Book 3 ans

HKDSE CHEMISTRY CHEMI STRY – A Modern View (Chemistry) Coursebook 3 Suggested answers Chapter Chapter 25 Simple molecular molecular substances substances with non-octet non-octet

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structures and shapes of simple molecules  Class Practice

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Chapter Exercise

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Chapter 2 Chapter 26 Bond po polar larity ity 

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Chapte Chapterr 27 Interm Intermole olecu cular lar forces forces 

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Chapter 28 Chapter 28 Structures Structures and properties properties of molecular molecular crystals crystals 

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Chapte Chapterr 29 29 Chemic Chemical al cell cells s in daily daily life 

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Chapter 30 Chapter 30 Simp Simple le che chemi mica call cell cells s 

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Chap Chapte terr 31 31 Redo Redox x rea react ctio ions ns © Aristo Educational Press Ltd. 2010

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Chapter 32 Chapter 32 Redox Redox reactio reactions ns in chemic chemical al cells cells 

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Chapter 3 Chapter 33 Elec lectro trolysis 

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Chapter Chapter 34 Importance Importance of redox redox reactio reactions ns in modern modern ways ways of living living 

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Chapter 35 Chapter 35 Energy Energy change changes s in chemic chemical al rea reactio ctions ns 

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Chapter 36 Chapter 36 Stand Standard ard enthalp enthalpy y chang change e of combusti combustion, on, neutra neutraliz lizatio ation, n, solution and formation  Class Practice 

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Chapter 3 Chapter 37 Hess’s Law 

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© Aristo Educational Press Ltd. 2010

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Chapter 32 Chapter 32 Redox Redox reactio reactions ns in chemic chemical al cells cells 

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Chapter 3 Chapter 33 Elec lectro trolysis 

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Chapter Chapter 34 Importance Importance of redox redox reactio reactions ns in modern modern ways ways of living living 

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Chapter 35 Chapter 35 Energy Energy change changes s in chemic chemical al rea reactio ctions ns 

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Chapter 36 Chapter 36 Stand Standard ard enthalp enthalpy y chang change e of combusti combustion, on, neutra neutraliz lizatio ation, n, solution and formation  Class Practice 

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Chapter 3 Chapter 37 Hess’s Law 

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HKDSE CHEMISTRY  A Modern View (Chemistry) (Chemist ry)

Coursebook 3

Chapter Chapter 25 Simple Simple molecular molecular substan substances ces with with non-octet non-octet structu structures res and 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 lone No. of Shape pairs bond pairs

CH4 NH3

No. of Spatial electron arrangement pairs of electron pairs 4 Tetrahedral 4 Tetrahedral

0 1

4 3

H2O

4

2

2

Tetrahedral


Tetrahedral Trigonal pyramidal V-shaped

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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. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

eight eight six 10 12 non-octet linear, 180° trigonal planar, 120° tetrahedral, 109.5° trigonal bipyramidal, 90°, 120° octahedral, 90° 2, 2 1, 3 shapes D B C A D A

21. (a) A: ; B: (b) A: WZ 3 B: YZ 3 (c) A. The central atom (W ) has only six electrons in its outermost shell. 22. (a) X : Trigonal pyramidal Y : Trigonal bipyramidal (b) 107° (c) 90°(for Cl(axial) −P−Cl(equatorial) bond angles) and 120° (for Cl−P−Cl bond angles within the plane of triangle)

23. (a) (b)


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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°.


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Chapter 26 Bond polarity Class Practice A26.1

(a) (b)

(c) A26.2 (a) NCl3 has three polar N−Cl 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 B−Cl 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 NBr 3 and NF3 are pointing to different directions. The situations are shown below:

In a non-uniform electrostatic field, the nitrogen end of NBr 3 will point to the positive pole while the nitrogen end of NF3 will point to the negative pole.


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Chapter 26 Bond polarity Chapter Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

attract polar non-polar larger, less, smaller polar, cancel out polar, symmetrically deflection C B C B C 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 (F−F) is non-polar. 2. Polar X−F bonds may cancel out their individual dipole moments to give © Aristo Educational Press Ltd. 2010

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a non-polar molecule e.g. CF4, SF6, etc.


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Chapter 27 Intermolecular forces 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)

A27.2 1. (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


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(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.


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Chapter 27 Intermolecular forces Chapter Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

molecules, electrostatic an der Waals’ forces, ydrogen bonding Dipole-dipole forces Dispersion forces, induced molecular, surface, olarity hydrogen, electrons energy surface, iscosity hydrogen bonds high proteins, deoxyribonucleic acids D A A A B 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


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−

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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.


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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). (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. A28.4 1. (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.


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Coursebook 3

Chapter 28 Structures and properties of molecular crystals Chapter Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

order four, hydrogen oxygen open hydrogen carbon, hollow spherical, benzene, close, strong, hard, insulator cylindrical, melting, solvents, carbon, tensile, conductors D B D B A C

15. (a) a b

a b

a

b

* Label all other bonds accordingly.

(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) (b) (c) (d)

It is an element as it is composed of carbon atoms only. 12 pentagons and 0 hexagon Other than pentagons, there are hexagons over the surface of C60. 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 strong © Aristo Educational Press Ltd. 2010

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Coursebook 3

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.


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Coursebook 3

Part VI Microscopic world II Part Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

C A C A C B A C C A

11. (a) WY 2 (b) The electron diagram of WY 2 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.


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(d) (e) (e) It is is non-po non-polar lar as the the effe effects cts of of all polar polar S−F bonds cancel each other. 14. (a) (i) Non-polar (ii) Non-polar (iii) iii) Polar lar (b) (b) Iodi Iodine ne is nonnon-p polar olar.. It diss dissol olve vess well well in tetr tetrac achl hlor orom omet etha hane ne beca becaus usee 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 (dispers (dispersion ion forces) forces) between between tetrachlo tetrachlorome romethane thane molecules. molecules. On the other 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) Tetrachlo etrachlorome romethan thanee cannot cannot dissolve dissolve in water water.. (d) Two separa separate te layers layers will be seen. The The yellow yellow colour colour of the aqueo aqueous us layer layer fades while the tetrachloromethane layer turns violet. 15. 15. (a) (a) In ice, ice, water ater mole molecu cule less are are arra arrang nged ed order orderly ly in a regu regula larr ope open netw networ ork k 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. only. Thus, more energy is needed to separate water molecules in the boiling process. (c) (c) Alth Althou ough gh both compo compoun unds ds can form form hydro hydroge gen n bond bonds, s, CH3CH2CH2CH2OH has a larger size than CH3CH2OH and it has larger dispersion forces between molecu molecules les.. Th Thus, us, mo more re energ energy y is needed needed to separ separate ate CH3CH2CH2CH2OH molecules in the boiling process. (d) Hexane Hexane is non-polar non-polar.. It cannot cannot dissolve dissolve in water water which which is polar. 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 Dispersion force Dispersion 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

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17. (a)

(b)

(c)

(d) (e) The The roots of the plant plantss can absorb absorb the fertil fertilize izerr dissolve dissolved d in water water from 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) Am Ammonia monia has has a higher boiling boiling point point than nitrogen. nitrogen. Am Ammoni moniaa molecule moleculess are attrac attracted ted by hyd hydrog rogen en bonds bonds while while nitrog nitrogen en molecu molecules les are attrac attracted ted by weaker dispersion forces. More energy is needed in boiling ammonia. 19. 19. (a) Either Either by by using using ver very y high high press pressure uress to comp compre ress ss the the hydro hydroge gen n gas or or very very low temperatures temperatures to turn the hydrogen gas into a liquid. (b) (b) Dispe ispers rsio ion n forc forces es (c) Hydroge Hydrogen n in molecu molecular lar form form can can be release released d easily easily and used as as fuel. fuel.


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Coursebook 3

Chapte Chapterr 29 Chemic Chemical al cells cells in dail daily y life life Class Practice A29.1 A little 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 Its elec electr trol olyt ytee is potas potassi sium um hydr hydrox oxid ide, e, whic which h is alkaline. Its positive electrode consists of manganese(IV) oxide (mixed with a little powdered graphite). graphite). 2. Heari Hearing ng aids aids are small small and and they they need need very sma small ll cells. cells. Silv Silver er oxide oxide cells cells (shap (shaped ed like a button) would serve the purpose. 3. Silve Silverr and its its comp compoun ounds ds (e.g. (e.g. silver silver oxid oxide) e) are are expens expensive ive.. 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) A flat flat dis disch char arge ge curv curve. e. (b) (b) Alkal lkalin inee mang angane anese cell cell/s /sil ilve verr oxid oxidee cell cell/n /nic icke kell-ca cadm dmiu ium m rech rechar arge geab able le cell/lith cell/lithiumium-ion ion recharge rechargeable able cell/nick cell/nickel-m el-metal etal hydride hydride recharge rechargeable able cell (any ONE)


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Coursebook 3

Chapter 29 Chemical cells in daily life Chapter Exercise 1. 2. 3. 4. 5. 6. 7. 8.

9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

electricity electrolyte, paste inc-carbon, lkaline manganese, ilver oxide, ithium primary nickel-cadmium rechargeable, lithium-ion rechargeable, nickel-metal hydride rechargeable, lead-acid accumulator zinc-carbon, alkaline manganese, nickel-metal hydride rechargeable, 3.7 V, Leadacid accumulator, 12 V voltage alkaline manganese (a) size of cell (b) shape of cell (c) type of cell nvironment, mercury, rechargeable, recycling D A B B A D C D B C A B C B B 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; (−) zinc Electrolyte: ammonium chloride 27. (a) B New 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 © Aristo Educational Press Ltd. 2010

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(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) C Alkaline manganese cells cannot be recharged. During recharging, heat and hydrogen may be produced and hence lead to explosion. (e) D Small children may mistake small-sized button cells as sweets. 28. (a) Silver oxide cell/lithium button cell These 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 cell It 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 © Aristo Educational Press Ltd. 2010

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Coursebook 3

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 lowend 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) −


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Chapter 30 Simple chemical cells 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. A30.2 Mg/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.


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Chapter 30 Simple chemical cells Chapter Exercise 1. 2. 3. 4. 5. 6.

7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

metals, electrolyte, metals, wire negative, positive electromotive force (or e.m.f.), higher voltmeter, igital, ultimeter voltage (or e.m.f.), Electrochemical Series (a) similar (b) lose, cations, aqueous solution (c) reactivity, air, dilute acids, losing salt bridge, complete, charges porous pot D C C C B A C C A

18. (a) Dilute sulphuric acid (or any other dilute acid, or an acidic salt solution like ammonium chloride.) (b) Positive electrode: copper electrode Negative 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)

(b) (c)

(d) (e)

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 halfcells. 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. (ii) The iron electrode slowly dissolves and the green iron(II) solution becomes darker. Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) From iron electrode to copper electrode (or from left to right).


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HKDSE CHEMISTRY  A Modern View (Chemistry) 2+

Coursebook 3

−

20. (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. 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) (b) (c) (d) (e) (f)

22.

−

A metal couple is a combination of two different metals connected to an external circuit. The filter paper separates the two metal strips, preventing a short circuit. Besides, the salt solution allows movement of ions to complete the circuit. From copper to silver. The voltmeter would give a positive reading. Removing the filter paper would cause a short circuit, thus the voltmeter reading drops to zero. The voltage would be higher. This is because magnesium is in a higher position than copper in the Electrochemical Series. Connect the metals aluminium, copper, iron, lead, magnesium and zinc one by one in turn to the circuit as shown below:

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. 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.


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Chapter 31 Redox reactions 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 oxygen electron oxidation number 2.

Oxidation +O e increases

Reduction O +e decreases

(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


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3.

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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) (ii) (iii) (b) (i) (ii) (iii)

−

−

Cl2(g) + 2e → 2Cl (aq) − − 2Br (aq) → Br 2(aq) + 2e − − Cl2(g) + 2Br (aq) → 2Cl (aq) + Br 2(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)

A31.5 (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) A31.8 1. Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) 2. (a) Zn to Cu (b) Mg to Ag (c) Zn to Pb


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Chapter 31 Redox reactions Chapter Exercise 1. 2. 3.

8.

transfer loses, gains (a) oxidizes, accepting (b) reduces, 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, NO2(g), oxidizing, SO2(g)

9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

B C D D C C C C D B B B D D B C B C C A

4.

5. 6. 7.


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29. (a) (b) (c) (d) (e) (f) (g) (h) (i)

Coursebook 3

0 +2 +3 −3 +4 +6 +7 +5 +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) Cr2 O72 (aq) + 14H+(aq) + 6e → 2Cr 3+(aq) + 7H2O(l) − − + 3 × (2Br (aq) → Br 2(aq) + 2e ) ____________________________________________________________ − − Cr 2O72 (aq) + 14H+(aq) + 6Br (aq) → 2Cr 3+(aq) + 3Br 2(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.


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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.


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Chapter 32 Redox reactions in chemical cells 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 Br 2(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) → Br 2(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) + 5Br 2(aq) + 8H2O(l) A32.3 1. 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. © Aristo Educational Press Ltd. 2010

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Chapter 32 Redox reactions in chemical cells Chapter Exercise 1. 2. 3. 4.

5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

oxidation, cathode negative, 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, oxygen oxidant, catalyst B A D D D D C C B

17. (a) Dilute sulphuric acid − − (b) Cr2 O72 (aq) + 14H+(aq) + 6e → 2Cr 3+(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. −

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, 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.

18. (a) (b) (c) (d)

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. © Aristo Educational Press Ltd. 2010

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2+

(e) (1) In the Daniell cell, Cu (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) (b) (c) (d) (e) (f) (g)

Hot, concentrated potassium hydroxide solution. As a catalyst. − − H2(g) + 2OH (aq) → 2H2O(l) + 2e Oxidation. It is because the oxidation number of H changes from 0 to +1. − − 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. (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 timeconsuming.


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Chapter 33 Electrolysis 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 A33.2 − − H+(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 A33.4 1. (a)

2.

(b) (a) (b) (c) (d) (e)

−

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. The blue colour of the solution becomes paler. chlorine sodium amalgam oxygen ionizes (dissolves) Cu2+(aq) (i)

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. A33.6 1. 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. © Aristo Educational Press Ltd. 2010

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Chapter 33 Electrolysis Chapter Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

decomposition chemical, electricity cathode, anode positive ions, cathode, anode, cathode, gain, anode, lose, anode, cathode E.C.S., concentration, electrodes electroplating, electroplated, corrosion cathode anode thicken lkalis, heavy metal, rganic, health Environmental Protection C A B B A D A 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) → Br 2(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


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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)


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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 Anodizing Can be applied to most metal items (e.g. Can only be applied to metal items made iron or copper items) and non-metal items. of aluminium. The item to be electroplated is connected Aluminium item to be anodized is to the cathode during electroplating. connected to the anode during anodizing. A thin layer of another metal (e.g. nickel A thin layer of protective aluminium oxide coating on iron) is coated on the item. is formed on the aluminium item. The item acquires the colour of the coating The oxide layer is too thin to be seen and metal (e.g. the golden colour of gold the aluminium item retains its own silvery plating). The item cannot be further dyed colour. The oxide layer can be dyed to a to another colour. new colour if necessary.


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Chapter 34 Importance of redox reactions in modern ways of living Class Practice A34.1 1. (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) A34.2 Lithium 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.


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Chapter 34 Importance of redox reactions in modern ways of living Chapter Exercise 1. 2. 3. 4. 5. 6.

hotosynthesis, etabolism ermentation, ntioxidants, reathalyzers, leaching Alkaline Fuel Cells, Proton Exchange Membrane Fuel Cells, Phosphoric Acid Fuel Cells, Molten Carbonate Fuel Cells, Solid Oxide Fuel Cells lithium, lithium-ion rechargeable, lithium-ion polymer rechargeable oxidized, reduced lithium-carbon compound, metal oxide, lithium, organic solvent, discharging +

−

CoO2 + xLi + xe

Li xCoO2

⇌

charging discharging −

Li xC6 ⇌ 6C + xLi+ + xe charging discharging

CoO2 + Li xC6 ⇌ Li xCoO2 + 6C charging

7. 8. 9. 10. 11. 12.

harge density, oltage, rain capacity, recharged, nvironmentally preferred A A D A 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: 2K 2Cr 2O7(aq) + 3CH3CH2OH(aq) + 8H2SO4(aq) → 2Cr 2(SO4)3(aq) + 3CH3COOH(aq) + 2K 2SO4(aq) + 11H2O(l)


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(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 40°C (the melting point of phosphoric acid) before the PAFC can function. (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). (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)


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Part VII Redox reactions, chemical cells, and electrolysis Part Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

A B D B B B C B D A C A D A D

16. (a) (b) (c) (d) (e)

(f)

manganese(IV) oxide and powdered graphite; ammonium chloride paste − Zn(s) → Zn2+(aq) + 2e − 2MnO2(s) + 2H+(aq) + 2e → Mn2O3(aq) + H2O(l) From zinc to copper. 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. 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 zinccarbon 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 reducing © Aristo Educational Press Ltd. 2010

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Coursebook 3

agents like carbon. At 850°C, 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) K2Cr 2O7(aq) + H2SO4(aq) + 3SO2(g) → Cr 2(SO4)3(aq) + K 2SO4(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)


40


21. (a) (b) (c) (d) (e)

+

Coursebook 3

−

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. Methanol is toxic and is highly flammable. 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 (Mg 2+(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) −


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Coursebook 3

Chapter 35 Energy changes in chemical reactions 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.


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Coursebook 3

Chapter 35 Energy changes in chemical reactions Chapter Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

pressure gives out takes in products, reactants lower ombustion, recipitation, neutralization oil fractions take in, give out bond-forming, 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. 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) b) (c) (d)

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) C−C, C−H and O=O C=O and O−H The enthalpy change in bond-forming processes is larger than the enthalpy change in bond-breaking processes.


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Coursebook 3

Chapter 36 Standard enthalpy change of combustion, neutralization, solution and formation Class Practice A36.1

11.5

Heat change = −1371 ×

12.0 × 2 + 1.0 × 6 + 16.0 Hence, the heat evolved is 342.75 kJ.

kJ = −342.75 kJ

A36.2 1. (a)

Let Y be the mass of hydrogen in the mixture. 50 − Y Y × −285.8 + × −890.4 = −4725 1.0 × 2 12.0 + 1.0 × 4 Y = 22.3 Thus, 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. 50 − X X × −285.8 + × −890.4 = −5000 1.0 × 2 12.0 + 1.0 × 4 X = 25.4 Thus, the percentage by mass of hydrogen in the mixture 25.4 = × 100% 50 = 50.8%

2.

(a) CH3CH2COOH (b) NH3 (c) Heat required for the ionization of CH3CH2COOH − − = −49.9 −(−57.1) kJ mol 1 = 7.2 kJ mol 1 Heat required for the ionization of NH3 − − = −52.2 −(−57.1) kJ mol 1 = 4.9 kJ mol 1 Estimated Δ H neut for CH3CH2COOH and NH3 − − = −57.1 + 7.2 + 4.9 kJ mol 1 = −45.0 kJ mol 1

A36.3 1. (a)

(b) (c) (d) 2.

∆ H = −92.3 ×

−

−

1 O2(g) → 8CO2(g) + 9H2O(l) 2

∆ H c = −5512

(c)

−

2 kJ mol 1 = −477.2 kJ mol 1 −1 ∆ H = +1.9 kJ mol −1 −1 ∆ H = −110.5 × 2 + (−393.5) kJ mol = −614.5 kJ mol ∆ H = −238.6 ×

(a) C8H18(l) + 8

(b)

−

2 kJ mol 1 = −184.6 kJ mol 1

−

kJ mol 1

CH3COOH(aq) + NH3(aq) → CH3COONH4(aq) − Δ H neut = −51.5 kJ mol 1 − KBr(s) + aq → K +(aq) + Br (aq) − Δ H soln = +20.0 kJ mol 1


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(d)

Coursebook 3

2B(s) + 3H2(g) → B2H6(g) − Δ H f = +36.0 kJ mol 1

A36.4 Heat transferred to water = mw × c × (T 2 − T 1) − − = (500)(1) g × 4.2 J g 1 K 1 × [(50.1 + 273) −(25.3 + 273)] K = 52 080 J Number of moles of methanol burnt 2.65 g = = 0.0828 mol 12.0 + 1.0 × 4 + 16.0 g mol −1

Heat released per mole of methanol burnt 52 080 J − − = = 62 900 J mol 1 = 629 kJ mol 1 0.0828 mol − Hence, the enthalpy change of combustion of methanol is −629 kJ mol 1. A36.5 Heat transferred to solution = m × c × (T 2 −T 1) − − = (75.0 + 50.0)(1) g × 4.2 J g 1 K 1 × [(27.0 + 273) −(25.0 + 273)] K = 1050 J Number of moles of water formed 50.0 − = 0.5 mol dm 3 × dm3 1000 = 0.025 mol (Sodium hydroxide is the limiting reactant.) 1050 − − − Δ H neut = − J mol 1 = −42 000 J mol 1 = −42 kJ mol 1 0.025 A36.6 1. Heat transferred to solution = m × c × (T 2 − T 1) − − = 45.0 g × 4.2 J g 1 K 1 × 4.8 K = 907.2 J 907.2 − Δ H soln = − 1.06 J mol 1

6.9 + 35.5 − − = −36 288 J mol 1 (−36.3 kJ mol 1) 2.

(a) Heat transferred to solution = m × c × (T 2 − T 1) − − = (50)(1) g × 4.2 J g 1 K 1 × 3.89 K = 816.9 J 10 − Mass of HCl dissolved = 10 g min 1 × min = 1.67 g 60 1.67 Number of moles of HCl dissolved = mol = 0.0458 mol 1 +35.5 Heat released per mole of HCl dissolved 816.9 J − = = 17 836 J mol 1 0.0458 mol Hence, the enthalpy change of solution of hydrogen chloride is −17 836 J − mol 1. (b) To make sure that the solution is at infinite dilution.


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Coursebook 3

Chapter 36 Standard enthalpy change of combustion, neutralization, solution and formation Chapter Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

1, 101 325, 298, concentration moles combustion neutralization,water solution, heat, infinite dilution formation, substance, standard conditions heat loss, standard conditions Bomb heat, specific heat capacity, experiment A B D C B C

16. (a) Energy is consumed to break the strong ionic bonds in aluminium oxide. 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) Δ H f = −1675 kJ mol 1 2

17. (a) Energy needed = mcΔT − − − = 400 cm3 × 1 g cm 3 × 4.2 J g 1 K 1 × [(70 + 273) − (20 + 273)] K = 84 000 J (84 kJ) 84 No. of moles of propan-1-ol needed = mol = 0.0418 mol 2010 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.


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Coursebook 3

18. (a) H2SO4(aq) + 2KOH(aq) → K 2SO4(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. 1 1 (c) H2SO4(aq) + KOH(aq) → K 2SO4(aq) + H2O(l) 2 2 65.00 (d) No. of moles of H2SO4(aq) used = 0.50 × mol = 0.0325 mol 1000 50.00 No. of moles of KOH(aq) used = 0.90 × mol = 0.0450 mol 1000 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 cm 3 × 4.2 J g 1 K 1 × [(30.3 + 273) − (25.0 + 273)] K = 2560 J (2.56 kJ) Standard enthalpy change of neutralization between sulphuric acid and potassium hydroxide 2.56 − − =− kJ mol 1 = −56.9 kJ mol 1 0.0450 19. (a) Exothermic (b) Energy absorbed by water = mcΔT − − − = 850 cm3 × 1.0 g cm 3 × 4.2 J g 1 K 1 × [(28.5 + 273) − (23.0 + 273)] K = 19 635 J (19.635 kJ) (c) Standard enthalpy change of solution of NaOH 19.635 − − = − 80.0 kJ mol 1 = −9.82 kJ mol 1 40.0


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Coursebook 3

Chapter 37 Hess’s Law Class Practice A37.1 (a)

SHAPE

\* ∆ H

SO2(g) + O2(g)

+297 kJ mol−1

MERGEFORMAT

SO3(g)

−396.0

kJ mol−1

S(s) + O2(g) By applying Hess’s Law, −1 −1 ∆ H = +297.0 + (−396.0) kJ mol = −99.0 kJ mol (b) S(s) +O2(g) y p l a h t n E

SO2((g) + +297.0 kJ mol−1

O2(g) −99.0

kJ mol−1

−396.0

kJ mol−1

SO3(g)

Reaction coordinate A37.2 (a) By applying Hess’s Law, ∆ H 1 = ∆ H f + ∆ H 2 −1 −1 ∆ H f = (−394.0) − (−283.0) kJ mol = −111.0 kJ mol


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Coursebook 3

(b) C(s)) +O2((g) +O2(g) y p l a h t n E

−111.0

kJ mol−1

CO(g) O + O2(g) −283.0

−394.0

kJ mol−1

kJ mol−1 CO2(g)

Reaction coordinate A37.3

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

C2H5OH(l)

∆ H

f

+ 2O2(g) −394.0

kJ mol−1 × 2 + 3O2(g)

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

−1371

+ O2(g) −286.0

kJ mol−1

kJ mol−1 × 3

2CO2(g) + 3H2O(l) −

2 × (−394.0) + 3 × (−286.0) −(−1371) kJ mol 1 − = −275.0 kJ mol 1 ∆ H f =

A37.4 (a) CaCO3(s) → CaO(s) + CO2(g) (b) By applying Hess’s Law, ∆ H = Σ∆ H f (products) – Σ∆ H f (reactants) ∆ H = ∆ H f [CaO(s)] + ∆ H f [CO2(g)] – ∆ H f [CaCO3(s)] − = (−635.0) + (−395.0) – (−1207) kJ mol 1 − = +177.0 kJ mol 1


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Coursebook 3

A37.5

1.

3C(s) + 4H2(g) +

1

O2(g) → C3H7OH(l) 2 By applying Hess’s Law, ∆ 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) + 4 × (−286.0) −(−1004) kJ mol 1 − = −1322 kJ mol 1 Enthalpy change of formation of 100 g liquid propanol 100 − = −1322 kJ mol 1 × mol 12.0 × 3 + 1.0 × 8 + 16.0 = −2203 kJ

2. ∆ H

2NaHCO3(s)

∆ H

f

[ NaHCO3(s)]

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

∆ H

f

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

2

∆ H

c

[H2(g)]

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

[ Na2CO3(s)]

∆ H

c

[C(s)]

Na2CO3(s) + H2(g) + C(s) + O2(g) 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 mol 1 − = +85 kJ mol 1


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Coursebook 3

Chapter 37 Hess’s Law Chapter Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

enthalpy, route lower, higher, enthalpy, endothermic, exothermic products, reactants reactants, products D C A B B A

11. (a)

1 2

H2(g) +

1 2

Cl2(g) → HCl(g)

(b) N2(g) + 2H2(g) + Cl2(g) ∆ H

∆ H

3

2

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

∆ H

f

NH4Cl(s)

[HCl(g)]

∆ H

1

NH3(g) + HCl(g) ∆ H f [HCl(g)]

= – ∆ H 3 + ∆ H 2 – ∆ H 1 92.2 629.0 − = –(– ) + (– ) – (–176.1) kJ mol 1 2 2 −1 = –92.3 kJ mol 12. (a) C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(l) −1 ∆ H c[C3H6O(l)] = –1790 kJ mol (b) ∆ H

1

C3H6O(l) + 4O2(g) ∆ H

f

3CO2(g) + 3H2O(l) + 4O2(g)

[C3H6O(l)]

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


+ 4O2(g) ∆ H

2

51


Coursebook 3

∆ H f [C3H6O(l)]

= ∆ H 2 – ∆ H 1 − = 3 × (–394.0) + 3 × (–286.0) – (–1790) kJ mol 1 − = –250 kJ mol 1 13. (a)

∆ H f [H2O(g)] = (–286.0) + (+41.1) kJ

−

−

mol 1 = –244.9 kJ mol 1

6720 − − = – kJ mol 1 = –3360 kJ mol 1 2 (c) For the reaction, 3 3 1 Al(s) + NH4 NO3(s) → N2(g) + 3H2O(g) + Al2O3(s) 2 2 2 By applying Hess’s Law, ∆ 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 [NH4 NO3(s)] 2 3 1 3 –1015 = × 0 + 3 × (–244.9) + × (–3360) – 0 – × ∆ H f [NH4 NO3(s)] 2 2 2 −1 ∆ H f [NH4 NO3(s)] = –933.1 kJ mol (b)

∆ H f [Al2O3(s)]

14. (a) ∆ H c

C2H4(g) + 3O2(g)

[C2H4(g)]

2CO2(g) + 2H2O(l)

∆ H

∆ H

2

1

2C(s) + 2H2(g) + 3O2(g) By applying Hess’s Law, ∆ H c[C2H4(g)] = – ∆ H 2 + ∆ H 1 − = –(+52.3) + 2 × (–394.0) + 2 × (–286.0) kJ mol 1 − = –1412 kJ mol 1 (b) ∆ H

1

C2H4(g) + H2(g) ∆ H

2

3O2(g)

C2H6(g) 3O2(g)

∆ H

3

2CO2(g) + 3H2O(l) By applying Hess’s Law, ∆ H 1 = ∆ H 2 – ∆ H 3 − = [–1412 + (–286.0)] – (–84.6) kJ mol 1 − = –1613 kJ mol 1


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Coursebook 3

Part VIII Chemical reactions and energy Part Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

D D D C B B A D A C D

12. (a) C6H12O6 + 6O2 → 6CO2 + 6H2O (b) Molar enthalpy change of combustion of glucose 15.7

= – 12.0 × 6

−

kJ mol 1

1 + 1.0 × 12 + 16.0 ×

6

−1

= –2826 kJ mol (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 m × 2826 × 30% = 330 180 m = 70.1 70.1 g of glucose is needed for providing the energy required. 70.1 g = 200 g ∴ mass of cereal needed = 35% 13. (a) Energy released by burning (CH3)2 NNH2 1694 − = kJ g 1 12.0 × 2 + 1.0 × 8 + 14.0 × 2 − = 28.2 kJ g 1 Energy released by burning CH3OH 726 − = kJ g 1 12.0 × 1 + 1.0 × 4 + 16.0 − = 22.7 kJ g 1 Energy released by burning C8H18 5590 − = kJ g 1 12.0 × 8 + 1.0 × 18 − = 49.0 kJ g 1 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.


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Coursebook 3

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 cm 3 × 4.2 J g 1 K 1 × [(22.2 + 273) – (20.0 + 273)] K = 184.8 J Enthalpy change of precipitation of magnesium carbonate 184.8 − = – 10.0 J mol 1 × 1 .0 1000 − − = –18 480 J mol 1 (18.5 kJ mol 1) For 30.0 cm3 of 1.0 M solutions, Energy released 30.0 = 18.5 × × 1.0 kJ 1000 = 0.555 kJ (555 J) mc∆T = 555 (30.0 + 30.0) × 1.0 × 4.2 × ∆T = 555 o ∆T = 2.20 K ( C) (d) For 30.0 cm3 of 1.2 M solutions, Energy released 30.0 = 18.5 × × 1.2 kJ 1000 = 0.666 kJ (666 J) mc∆T = 666 (30.0 + 30.0) × 1.0 × 4.2 × ∆T = 666 o ∆T = 2.64 K ( C) (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 =

50.0

× 0.04 mol = 0.002 mol 1000 (b) Mg(s) + Cu2+(aq) → Cu(s) + Mg2+(aq) − 1050 − − − J mol 1 = –525 000 J mol 1 = –525 kJ mol 1 ∆ H = 0.002 − (c) Mg(s) + Cu2+(aq) → Cu(s) + Mg2+(aq) ∆ H = –525 kJ mol 1


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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 =

50.0

× 0.05 mol = 0.0025 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 0.0025 According to the equation, the standard enthalpy change of the above displacement reaction involves the complete reaction of 2 moles of Ag+(aq). Therefore, −1 −1 ∆ H = –74 × 2 kJ mol = –148 kJ mol −1 (c) Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) ∆ H = –148 kJ mol (d) ∆ H

1

2+

Cu(s) + Mg2+(aq)

Mg(s) + Cu (aq) ∆ H

2Ag+(aq)

+

2Ag (aq) ∆ H

2

2+

2+

Mg (aq) + Cu (aq) + 2Ag(s) ∆ H

= ∆ H 1 + ∆ H 2 − = (–525) + (–148) kJ mol 1 − = –673 kJ mol 1

17. (a) Hydrochloric acid (b)

∆ H

Mg(s) + H2O(l)

MgO(s) + H2(g) ∆ H

1

2H+(aq) ∆ H

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

2

∆ H = ∆ H 1 – ∆ H 2


55


Coursebook 3

(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 mol 1 − = –11 840 kJ mol 1 (c) Energy needed = mc∆T = (600 × 1000) g × 4.2 J g –1 K –1 × [(25 + 273) – (0 + 273)] K = 63 000 kJ No. of moles of arachidonic acid needed 63 000 = mol = 5.32 mol 11 840 Mass of arachidonic acid needed = 5.32 × (12.0 × 20 + 1.0 × 32 + 16.0 × 2) g = 1617 g 19. (a) (b) (c) (d)

Positive value 15 s 40 s 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 10 = mol 39.1 + 14.0 + 16.0 × 3 = 0.0989 mol Heat 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 J Standard enthalpy change of solution of potassium nitrate 3465 − =+ J mol 1 0.0989 − − = +35 035 J mol 1 (+35 kJ mol 1)


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Coursebook 3

20. (a)

thermometer lid

expanded polystyrene cup

beaker

Fe(s) + H2SO4(aq)

cotton wool (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 mol Methanol CH3OH –26.34 × (12.0 + 1.0 × 4 + 16.0) = –843 Ethanol 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) = –2010 Butan-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 mol 1 22. N2(g) + 3H2(g) O2(g)

∆ H

2NH3(g)

∆ H

∆ H

4

1

2NO2(g) + 3H2O(g)

2NO(g) + 3H2(g)

O2(g)

3O2(g)

∆ H

∆ H

3

2

1O2(g)

2NO2(g) + 3H2(g) By applying Hess’s Law, ∆ H =

–(–180.0) + (–112.0) +


3

1 − (–572.0) – (–1396) kJ mol 1 2 2 57

1

Book 3 ans

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