another metal (e. the golden colour of gold the aluminium item retains its own silvery plating). The item acquires the colour of the coating The oxide layer is too thin to be seen and metal (e.g. Differences: Electroplating Anodizing Can be applied to most metal items (e.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 26. new colour if necessary.g. Can only be applied to metal items made iron or copper items) and non-metal items. The item cannot be further dyed colour. 2010 35 . is formed on the aluminium item. If the seal of a cell with a lithium metal anode is broken. so it is more suitable for on-site energy production for a block of flats.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 34 Importance of redox reactions in modern ways of living Class Practice A34. bad smell. In other words. and a lot of NOx and SO2. like other alkali metals (sodium. producing smoke. causing hydrogen and alkaline solution to leak out.) reacts vigorously with water to produce hydrogen and a corrosive. strongly alkaline solution LiOH. a diesel generator consumes more fuel to produce the same quantity of heat and electricity as compared to a fuel cell. potassium. (a) At cathode: O2(g) + 4H+(aq) + 4e → 2H2O(l) − At anode: 2H2(g) → 4H+(aq) + 4e (b) Overall equation of the cell reaction: 2H2(g) + O2(g) → 2H2O(l) A34. Hydrogen may cause explosion and the alkaline solution can cause severe skin burns. water or even moisture in the air may react with lithium. © Aristo Educational Press Ltd. − 2.1 1.2 Lithium metal. This again is better for on-site power production. (1) A diesel generator has a lower efficiency than a fuel cell system. etc. (2) A diesel generator causes pollution to the environment. (3) A diesel generator is very noisy while a fuel cell operates quietly. A fuel cell system is clean and the exhaust is non-polluting. 2010 36 . 14. lithium-ion rechargeable. 8. hotosynthesis. the only exhaust is water). (b) A fuel cell is better than a secondary cell in the way that recharging secondary cells usually takes hours (2 hours for lithium-ion cells. A fuel cell can in theory work forever because it can continue to work as long as the fuel is replenished. Besides. methanol for DMFC) can be easily carried around whereas electrical power for recharging secondary cells may not be available in remote locations. metal oxide. Proton Exchange Membrane Fuel Cells. (2) Environmentally friendly (for hydrogen fuel cells. harge density. Molten Carbonate Fuel Cells. Phosphoric Acid Fuel Cells. nvironmentally preferred A A D A A 13. ntioxidants. recharged.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 34 Importance of redox reactions in modern ways of living Chapter Exercise 1. 9. lithium. 5. the dichromate is reduced and changes from orange to green: 2K2Cr2O7(aq) + 3CH3CH2OH(aq) + 8H2SO4(aq) → 2Cr2(SO4)3(aq) + 3CH3COOH(aq) + 2K2SO4(aq) + 11H2O(l) © Aristo Educational Press Ltd. A fuel cell is not a secondary cell because it cannot be recharged by using an external power source. refilling the fuel of a fuel cell may just take seconds. lithium-ion polymer rechargeable oxidized. (a) It is commonly used to measure the amount of ethanol present in the breath of drivers to help police collect evidence for drink driving prosecutions. 11. reduced lithium-carbon compound. 2010 37 . 6. 3. 12. rain capacity. leaching Alkaline Fuel Cells. organic solvent. (c) (1) High efficiency. 10. (3) Convenient (easy to refill the fuel). a small bottle of fuel (say. reathalyzers. If alcohol is present in a driver’s breath. (a) The statement is half correct and half incorrect. 4. Instead. oltage. 2. and about 8 hours for nickel-cadmium cells). discharging CoO2 + xLi + xe discharging charging + − ⇌ − LixCoO2 charging LixC6 ⇌ 6C + xLi+ + xe discharging charging CoO2 + LixC6 ⇌ LixCoO2 + 6C 7. etabolism ermentation. (b) A conventional breathalyzer contains a solution of acidified potassium dichromate. Solid Oxide Fuel Cells lithium. (2) The EC/IR intoximeter is more environmentally friendly (no need to dispose of spent chemicals). (e) Generating great power/providing power at a relatively low cost/high efficiency/non-polluting/providing hot water or heat on-site (any TWO) 16. electrolyte to leak and the cell may even explode. − Lithium-ion cells do not contain/contain less heavy metals that pollute the environment like zinc-carbon cells or nickel-cadmium cells. 2010 38 . − The unit cost of lithium-ion cells is currently several times higher than that of other cell types. Advantages of lithium-ion cells: − Lithium-ion cells are rechargeable and can provide high voltage (3. A higher concentration of ethanol produces a larger current. − Lithium-ion cells are comparatively lightweight (high charge density). (d) The exhaust of this fuel cell is steam (at 150°C to 200°C). It can be used to heat up water (for hot water supply) or heat up the air (for warming air in winter). using a lithium-ion cell may cause the cell to heat up. This makes the charging process more economical and more environmentally friendly.7 V) and large current.6 V. − Lithium-ion cells must be recharged with a dedicated charger. − After recharging for more than 1200 cycles. (a) At cathode: O2(g) + 4H+(aq) + 4e → 2H2O(l) − At anode: H2(g) → 2H+(aq) + 2e (b) 2H2(g) + O2(g) → 2H2O(l) (c) To start up a cold PAFC.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (c) The alcohol in the breath of the driver. linear reading of the breath alcohol concentration of a driver. so discharging fully before recharging is unnecessary. which may not be good enough for legal procedures. They may also explode if overheated or if charged to an excessively high voltage. is used as the fuel in an ethanol fuel cell. The colour change of the conventional breathalyzer can only give an approximate data. when blown into the intoximeter. (e) An infrared (IR) sensor 15. it must first be heated to a temperature well above 40°C (the melting point of phosphoric acid) before the PAFC can function. (Any THREE) − © Aristo Educational Press Ltd. (d) (1) An EC/IR intoximeter can give a continuous. and if the charger is not available. − Lithium-ion cells have no memory effect. the lithium-ion cell cannot be recharged. thus the intoximeter can give a reading that is proportional to the breath alcohol concentration of the driver. (Any THREE) Disadvantages of lithium-ion cells: − Lithium-ion cells cannot be used for appliances that need a voltage lower than 3. Zinc loses electrons more readily than copper. of Zn = +2). so O2 is preferentially discharged. 3. 6. MnO2 (O. 15. − − − Besides.N. 8. of Zn = 0) is oxidized to ZnO (O. Zn (O. F is a weaker reducing agent than O2 . calcium fluoride added does not affect the result of the electrolysis. of Mn = +4) is reduced to Mn2O3 (O.N. A B D B B B C B D A C A D A D manganese(IV) oxide and powdered graphite. Oxidation occurs at the anode. Thus. so electrons flow from zinc to copper in the external circuit.N. 14. ammonium chloride paste − Zn(s) → Zn2+(aq) + 2e − 2MnO2(s) + 2H+(aq) + 2e → Mn2O3(aq) + H2O(l) From zinc to copper. 13. (a) To lower the melting point so that energy and cost for heating up the ore can be saved. 12. so Al3+ is preferentially discharged. (b) Ca2+ is a weaker oxidizing agent than Al3+. 2. Therefore. of Mn = +3). 4. − − 16. it acts as an electrolyte to complete the circuit by movement of ions. The lemon juice (citric acid solution) contains mobile H+(aq) ions. (a) (i) Zn(s) + 2OH (aq) → ZnO(s) + H2O(l) + 2e − − 2MnO2(s) + H2O(l) + 2e → Mn2O3(s) + 2OH (aq) (ii) Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s) (b) (i) Electrode A is the cathode. (c) The alkaline manganese cell is often preferred to the less expensive zinccarbon cell because the alkaline manganese cell has a much longer service life/higher charge capacity.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Part VII Redox reactions. (d) The silver oxide cell is commonly used in electronic or electrical devices which need a small-sized cell with a steady voltage. 9. and electrolysis Part Exercise 1. 11.N. (ii) Electrode B is the anode. chemical cells. 2010 39 . 5. Reduction occurs at the cathode. 18. 10. (f) Zinc metal is the anode as it is oxidized to Zn2+(aq) ions in the cell reaction. − (c) Al3+(l) + 3e →Al(l) − − (d) 2O2 (l) → O2(g) + 4e (e) Aluminium is too high in the reactivity series to be reduced by other reducing © Aristo Educational Press Ltd. (a) (b) (c) (d) (e) 17. 7. (f) At 850°C. The lead plate is the anode because lead is oxidized to lead(II) sulphate (oxidation number of Pb increases from 0 to +2). (h) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l) (i) The limewater will turn milky but the dichromate solution will have no observable change. They will be both removed by bottle B. The car battery can produce a large current and may cause electric sparks and heat burns if short-circuited. − (c) Pb(s) + PbO2(s) + 4H+(aq) + 2SO42 (aq) ⇌ 2PbSO4(s) + 2H2O(l) (d) The car battery is a secondary cell because the overall cell reaction (as shown above) is reversible. The lead plate is the negative electrode as it gives out electrons for oxidation to take place.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 agents like carbon. It can be reversed by applying an external voltage across the electrodes. (g) The limewater turns milky. (2) Beware of acid burns by sulphuric acid. the oxidation number of S increases from +4 to +6. 19. (Any TWO) © Aristo Educational Press Ltd. 20. This is because both SO2 and CO2 are acidic gases. (b) The lead plate coated with lead(IV) oxide is the positive electrode as it takes in electrons for reduction to take place. (f) (1) Wear gloves and goggles. (3) Beware of electric shock. (e) The car battery has a voltage of 12 V. This is achieved by combining six cells in series. (3) Beware of heat burns. (d) The colour of the acidified potassium dichromate solution changes from orange to green. (e) K2Cr2O7(aq) + H2SO4(aq) + 3SO2(g) → Cr2(SO4)3(aq) + K2SO4(aq) + H2O(l) (f) This is a redox. and the oxidation number of S decreases from +6 to +4. (2) Perform the experiment in a fume cupboard (in case of leakage of toxic sulphur dioxide). and the oxidation number of Cr decreases from +6 to +3. 2010 40 . (b) 2H2SO4(l) + C(s) → CO2(g) + 2SO2(g) + 2H2O(l) (c) This is a redox. the carbon electrode burns in the oxygen formed at the anode to give carbon dioxide. (a) The lead plate coated with lead(IV) oxide is the cathode because the lead(IV) oxide is reduced to lead(II) sulphate (oxidation number of Pb decreases from +4 to +2). (j) (1) Wear gloves and goggles (concentrated sulphuric acid is highly corrosive). During the reaction. (a) Colourless bubbles are formed from the carbon. the oxidation number of C increases from 0 to +4. During the reaction. No gas will enter bottle A. Then the battery can be recharged. HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 21. magnesium combining with oxygen undergoes oxidation while copper(II) oxide losing oxygen undergoes reduction. − Oxidation reactions involve addition of oxygen while reduction reactions involve removal of oxygen. − For example. Methanol is toxic and is highly flammable. Mg(s) + CuO(s) → MgO(s) + Cu(s) − Oxidation reactions involve loss of electron(s) while reduction reactions involve gain of electron(s). magnesium undergoes oxidation due to the increase in its oxidation number from 0 (Mg(s)) to +2 (Mg2+(aq)) while copper(II) ion undergoes reduction due to the decrease in its oxidation number from +2 (Cu2+(aq)) to 0 (Cu(s)). This is ideal for powering electronic products. (a) (b) (c) (d) (e) O2(g) + 4H (in PEM) + 4e → 2H2O(l) 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l) Methanol is easier and safer to carry and handle than hydrogen gas. magnesium losing 2 electrons undergoes oxidation while copper(II) ion gaining 2 electrons undergoes reduction. − For example. which is more environmentally friendly. it is not suitable for powering automobiles. (f) DMFCs can be made into a small size and can produce a small current for a long time. + − 22. DMFCs produce the greenhouse gas. 2010 41 . it cannot provide a large current. Therefore. carbon dioxide as one of its exhaust. Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) − Oxidation reactions involve the increase in the oxidation number while reduction reactions involve the decrease in the oxidation number. Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) © Aristo Educational Press Ltd. However. − For example. Hydrogen fuel cells only produce water as the only exhaust. two moles of gaseous reactants (H2 and Cl2) react to give two moles of product (HCl). In the reaction. Energy is released when water molecules come close together during condensation. © Aristo Educational Press Ltd. (c) Endothermic. Energy is needed to break the covalent bond between chlorine atoms. (b) Endothermic. Energy is needed to overcome the intermolecular forces (hydrogen bonds) between water molecules. 2010 42 . A35. (e) Exothermic.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 35 Energy changes in chemical reactions Class Practice A35. Energy is needed to overcome the attraction between the nucleus of the sodium atom and its outermost shell electron. Heat change at constant volume is equal to the change in internal energy.1 Yes.2 (a) Exothermic. The total gas volume remains constant. (d) Endothermic. Energy is released when the nucleus of chlorine atom attracts the incoming electron. hydrogen bond) between water molecules.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 35 Energy changes in chemical reactions Chapter Exercise 1. 11. 9. give out bond-forming. 14. 10. pressure gives out takes in products. 8. The change in internal energy is equal to the heat change at constant volume. 16. 13. set-up (C) releases less heat.e. neutralization oil fractions take in. 7. (c) (C). energy is needed to overcome the intermolecular forces (i. 6. (A) 18. 2. bond-breaking smaller D A A D B D (a) Set-up (A) releases more heat as the reaction is carried out in a closed system. 4. Set-up (B) is open to the atmosphere. (a) Endothermic process (b) The energy needed for overcoming the ionic bonds in potassium sulphate and intermolecular forces between water molecules cannot be compensated by the energy released from the formation of chemical bonds between water molecules and potassium and sulphate ions. set-up (C) has less magnesium metal (the limiting reactant). (c) Cooling of hot engines 19. recipitation. (c) Treatment of athletes’ injuries 20. Some energy is used as the work done on the surroundings. 2010 43 . H C=O and O− H The enthalpy change in bond-forming processes is larger than the enthalpy change in bond-breaking processes. reactants lower ombustion. 15. 3. 17. 12. (b) For the same amount of acid used. (a) Endothermic process (b) During the process of evaporation. © Aristo Educational Press Ltd. Therefore. (B). (a) b) (c) (d) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) C− C− and O=O C. 5. 2 −(− 57.75 kJ.9 kJ mol 1 = − 45.3 × 2 kJ mol 1 = − 184.1) kJ mol 1 = 4.9 −(− 57.0 kJ mol 1 A36.0 × 2 12.0 × 4 Y = 22. (a) ∆H = − 92. (b) Let X be the mass of hydrogen in the mixture. 1 (a) C8H18(l) + 8 O2(g) → 8CO2(g) + 9H2O(l) 2 ∆H c = − 5512 kJ mol 1 (b) CH3COOH(aq) + NH3(aq) → CH3COONH4(aq) − ΔH neut = − 51.0 × 2 + 1.3 g and 27. Y 50 − Y × − 285.2 kJ mol 1 Heat required for the ionization of NH3 − − =− 52.5 kJ mol 1 2.2 1.4 = × 100% 50 = 50.0 + 1.8% 2.8 + × − 890.8 + × − 890.5) kJ mol 1 = − 614. solution and formation Class Practice A36.1 11. (a) Let Y be the mass of hydrogen in the mixture. X 50 − X × − 285.75 kJ 12. the heat evolved is 342.0 × 6 + 16.9 kJ mol 1 Estimated ΔH neut for CH3CH2COOH and NH3 − − =− 57. neutralization. the percentage by mass of hydrogen in the mixture 25.6 kJ mol 1 (b) ∆H = − 238.5 × 2 + (− 393.2 kJ mol 1 − (c) ∆H = +1.3 − − 1. 2010 44 − − − .6 × 2 kJ mol 1 = − 477.0 × 2 12.4 = − 5000 1. (a) CH3CH2COOH (b) NH3 (c) Heat required for the ionization of CH3CH2COOH − − =− 49.5 kJ = − 342.9 kJ mol 1 − − (d) ∆H = − 110.3 Thus.0 Hence. Heat change = − 1371 × A36.7 g respectively.0 × 4 X = 25.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 36 Standard enthalpy change of combustion.2 + 4.5 kJ mol 1 − (c) KBr(s) + aq → K+(aq) + Br (aq) − ΔH soln = +20.0 kJ mol 1 © Aristo Educational Press Ltd.1 + 7. the mass of hydrogen and methane are 22.4 Thus.4 = − 4725 1.1) kJ mol 1 = 7.0 + 1. 0 + 273) −(25.0 kJ mol 1 A36. Heat transferred to solution = m × c × (T2 −T1) − − = 45.06 J mol 1 6.025 A36.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (d) 2B(s) + 3H2(g) → B2H6(g) − ΔH f = +36. the enthalpy change of combustion of methanol is − 629 kJ mol 1. (b) To make sure that the solution is at infinite dilution.0 g × 4.0 g mol −1 Heat released per mole of methanol burnt 52 080 J − − = = 62 900 J mol 1 = 629 kJ mol 1 0.67 Number of moles of HCl dissolved = mol = 0.0 × 4 + 16.65 g = = 0.2 J g 1 K 1 × 4.2 J g 1 K 1 × 3.9 + 35. A36.0)(1) g × 4.0458 mol 1 +35.0 − = 0.025 mol (Sodium hydroxide is the limiting reactant.5 mol dm 3 × dm3 1000 = 0.0828 mol 12.5 − − = − 288 J mol 1 (− 36 36.0 + 1.89 K = 816.5 Heat released per mole of HCl dissolved 816.4 Heat transferred to water = mw × c × (T2 −T1) − − = (500)(1) g × 4.5 Heat transferred to solution = m × c × (T2 −T1) − − = (75.67 g 60 1.1 + 273) −(25.2 − ΔH soln = − 1.3 + 273)] K = 52 080 J Number of moles of methanol burnt 2.9 J 10 − Mass of HCl dissolved = 10 g min 1 × min = 1. 2010 45 .0 + 50.0828 mol − Hence. © Aristo Educational Press Ltd.2 J g 1 K 1 × [(27.3 kJ mol 1) 2.2 J g 1 K 1 × [(50.0 + 273)] K = 1050 J Number of moles of water formed 50.6 1. the enthalpy change of solution of hydrogen chloride is − 836 J 17 − mol 1. (a) Heat transferred to solution = m × c × (T2 −T1) − − = (50)(1) g × 4.) 1050 − − − ΔH neut = − J mol 1 = − 000 J mol 1 = − kJ mol 1 42 42 0.8 K = 907.0458 mol Hence.2 J 907.9 J − = = 17 836 J mol 1 0. concentration moles combustion neutralization. 10. standard conditions heat loss. 8. neutralization.51 g (b) A larger mass is needed.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 36 Standard enthalpy change of combustion.2 J g 1 K 1 × [(70 + 273) − (20 + 273)] K = 84 000 J (84 kJ) 84 No. 1 (b) 2Al(s) + 1 O2(g) → Al2O3(s) 2 + 3350 − − (c) − kJ mol 1 = − 1675 kJ mol 1 2 1 − (d) 2Al(s) + 1 O2(g) → Al2O3(s) ΔHf = − 1675 kJ mol 1 2 17.0418 × (12. 15. infinite dilution formation.0) g = 2.water solution. 12. 2. solution and formation Chapter Exercise 1. 5. heat. 1. © Aristo Educational Press Ltd.0418 mol 2010 Mass of propan-1-ol needed = 0. 11.0 × 3 + 1. 101 325. 7. 298. (a) Energy needed = mcΔT − − − = 400 cm3 × 1 g cm 3 × 4.0 × 8 + 16. (a) Energy is consumed to break the strong ionic bonds in aluminium oxide. 3. specific heat capacity. experiment A B D C B C 16. (c) There is heat loss to surroundings during the heating process. 9. 2010 46 . 4. 6. 14. substance. standard conditions Bomb heat. 13. of moles of propan-1-ol needed = mol = 0. 2010 47 . Thus.0325 mole of H2SO4(aq) requires 0.3 + 273) − (25. (a) H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l) (b) Standard enthalpy change of neutralization is the enthalpy change when 1 mole of water is formed from neutralization between an acid and an alkali under standard conditions.0 g cm 3 × 4.82 kJ mol 1 40.56 − − =− kJ mol 1 = − 56.00 (d) No.90 × mol = 0. Hence.0 © Aristo Educational Press Ltd.00 + 50.0450 mol 1000 From the equation. 1 1 (c) H2SO4(aq) + KOH(aq) → K2SO4(aq) + H2O(l) 2 2 65.0450 19. of moles of H2SO4(aq) used = 0.0 + 273)] K = 19 635 J (19.50 × mol = 0.9 kJ mol 1 0.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 18. (a) Exothermic (b) Energy absorbed by water = mcΔT − − − = 850 cm3 × 1.0325 mol 1000 50.00 No.0 + 273)] K = 2560 J (2.635 kJ) (c) Standard enthalpy change of solution of NaOH 19. 0.00) cm3 × 1.0650 mole of KOH(aq) for complete neutralization.2 J g 1 K 1 × [(28.0 kJ mol 1 = − 9.0 g cm 3 × 4.635 − − = − 80. of moles of KOH(aq) used = 0. KOH is the limiting reactant.56 kJ) Standard enthalpy change of neutralization between sulphuric acid and potassium hydroxide 2. Energy released during neutralization = mcΔT − − − = (65.5 + 273) −(23.2 J g 1 K 1 × [(30. 1 mole of H2SO4(aq) requires 2 moles of KOH(aq) for complete neutralization. 0 kJ mol−1 − 396. ∆H 1 = ∆H f + ∆H 2 − − ∆H f = (− 394.0) kJ mol 1 = − 99.1 (a) SHAPE SO2(g) + O2(g) ∆H \* SO3(g) MERGEFORMAT +297 kJ mol−1 S(s) + O2(g) − 396.0 + (− 396.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 37 Hess’s Law Class Practice A37. 2010 48 .0 kJ mol 1 (b) S(s) +O2(g) Enthalpy SO2(g) + ( O2(g) − 99.0 kJ mol−1 +297.0) −(− 283.2 (a) By applying Hess’s Law.0) kJ mol 1 = − 111. − − ∆H = +297.0 kJ mol−1 By applying Hess’s Law.0 kJ mol 1 © Aristo Educational Press Ltd.0 kJ mol−1 SO3(g) Reaction coordinate A37. ∆H = Σ∆H f (products) – Σ∆H f (reactants) ∆H = ∆H f [CaO(s)] + ∆H f [CO2(g)] – ∆H f [CaCO3(s)] − = (− 635.0 kJ mol−1 CO2(g) − 394.0 kJ mol−1 × 2 2CO2(g) + 3H2(g) +O2(g) + O2(g) − 286.0 kJ mol 1 − © Aristo Educational Press Ltd.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (b) C(s) +O2(g) +O2(g) ) ( Enthalpy − 111.0) + 3 × (− 286.4 (a) CaCO3(s) → CaO(s) + CO2(g) (b) By applying Hess’s Law.0) – (− 1207) kJ mol 1 − = +177.0) −(− 1371) kJ mol 1 − 1 =− 275.0 kJ mol−1 Reaction coordinate A37.0 kJ mol−1 × 3 + 3O2(g) − 1371 kJ mol−1 ∆H f C2H5OH(l) 2CO2(g) + 3H2O(l) ∆H f = 2 × (− 394.0 kJ mol−1 CO(g) + O2(g) O − 283. 2010 49 .3 2C(s) + 3H2(g) +O2(g) + 2O2(g) − 394.0 kJ mol A37.0) + (− 395. 0) kJ mol 1 − = +85 kJ mol 1 © Aristo Educational Press Ltd.0) + (− 1131) + (− 394. ∆H = − × ∆H f [NaHCO3(s)] + ∆H f [Na2CO3(s)] + ∆H c [C(s)] + ∆H c [H2(g)] 2 − = − × (− 2 948.0 × 8 + 16. 2NaHCO3(s) ∆H Na2CO3(s) + H2O(g) + CO2(g) ∆H f [NaHCO3(s)] × 2 ∆H c [H2(g)] 2Na(s) + H2(g) + 2C(s) + 3O2(g) Na2CO3(s) + CO2(g) + H2(g) + O2(g) ∆H f [Na2CO3(s)] ∆H c [C(s)] Na2CO3(s) + H2(g) + C(s) + O2(g) By applying Hess’s Law. 2010 50 .0) + 4 × (− 286.5 1.0) −(− 1004) kJ mol 1 − =− 1322 kJ mol 1 Enthalpy change of formation of 100 g liquid propanol 100 − =− 1322 kJ mol 1 × mol 12. ∆H f = Σ∆H c(reactants) −Σ∆H c(products) = 3 × ∆ H c [C(s)] + 4 × ∆H c [H2(g)] −∆H c [C3H7OH(l)] − = 3 × (− 394.0 × 3 + 1.0 =− 2203 kJ 2.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 A37. 1 3C(s) + 4H2(g) + O2(g) → C3H7OH(l) 2 By applying Hess’s Law.0) + (− 286. 9. 10. enthalpy. 5. 2010 51 . 3. reactants reactants. enthalpy. exothermic products. 2. higher. (a) C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(l) − ∆H c[C3H6O(l)] = –1790 kJ mol 1 (b) C3H6O(l) + 4O2(g) ∆H f [C3H6O(l)] ∆H1 3CO2(g) + 3H2O(l) + 4O2(g) ∆H2 11.0 − = –(– ) + (– ) – (–176.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Chapter 37 Hess’s Law Chapter Exercise 1. 7.2 629. (a) (b) ∆H f [HCl(g)] ] + 4O2(g) 3C(s) + 3H2(g) +O2(g) © Aristo Educational Press Ltd. 4. endothermic. products D C A B B A 1 1 H2(g) + Cl2(g) → HCl(g) 2 2 N2(g) + 2H2(g) + Cl2(g) ∆H3 NH3(g) + H2(g) + Cl2(g) ∆H2 NH4Cl(s) ∆H1 NH3(g) + HCl(g) ∆H f [HCl(g)] = –∆H3 + ∆H2 – ∆H1 92.1) kJ mol 1 2 2 − 1 = –92. 8. 6.3 kJ mol 12. route lower. 1) kJ mol 1 = –244.9 kJ mol 1 6720 − − (b) ∆H f [Al2O3(s)] = – kJ mol 1 = –3360 kJ mol 1 2 (c) For the reaction. (a) ∆H f [H2O(g)] = (–286. 2010 52 .0) – (–1790) kJ mol 1 − = –250 kJ mol 1 13.6) kJ mol 1 − = –1613 kJ mol 1 © Aristo Educational Press Ltd.0) + 3 × (–286. ∆H = Σ∆H f (products) – Σ∆H f (reactants) 3 1 ∆H = × ∆H f [N2(g)] + 3 × ∆H f [H2O(g)] + × ∆H f [Al2O3(s)] 2 2 3 – ∆H f [Al(s)] – × ∆H f [NH4NO3(s)] 2 3 1 3 –1015 = × 0 + 3 × (–244. 3 3 1 Al(s) + NH4NO3(s) → N2(g) + 3H2O(g) + Al2O3(s) 2 2 2 By applying Hess’s Law.3) + 2 × (–394.9) + × (–3360) – 0 – × ∆H f [NH4NO3(s)] 2 2 2 − 1 ∆H f [NH4NO3(s)] = –933.0) + 2 × (–286. ∆H1 = ∆H2 – ∆H3 − = [–1412 + (–286.0) kJ mol 1 − = –1412 kJ mol 1 (b) C2H4(g) + H2(g) ∆H2 3O2(g) ∆H1 3O2(g) C2H6(g) ∆H3 ∆Hc [C2H4(g)] 2CO2(g) + 2H2O(l) ∆H1 − − 2CO2(g) + 3H2O(l) By applying Hess’s Law. ∆Hc[C2H4(g)] = –∆H2 + ∆H1 − = –(+52. (a) C2H4(g) + 3O2(g) ∆H2 2C(s) + 2H2(g) + 3O2(g) By applying Hess’s Law.0)] – (–84.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 ∆H f [C3H6O(l)] = ∆H2 – ∆H1 − = 3 × (–394.1 kJ mol 14.0) + (+41. 0 × 8 + 1. (a) C6H12O6 + 6O2 → 6CO2 + 6H2O (b) Molar enthalpy change of combustion of glucose 15. C8H18 releases the largest amount of energy per gram of substance burnt.0 − = 22.0 × 2 + 1. (d) Energy needed for walking a distance of 1.0 × 6 − = –2826 kJ mol 1 (c) Some energy is used for supporting other body activities. 2010 53 .1 ∴ mass of cereal needed = g = 200 g 35% 13.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 Part VIII Chemical reactions and energy Part Exercise 1.5 km = 220 × 1.0 kJ g 1 Hence. 2.0 × 4 + 16.1 70. 5.1 g of glucose is needed for providing the energy required. 3.0 × 8 + 14. 8. D D D C B B A D A C D 12.0 × 1 + 1. 6. 70.7 − 1 =– kJ mol 1 12.7 kJ g 1 Energy released by burning C8H18 5590 − = kJ g 1 12. (b) C8H18 is the best for use in motor cars as it can release the largest amount of energy for the same mass of fuel carried. (a) Energy released by burning (CH3)2NNH2 1694 − = kJ g 1 12.0 × 18 − = 49.0 × 12 + 16.0 × 2 − = 28. 10.0 × 6 + 1. 7.5 kJ = 330 kJ Let m be the mass of glucose needed for providing the energy required m × 2826 × 30% = 330 180 m = 70. © Aristo Educational Press Ltd. 4.2 kJ g 1 Energy released by burning CH3OH 726 − = kJ g 1 12. 11. 9. 2 J g 1 K 1 × [(22.2 × ∆T = 666 ∆T = 2.555 kJ (555 J) mc∆T = 555 (30.002 − (c) Mg(s) + Cu2+(aq) → Cu(s) + Mg2+(aq) ∆H = –525 kJ mol 1 © Aristo Educational Press Ltd.2 × ∆T = 555 ∆T = 2.2 + 273) – (20.0 cm3 of 1.0 + 30. 2010 54 .8 − = – 10.0 × 4.0 + 273)] K = 184.0 × 4.20 K (oC) (d) For 30.0) × 1.2 M solutions.0 = 18.0 = 18. (a) Number of moles of Cu2+ used = 50.0 × 0.5 × × 1.0 M solutions. (a) Mg(NO3)2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaNO3(aq) (b) Exothermic.0 + 30. Energy released 30.8 J Enthalpy change of precipitation of magnesium carbonate 184.0) × 1.0 kJ 1000 = 0.0 J mol 1 × 1.0 + 10. The reaction released heat that raised the temperature of the reaction mixture.0 1000 − − = –18 480 J mol 1 (18.04 mol = 0.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 14.5 kJ mol 1) For 30. Energy released 30.666 kJ (666 J) mc∆T = 666 (30.002 mol 1000 (b) Mg(s) + Cu2+(aq) → Cu(s) + Mg2+(aq) − 1050 − − − ∆H = J mol 1 = –525 000 J mol 1 = –525 kJ mol 1 0.0 g cm 3 × 4.0 cm3 of 1. (c) Energy released in the first experiment − − − = (10.2 kJ 1000 = 0. 15.0) cm3 × 1.5 × × 1.64 K (oC) (e) The third experiment is the most accurate as it involves a larger temperature change and so any reading errors can be minimized. 0025 According to the equation.0 × 0. 2010 ∆H MgO(s) + H2(g) 2H+(aq) ∆H2 55 .0025 mol 1000 (b) Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) Enthalpy change for each mole of Ag+(aq) reacted − 185 − − − = J mol 1 = –74 000 J mol 1 = –74 kJ mol 1 0. (a) Hydrochloric acid (b) Mg(s) + H2O(l) ∆H1 Mg2+(aq) + H2(g) + H2O(l) ∆H = ∆H1 – ∆H2 © Aristo Educational Press Ltd. − − ∆H = –74 × 2 kJ mol 1 = –148 kJ mol 1 − (c) Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) ∆H = –148 kJ mol 1 (d) ∆H1 Mg(s) + Cu (aq) 2Ag (aq) + 2+ Cu(s) + Mg2+(aq) ∆H ∆H2 2Ag+(aq) Mg2+(aq) + Cu2+(aq) + 2Ag(s) ∆H = ∆H1 + ∆H2 − = (–525) + (–148) kJ mol 1 − = –673 kJ mol 1 17.05 mol = 0. the standard enthalpy change of the above displacement reaction involves the complete reaction of 2 moles of Ag+(aq).HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (d) Enthalpy Mg(s) + Cu2+(aq) ∆H = �25 kJ mol(− Cu(s) + Mg2+(aq) Reaction coordinate 16. (a) Number of moles of Ag+ used = 50. Therefore. (e) Number of moles of solute used 10 = mol 39.0 + 16.0989 − − = +35 035 J mol 1 (+35 kJ mol 1) © Aristo Educational Press Ltd.0 × 2) g = 1617 g 19.0 g cm–3 × 4.2 J g–1 K–1 × [(25 + 273) – (0 + 273)] K = 63 000 kJ No.0 × 3 = 0. When potassium nitrate was added at 15 s. (a) C20H32O2 + 27O2 → 20CO2 + 16H2O − (b) ∆H c = 20 × (–395. all solute had been dissolved and the temperature of the solution reached a minimum. 2010 56 .0 × 32 + 16.2 J g–1 K–1 × [(23.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 (c) She is correct.0 + 273) – (17. Ca(s) + 2H2O(l)  Ca(OH)2(aq) + H2(g) 18.1 + 14.5 + 273)] K = 3465 J Standard enthalpy change of solution of potassium nitrate 3465 − =+ J mol 1 0. the temperature of the solution rose. In fact. calcium reacts with water to give calcium hydroxide instead of calcium oxide. of moles of arachidonic acid needed 63 000 = mol = 5. As heat was acquired from the surroundings.32 × (12. At 40 s. (a) (b) (c) (d) Positive value 15 s 40 s The water temperature was constant at 23oC before the addition of potassium nitrate.0) – (–636.32 mol 11 840 Mass of arachidonic acid needed = 5.0) kJ mol 1 − = –11 840 kJ mol 1 (c) Energy needed = mc∆T = (600 × 1000) g × 4.0989 mol Heat absorbed during dissolution = 150 cm3 × 1. heat was absorbed for the dissolution and the temperature of the solution fell.0 × 20 + 1.0) + 16 × (–286. 0) – (–1396) kJ mol 1 2 2 © Aristo Educational Press Ltd. (iii) Lowered.0 × 8 + 16. lid beaker 21.0) = –843 Ethanol CH3CH2OH –29. less heat will be released.34 × (12. As the actual specific heat capacity is smaller. the temperature rise during measurement will be lowered and so is the calculated value. (a) − Name Formula Standard enthalpy change of combustion / kJ mol 1 Methanol CH3OH –26.0) + (–572.80 × (12.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 20. (ii) Raised.0 × 4 + 16. 2NO2(g) + 3H2O(g) O2(g) 1O2(g) 3 1 − ∆H = –(–180. − (c) About 3200–3330 kJ mol 1 22.0) = –2010 Butan-1-ol CH3CH2CH2CH2OH –36. As less iron reacts. 2010 57 . This results in a smaller temperature rise and thus the calculated value will be lowered.0) = –1371 Propan-1-ol CH3CH2CH2OH –33.0) + (–112.0 × 3 + 1.50 × (12.0 × 10 + 16. As some heat is lost.0 × 4 + 1.0) = –2673 (b) The larger alcohols have higher standard enthalpy change of combustion. N2(g) + 3H2(g) O2(g) ∆H 2NH3(g) 3O2(g) ∆H 1 ∆H 4 2NO(g) + 3H2(g) ∆H 2 ∆H 3 2NO2(g) + 3H2(g) By applying Hess’s Law.0 × 2 + 1.0 + 1. (a) thermometer expanded polystyrene cup Fe(s) + H2SO4(aq) cotton wool (b) (i) Lowered.0 × 6 + 16. there will be a higher temperature rise and so is the calculated value.12 × (12. 0) + 3 × (–286. (a) ∆H = 3 × (–395. Since 2SO2(g) + 4HCl(g) → 2SOCl2(l) + 2H2O(l) − − ∆H = 2 × (− 10.0) – (–2542) kJ mol 1 − = +104.HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 3 = –92.6 + (− 651.4) + (+202.0 kJ mol 1 − © Aristo Educational Press Ltd.6 kJ mol 1 2PCl3(l) + O2(g) → 2POCl3(l) − − ∆H = 2 × (− 325.3) kJ mol 1 = − 20.0 kJ mol − 1 23.0) – (–1574) kJ mol 1 − = –755.4 kJ mol 1 2Cl2(g) + 2H2O(l) → 4HCl(g) + O2(g) − ∆H = +202.4) + (− 613.4 kJ mol 1 2P(s) + 3Cl2(g) → 2PCl3(l) − − ∆H = 2 × (− 306.7) kJ mol 1 = − 651.7) kJ mol 1 = − 613.0 kJ mol 1 (b) Enthalpy change of formation of C4H6(g) is defined by the following equation: 4C(s) + 3H2(g) → C4H6(g) − ∆H = 4 × (–395.0) kJ mol 1 − = +42.6) kJ mol 1 = − 1083 kJ mol 1 24. 2010 58 .0 kJ mol 1 (c) The chemical equation for the conversion is: 2C2H4(g) → C4H6(g) + H2(g) − ∆H = 2 × (–1393) – (–2542) – (–286.0) + 4 × (–286.6 kJ mol 1 Summing up the above equations: 2SO2(g) + 4HCl(g) + 2PCl3(l) + O2(g) + 2P(s) + 3Cl2(g) + 2Cl2(g) + 2H2O(l) → 2SOCl2(l) + 2H2O(l) + 2POCl3(l) + 2PCl3(l) + 4HCl(g) + O2(g) − − ∆H = − 20.