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

Page

Number

Class Practice 1

Chapter Exercise 2

Chapter 26 Bond polarity

Class Practice 4

Chapter Exercise 5

Chapter 27 Intermolecular forces

Class Practice 7

Chapter Exercise 9

Chapter 28 Structures and properties of molecular crystals

Class Practice 11

Chapter Exercise 12

Part Exercise 14

Chapter 29 Chemical cells in daily life

Class Practice 17

Chapter Exercise 18

Chapter 30 Simple chemical cells

Class Practice 21

Chapter Exercise 22

© Aristo Educational Press Ltd. 2010

Chapter 31 Redox reactions

Class Practice 24

Chapter Exercise 26

Chapter 32 Redox reactions in chemical cells

Class Practice 29

Chapter Exercise 30

Chapter 33 Electrolysis

Class Practice 32

Chapter Exercise 33

Chapter 34 Importance of redox reactions in modern ways of living

Class Practice 36

Chapter Exercise 37

Part Exercise 39

Chapter 35 Energy changes in chemical reactions

Class Practice 42

Chapter Exercise 43

Chapter 36 Standard enthalpy change of combustion, neutralization,

solution and formation

Class Practice 44

Chapter Exercise 46

Chapter 37 Hess’s Law

Class Practice 48

Chapter Exercise 51

Part Exercise 53


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.

A25.2Molecule No. of

electron pairs

Spatial arrangement ofelectron pairs

No. of lone pairs

No. of bond pairs

Shape

CH4 4 Tetrahedral 0 4 TetrahedralNH3 4 Tetrahedral 1 3 Trigonal

pyramidalH2O 4 Tetrahedral 2 2 V-shaped

© Aristo Educational Press Ltd. 2010 1


Chapter 25 Simple molecular substances with non-octet structures and

shapes of simple molecules

Chapter Exercise

1. eight2. eight3. six4. 105. 126. non-octet7. linear, 180°8. trigonal planar, 120°9. tetrahedral, 109.5°10. trigonal bipyramidal, 90°, 120°11. octahedral, 90°12. 2, 213. 1, 314. shapes15. D16. B17. C18. A19. D20. A

21. (a) A: ; B: (b) A: WZ3

B: YZ3

(c) A. The central atom (W) has only six electrons in its outermost shell.

22. (a) X: Trigonal pyramidalY: Trigonal bipyramidal

(b) 107°(c) 90°(for Cl(axial) PCl(equatorial) bond angles) and 120° (for ClPCl bond angles

within the plane of triangle)

23. (a)

(b)



24. (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, all the O=S=O bond angles are about 120°.




Class Practice

A26.1

(a)(b)

(c)

A26.2(a) NCl3 has three polar NCl bonds and is trigonal pyramidal in shape. As there is a

resultant dipole moment arising from the three polar bonds, the molecule is polar. BCl3 has three polar BCl bonds and is trigonal planar in shape. As the dipole moments of the three polar bonds cancel out each other, the molecule is non-polar.

(b) As the order of electronegativity is F > N > Br, the resultant dipole moments of NBr3 and NF3 are pointing to different directions. The situations are shown below:

In a non-uniform electrostatic field, the nitrogen end of NBr3 will point to the positive pole while the nitrogen end of NF3 will point to the negative pole.




Chapter Exercise

1. attract2. polar3. non-polar4. larger, less, smaller5. polar, cancel out6. polar, symmetrically 7. deflection8. C9. B10. C11. B12. C13. A

14. (a) The electronegativity values of elements do not follow the trend of changing masses.

(b) It increases.(c) It decreases.(d) 2.5–3.5(e) 0–2.8(f) C and F(g) Cl and Br

15. (a)

(b)(c)

16. (a) Y, Z, X.

(b)

(c)

(d)

17. (a) Mistakes:1. ‘covalent bond involving fluorine is polar in nature’2. ‘all compounds of fluorine must be polar’

(b) 1. The covalent bond in a fluorine molecule (FF) is non-polar.



2. Polar XF bonds may cancel out their individual dipole moments to give a non-polar molecule e.g. CF4, SF6, etc.




Class Practice

A27.1(a) When the electron distribution around the two iodine nuclei is uneven,

instantaneous dipole is formed in which the side with more electrons will carry a partial negative charge.

(b)

A27.21. (a) Non-polar molecules

(b) Dispersion forces(c) (i) The statement is correct as butane has a higher boiling point than that

of propane. 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.

(ii) The statement is incorrect as butane has a higher boiling point than that of 2-methylpropane. Butane (straight-chain hydrocarbon) has a long, thin shape. This contributes to a larger contact surface area between butane molecules, resulting in larger dispersion forces.

2. ClF and CH2Cl2 are polar in nature and their molecules are attracted by both dipole-dipole forces and dispersion forces. On the other hand, F2 and Cl2 are non-polar in nature and their molecules are attracted by dispersion forces only. As a result, ClF and CH2Cl2 have higher boiling points than those of F2 and Cl2.Since the molecular size of CH2Cl2 is larger than that of ClF, the dispersion forces between CH2Cl2 molecules are larger. So CH2Cl2 has a higher boiling point.Similarly, since the molecular size of Cl2 is larger than that of F2, Cl2 has a higher boiling point.

3. (a) CH3F is polar in nature and its molecules are attracted by both dipole-dipole forces and dispersion forces. On the other hand, C2H6 is non-polar in nature and its molecules are attracted by dispersion forces only. As a result, CH3F has a higher boiling point than that of C2H6.

(b) Although Cl2 is non-polar and its molecules are attracted by dispersion forces only, the larger dispersion forces in Cl2 outweigh the dipole-dipole forces in HCl. Thus, Cl2 has a higher boiling point.

A27.3(a)


CH3OH

7


(b)

(c)

A27.4(a) In glucose, 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. As a result, glucose is very soluble in water. On the other hand, 1,2,4-trichlorobenzene cannot form any hydrogen bond with water.

(b) Glucose molecules are attracted to each other by extensive hydrogen bonds. A considerable amount of energy is needed for separating glucose molecules in melting. On the other hand, 1,2,4-trichlorobenzene molecules are attracted by dipole-dipole forces. Less energy is needed to melt 1,2,4-trichlorobenzene.

A27.5(a) In diamond, carbon atoms are held together by strong covalent bonds and much

energy is needed for separating the atoms during melting. In ice, water molecules are held together by weak intermolecular forces (hydrogen bonds and van der Waals’ forces), 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. Therefore, more energy is needed to separate water molecules during boiling.

(c) Hydrogen chloride molecules are held together by van der Waals’ forces but hydrogen fluoride molecules are held together by hydrogen bonds. As the intermolecular forces in hydrogen fluoride are stronger than those in hydrogen chloride, it is more difficult for hydrogen fluoride molecules to escape into the atmosphere.




Chapter Exercise

1. molecules, electrostatic2. an der Waals’ forces, ydrogen bonding3. Dipole-dipole forces4. Dispersion forces, induced5. molecular, surface, olarity6. hydrogen, electrons7. energy8. surface, iscosity9. hydrogen bonds10. high11. proteins, deoxyribonucleic acids12. D13. A14. A15. A16. B17. A

18. (a) GroupVIII/0(b) Dispersion forces(c) The intermolecular forces arise from the uneven distribution of electrons

within the atoms.(d) The boiling point increases down the group.(e) The larger the atom, the higher is the chance of uneven distribution of

electrons. This gives rise to a more prominent instantaneous dipole and larger dispersion forces.

19. (a) Dipole-dipole forces and dispersion forces(b) Dispersion forces(c) Since the molecular sizes of HCl and F2 do not differ much, the dispersion

forces among their molecules are similar in strength. However, the presence of dipole-dipole forces in HCl strengthens the attractions among molecules of HCl and so HCl has a higher boiling point.

20. (a) A and B. Their molecules are held together by hydrogen bonds.(b) Lower. As there is no hydrogen bond between molecules of C, less energy is

needed to separate the molecules in boiling.

21. Ethanol and ethane-1,2-diol have higher boiling points as their molecules are held together by hydrogen bonds. Among them, each ethane-1,2-diol molecule can form two hydrogen bonds per molecule while each ethanol molecule can form one hydrogen bond per molecule. Therefore, ethane-1,2-diol has a higher boiling point than ethanol. 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. It has the lowest boiling point.

22. The melting point and boiling point of water are much higher than those of simple molecular substances without hydrogen bonds. This is because



much energy is needed to overcome the strong hydrogen bonds between water molecules and separate them.

The existence of extensive hydrogen bonds makes the surface tension of water exceptionally high. That is, water molecules are held strongly together to form an unstretchable surface.

Regarding to its viscosity, water is more viscous than most molecular liquids. This is because the strong hydrogen bonds hold water molecules together and do not allow them to move past one another easily.




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).(c) To overcome a certain amount of hydrogen bonds and separate the water

molecules.(d) During boiling, all hydrogen bonds have to be broken before the water molecules

can escape as steam.

A28.2(a) Diamond and graphite were discovered before fullerene.(b) Buckminsterfullerene(c) Covalent bond(d) Dispersion forces (or van der Waals’ forces)

A28.3(a) Both of them have molecules held together by dispersion forces.(b) Buckminsterfullerene is spherical in shape but carbon nanotube is cylindrical. In

buckminsterfullerene, pentagonal patterns of atoms are found between the hexagonal patterns. In a carbon nanotube, pentagonal patterns of atoms are found only near the two ends of molecules.

A28.41. (a) In diamond, carbon atoms are held strongly by covalent bonds but the

buckminsterfullerene molecules are held by weak dispersion forces.(b) Graphite has delocalized electrons in its structure but diamond and

buckminsterfullerene do not.(c) As buckminsterfullerene molecules are held by dispersion forces, less

energy is needed during melting. In melting diamond and graphite, strong covalent bonds have to be overcome.

2. (a) Carbon nanotubes are electrically conductive because of the presence of delocalized electrons in their structures.

(b) Carbon nanotubes have very high tensile strength.




Chapter Exercise

1. order2. four, hydrogen3. oxygen4. open5. hydrogen6. carbon, hollow7. spherical, benzene, close, strong, hard, insulator8. cylindrical, melting, solvents, carbon, tensile, conductors9. D10. B11. D12. B13. A14. C

15. (a)

(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. (a) Ice has a lower density than water.(b) Ice floats on water. This prevents heat loss from water and helps maintain

the water temperature.(c) Ice possesses an open structure in which H2O molecules are separated

farther and there are more spaces between molecules.

17. (a) It is an element as it is composed of carbon atoms only.(b) 12 pentagons and 0 hexagon(c) Other than pentagons, there are hexagons over the surface of C60.(d) C60 has a larger molecular size than C20. Hence, C60 has larger dispersion

forces between its molecules and thus a higher boiling point than that of C20.

18. (a) X Carbon nanotube; Y Diamond; Z Buckminsterfullerene(b) Carbon nanotubes are good electrical conductors due to the movement of

delocalized electrons. Furthermore, each carbon atom in a carbon nanotube is covalently-bonded to three neighbouring carbon atoms. Because of the


* Label all other bonds accordingly.

a

a

b

b

b a

12


strong covalent bonds holding carbon atoms along the tube axis, carbon nanotubes can withstand high tension.

(c) Some metal (e.g. potassium) atoms(d) No. There are no strong covalent bonds holding the carbon atoms along a

specific direction.



Part VI Microscopic world II

Part Exercise

1. C2. A3. C4. A5. C6. B7. A8. C9. C10. A

11. (a) WY2

(b) The electron diagram of WY2 is:

There are two double bonds surrounding the central atom W. The molecule should adopt a linear shape to minimize the repulsion between the two groups of electrons.

(c) X and Z.

Due to the repulsion between the lone pair of electrons and the three bond pairs of electrons, the molecule adopts a trigonal pyramidal shape with a bond angle of about 107°.

12. (a)(b) C–B–B–C

(c)

13. (a)

(b) SF2 105°; SF6 90°(c) SF2 is polar while SF6 is non-polar.



(d)(e) It is non-polar as the effects of all polar SF bonds cancel each other.

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 Hydrogen bondNH3 Hydrogen bondCl2 Dispersion force

CH3CH2CH2CH2CH3 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.

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.




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.

A29.21. 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 2.79 104 kWh (=180 103 1.55 103

kWh) electricity. To produce one ‘unit’ of electricity (1 kWh), about 3585 (=

) cells are needed. Therefore, the approximate cost is HK$30

3585 = 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)




Chapter Exercise

1. electricity2. electrolyte, paste3. inc-carbon, lkaline manganese, ilver oxide, ithium primary4. nickel-cadmium rechargeable, lithium-ion rechargeable, nickel-metal hydride

rechargeable, lead-acid accumulator5. zinc-carbon, alkaline manganese, nickel-metal hydride rechargeable, 3.7 V, Lead-

acid accumulator, 12 V6. voltage7. alkaline manganese8. (a) size of cell

(b) shape of cell(c) type of cell

9. nvironment, mercury, rechargeable, recycling10. D11. A12. B13. B14. A15. D16. C17. D18. B19. C20. A21. B22. C23. B24. B25. C

26. (a) Primary cells are not rechargeable whereas secondary cells can be recharged and used again.

(b) Primary cells: zinc-carbon cells, alkaline manganese cells, silver oxide cells, lithium primary cells (any TWO)Secondary cells: nickel-cadmium rechargeable cells, lithium-ion rechargeable cells, 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, and can work even when the voltage and current are not steady.

(d) Electrodes: (+) graphite; () zincElectrolyte: ammonium chloride

27. (a) BNew cells may have a higher voltage than the old cells, thus charging the old cells during operation. Charging old zinc-carbon cells may cause cells to overheat or even explode.

(b) A The acidic/alkaline electrolyte of the cells may corrode the metal case



(negative electrode) of the cells. The electrolyte leaks out when holes appear in the metal case. The acidic/alkaline electrolyte can damage metal contacts/parts of the appliance. The electrolyte can also cause skin burns.

(c) C Heat of fire may cause chemicals in cells to expand. Most cells are tightly sealed to prevent leakage, so the expansion of chemicals under pressure may lead to explosion. Chemicals inside cells may be flammable (hydrogen produced at electrodes during discharging), or oxidizing (depolarizer added) and can hence cause fire to burn more fiercely.

(d) CAlkaline manganese cells cannot be recharged. During recharging, heat and hydrogen may be produced and hence lead to explosion.

(e) DSmall children may mistake small-sized button cells as sweets.

28. (a) Silver oxide cell/lithium button cellThese cells are small enough to fit into a watch. Since a watch consumes little energy, a silver oxide cell can power a watch for about 1–2 years, while a lithium button cell can power a watch for about 5 years.

(b) Lithium-ion rechargeable cellIt has a high charge capacity but lightweight, and has a high voltage. It can be recharged for about 1200 cycles and a life span of about 3 years, and has no memory effect. It is most suitable for daily recharge irrespective of whether it has been fully discharged, and meets the demands for steady high-current, high-voltage of mobile phones.

(c) Zinc-carbon cell/alkaline manganese cell/nickel-metal hydride rechargeable cell (any ONE)Zinc-carbon cell is cheap./Alkaline manganese cell is more expensive but more durable./Nickel-metal hydride rechargeable cell is more environmentally friendly. (Any ONE)

(d) Zinc-carbon cell/alkaline manganese cell/nickel-metal hydride rechargeable cell (any ONE)Zinc-carbon cell is cheap./Alkaline manganese cell is more expensive but more durable./Nickel-metal hydride rechargeable cell is more environmentally friendly. (Any ONE)

29. (a) A silver oxide cell is normally made into a button-sized cell. It is lightweight and small so is suitable for use in watches. Most other cell types are normally larger in size and too heavy for use in watches.

(b) Calculators are always used when there is light. Watches, however, must continue to run even in the dark, so watches powered by solar cells are rare. Measures must be taken to prevent the watch to stop when there is no light.

(c) A solar cell powered watch must have another secondary chemical cell inside. The solar cell produces electricity when there is light. The secondary cell stores the excess electrical energy and powers the watch when there is no light.

(d) The solar cell is not a chemical cell. The energy change involved in a solar cell is a physical change instead of a chemical change. The electrical energy is not stored in the form of chemical energy.

30. (a) Lithium-ion cells are rechargeable. They can provide a large and steady current, and are lightweight. These are advantageous when they are used in



portable electronic devices like mobile phones, notebook computers and digital cameras.

(b) Lithium-ion cells are expensive (each costs at least HK$200). For use in low-end electrical appliances like portable radios, hand torches (flashlights) and motorized toys, standard-sized and low-priced zinc-carbon cells or alkaline manganese cells are more affordable and more readily available.

31. Advantages of a lead-acid accumulator: It provides a large current and a steady voltage. It is rechargeable, so can be used for many years without the need of

replacement.Disadvantages of a lead-acid accumulator: It is heavy and large in size, difficult to carry. It is a wet cell, turning the cell upside down may cause leakage of acid.Suitability of using it as a car battery: It is very suitable for use as a car battery, since starting the car needs a large

and steady current. Besides, frequent replacement of battery is inconvenient particularly when the car is travelling a long journey between cities.

The disadvantages do not affect its suitability for use as a car battery. The car is powerful enough to carry the battery along. The acid electrolyte does not leak out unless the car overturns.

32. Dry cells contain toxic chemicals which may pollute the environment. These may include mercury in low-priced zinc-carbon cells, and cadmium in nickel-cadmium rechargeable cells.

Other remaining chemicals in spent dry cells, when disposed of, are wasted and cause burden to landfill sites.

To minimize pollution, the following measures can be taken: Use mains electricity instead of cells whenever possible. Do not use mercury-containing zinc-carbon cells or alkaline manganese

cells. Use rechargeable cells instead of disposable cells. Use nickel-metal hydride rechargeable cells or lithium-ion rechargeable

cells instead of nickel-cadmium rechargeable cells. Return expired rechargeable cells to the manufacturer or the Environmental

Protection Department for recycling of the remaining chemicals.(Any FOUR)




Class Practice

A30.1(a) Copper to silver. Copper is more reactive than silver, thus it loses electrons more readily.(b) Copper foil(c) Yes. Electrons flow from zinc to copper. Zinc is more reactive than copper, thus

it loses electrons more readily.

A30.2Mg/Ag > Zn/Ag > Fe/Ag > Pb/Ag > Cu/Ag

A30.3(a) Negative electrode. It is because copper is higher than silver in the E.C.S. Copper

would lose electrons to silver when they are connected together.(b) From left to right. It is because electrons flow from copper rod to silver rod in the

external circuit.(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, the silver ions can come

into contact with the copper electrode and a direct displacement reaction occurs on the copper surface. A grey deposit forms on the copper rod and no voltage can be recorded in the multimeter.




Chapter Exercise

1. metals, electrolyte, metals, wire2. negative, positive3. electromotive force (or e.m.f.), higher4. voltmeter, igital, ultimeter5. voltage (or e.m.f.), Electrochemical Series6. (a) similar

(b) lose, cations, aqueous solution(c) reactivity, air, dilute acids, losing

7. salt bridge, complete, charges8. porous pot9. D10. C11. C12. C13. B14. A15. C16. C17. A

18. (a) Dilute sulphuric acid (or any other dilute acid, or an acidic salt solution like ammonium chloride.)

(b) Positive electrode: copper electrodeNegative electrode: zinc electrode

(c) From zinc electrode to copper electrode (or from left to right).(d) (i) Bubbles appear at the copper electrode.

(ii) Bubbles appear at the zinc electrode and the zinc electrode dissolves gradually.

(e) (i) The copper electrode is quickly covered with a brown deposit.(ii) The zinc electrode dissolves gradually.

19. (a) To complete the circuit by allowing ions to move from one half-cell to another half-cell, and to balance the charges in the solutions of the two half-cells.

(b) Positive electrode: copper electrodeNegative electrode: iron electrode

(c) (i) A brown solid deposit forms on the copper electrode and the blue colour of the copper(II) solution becomes paler.

(ii) The iron electrode slowly dissolves and the green iron(II) solution becomes darker.

(d) Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)(e) From iron electrode to copper electrode (or from left to right).



20. (a) Zn(s) Zn2+(aq) + 2e

(b) Copper containerCu2+(aq) + 2e Cu(s)

(c) The porous pot is used to separate the two electrolyte solutions and prevent them from direct mixing. Besides, it allows ions to pass through to complete the circuit.

(d) There would be a displacement reaction occurring at the surface of the zinc electrode immediately: Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq). Once the zinc rod is coated with copper, both electrodes become the same (copper electrodes) and the electrolyte is also the same, so the voltage of the cell drops to zero.

(e) Yes. The zinc strip loses electrons and these electrons pass through the external circuit to the copper container. The zinc strip is therefore the negative pole and the copper container is the positive pole.

21. (a) A metal couple is a combination of two different metals connected to an external circuit.

(b) The filter paper separates the two metal strips, preventing a short circuit. Besides, the salt solution allows movement of ions to complete the circuit.

(c) From copper to silver.(d) The voltmeter would give a positive reading.(e) Removing the filter paper would cause a short circuit, thus the voltmeter

reading drops to zero.(f) The voltage would be higher. This is because magnesium is in a higher

position than copper in the Electrochemical Series.

22. Connect the metals aluminium, copper, iron, lead, magnesium and zinc one by one in turn to the circuit as shown below:

The metal strip under test and the silver sheet is separated by a piece of filter paper moistened with sodium chloride solution.

Measure and record the voltage read out from the digital voltmeter for each metal tested.

Since the same reference electrode (silver) is used, the Electrochemical Series of these metals can be obtained by arranging the metals in the order of voltage reading.

The higher the voltage reading, the higher position the metal is in the Electrochemical Series.


clip

metal strip

under test

filter paper moistened with sodium

chloride solution

silver sheet

23



Class Practice

A31.1(a) It involves a transfer of electrons (from Fe(s) to Cu2+(aq)).(b) Fe(s) is being oxidized. Fe(s) loses electrons to others.(c) Cu2+(aq) is the oxidizing agent. Cu2+(aq) gains electrons (or Cu2+(aq) is reduced).

A31.2(a) O.N. of S = 0(b) (+2) 3 + (O.N. of N) 2 = 0

O.N. of N = 3(c) (+1) 2 + (O.N. of S) + (2) 4 = 0

O.N. of S = +6(d) (+1) 2 + (O.N. of S) + (2) 3 = 0

O.N. of S = +4(e) (+1) 2 + (O.N. of Cr) + (2) 4 = 0

O.N. of Cr = +6(f) For NH4

+,(O.N. of N) + (+1) 4 = +1 O.N. of N = 3

(g) (+1) + (O.N. of C) + (2) 3 = 1 O.N. of C = +4

(h) (O.N. of Cr) 2 + (2) 7 = 2 O.N. of Cr = +6

(i) (O.N. of N) + (1) 3 = 0 O.N. of N = +3

A31.3

1.Defined in terms of Oxidation Reduction

oxygen +O Oelectron e +e

oxidation number increases decreases

2. (a) (i) Redox(ii) O2

(iii) CH4

(iv) Carbon(b) (i) Not redox

(ii), (iii) & (iv): not applicable(c) (i) Redox

(ii) Cl2

(iii) FeSO4 (or Fe2+)(iv) Iron



3. Both statements are correct. However, the second statement is not a correct explanation of the first. A correct explanation would be: ‘The oxidation number of nitrogen in ammonia can be increased when ammonia reacts with a strong oxidizing agent.’

A31.4(a) (i) Cl2(g) + 2e 2Cl(aq)

(ii) 2Br(aq) Br2(aq) + 2e

(iii) Cl2(g) + 2Br(aq) 2Cl(aq) + Br2(aq)(b) (i) MnO4

(aq) + 8H+(aq) + 5e(aq) Mn2+(aq) + 4H2O(l)(ii) Fe2+(aq) Fe3+(aq) + e

(iii) MnO4(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

A31.5(a) K+(aq)(b) F(aq)(c) Fe2+(aq)

A31.6(a) Ag(s) + 2HNO3(aq) AgNO3(aq) + NO2(g) + H2O(l)(b) Ag(s) + 2H+(aq) + NO3

(aq) Ag+(aq) + NO2(g) + H2O(l)

A31.7(a) No reaction.(b) C(s) + H2SO4(l) CO2(g) + 2SO2(g) + 2H2O(l)

A31.81. Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s)2. (a) Zn to Cu

(b) Mg to Ag(c) Zn to Pb




Chapter Exercise

1. transfer2. loses, gains3. (a) oxidizes, accepting

(b) reduces, donating4. (a) 0

(b) ionic charge(c) 0(d) ionic charge

5. (a) increases(b) decreases

6. Electrochemical7. (a) weak

(b) increases(c) strong(d) oxidizing(e) reducing(f) strong(g) decreases(h) weak

8. oxidizing, NO2(g), oxidizing, SO2(g)

9. B10. C11. D12. D13. C14. C15. C16. C17. D18. B19. B20. B21. D22. D23. B24. C25. B26. C27. C28. A



29. (a) 0(b) +2(c) +3(d) 3(e) +4(f) +6(g) +7(h) +5(i) +1

30. (a) Redox reaction. Oxidizing agent: O2; reducing agent: NO; nitrogen monoxide is oxidized.

(b) Redox reaction. Oxidizing agent: HNO3; reducing agent: C; carbon is oxidized.

(c) Not a redox reaction.(d) Redox reaction. Oxidizing agent: CuSO4; reducing agent: Zn; zinc is

oxidized.(e) Redox reaction. Oxidizing agent: Cl2; reducing agent: Cl2; chlorine is

oxidized. (Chlorine is both oxidized and reduced in this reaction.)

31. (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, and a yellow solution is formed.

(ii) The orange dichromate solution becomes green, and bromine is formed which is yellow. The resultant solution looks yellowish-green.

32. (a) Colourless bubbles are evolved from the copper turnings, and the solution turns blue.

(b) 3Cu(s) + 8HNO3(aq) 3Cu(NO3)2(aq) + 2NO(aq) +4H2O(l)(c) Dilute nitric acid is the oxidizing agent. It is reduced as the oxidation

number of nitrogen changes from +5 to +2. Copper is the reducing agent. It is oxidized as the oxidation number of copper changes from 0 to +2.

(d) If the plug of cotton wool is removed, brown fumes appear at the mouth of the test tube.

(e) 2NO(g) + O2(g) 2NO2(g)(f) The cotton wool plug prevents oxygen in air from entering the test tube, so

the nitrogen monoxide formed will not be converted to nitrogen dioxide.



33. (a) Bubbles form immediately, and green fumes are evolved.(b) 2MnO4

(aq) + 16H+(aq) + 10Cl(aq) 2Mn2+(aq) + 5Cl2(aq) + 8H2O(l)(c) Permanganate ion is the oxidizing agent. It is reduced as the oxidation

number of manganese changes from +7 to +2. Hydrochloric acid is the reducing agent. It is oxidized as the oxidation number of chlorine changes from –1 to 0.

(d) The stopcock should be closed. Otherwise, the chlorine gas would leave the flask through the funnel rather than through the side arm to react with the potassium iodide.

(e) The water in the moistened broken porcelain chips can absorb unreacted hydrochloric acid vapour (hydrogen chloride gas), which is highly soluble in water.

(f) The potassium iodide crystals turn black/dark brown.(g) Cl2 + 2I 2Cl + I2

(h) Chlorine is the oxidizing agent. It is reduced as the oxidation number of chlorine changes from 0 to –1. Potassium iodide is the reducing agent. It is oxidized as the oxidation number of iodine changes from –1 to 0.

(i) The experiment should be carried out in the fume cupboard as chlorine is a poisonous gas.




Class Practice

A32.1(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)

A32.2(a) Electrons flow from electrode U to electrode V in the external circuit. It is

because the oxidizing agent MnO4(aq) will gain electrons while the reducing

agent Br(aq) will lose electrons.(b) Oxidation takes place at electrode U. Hence, U is the anode and V is the cathode.(c) V is the positive electrode and U is the negative electrode. Br(aq) loses electrons

at electrode U and is oxidized to Br2(aq). Electrons thus move in the external circuit from U to V. Therefore, U is the negative electrode.

(d) At electrode U: solution around electrode turns yellow/brown.At electrode V: solution around electrode becomes pale purple (partially decolorized).

(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)

A32.31. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)2. (a) A paste of ammonium chloride is used so that ions are mobile and can

conduct electricity.(b) If the cell is dry, the ions inside are no longer mobile. Electricity cannot

pass through and so the cell will not produce electrical energy.(c) No. As shown by the half equation, zinc participates in the electrode

reaction.(d) Yes. As shown by the half equation, carbon does not participate in the

electrode reaction.(e) At the zinc cup (the anode), zinc dissolves to form zinc ions. A metal-metal

ion system is set up.

A32.4(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 poses a lower risk of explosion than hydrogen. (Any ONE)(c) Methanol is flammable, if carelessly handled, it may catch fire. Furthermore,

methanol is a colourless liquid like water, yet it is highly poisonous. If it is not stored or labelled properly, there is a danger of accidental poisoning.




Chapter Exercise

1. oxidation, cathode2. negative, cathode3. Zn(s) Zn2+(aq) + 2e

2MnO2(s) + 2H+(aq) + 2e Mn2O3(s) + H2O(l)4. secondary

Pb (s) + SO42(aq) ⇌ PbSO4(s) + 2e

PbO(s) + 4H+(aq) + SO42(aq) + 2e ⇌ PbSO4(s) + H2O(l)

5. primary6. hydrogen, oxygen7. oxidant, catalyst8. B9. A10. D11. D12. D13. D14. C15. C16. B

17. (a) Dilute sulphuric acid(b) Cr2O7

2(aq) + 14H+(aq) + 6e 2Cr3+(aq) + 7H2O(l)(c) The colour of the acidified potassium dichromate solution slowly changes

from orange to green.(d) Fe2+(aq) Fe3+(aq) + e

(e) The colour of the iron(II) sulphate solution slowly changes from pale green to yellow.

(f) Electrons flow from the right electrode to the left electrode in the external circuit.

18. (a) 2MnO2(s) + 2H+(aq) + 2e Mn2O3(s) + H2O(l)(b) Zn(s) Zn2+(aq) + 2e

(c) Ammonium chloride(d) After a zinc-carbon cell has been used for a certain time, all the NH4

+(aq) ions will be used up. No more H+(aq) ions are available. Therefore, the reaction at the cathode cannot take place. The cell will stop functioning.

(e) During storage, the zinc casing of the cell will react with H+(aq) ions present in the electrolyte. Therefore, zinc can be used up even when the cell is not being used.

19. (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, they are the same.(e) (1) In the Daniell cell, Cu2+(aq) and Zn(s) are separated by a porous



partition. There is no direct mixing of the chemicals. Solid copper deposits on the surface of the copper container. In the beaker, the Cu2+

(aq) and Zn(s) are in direct contact and solid copper deposits directly on the surface of the zinc rod.

(2) In the Daniell cell, the reaction generates energy mainly in the form of electricity. In the beaker, the reaction generates energy mainly in the form of heat.

20. (a) Hot, concentrated potassium hydroxide solution.(b) As a catalyst.(c) H2(g) + 2OH(aq) 2H2O(l) + 2e

(d) Oxidation. It is because the oxidation number of H changes from 0 to +1.(e) O2(g) + 2H2O(l) + 4e 4OH(aq)(f) 2H2(g) + O2(g) 2H2O(l)(g) The electrons flow from the left hand side to the right hand side (clockwise)

in the external circuit.(h) The fuel of fuel cells can be refilled easily. They can be used basically

forever (by refilling the fuel). Conventional cells either run out (primary cells) and need to be discarded, or need recharging (secondary cells) which is time-consuming.




Class Practice

A33.1(a) (i) At cathode: magnesium; at anode: chlorine

(ii) No electrolysis takes place. (b) Molten magnesium chloride contains mobile ions that conduct electricity while

solid magnesium chloride does not conduct electricity.(c) For (a)(i),

At cathode: Mg2+(l) + 2e Mg(l)At anode: 2Cl(l) Cl2(g) + 2e

A33.2H+(aq), OH(aq), cation, metal, H+(aq), hydrogen, OH(aq), oxygen, halide ions, concentration

A33.3

(a) Cations: Na+(aq), H+(aq); anions: OH(aq)(b) (i) Hydrogen

(ii) Oxygen(c) Volume ratio of hydrogen : oxygen = 2 : 1

A33.41. (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.

(b) The blue colour of the solution becomes paler.2. (a) chlorine

(b) sodium amalgam(c) oxygen(d) ionizes (dissolves)(e) Cu2+(aq)

A33.5(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.

A33.61. Remove the metal ions from effluents by chemical treatment, if it is economical

to do so.2. Acidic effluents usually contain sulphuric acid. The calcium sulphate formed is

only slightly soluble in water, so it would prevent the neutralization reaction from going on. Besides, it is difficult to remove the insoluble calcium sulphate.




Chapter Exercise

1. decomposition2. chemical, electricity3. cathode, anode4. positive5. ions, cathode, anode, cathode, gain, anode, lose, anode, cathode6. E.C.S., concentration, electrodes7. electroplating, electroplated, corrosion8. cathode9. anode10. thicken11. lkalis, heavy metal, rganic, health12. Environmental Protection13. C14. A15. B16. B17. A18. D19. A20. B

21. (a) Carbon(b) Lead(II) ions and bromide ions(c) At the cathode: A silver colour appears.

At the anode: Some brown gas bubbles appear.(d) At the cathode: Pb2+(l) + 2e Pb(l)

At the anode: 2Br(l) Br2(g) + 2e

22. (a) Ions present: H+, OH, SO42.

Product at cathode: H2; product at anode: O2

(b) Ions present: H+, Na+, OH, Cl.Product at cathode: H2; product at anode: O2

(c) Ions present: H+, Na+, OH, Cl.Product at cathode: H2; product at anode: Cl2

(d) Ions present: H+, Cu2+, OH, Cl.Product at cathode: Cu; product at anode: O2

(e) Ions present: H+, Cu2+, OH, SO42.

Product at cathode: Cu; product at anode: O2



23. (a) Graphite (carbon) electrodes should be used. During electrolysis of halides, if platinum electrodes are used, they can be easily corroded by the respective halogens produced at the anode. Carbon is used as electrodes for electrolysis of halides as there is no direct reaction between carbon and the halogens.

(b) The left electrode. During electrolysis, the electrode connected to the positive pole of the d.c. source is the anode.

(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. When an inert electrode (graphite or platinum) is used, OH(aq) is preferentially discharged as it is a stronger reducing agent than Cl(aq). When copper metal is used as the anode, Cu(s) is preferentially discharged. Cu(s) is a stronger reducing agent than OH(aq) and Cl(aq).

(e) The copper anode becomes smaller.(f) There would be colourless bubbles (of oxygen) appearing on the surface of

the graphite anode.

24. (a) At electrode D: 2Cl(aq) Cl2(g) + 2e

At electrode E:2H+(aq) + 2e H2(g)

(b) Judging from the equations in (a), when 2 moles of electrons pass through the electrolytic cell, 1 mole of chlorine is liberated at the anode and at the same time 1 mole of hydrogen is liberated at the cathode. The theoretical ratio of the volumes of gases collected over the electrodes should be 1:1.

(c) The relative volume of gases shown in the diagram is reasonable. Since hydrogen is insoluble in water whereas chlorine is quite soluble in water, the actual volume of chlorine collected is less than the volume of hydrogen collected.

(d) At the anode, when the chloride ions are discharged, chlorine gas is produced. Since chlorine water is acidic and bleaching, the solution near the anode turns red and then colourless.At the cathode, when hydrogen ions are discharged, an excess of hydroxide ions are present. The solution near the cathode turns blue.

25. (a) The three mistakes are: (1) The door handle should be connected to the negative terminal

(cathode).(2) The piece of nickel metal should be connected to the positive terminal

(anode).(3) The electrolyte should be an aqueous solution of nickel(II) sulphate

instead of dilute sulphuric acid.(b) Two methods to shorten the time needed for electrolysis:

(1) Use a larger current (higher voltage).(2) Increase the surface area of the piece of nickel metal connected to the

anode.(3) Increase the concentration of the electrolyte.(Any TWO)



26. Similarities: Both electroplating and anodizing are industrial processes. Both processes involve electrolysis with the purpose of beautifying and/or

strengthening the surface of metal items.Differences:

Electroplating AnodizingCan be applied to most metal items (e.g. iron or copper items) and non-metal items.

Can only be applied to metal items made of aluminium.

The item to be electroplated is connected to the cathode during electroplating.

Aluminium item to be anodized is connected to the anode during anodizing.

A thin layer of another metal (e.g. nickel coating on iron) is coated on the item.

A thin layer of protective aluminium oxide is formed on the aluminium item.

The item acquires the colour of the coating metal (e.g. the golden colour of gold plating). The item cannot be further dyed to another colour.

The oxide layer is too thin to be seen and the aluminium item retains its own silvery colour. The oxide layer can be dyed to a new colour if necessary.




Class Practice

A34.11. (1) A diesel generator has a lower efficiency than a fuel cell system. In other

words, a diesel generator consumes more fuel to produce the same quantity of heat and electricity as compared to a fuel cell.

(2) A diesel generator causes pollution to the environment, producing smoke, bad smell, and a lot of NOx and SO2. A fuel cell system is clean and the exhaust is non-polluting, so it is more suitable for on-site energy production for a block of flats.

(3) A diesel generator is very noisy while a fuel cell operates quietly. This again is better for on-site power production.

2. (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)

A34.2Lithium metal, like other alkali metals (sodium, potassium, etc.) reacts vigorously with water to produce hydrogen and a corrosive, strongly alkaline solution LiOH. If the seal of a cell with a lithium metal anode is broken, water or even moisture in the air may react with lithium, causing hydrogen and alkaline solution to leak out. Hydrogen may cause explosion and the alkaline solution can cause severe skin burns.




Chapter Exercise

1. hotosynthesis, etabolism2. ermentation, ntioxidants, reathalyzers, leaching3. Alkaline Fuel Cells, Proton Exchange Membrane Fuel Cells, Phosphoric Acid

Fuel Cells, Molten Carbonate Fuel Cells, Solid Oxide Fuel Cells4. lithium, lithium-ion rechargeable, lithium-ion polymer rechargeable5. oxidized, reduced6. lithium-carbon compound, metal oxide, lithium, organic solvent,

discharging

CoO2 + xLi+ + xe ⇌ LixCoO2 charging

discharging

LixC6 ⇌ 6C + xLi+ + xe

charging discharging

CoO2 + LixC6 ⇌ LixCoO2 + 6C charging

7. harge density, oltage, rain capacity, recharged, nvironmentally preferred8. A9. A10. D11. A12. A

13. (a) The statement is half correct and half incorrect. A fuel cell can in theory work forever because it can continue to work as long as the fuel is replenished. A fuel cell is not a secondary cell because it cannot be recharged by using an external power source.

(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, and about 8 hours for nickel-cadmium cells). Instead, refilling the fuel of a fuel cell may just take seconds. Besides, a small bottle of fuel (say, methanol for DMFC) can be easily carried around whereas electrical power for recharging secondary cells may not be available in remote locations.

(c) (1) High efficiency.(2) Environmentally friendly (for hydrogen fuel cells, the only exhaust is

water). (3) Convenient (easy to refill the fuel).

14. (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.

(b) A conventional breathalyzer contains a solution of acidified potassium dichromate. If alcohol is present in a driver’s breath, 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)



(c) The alcohol in the breath of the driver, when blown into the intoximeter, is used as the fuel in an ethanol fuel cell. A higher concentration of ethanol produces a larger current, thus the intoximeter can give a reading that is proportional to the breath alcohol concentration of the driver.

(d) (1) An EC/IR intoximeter can give a continuous, linear reading of the breath alcohol concentration of a driver. The colour change of the conventional breathalyzer can only give an approximate data, which may not be good enough for legal procedures.

(2) The EC/IR intoximeter is more environmentally friendly (no need to dispose of spent chemicals).

(e) An infrared (IR) sensor

15. (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, it must first be heated to a temperature well above

40C (the melting point of phosphoric acid) before the PAFC can function.(d) The exhaust of this fuel cell is steam (at 150C to 200C). It can be used to

heat up water (for hot water supply) or heat up the air (for warming air in winter).

(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. Advantages of lithium-ion cells: Lithium-ion cells are rechargeable and can provide high voltage (3.7 V)

and large current. Lithium-ion cells are comparatively lightweight (high charge density). Lithium-ion cells do not contain/contain less heavy metals that pollute the

environment like zinc-carbon cells or nickel-cadmium cells. Lithium-ion cells have no memory effect, so discharging fully before

recharging is unnecessary. This makes the charging process more economical and more environmentally friendly.

(Any THREE)Disadvantages of lithium-ion cells: Lithium-ion cells cannot be used for appliances that need a voltage lower

than 3.6 V. The unit cost of lithium-ion cells is currently several times higher than that

of other cell types. Lithium-ion cells must be recharged with a dedicated charger, and if the

charger is not available, the lithium-ion cell cannot be recharged. After recharging for more than 1200 cycles, using a lithium-ion cell may

cause the cell to heat up, electrolyte to leak and the cell may even explode. They may also explode if overheated or if charged to an excessively high voltage.

(Any THREE)



Part VII Redox reactions, chemical cells, and electrolysis

Part Exercise

1. A2. B3. D4. B5. B6. B7. C8. B9. D10. A11. C12. A13. D14. A15. D

16. (a) manganese(IV) oxide and powdered graphite; ammonium chloride paste(b) Zn(s) Zn2+(aq) + 2e

(c) 2MnO2(s) + 2H+(aq) + 2e Mn2O3(aq) + H2O(l)(d) From zinc to copper.(e) Zinc loses electrons more readily than copper, so electrons flow from zinc

to copper in the external circuit. The lemon juice (citric acid solution) contains mobile H+(aq) ions. Thus, it acts as an electrolyte to complete the circuit by movement of ions.

(f) Zinc metal is the anode as it is oxidized to Zn2+(aq) ions in the cell reaction.

17. (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. MnO2 (O.N. of Mn = +4) is reduced to Mn2O3 (O.N. of Mn = +3). Reduction occurs at the cathode.

(ii) Electrode B is the anode. Zn (O.N. of Zn = 0) is oxidized to ZnO (O.N. of Zn = +2). Oxidation occurs at the anode.

(c) The alkaline manganese cell is often preferred to the less expensive zinc-carbon cell because the alkaline manganese cell has a much longer service life/higher charge capacity.

(d) The silver oxide cell is commonly used in electronic or electrical devices which need a small-sized cell with a steady voltage.

18. (a) To lower the melting point so that energy and cost for heating up the ore can be saved.

(b) Ca2+ is a weaker oxidizing agent than Al3+, so Al3+ is preferentially discharged. Besides, F is a weaker reducing agent than O2, so O2 is preferentially discharged. Therefore, calcium fluoride added does not affect the result of the electrolysis.

(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 © Aristo Educational Press Ltd. 2010 39


reducing agents like carbon.(f) At 850C, the carbon electrode burns in the oxygen formed at the anode to

give carbon dioxide.

19. (a) Colourless bubbles are formed from the carbon.(b) 2H2SO4(l) + C(s) CO2(g) + 2SO2(g) + 2H2O(l)(c) This is a redox. During the reaction, the oxidation number of C increases

from 0 to +4, and the oxidation number of S decreases from +6 to +4.(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. During the reaction, the oxidation number of S increases

from +4 to +6, and the oxidation number of Cr decreases from +6 to +3.(g) The limewater turns milky.(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. This is because both SO2 and CO2 are acidic gases. They will be both removed by bottle B. No gas will enter bottle A.

(j) (1) Wear gloves and goggles (concentrated sulphuric acid is highly corrosive).

(2) Perform the experiment in a fume cupboard (in case of leakage of toxic sulphur dioxide).

(3) Beware of heat burns.

20. (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). The lead plate is the anode because lead is oxidized to lead(II) sulphate (oxidation number of Pb increases from 0 to +2).

(b) The lead plate coated with lead(IV) oxide is the positive electrode as it takes in electrons for reduction to take place. The lead plate is the negative electrode as it gives out electrons for oxidation to take place.

(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. It can be reversed by applying an external voltage across the electrodes. Then the battery can be recharged.

(e) The car battery has a voltage of 12 V. This is achieved by combining six cells in series.

(f) (1) Wear gloves and goggles.(2) Beware of acid burns by sulphuric acid.(3) Beware of electric shock. The car battery can produce a large current

and may cause electric sparks and heat burns if short-circuited.(Any TWO)



21. (a) O2(g) + 4H+(in PEM) + 4e 2H2O(l) (b) 2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(l)(c) Methanol is easier and safer to carry and handle than hydrogen gas.(d) Methanol is toxic and is highly flammable.(e) DMFCs produce the greenhouse gas, carbon dioxide as one of its exhaust.

Hydrogen fuel cells only produce water as the only exhaust, which is more environmentally friendly.

(f) DMFCs can be made into a small size and can produce a small current for a long time. This is ideal for powering electronic products. However, it cannot provide a large current. Therefore, it is not suitable for powering automobiles.

22. Oxidation reactions involve addition of oxygen while reduction reactions involve removal of oxygen.

For example, magnesium combining with oxygen undergoes oxidation while copper(II) oxide losing oxygen undergoes reduction.Mg(s) + CuO(s) MgO(s) + Cu(s)

Oxidation reactions involve loss of electron(s) while reduction reactions involve gain of electron(s).

For example, magnesium losing 2 electrons undergoes oxidation while copper(II) ion gaining 2 electrons undergoes reduction.

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. For example, 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)). Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)




Class Practice

A35.1 Yes. In the reaction, two moles of gaseous reactants (H2 and Cl2) react to give two moles of product (HCl). The total gas volume remains constant. Heat change at constant volume is equal to the change in internal energy.

A35.2(a) Exothermic. Energy is released when water molecules come close together

during condensation.(b) Endothermic. Energy is needed to overcome the intermolecular forces (hydrogen

bonds) between water molecules.(c) Endothermic. Energy is needed to break the covalent bond between chlorine

atoms.(d) Endothermic. Energy is needed to overcome the attraction between the nucleus

of the sodium atom and its outermost shell electron.(e) Exothermic. Energy is released when the nucleus of chlorine atom attracts the

incoming electron.




Chapter Exercise

1. pressure2. gives out3. takes in4. products, reactants5. lower6. ombustion, recipitation, neutralization7. oil fractions8. take in, give out9. bond-forming, bond-breaking10. smaller11. D12. A13. A14. D15. B16. D17. (a) Set-up (A) releases more heat as the reaction is carried out in a closed

system. The change in internal energy is equal to the heat change at constant volume. Set-up (B) is open to the atmosphere. Some energy is used as the work done on the surroundings.

(b) For the same amount of acid used, set-up (C) has less magnesium metal (the limiting reactant). Therefore, set-up (C) releases less heat.

(c) (C), (B), (A)

18. (a) Endothermic process(b) During the process of evaporation, energy is needed to overcome the

intermolecular forces (i.e. hydrogen bond) between water molecules.(c) Cooling of hot engines

19. (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.

(c) Treatment of athletes’ injuries

20. (a) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)b) CC, CH and O=O(c) C=O and OH(d) The enthalpy change in bond-forming processes is larger than the enthalpy

change in bond-breaking processes.



Chapter 36 Standard enthalpy change of combustion, neutralization, solution and formation

Class Practice

A36.1

Heat change = 1371 kJ = 342.75 kJ

Hence, the heat evolved is 342.75 kJ.

A36.21. (a) Let Y be the mass of hydrogen in the mixture.

285.8 + 890.4 = 4725

Y = 22.3Thus, the mass of hydrogen and methane are 22.3 g and 27.7 g respectively.

(b) Let X be the mass of hydrogen in the mixture.

285.8 + 890.4 = 5000

X = 25.4Thus, the percentage by mass of hydrogen in the mixture

= 100%

= 50.8%

2. (a) CH3CH2COOH(b) NH3

(c) Heat required for the ionization of CH3CH2COOH = 49.9 (57.1) kJ mol1 = 7.2 kJ mol1

Heat required for the ionization of NH3 = 52.2 (57.1) kJ mol1 = 4.9 kJ mol1

Estimated ΔH neut for CH3CH2COOH and NH3

= 57.1 + 7.2 + 4.9 kJ mol1 = 45.0 kJ mol1

A36.31. (a) H = 92.3 2 kJ mol1 = 184.6 kJ mol1

(b) H = 238.6 2 kJ mol1 = 477.2 kJ mol1

(c) H = +1.9 kJ mol1

(d) H = 110.5 2 + (393.5) kJ mol1 = 614.5 kJ mol1

2. (a) C8H18(l) + 8 O2(g) 8CO2(g) + 9H2O(l)

H c = 5512 kJ mol1

(b) CH3COOH(aq) + NH3(aq) CH3COONH4(aq)ΔH neut

= 51.5 kJ mol1

(c) KBr(s) + aq K+(aq) + Br(aq)ΔH

soln = +20.0 kJ mol1



(d) 2B(s) + 3H2(g) B2H6(g)ΔH f

= +36.0 kJ mol1

A36.4Heat transferred to water = mw c (T2 T1)= (500)(1) g 4.2 J g1 K1 [(50.1 + 273) (25.3 + 273)] K = 52 080 JNumber of moles of methanol burnt

= = 0.0828 mol

Heat released per mole of methanol burnt

= = 62 900 J mol1 = 629 kJ mol1

Hence, the enthalpy change of combustion of methanol is 629 kJ mol1.

A36.5Heat transferred to solution = m c (T2 T1)= (75.0 + 50.0)(1) g 4.2 J g1 K1 [(27.0 + 273) (25.0 + 273)] K = 1050 JNumber of moles of water formed

= 0.5 mol dm3 dm3

= 0.025 mol (Sodium hydroxide is the limiting reactant.)

ΔH neut

= J mol1 = 42 000 J mol1 = 42 kJ mol1

A36.61. Heat transferred to solution = m c (T2 T1)

= 45.0 g 4.2 J g1 K1 4.8 K = 907.2 J

ΔH soln = J mol1

= 36 288 J mol1 (36.3 kJ mol1)

2. (a) Heat transferred to solution = m c (T2 T1)= (50)(1) g 4.2 J g1 K1 3.89 K = 816.9 J

Mass of HCl dissolved = 10 g min1 min = 1.67 g

Number of moles of HCl dissolved = mol = 0.0458 mol

Heat released per mole of HCl dissolved

= = 17 836 J mol1

Hence, the enthalpy change of solution of hydrogen chloride is 17 836 J mol1.

(b) To make sure that the solution is at infinite dilution.



Chapter 36 Standard enthalpy change of combustion, neutralization, solution and formation

Chapter Exercise

1. 1, 101 325, 298, concentration2. moles3. combustion4. neutralization,water5. solution, heat, infinite dilution6. formation, substance, standard conditions7. heat loss, standard conditions8. Bomb9. heat, specific heat capacity, experiment10. A11. B12. D13. C14. B15. C

16. (a) Energy is consumed to break the strong ionic bonds in aluminium oxide.

(b) 2Al(s) + 1 O2(g) Al2O3(s)

(c) kJ mol1 = 1675 kJ mol1

(d) 2Al(s) + 1 O2(g) Al2O3(s) ΔHf = 1675 kJ mol1

17. (a) Energy needed = mcΔT = 400 cm3 1 g cm3 4.2 J g1 K1 [(70 + 273) (20 + 273)] K = 84 000 J (84 kJ)

No. of moles of propan-1-ol needed = mol = 0.0418 mol

Mass of propan-1-ol needed = 0.0418 (12.0 3 + 1.0 8 + 16.0) g = 2.51 g

(b) A larger mass is needed.(c) There is heat loss to surroundings during the heating process.



18. (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.

(c) H2SO4(aq) + KOH(aq) K2SO4(aq) + H2O(l)

(d) No. of moles of H2SO4(aq) used = 0.50 mol = 0.0325 mol

No. of moles of KOH(aq) used = 0.90 mol = 0.0450 mol

From the equation, 1 mole of H2SO4(aq) requires 2 moles of KOH(aq) for complete neutralization. Thus, 0.0325 mole of H2SO4(aq) requires 0.0650 mole of KOH(aq) for complete neutralization. Hence, KOH is the limiting reactant.Energy released during neutralization = mcΔT = (65.00 + 50.00) cm3 1.0 g cm3 4.2 J g1 K1 [(30.3 + 273) (25.0 + 273)] K = 2560 J (2.56 kJ)Standard enthalpy change of neutralization between sulphuric acid and potassium hydroxide

= kJ mol1 = 56.9 kJ mol1

19. (a) Exothermic(b) Energy absorbed by water

= mcΔT = 850 cm3 1.0 g cm3 4.2 J g1 K1 [(28.5 + 273) (23.0 + 273)] K = 19 635 J (19.635 kJ)

(c) Standard enthalpy change of solution of NaOH

= kJ mol1 = 9.82 kJ mol1




Class Practice

A37.1 (a) SHAPE \* MERGEFORMAT

By applying Hess’s Law, H = +297.0 + (396.0) kJ mol1 = 99.0 kJ mol1

(b)

A37.2(a) By applying Hess’s Law,

H 1 = H f + H 2

H f = (394.0) (283.0) kJ mol1 = 111.0 kJ mol1


SO2(g) + O2(g)SO3(g)

+297 kJ mol1 396.0 kJ mol1

S(s) + O2(g)

H

SO2(g) +

O2(g)

SO3(g)

+297.0 kJ mol1 396.0 kJ mol1

99.0 kJ mol1

Reaction coordinate

Ent

halp

y

S(s) +O2(g)

48


(b)

A37.3

H f = 2 (394.0) + 3 (286.0) (1371) kJ mol1

= 275.0 kJ mol1

A37.4 (a) CaCO3(s) CaO(s) + CO2(g)(b) By applying Hess’s Law,

H = ΣH f (products) – ΣH f (reactants)H = H f [CaO(s)] + H f [CO2(g)] – H f [CaCO3(s)]

= (635.0) + (395.0) – (1207) kJ mol1

= +177.0 kJ mol1


C(s) +O2(g) +O2(g)

111.0 kJ mol1

CO(g) + O2(g)

283.0 kJ mol1

394.0 kJ mol1

CO2(g)

Reaction coordinate

Ent

halp

y

H f2C(s) + 3H2(g) +O2(g)C2H5OH(l)

2CO2(g) + 3H2(g) +O2(g)

2CO2(g) + 3H2O(l)

+ 2O2(g)

394.0 kJ mol1 2

+ O2(g)

286.0 kJ mol1 3

+ 3O2(g)

1371 kJ mol1

49

Na2CO3(s) + H2O(g) + CO2(g)


A37.5

1. 3C(s) + 4H2(g) + O2(g) C3H7OH(l)

By applying Hess’s Law,H

f = ΣH c(reactants) ΣH c(products)

= 3 H c [C(s)] + 4 H c [H2(g)] H c [C3H7OH(l)]= 3 (394.0) + 4 (286.0) (1004) kJ mol1

= 1322 kJ mol1

Enthalpy change of formation of 100 g liquid propanol

= 1322 kJ mol1 mol

= 2203 kJ

2.

By applying Hess’s Law,H = 2 H

f [NaHCO3(s)] + H f [Na2CO3(s)] + H

c [C(s)] + H c [H2(g)]

= 2 (948.0) + (1131) + (394.0) + (286.0) kJ mol1 = +85 kJ mol1


2NaHCO3(s)

2Na(s) + H2(g) + 2C(s) + 3O2(g)

H f [NaHCO3(s)] 2

Na2CO3(s) + CO2(g) + H2(g) + O2(g)

H c [H2(g)]

H f [Na2CO3(s)]

Na2CO3(s) + H2(g) + C(s) + O2(g)

H c [C(s)]

H

50



Chapter Exercise

1. enthalpy, route2. lower, higher, enthalpy, endothermic, exothermic3. products, reactants4. reactants, products5. D6. C7. A8. B9. B10. A

11. (a) H2(g) + Cl2(g) HCl(g)

(b)

H f [HCl(g)]

= –H3 + H2 – H1

= –(– ) + (– ) – (–176.1) kJ mol1

= –92.3 kJ mol1

12. (a) C3H6O(l) + 4O2(g) 3CO2(g) + 3H2O(l)H c[C3H6O(l)] = –1790 kJ mol1

(b)

H f [C3H6O(l)] = H2 – H1


NH3(g) + H2(g) + Cl2(g)

H3

NH4Cl(s)

H f [HCl(g)]

N2(g) + 2H2(g) + Cl2(g)

H1

NH3(g) + HCl(g)

H2

3C(s) + 3H2(g) +O2(g)

H f [C3H6O(l)]

C3H6O(l) + 4O2(g)

H1

H2

3CO2(g) + 3H2O(l)

+ 4O2(g) + 4O2(g)

51


= 3 (–394.0) + 3 (–286.0) – (–1790) kJ mol1

= –250 kJ mol1

13. (a) H f [H2O(g)] = (–286.0) + (+41.1) kJ mol1 = –244.9 kJ mol1

(b) H f [Al2O3(s)] = – kJ mol1 = –3360 kJ mol1

(c) For the reaction,

Al(s) + NH4NO3(s) N2(g) + 3H2O(g) + Al2O3(s)

By applying Hess’s Law,H = ΣH

f (products) – ΣH f (reactants)

H = H f [N2(g)] + 3 H

f [H2O(g)] + H f [Al2O3(s)]

– H f [Al(s)] – H

f [NH4NO3(s)]

–1015 = 0 + 3 (–244.9) + (–3360) – 0 – H f [NH4NO3(s)]

H f [NH4NO3(s)] = –933.1 kJ mol1

14. (a)

By applying Hess’s Law,Hc[C2H4(g)] = –H2 + H1

= –(+52.3) + 2 (–394.0) + 2 (–286.0) kJ mol1

= –1412 kJ mol1

(b)

By applying Hess’s Law,H1 = H2 – H3

= [–1412 + (–286.0)] – (–84.6) kJ mol1

= –1613 kJ mol1


2C(s) + 2H2(g) + 3O2(g)

C2H4(g) + 3O2(g)

H1H2

2CO2(g) + 2H2O(l)Hc [C2H4(g)]

3O2(g)

2CO2(g) + 3H2O(l)

C2H4(g) + H2(g)H1

H2

C2H6(g)

3O2(g) H3

52


Part VIII Chemical reactions and energy

Part Exercise

1. D2. D3. D4. C5. B6. B7. A8. D9. A10. C11. D

12. (a) C6H12O6 + 6O2 6CO2 + 6H2O(b) Molar enthalpy change of combustion of glucose

= – kJ mol1

= –2826 kJ mol1

(c) Some energy is used for supporting other body activities.(d) Energy needed for walking a distance of 1.5 km = 220 1.5 kJ = 330 kJ

Let m be the mass of glucose needed for providing the energy required

2826 30% = 330

m = 70.170.1 g of glucose is needed for providing the energy required.

mass of cereal needed = g = 200 g

13. (a) Energy released by burning (CH3)2NNH2

= kJ g1

= 28.2 kJ g1

Energy released by burning CH3OH

= kJ g1

= 22.7 kJ g1

Energy released by burning C8H18

= kJ g1

= 49.0 kJ g1

Hence, C8H18 releases the largest amount of energy per gram of substance burnt.

(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.



14. (a) Mg(NO3)2(aq) + Na2CO3(aq) MgCO3(s) + 2NaNO3(aq)(b) Exothermic. The reaction released heat that raised the temperature of the

reaction mixture.(c) Energy released in the first experiment

= (10.0 + 10.0) cm3 1.0 g cm3 4.2 J g1 K1 [(22.2 + 273) – (20.0 + 273)] K= 184.8 JEnthalpy change of precipitation of magnesium carbonate

= – J mol1

= –18 480 J mol1 (18.5 kJ mol1)For 30.0 cm3 of 1.0 M solutions,Energy released

= 18.5 1.0 kJ

= 0.555 kJ (555 J)mcT = 555(30.0 + 30.0) 1.0 4.2 T = 555T = 2.20 K (oC)

(d) For 30.0 cm3 of 1.2 M solutions,Energy released

= 18.5 1.2 kJ

= 0.666 kJ (666 J)mcT = 666(30.0 + 30.0) 1.0 4.2 T = 666T = 2.64 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.

15. (a) Number of moles of Cu2+ used = 0.04 mol = 0.002 mol

(b) Mg(s) + Cu2+(aq) Cu(s) + Mg2+(aq)

H = J mol1 = –525 000 J mol1 = –525 kJ mol1

(c) Mg(s) + Cu2+(aq) Cu(s) + Mg2+(aq) H = –525 kJ mol1

(d)


Reaction coordinate

Enthalpy

Mg(s) + Cu2+(aq)

Cu(s) + Mg2+(aq)

H = –525 kJ mol1

54


16. (a) Number of moles of Ag+ used = 0.05 mol = 0.0025 mol

(b) Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)Enthalpy change for each mole of Ag+(aq) reacted

= J mol1 = –74 000 J mol1 = –74 kJ mol1

According to the equation, the standard enthalpy change of the above displacement reaction involves the complete reaction of 2 moles of Ag+(aq). Therefore, H = –74 2 kJ mol1 = –148 kJ mol1

(c) Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) H = –148 kJ mol1

(d)

H = H1 + H2

= (–525) + (–148) kJ mol1

= –673 kJ mol1

17. (a) Hydrochloric acid(b)

H = H1 – H2


Mg2+(aq) + Cu2+(aq) + 2Ag(s)

2Ag+(aq)

Mg(s) + Cu2+(aq)

H1

H2

Cu(s) + Mg2+(aq)

H 2Ag+(aq)

Mg2+(aq) + H2(g) + H2O(l)

Mg(s) + H2O(l)

H1

H2

MgO(s) + H2(g)

H

2H+(aq)

55


(c) She is correct. In fact, calcium reacts with water to give calcium hydroxide instead of calcium oxide.Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g)

18. (a) C20H32O2 + 27O2 20CO2 + 16H2O(b) H c

= 20 (–395.0) + 16 (–286.0) – (–636.0) kJ mol1 = –11 840 kJ mol1

(c) Energy needed = mcT= (600 1000) g 4.2 J g–1 K–1 [(25 + 273) – (0 + 273)] K= 63 000 kJNo. of moles of arachidonic acid needed

= mol = 5.32 mol

Mass of arachidonic acid needed = 5.32 (12.0 20 + 1.0 32 + 16.0 2) g= 1617 g

19. (a) Positive value(b) 15 s(c) 40 s(d) The water temperature was constant at 23oC before the addition of

potassium nitrate. When potassium nitrate was added at 15 s, heat was absorbed for the dissolution and the temperature of the solution fell. At 40 s, all solute had been dissolved and the temperature of the solution reached a minimum. As heat was acquired from the surroundings, the temperature of the solution rose.

(e) Number of moles of solute used

= mol

= 0.0989 molHeat absorbed during dissolution = 150 cm3 1.0 g cm–3 4.2 J g–1 K–1 [(23.0 + 273) – (17.5 + 273)] K= 3465 JStandard enthalpy change of solution of potassium nitrate

= + J mol1

= +35 035 J mol1 (+35 kJ mol1)



20. (a)

(b) (i) Lowered. As some heat is lost, the temperature rise during measurement will be lowered and so is the calculated value.

(ii) Raised. As the actual specific heat capacity is smaller, there will be a higher temperature rise and so is the calculated value.

(iii) Lowered. As less iron reacts, less heat will be released. This results in a smaller temperature rise and thus the calculated value will be lowered.

21. (a)Name Formula Standard enthalpy change of combustion / kJ mol1

Methanol CH3OH –26.34 (12.0 + 1.0 4 + 16.0) = –843Ethanol CH3CH2OH –29.80 (12.0 2 + 1.0 6 + 16.0) = –1371

Propan-1-ol CH3CH2CH2OH –33.50 (12.0 3 + 1.0 8 + 16.0) = –2010Butan-1-ol CH3CH2CH2CH2OH –36.12 (12.0 4 + 1.0 10 + 16.0) = –2673

(b) The larger alcohols have higher standard enthalpy change of combustion.(c) About 3200–3330 kJ mol1

22.

By applying Hess’s Law,

H = –(–180.0) + (–112.0) + (–572.0) – (–1396) kJ mol1

= –92.0 kJ mol1


thermometer

expanded

polystyrene cup

lid

beaker

cotton wool

Fe(s) + H2SO4(aq)

2NO(g) + 3H2(g) 2NO2(g) + 3H2O(g)

N2(g) + 3H2(g)

O2(g)

2NO2(g) + 3H2(g)

2NH3(g) H

H 1

H 2 H

3

H 4

3O2(g)

1O2(g)O2(g)

57


23. Since2SO2(g) + 4HCl(g) 2SOCl2(l) + 2H2O(l)

H = 2 (10.3) kJ mol1 = 20.6 kJ mol1

2PCl3(l) + O2(g) 2POCl3(l)H = 2 (325.7) kJ mol1 = 651.4 kJ mol1

2P(s) + 3Cl2(g) 2PCl3(l)H = 2 (306.7) kJ mol1 = 613.4 kJ mol1

2Cl2(g) + 2H2O(l) 4HCl(g) + O2(g)H = +202.6 kJ mol1

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 = 20.6 + (651.4) + (613.4) + (+202.6) kJ mol1 = 1083 kJ mol1

24. (a) H = 3 (–395.0) + 4 (–286.0) – (–1574) kJ mol1

= –755.0 kJ mol1

(b) Enthalpy change of formation of C4H6(g) is defined by the following equation:4C(s) + 3H2(g) C4H6(g)H = 4 (–395.0) + 3 (–286.0) – (–2542) kJ mol1

= +104.0 kJ mol1

(c) The chemical equation for the conversion is:2C2H4(g) C4H6(g) + H2(g)H = 2 (–1393) – (–2542) – (–286.0) kJ mol1

= +42.0 kJ mol1


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