Suggested Teaching Schemeintranet.stmgss.edu.hk/~ted/ccy/new_edition/Book3Canswer.pdf · • Standard enthalpy changes of combustion of some common substances • Practice 34.7 34

�

Suggested Teaching Scheme

The following suggested teaching schemes are for teachers’ reference only. Teachers may revise them based on the time-tabling arrangement of their own schools.

Scheme 1: Chemistry to be studied in Secondary 3, 4, 5 and 6In many schools, the Chemistry curriculum is studied in Secondary 3, 4, 5 and 6. Although the distribution of periods varies from school to school, the total number of periods for the curriculum is generally around 4�6. A possible distribution of periods is as follows:

A possible distribution of periods for S3, S4, S5 and S6

S3 S4 S5 S6

Number of teaching weeks per year 28 28 28 �6

Number of periods per week 2 5 5 5

Total number of periods per year 56 �40 �40 80

Total number of periods for the curriculum 4�6

Suggested teaching scheme for the curriculum

Level ContentSuggested number

of period(s)

S3(56 periods)

Topic � Planet Earth �2

Topic 2 Microscopic World I 44

S4(�40 periods)

Revision on laboratory safety �

Topic 3 Metals 39

Topic 4 Acids and Bases 45

Topic 5 Redox Reactions, Chemical Cells and Electrolysis 4�

Topic 6 Microscopic World II �4

S5(�40 periods)


Topic 7 Fossil Fuels and Carbon Compounds 32

Topic 8 Chemistry of Carbon Compounds 45

Topic 9 Chemical Reactions and Energy �3

Topic �0 Rate of Reaction �6

Topic �� Chemical Equilibrium �8

Topic �2 Patterns in the Chemical World �5

S6(80 periods)


Topic �3 Industrial Chemistry 39

Topic �4 Materials Chemistry 39

Topic �5 Analytical Chemistry 40

Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.

Only 2 out of 3 Only 2 out of 3


2

Topic 9 Chemical Reactions and Energy

Scheme 2: Chemistry to be studied in Secondary 4, 5 and 6In some schools, the Chemistry curriculum is studied in Secondary 4, 5 and 6. The total number of periods for the curriculum is generally around 360. A possible distribution of periods is as follows:

A possible distribution of periods for S4, S5 and S6

S4 S5 S6

Number of teaching weeks per year 28 28 �6

Number of periods per week 5 5 5

Total number of periods per year �40 �40 80

Total number of periods for the curriculum 360

Suggested teaching scheme for the curriculum

Level ContentSuggested number

of period(s)

S4(�40 periods)

Topic � Planet Earth 8

Topic 2 Microscopic World I 3�

Topic 3 Metals 32

Topic 4 Acids and Bases 36

Topic 5 Redox Reactions, Chemical Cells and Electrolysis 33

S5(�40 periods)


Topic 6 Microscopic World II �3

Topic 7 Fossil Fuels and Carbon Compounds 29

Topic 8 Chemistry of Carbon Compounds 4�

Topic 9 Chemical Reactions and Energy �2

Topic �0 Rate of Reaction �5

Topic �� Chemical Equilibrium �6

Topic �2 Patterns in the Chemical World �3

S6(80 periods)


Topic �3 Industrial Chemistry 39

Topic �4 Materials Chemistry 39

Topic �5 Analytical Chemistry 40

Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.

Only 2 out of 3 Only 2 out of 3

3


Suggested number of periods for Topic 9

Chemistry forTotal number

of periodsSuggested number of periods for each unit

S3–S6(Scheme �)

�6Unit 34 Energy changes in chemical reactionsUnit 35 Hess’s Law and its applications

97

S4–S6(Scheme 2)


96

S3–S6(Scheme �)


97

S4–S6(Scheme 2))


85

4


Teaching Plan

Chemical Reactions and Energy

Unit 34Energy changes in chemical reactions

Unit 35Hess’s Law and its applications

Chemical reactions are accompanied by energy changes, which often appear in the form of heat. In fact, energy taken in or released by a chemical system may take different forms.

Unit 34 first provides knowledge of the quantitative treatment of heat changes based on heat capacity calculations. Then the basic concepts of chemical energetics and enthalpy terms are introduced, including enthalpy change of reaction and standard conditions, with particular reference to formation, combustion, neutralization and solution.

Students are also required to find enthalpy changes using simple calorimetric method and calculate enthalpy changes from experimental results.

Unit 35 focuses on Hess’s Law and its applications. Students learn to use Hess’s law to determine enthalpy changes which cannot be determined by experiment directly, such as the enthalpy change of formation of metal oxides or metal carbonates. They are also required to construct simple enthalpy change cycles and enthalpy level diagrams, and perform calculations involving such cycles and relevant energy terms.

Organization of the topic

5

Teaching Plan

Unit 34 Energy changes in chemical reactions

Continued on next page

Section Key point(s)Suggested task(s) for

studentsRemark

Total number of periods = xx (Scheme 1), total number of periods = xx (Scheme 2)

34.� What is energy? • Different forms of energy

• Unit of energy

34.2 Specific heat capacity

• Calculations involving specific heat capacity and heat capacity

• Practice 34.�

34.3 The system and the surroundings

• Law of conservation of energy

34.4 Internal energy of a system

• Introducing the term ‘enthalpy change’

• Difference between enthalpy change and inernal energy change is usually small in most cases


34.5 Enthalpy change of an exothermic reaction

• What an exothermic reaction is

• Enthalpy level diagram of an exothermic reaction

• Activity 34.� — Classifying whether reactions are exothermic or endothermic

• Do you know — Heat packs

34.6 Enthalpy change of an endothermic reaction

• What an endothermic reaction is

• Enthalpy level diagram of an endothermic reaction

• Practice 34.2


34.7 Enthalpy changes during physical and chemical changes

• Enthalpy change during the melting of ice

• Enthalpy change during the combustion of methane

• Practice 34.3 • Supercooling• Do you know — Cold packs

34.8 Thermochemical equations

• Writing and interpreting thermochemical equations

• Practice 34.4 • Greenhouse effect

6



studentsRemark


34.9 Explaining energy changes — breakage and formation of chemical bonds

• Exothermic reaction — the amount of energy released in the bond-forming step is greater than the amount of energy used in the bond-breaking step

• Endothermic reaction — the amount of energy released in the bond-forming step is less than the amount of energy used in the bond-breaking step

• Practice 34.5 • Energy changes associated with breaking and forming of chemical bonds

34.�0 Standard conditions for measuring enthalpy changes

• What the standard conditions are

• Practice 34.6

34.�� Standard enthalpy change of reaction

• Definition of standard enthalpy change of reaction


34.�2 Standard enthalpy change of formation

• Definition of standard enthalpy change of formation

• Enthalpy level diagram representing standard enthalpy change of formation

• Standard enthalpy changes of formation of some common substances

• Making use of enthalpy change of formation to compare the stability of a compound and its constituent elements


7

Teaching Plan


studentsRemark

34.�3 Standard enthalpy change of combustion

• Definition of standard enthalpy change of combustion

• Enthalpy level diagram representing standard enthalpy change of combustion

• Standard enthalpy changes of combustion of some common substances

• Practice 34.7

34.�4 Standard enthalpy change of neutralization

• Definition of standard enthalpy change of neutralization

• Standard enthalpy change of neutralization involving a strong acid and a strong alkali

• Standard enthalpy change of neutralization involving either a weak acid or a weak alkali or both

• Refer to the following

website for the animations on the heat changes when mixing different acids and alkalis:

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/animations/Rxnheat2.htm

(accessed July 20�4)


34.�5 Determining enthalpy changes of chemical reactions

• Experiment and calculation for determining the enthalpy change of neutralization between hydrochloric acid and sodium hydroxide solution

• Sources of inaccuracy in the experiment

• Determining the heat capacity of the polystyrene calorimeter

• Activity 34.2 — Determining the

enthalpy change of a reaction


enthalpy changes of neutralization

• Discussion • Practice 34.8

• Do you know — Measuring chemical

energy in food• Determining the

enthalpy change of neutralization of the reaction between hydrochloric acid and sodium hydroxide solution


8



studentsRemark

34.�6 Determining enthalpy changes of combustion

• Determining the enthalpy change of combustion of ethanol


enthalpy changes of combustion of some alcohols

• Discussion • Practice 34.9• Chemistry Magazine — The energy values of

fuels

9

Teaching Plan

Unit 35 Hess’s Law and its applications



studentsRemark


35.� Hess’s Law • Definition of Hess’s Law• Introducing Hess’s Law

via the oxidation of nitrogen to nitrogen dioxide

• Enthalpy change cycle relating the enthalpy changes of processes involved in the reaction between gaseous nitrogen and oxygen to form nitrogen dioxide gas

• Practice 35.� • Hess’s Law

35.2 Using Hess’s Law to determine enthalpy changes that cannot be easily obtained by experiment

• Applying Hess’s Law and making use of the enthalpy changes of Haber process (making ammonia through the reaction between nitrogen and hydrogen) and the formation of ammonia through the reaction between hydrazine and hydrogen, determine the enthalpy change of formation of hydrazine

• Enthalpy change cycle used for determining ΔHΔHO

r [N2H4(g)]

• Practice 35.2

�0



studentsRemark

35.3 Determining the enthalpy change of formation of magnesium hydroxide from enthalpy changes of other reactions

• Determining the enthalpy change of the reaction between magnesium and hydrochloric acid

• Determining the enthalpy change of the reaction between magnesium hydroxide and hydrochloric acid

• Calculating the enthalpy change of formation of magnesium hydroxide

• Activity 35.� — Determining the

enthalpy change of formation of magnesium oxide


enthalpy change of thermal decomposition of potassium hydrogencarbonate

• Practice 35.3

• Determining the

enthalpy change of formation of magnesium oxide


35.4 Determining the standard enthalpy change of formation of a compound from standard enthalpy changes of combustion

• ΔHOf [compound]

= ∑ ΔHOc[constituent

elements] – ΔHO

c[compound]

• Practice 35.4

35.5 Determining the standard enthalpy change of a reaction from standard enthalpy changes of formation

• ΔHOr

= ∑ ΔHOf [products]

– ∑ ΔHOf [reactants]

• Practice 35.5

��

Teaching Notes

Teaching Notes


page 9N1

Uses of chemical energy in daily life

Self-heating can

Nestlé U.K. has launched a self-heating can of prepared coffee. The can has a conventional ring-pull top. Heating is activated by pushing in a button on the base of the can. This produces a chemical reaction between water and quicklime that are contained in separate compartments in the can base. After 3 minutes, the coffee is heated to 60 °C.

CaO(s) + H2O(l) Ca(OH)2(aq) ΔH = –82 kJ mol–�

To make room for the heat-generating chamber, a can of 330 cm3 can only hold 2�0 cm3 of drink.

Solid fuels in portable stoves

The three main types of portable stoves are gas / liquid fuel stoves, solid fuel stoves and wood burning stoves.

Ethanol can be turned into a solid fuel by pouring it into a saturated calcium ethanoate solution. A precipitate readily forms as the solubility of calcium ethanoate is reduced. Ethanol is trapped in the network of solid calcium ethanoate, forming a gel. When ignited, ethanol in the gel burns. Similar gels (e.g. SternoTM) are used as a source of fuel in portable cooking stoves, e.g. Chinese hot pots. Other brands may use hexamethylenetetramine and trioxane as solid fuels.

The great benefit of these types of stoves is that the fuel and stove are both very lightweight and cheap. The downsides are that the heat output is difficult to regulate and they do not burn as hot as gas / liquid fuel stoves.

�2


Instant cold-packs

The liquid inside the cold-pack is water. In the water is another plastic bag or tube containing ammonium nitrate. When you hit the cold-pack, it breaks the tube so that the water mixes with the chemical. An endothermic process occurs. H2O(l) NH4NO3(s) NH4

+(aq) + NO3–(aq) ΔH = +25.5 kJ mol–�

The pack will begin to get cold almost immediately and should remain cold for approximately 20 minutes.

page 12N2

Enthalpy changes for changes of states

Melting and freezing

The enthalpy change of melting of a substance is the enthalpy change when one mole of the substance melts.

The enthalpy change of freezing of a substance is the enthalpy change when one mole of the substance freezes.

These changes do not take place under standard conditions. Enthalpy changes of melting and freezing are quoted at � atmosphere and the temperature at which the change of state takes place, i.e. the melting temperature. The enthalpy change of freezing is the negative of the enthalpy change of melting.

ΔH freezing = –ΔH melting

Vaporization and condensation

When a liquid is heated, the average energy of the molecules increases. Some molecules, with energy above average, are able to break away from the attraction of other molecules and enter the vapour phase.

The enthalpy change of vaporization of a substance is the difference in enthalpy between the vapour and the liquid state per mole of the substance.

ΔH vaporization = H vapour – H liquid

The enthalpy changes of vaporization are quoted at � atmosphere and the temperature at which the change of state takes place.

Condensation is the reverse of vaporization.

ΔH condensation = –ΔH vaporization

∆H vaporization

∆H melting

∆H condensation

= –∆H vaporization

∆H freezing

= –∆H melting

vapour

liquid

solid

Enth

alpy

�3

Teaching Notes

page 33N4

An improved apparatus: the flame calorimeter

An improved apparatus: the flame calorimeterThe simple calorimeter shown in Fig. 34.30 is not very efficient and accurate because energy released from combustion of the fuel is lost in heating the apparatus and surroundings. A more effective apparatus, the flame calorimeter, is shown in Fig. 34.32. The chimney is made of copper, so that heat released from combustion can be transferred effectively to the water. The flame is enclosed so as to reduce heat loss. Oxygen is supplied to ensure complete combustion.

Fig. 34.32 A flame calorimeter

When using a flame calorimeter, we need to know its heat capacity, (i.e. the heat required to raise the temperature of the whole apparatus by � K). This can be calculated from measurements made using a fuel with known enthalpy change of combustion.

The calorimeter is filled with water and a small electric immersion heater is placed in it. This is switched on until the same temperature rise is produced as was produced by burning the ethanol. The heat losses in both experiments will cancel out. The heat produced by the heater can be measured by an electrical joulemeter. It can also be determined by measuring the voltage (V) and current (I) of the heater, and the time (t) for which it ran. Then the following expression is used:

Heat produced = V x I x t

where V = voltage of the heater I = current applied t = time in seconds

�4


When using the flame calorimeter, the amount of heat gain for it is found as follows:

Amount of heat gain for the calorimeter

= heat capacity of the calorimeter x ΔT

where ΔT is the temperature change.

page 36N5

Fuel efficiency

To compare the efficiency of different fuels, it is more useful to calculate enthalpy changes of combustion per gram or per millilitre of substance rather than per mole.

Enthalpy change of combustion

kJ mol–1 kJ g–1 kJ ml–1

Hydrogen –285.8 –�4�.8 –9.9*

Ethanol –� 367 –29.7 –23.4

Graphite –393.5 –32.8 –73.8

Methane –890.3 –55.5 –30.8*

Methanol –726.4 –22.7 –�7.9

Octane –5 470 –47.9 –33.6

Toluene –3 9�0 –42.3 –36.7

*These values of enthalpy of combustion for hydrogen and methane are the values under the temperature of 0°C and the pressure of �0000 psi (i.e. 680 atm).

�5

Teaching Notes

For applications where mass is important, as in rocket engines, hydrogen is ideal because its enthalpy change of combustion of per gram is the highest of any known fuel.

For applications where volume is important, as in cars, a mixture of hydrocarbons is most efficient because enthalpy changes of combustion per millilitre of hydrocarbons are relatively high. Octane and toluene are representative examples.

�6


Suggested Answers

page 1

� Liquefied butane

2 CaO(s) + H2O(l) Ca(OH)2(s)

3 Combustion reaction


Practice

P34.1 page 5

� Mass of water = 50 cm3 x �.00 g cm−3 = 50 g

Heat released by the fuel = m x c x ΔT = 50 g x 4.�8 J g−� K−� x (56.0–�6.0)K = 8 360 J

\ the fuel releases 8 360 J of heat.

2 252 x �03 J = 4.00 x �03 g x 3.50 J g−� K−� x ∆T

∆T = �8 K

\ the temperature of the solution rises �8 K.

P34.2 page 11

� a) Exothermic

b) Endothermic

c) Endothermic

d) Exothermic

2 The reaction between dry calcium oxide and water is exothermic.

Heat is released when water is added to the dry calcium oxide.

Thus the egg is cooked by the heat released.

�7

Suggested Answers

P34.3 page 13

a)

Enthalpy

CaO(s) + CO2(g)

CaCO3(s)

b)

Enthalpy

C3H8(g) + 5O2(g)

3CO3(g) + 4H2O(l)

P34.4 page 16

� a) H2O2(l)

heat released by the system∆H = –98.2 kJ mol–1

Enth

alpy

(kJ

mol

–1)

H2O(l) + O2(g)12

b) Multiplying both sides of the given equation by 2 gives the following equation:

2H2O2(l) 2H2O(l) + O2(g) ΔH = 2(–98.2 kJ) = –�96 kJ

When the reaction proceeds in the reverse direction, the magnitude of ΔH for the equation remains the same, but its sign changes.

i.e. 2H2O(l) + O2(g) 2H2O2(l) ΔH = +�96 kJ

2 Number of moles of ethanol burnt = 3.00 g

2 x �2.0 + 6 x �.0 + �6.0 g = 0.0652 mol

Amount of heat released = 0.0652 mol x � 368 kJ mol−�

= 89.2 kJ

�8


P34.5 page 17

The amount of energy released in the bond-forming step is less than the amount of energy used in the bond-forming step.

P34.6 page 18

a) Change in

b) Enthalpy

c) The enthalpy change in measured under standard conditions

i.e. at a temperature of 25 °C;

at a pressure of � atm.

d) That heat is released.

P34.7 page 23

� a) 2Al(s) + 32

O2(g) Al2O3(s) ΔHOf = −� 676 kJ mol−�

b) C3H7OH(l) + 92

O2(g) 3CO2(g) + 4H2O(l) ΔHOf = −2 02� kJ mol−�

2 a) ΔHOr

b) ΔHOf [SO2(g)] or ΔHO

c [S(s)]

c) ΔHOc

�9

Suggested Answers

P34.8 page 31–32

� a) Enthalpy change of neutralization of HCl(aq) and KOH(aq)

Total volume of the reaction mixture = 25.0 cm3 + 25.0 cm3

= 50.0 cm3

Mass of the reaction mixture = 50.0 g

Amount of heat released during neutralization = 50.0 g x 4.�8 J g–� K–� x 3.4 K = 7�� J

Number of moles of HCl reacted = number of moles of KOH reacted

= 0.500 mol dm–3 x 25.0

� 000 dm–3

= 0.0�25 mol

0.0�25 mole of HCl reacted with 0.0�25 mole of KOH to produce 0.0�25 mole of H2O.

Amount of heat released when � mole of water is formed = 7�� J0.0�25 mol

= 56 900 J mol–�

= 56.9 kJ mol–�

\ the enthalpy change of neutralization is −56.9 kJ mol–�.

Enthalpy change of neutralization of CH3COOH(aq) and KOH(aq)

Amount of heat released during neutralization = 50.0 g x 4.�8 J g–� K–� x 2.6 K = 543 J

Number of moles of CH3COOH reacted = number of moles of KOH reacted = number of moles of H2O formed = 0.0�25 mol

Amount of heat released when � mole of water is formed = 543 J0.0�25 mol

= 43 400 J mol–�

= 43.4 kJ mol–�

\ the enthalpy change of neutralization is −43.4 kJ mol–�.

b) Some energy is consumed when the weak ethanoic acid dissociates to give hydrogen ions before neutralization.

c) i) Possible sources of error:

• heat loss to the surroundings;

• heat capacity of the expanded polystyrene calorimeter, the thermometer and the stirrer not taken into account;

• thermometer not being precise enough; and

• density and specific heat capacity of the solution not being exactly the same as those of water.

20


ii) • Determine the heat capacities of the calorimeter and the thermometer. Take these values into account in the calculations.

• Use a digital thermometer of higher precision in temperature reading.

• Determine the densities and specific heat capacities of the reaction mixtures. Take these values into account in the calculations.

b) Amount of heat released during the reaction = m x c x ΔT = 25.0 g x 4.�8 J g−� K−� x 36.5 K = 3 8�0 J = 3.8� kJ

Number of moles of CuSO4 = �.00 mol dm−3 x 25.0

� 000 dm–3

= 0.0250 mol

Enthalpy change of reaction = −3.8� kJ0.0250 mol

= −�52 kJ mol−�

\ the enthalpy change of the reaction is −�52 kJ mol−�.

2 a)

Greatest temperature change of the reaction mixture = (54–�7.5) °C = 36.5 °C

2�

Suggested Answers

b) Amount of heat released during the reaction = m x c x ΔT = 25.0 g x 4.�8 J g−� K−� x 36.5 K = 3 8�0 J = 3.8� kJ

Number of moles of CuSO4 = �.00 mol dm−3 x 25.0

� 000 dm–3

= 0.0250 mol

Enthalpy change of reaction = −3.8� kJ0.0250 mol

= −�52 kJ mol−�

\ the enthalpy change of the reaction is −�52 kJ mol−�.

P34.9 page 35

a) To minimize heat loss to the surroundings.

b) To minimize the heat absorbed by the metal can.

c) Temperature change of water = (42.9 − 2�.0) °C = 2�.9 °C

Amount of heat released during combustion of propan-�-ol = 200.0 g x 4.�8 J g–� K–� x 2�.9 K = � 8300 J = �8.3 kJ

Number of moles of propan-�-ol burnt = 0.600 g60.0 g mol–�

= 0.0�00 mol

ΔHc = –�8.3 kJ0.0�00 mol–�

= −� 830 kJ mol–�

\ the enthalpy change of combustion of propan-�-ol is −� 830 kJ mol–�.

Discussion page 28

• Replace the expanded polystyrene calorimeter by a vacuum flask calorimeter.

• Determine the heat capacities of the expanded polystyrene calorimeter, the thermometer and the stirrer. Take these values into account in the calculations.

• Use a digital thermometer of higher precision in temperature reading.

• Determine the density and specific heat capacity of the solution. Take these values into account in the calculations.

• Carry the experiment under standard conditions.

22


Discussion page 35

� • Incomplete combustion occurs (this would cause black soot to be deposited on the bottom of the copper can).

• Heat loss to the surroundings.

• The experiment is not carried out under standard conditions.

• Heat capacities of the copper can, the thermometer and the stirrer are not taken into account.

• There is possible loss of liquid fuel / water via evaporation.

• The thermometer is not precise enough.

2 • Replace the copper can with a flame calorimeter to reduce heat loss.

• Oxygen is supplied to ensure complete combustion.

• Determine the heat capacities of the copper can, the thermometer and the stirrer. Take these values into account in calculations.

• Carry out the experiment under standard conditions.

• Cover the container of the liquid fuel when not in use to reduce loss due to evaporation.

• Use a digital thermometer of higher precision.

Chemistry Magazine page 36

The energy values of fuels

� a) 2H2(g) + O2(g) 2H2O(l)

b) No carbon dioxide is formed.

Carbon dioxide is a greenhouse gas.

2 Benefits

• Less carbon monoxide / unburnt hydrocarbons / particulates / sulphur dioxide / carcinogens emitted.

• Fuels made from crops are renewable.

• The exhaust gas produced does not contribute much to global warming (burning ethanol returns carbon dioxide which has only recently been removed from the atmosphere during the photosynthesis of plants).

• Fossil fuels have other uses.

• There is no risk of large scale pollution from exploitation of fossil fuels.

• Biodegradable

23

Suggested Answers

Drawbacks

• Requires a large area of agricultural land to produce enough food crops to meet the demand.

• May lead to diversion of investment from food production, resulting in increased food prices.

• The toxic effluent from distilleries causes water pollution.

• Ethanal ( H

O

CCH3 ) is also produced during combustion. Without adequate pollution control, it can harm vegetation, irritate the skin and eyes, and damage the lung at high concentrations.

• More energy is consumed to make ethanol, which releases less energy in use/ fossil fuels are used in the production of ethanol.

• Carbon dioxide emissions in manufacture.

• Reduce biodiversity.

• Lower energy density.

• May not be possible to obtain cheap ethanol in every region of the world.

pages 40–48Unit Exercise

� a) calorimetry

b) bond-forming

c) bond-breaking

d) exothermic

e) released to

f) bond-breaking

g) bond-forming

h) endothermic

i) taken in from

2 a) 4C(graphite) + 5H2(g) C4H�0(g)

b) C4H�0(g) + �32

O2(g) 4CO2(g) + 5H2O(l)

c) 6C(graphite) + 6H2(g) + 3O2(g) C6H�2O6(s)

d) C6H�2O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

24


3 a) i) The standard conditions are:

• at a temperature of 25°C (298 K);

• at a pressure of � atmosphere (atm);

• all the substances involved are in their standard state; i.e. the most stable physical state at 25 °C and � atm; and

• all the solutions involved have a concentration of � mol dm–3

ii) The enthalpy change when one mole of the substance is formed from its elements in their standard states.

b) i) H2(g) + �2

O2(g) H2O(l)

ii) The ΔHOf of liquid water and the ΔHO

c of hydrogen refer to the ΔH value of the same thermochemical equation, therefore they are identical.

4 D Gold has the lowest specific heat capacity. With the same amount of heat added, it will have the greatest temperature change.

5 C Amount of heat required = 40.0 g x 0.90 J g–� K–� x (32.3 – 20.0) K = 443 J

6 B According to the thermochemical equation, the explosion reaction of � mole of sucrose releases 2 �92 kJ of heat.

Number of moles of sucrose exploded = 5.�3 g342.0 g mol−� = 0.0�50 mol

Heat released from 5.�3 g of sucrose = 0.00�50 mol x 2 �92 kJ mol−� = 32.9 kJ

\ 32.9 kJ of heat are releases when one jelly-baby is exploded.

7 B

8 C When we reverse an equation, the magnitude of ΔH for the equation remains the same, but its sign changes.

\ 2CO2(g) + 3H2O(l) C2H5OH(g) + 3O2(g) ΔH = +� 368 kJ

When we multiply both sides of an equation by a factor, ΔH must also be multiplied by the same factor.

\ 4CO2(g) + 6H2O(l) 2C2H5OH(g) + 6O2(g) ΔH = 2(+� 368 kJ) = + 2 736 kJ

25

Suggested Answers

9 B Amount of heat required to warm the water = 25.0 g x 4.�8 J g−� K−� x (85.0 − 25.0) K = 6 270 J

According to the thermochemical equation, � mole of Mg reacts to give 353 kJ of heat.

number of moles of Mg needed = 6 270 J353 kJ mol−� = 0.0�78 mol

= 0.0�78 mol

Mass of Mg needed = 0.0�78 mol x 24.3 g mol−� = 0.433 g

0.433 g of magnesium is needed to supply the heat.

�0 B In the equation of choice A, 2 moles of sodium nitrate are formed.

In the left side of the equation of choice C, the standard states of nitrogen and oxygen should be N2(g) and O2(g).

In the left side of the equation of choice D, the standard state of sodium should be Na(s).

�� CAlkane

Enthalpy change of combustion (kJ mol–1)

Difference in enthalpy change of combustion (kJ mol–1)

C3H8 –2 2�9

C4H�0 –2 877

C5H�2 –3 509

C6H�4 –4 �63654

632

658

From the above data, it can be deduced that the difference of enthalpy change of combustion of two neighbouring alkanes is about 600 kJ mol−�

The difference of the enthalpy changes of combustion of another alkane and C6H�4 is [–4�63–(–6�25)] kJ mol–� = �962 kJ mol−�

� 962600

≈ 3

the difference of � 962 kJ mol−� is equivalent to 3–CH2– groups, therefore the alkane should be C9H20.

26


�2 C

Mixture

Number of moles of water formed

Standard enthalpy change of

neutralizationHeat released

Temperature rise

�25 cm3 of � mol dm−3 HCl(aq) + 25 cm3 of� mol dm−3 NH3(aq)

0.025 mol

< 57 kJ for � mole of water formed as the neutralization involves a weak alkali (NH3(aq))

< �.43 kJ

ΔT� < ΔT2,i.e. x < y

225 cm3 of � mol dm−3 HCl(aq) + 25 cm3 of� mol dm−3 NaOH(aq)

0.025 mol57 kJ for � mole of water formed

0.025 mol x 57 k mol–� = �.43 kJ

�3 D Evaporation of the alcohol during weighing makes the value for the number of moles of alcohol burnt higher than the actual one. This makes the results less exothermic than the reported value.

�4 B (�) Dissolution of ammonium nitrate is an endothermic process and is used in cold packs. When using the cold pack, ammonium nitrate dissolves in water and the temperature of water drops.

(3) Heat is absorbed in the sublimation of dry ice, therefore it is an endothermic process.

�5 Amount of heat required = m x c x ΔT

947 J = �25 g x c x �5.7 K c = 0.483 J g–� K–�

The specific heat capacity of stainless steel is 0.483 J g–� K–�.

�6 Answers for the HKDSE question are not provided.

�7 a)

27

Suggested Answers

b) Total volume of reaction mixture = 50.0 cm3 + 50.0 cm3 = �00.0 cm3

Mass of reaction mixture = �00.0 cm3 x �.00 g cm−3 = �00 g

Heat absorbed by the reaction mixture = �00 g x 4.�8 J g−� K−� x 3.2 K = � 340 J

Number of moles of the reacting CaCl2 = 2.00 mol dm−3 x 50.0� 000 dm3

= 0.�00 mol

Enthalpy change of the reaction = +� 340 J0.�00 mol

= +�3400 J mol−�

= +�3.4 kJ mol−�

c) The fall in temperature remains unchanged (i.e. it is still 3.2°C).

�8 Amount of heat released during hydrogenation = 500.0 g x 4.�8 J g–� K–� x �5.4 K = 32 200 J = 32.2 kJ

Number of moles of C6H8 used = �2.0 g80.0 g mol–�

= 0.�50 mol

Enthalpy change of hydrogenation = –32.2 kJ0.�50 mol

= –2�5 kJ mol–�

the enthalpy change of hydrogenation of cyclohexa-�,3-diene is –2�5 kJ mol–�.

b) Any two of the following:

• No heat loss to the surroundings.

• Heat capacities of the reaction vessel, the beaker and the thermometer are negligible.

• Cyclohexa-�,3-diene undergoes complete hydrogenation.

• The thermometer is of high precision.

�9 Number of moles of propane = �76.0 g(�2.0 x 3 + �.0 x 8) g mol–�)

= 4.00 mol

Heat evolved when propane are burnt in air = 2220 kJ mol−� x 4.00 mol = 8 880 kJ

20 a) The standard enthalpy change of neutralization ΔHc is the enthalpy change when an acid reacts with an alkali to form one mole of water under standard conditions.

b) i) Both neutralization reactions involve the reaction between a strong monobasic acid and NaOH(aq).

The same chemical change occurs in both cases.

H+(aq) + OH–(aq) H2O(l)

28


ii) Ethanoic acid is a weak acid.

In the course of neutralization, molecules of ethanoic acid consume energy for dissociation.

2� a) Standard enthalpy of combustion of a substance is the enthalpy change when one mole of the substance is completely burnt in oxygen under standard conditions.

b) i) Heat released during the combustion of pentan-�-ol = 250.0 g x 4.�8 J g K−� x (78.0 − 24.0) K = 56 400 J = 56.4 kJ

ii) Number of moles of pentan-�-ol that was burnt = �.76 g(�2.0 x 5 + �.0 x �2 + �6.0) g mol−�

= 0.0200 mol

iii) ΔHc[C5H��OH(l)] = −56.4 kJ0.0200 mol

= −2 820 kJ mol−�

22 a) i) q = m x c x ΔT

ii) Amount of heat required = 8.80 g x �.92 J g–� K–� x 9.5 K = �6� J

iii) Amount of heat required = ��.95 g x 0.96 J g–� K–� x 9.5 K = �09 J

iv) Molar enthalpy change for the mixing process

= – Total amount of heat required to rise the temperature of ethyl ethanoate and trichloromethane by 9.5 K0.0200 mol

= – (�6� + �09) J0.�00 mol

= –2700 J mol–�

= –2.70 kJ mol–�

b) Any one of the following:

• Each liquid has its own intermolecular forces within themselves before mixing.

• The liquids may react.

23 Answers for the HKDSE question are not provided.

24 a) Any one of the following:

• Use an expanded polystyrene cup.

• Insulate the beaker.

• Add a lid.

b) To ensure that the chemicals react completely.

29

Suggested Answers

c) Experiment 2.

The initial temperature is different.

d) The temperature change of this experiment is an outlier.

e) 7.0 °C

f) Enthalpy change = (�00.0 x 4.2 x 7.0) J = 2 940 J

g) Diagram A

The reaction between HCl(aq) and NaOH(aq) is exothermic.

The total enthalpy of the products is less than that of the reactants.

25 a) i)

Maximum temperature change for the reaction = (67.5 – 2�.5) °C = 46 °C

ii) Heat produced in the reaction = 50.0 g x 4.�8 J g−� K−� x 46.0 K = 9 6�0 J

iii) Number of moles of CuSO4 undergoing the reaction = �.00 mol dm−3 x 50.0� 000 dm3

= 0.0500 mol

ΔH = −9 6�0 J0.0500 mol

= −�92 kJ mol−�

30


b) i) To ensure a steady temperature.

ii) To perform cooling correction / to compensate the heat loss

(As there is heat loss to the surroundings when the system releases heat, therefore it is not possible to measure the accurate temperature rise at that moment. Extending the time for measuring temperature enables the use of extrapolation method to perform cooling correction, in order to compensate the heat loss and to deduce the maximum temperature change without heat loss to the surroundings.)

iii) Low heat capacity / good insulator of heat

iv) To ensure uniform temperature in the reaction mixture.

v) Burette / pipette

vi) Cover the polystyrene cup with a lid/ Improve insulation.

c) Zinc > lead > copper

The more exothermic / more negative the ΔH, the more reactive the metal.


Practice

P35.1 page 52

a) Route 1

Route 2

∆H = –441 kJ mol–1

∆Hx = ? ∆Hy = ?

SO2(s) + O2(g)12

S(s) + O2(g)32 SO3(g)

b) ΔHx = −297 kJ mol−�

ΔHy = −�44 kJ mol−�

3�

Suggested Answers

P35.2 page 57

� a) Rhombic sulphur is more stable.

b) The following data is given in the question:

(�) S(rhombic) + O2(g) SO2(g) ΔH = –296.0 kJ mol–�

(2) S(monoclinic) + O2(g) SO2(g) ΔH = –296.4 kJ mol–�

We need � mole of S(monoclinic) as the product. So, reverse equation (2) to give equation (2)’.

(2)’ SO2(g) S(monoclinic) + O2(g) ΔH = +296.4 kJ mol–�

Combine equations (�) and (2)’, to obtain the target equation.

(�) S(rhombic) + O2(g) SO2(g) ΔH = –296.0 kJ mol–�

(2)’ SO2(g) S(monoclinic) + O2(g) ΔH = +296.4 kJ mol–�

S(rhombic) S(monoclinic) ΔH = +0.4 kJ mol–�

\ the enthalpy change for the process is +0.4 kJ mol–�.

2 ∆HOr

∆HO = –98 kJ mol–1∆HO = –441 kJ mol–1

CH4(g) + 2Cl2(g) CH2Cl2(g) + 2HCl(g)

CH3Cl(g) + HCl(g) + Cl2(g)

By Hess’s Law,

ΔHOr = [(–98) + (–�04)] kJ mol–�

= –202 kJ mol–�

\ the standard enthalpy change of the reaction is –202 kJ mol–�.

P35.3 page 61

a) Amount of heat released when K2CO3 reacted with HCl = 32.2 g x 4.�8 J g–� K–� x 5.4 K = 727 J

Number of moles of K2CO3 reacted = 2.07 g�38.2 g mol–�

= 0.0�50 mol

Enthalpy change of the reaction = –727 J0.0�50 mol

= –48.5 kJ mol–�

32


b) Enthalpy change of decomposition of KHCO3(s) refers to the enthalpy change of the following process:

KHCO3(s) �2

K2CO3(s) + �2

H2O(l) + �2

CO2(g)

ΔH values for the following processes are known:

(�) KHCO3(s) + HCl(aq) KCl(aq) + H2O(l) + CO2(g) ΔH = +�9.6 kJ

(2) K2CO3(s) + 2HCl(aq) 2KCl(aq) + H2O(l) + CO2(g) ΔH = –48.5 kJ

Multiply equation (�) by 2, giving equation (�)’.

Reverse equation (2), giving equation (2)’.

Combine the two equations to obtain the target equation.

(�)’ 2KHCO3(s) + 2HCl(aq) 2KCl(aq) + 2H2O(l) + 2CO2(g) ΔH = +39.2 kJ

(2)’ 2KCl(aq) + H2O(l) + CO2(g) K2CO3(s) + 2HCl(aq) ΔH = +48.5 kJ

2KHCO3(s) K2CO3(s) + H2O(l) + CO2(g)

By Hess’s Law,

enthalpy change of decomposition of KHCO3(s) = �2

[(+39.2) + (+48.5)] kJ mol–�

= +43.9 kJ mol–�

\ the enthalpy change of decomposition of KHCO3(s) is +43.9 kJ mol–�.

P35.4 page 66

� a) ΔH2 represents the enthalpy change of formation of C2H4(g).

ΔH3 represents the enthalpy change of combustion of C2H4(g).

b) ΔH� represents the enthalpy change of combustion of C(graphite) and the enthalpy change of combustion of H2(g).

c) ΔHOf [C2H4(g)] = 2 x ΔHO

c [C(graphite)] + 2 x ΔHOc [H2(g)] – ΔHO

c [C2H4(g)] = [2(–394) + 2(–286) – (–� 4��)] kJ mol–�

= +5� kJ mol–�

\ the standard enthalpy change of formation of ethene is +5� kJ mol–�.

d) Carbon and hydrogen do not react to produce ethene under normal conditions.

2 ΔHOf [C�2H22O��(s)] refers to the standard enthalpy change of the following process:

�2C(graphite) + ��H2(g) + ��2

O2(g) C�2H22O��(s)

ΔHOf [compound] = ∑ ΔHO

c [constituent elements] – ΔHOc [compound]

ΔHOf [C�2H22O��(s)] = �2 x ΔHO

c [C(graphite)] + �� x ΔHOc [H2(g)] – ΔHO

c [C�2H22O��(s)] = [�2(–394) + ��(–286) – (–5 640)] kJ mol–�

= –2 230 kJ mol–�

\ the standard enthalpy change of formation of sucrose is –2 230 kJ mol–�.

33

Suggested Answers

P35.5 page 71–72

� a) 4CO2(g) + 5H2O(l)

∆HOr

C4H9OH(l) + 6O2(g)

4C(graphite) + 5H2(g) + 6 O2(g)12

4 x ∆HOf [CO2(g)] + 5 x ∆HO

f [H2O(l)]∆HOf [C4H9OH(l)]

b) By Hess’s Law, we can write

ΔHOf [C4H9OH(l)] + ΔHO

r = 4 x ΔHOf [CO2(g)] + 5 x ΔHO

f [H2O(l)]

By rearranging the equation, we have

ΔHOr = 4 x ΔHO

f [CO2(g)] + 5 x ΔHOf [H2O(l)] − ΔHO

f [C4H9OH(l)]

= [4(−394) + 5(−286) − (−327)] kJ = −2 680 kJ

\ the standard enthalpy change of the reaction is −2 680 kJ.

2 ΔHOr = ΔHO

f [product] − ∑ΔHOf [reactants]

= ΔHOf [CH3CH2CH3(g)] − ΔHO

f [cyclopropane]

ΔHOf [cyclopropane] = ΔHO

f [CH3CH2CH3(g)] − ΔHOr

= [(−�05) − (−�58)] kJ mol–�

= +53 kJ mol–�

pages 75–83Unit Exercise

� a) initial

b) final

c) ΔH�

d) ΔH2

e) ΔH3

f) constituent elements

g) compound

h) products

i) reactants

34


2 a) C8H�8(l) + 252 O2(g) 8CO2(g) + 9H2O(l)

b) ∆HOf [C8H18(l)]

C8H18(l) + O2(g)

8CO2(g) + 9H2O(l)

252

8C(graphite) + 9H2(g) + O2(g)252

c) ΔHOf [compound] = ∑ ΔHO

c [constituent elements] – ΔHOc [compound]

ΔHOf [C8H�8(l)] = 8 x ΔHO

c [C(graphite)] + 9 x ΔHOc [H2(g)] – ΔHO

c [C8H�8(l)] = [8(–394) + 9(–286) – (–5 470)] kJ mol–�

= –256 kJ mol–�

3 a) ∆HOr

CH4(g) + 2H2O(g) CO2(g) + 4H2(g)

C(graphite) + 4H2(g) + O2(g)

b) ΔHOf [H2(g)] = 0

c) ΔHOr = ∑ ΔHO

f [products] – ∑ ΔHOf [reactants]

ΔHOr = ΔHO

f [CO2(g)] – ΔHOf [CH4(g)] – 2 x ΔHO

f [H2O(g)] = [(–394) – (–74.8) – 2(–242)] kJ mol–�

= +�65 kJ mol–�

4 A

35

Suggested Answers

5 C Look at the target equation, we need � mole of NO2(g) as the reactant. So, reverse the first equation

and multiply it by �2

, giving equation (�)’.

(�)’ NO2(g) �2

N2(g) + O2(g ΔH = �2

(−68 kJ) = −34 kJ

Multiply the second equation by �2

, giving (2)’.

(2)’ �2

N2(g) + O2(g) �2

N2O4(g) ΔH = �2

(+�0 kJ) = +5 kJ

Combine the equations as shown below.

(�)’ NO2(g) �2

N2(g) + O2(g) ΔH = �2

(−68 kJ) = −34 kJ

(2)’ �2

N2(g) + O2(g) �2

N2O4(g) ΔH = �2

(+�0 kJ) = +5 kJ

NO2(g) �2

N2O4(g) ΔH = [(−34) + (+5)] kJ

= −29 kJ

\ the enthalpy change of the reaction is −29 kJ.

6 B

7 B ΔHOf [compound] = ∑ ΔHO

c [constituent elements] − ΔHOc [compound]

ΔHOf [CH4(g)] = ∑ ΔHO

c [C(graphite)] + 2 x ΔHOc [H2(g)] − ΔHO

c [CH4(g)] = [(−394) + 2(−286) − (−89�)] kJ mol−�

= –394–(2 x 286)+89� kJ mol−�

8 C ΔHOr = ∑ ΔHO

f [products] − ∑ ΔHOf [reactants]

= [Fe2O3(s)] − 2 x [FeO(s)] = [(−820) − 2(−270)] kJ = –820 + (270 x 2) kJ = −280 kJ



= 2 × [CO2(g)] + 3 x [H2O(l)] − [C2H6(g)] = [2(−��) + 3(−286) − (−84.7)] kJ = −995 kJ

\ the standard enthalpy change of the reaction is −995 kJ.

36


�0 B ΔHOr [CH3COCH3(l)] refers to the standard enthalpy change of the following process:

3C(graphite) + 3H2(g) + �2

O2(g) CH3COCH3(l)



ΔHOf [CH3COCH3(l)] = 3 x ΔHO


c [CH3COCH3(l)] = 3 x ΔHO

f [CO2(g)] + 3 x ΔHOf [H2O(l)] − ΔHO

c [CH3COCH3(l)] = [3(−394) + 3(−286) − (−� 786)] kJ mol−�

= −254 kJ mol−�

\ the standard enthalpy change of formation of propanone is −254 kJ mol−�.

�� B ΔHOr = ∑ ΔHO


= [Ca(OH)2(s)] − 2 × ΔHOf [H2O(l)]

\ to calculate ΔHOf [Ca(OH)2(s)], ΔHO

r and ΔHOf [H2O(l)] (i.e. ΔHO

c [H2(g)]) are required.

�2 a) i) ∆HOc

∆HO2∆HO

1

CH3OH(l) + O2(g)32 CO2(g) + 2H2O(l)

C(graphite) + 2H2(g) +2O2(g)

ii) The standard enthalpy change of formation of a substance, ΔHOf, is the enthalpy change when one mole of the substance is formed from its elements in their standard states.

The temperature and pressure under the standard conditions are 25 °C and � atmosphere respectively.

iii) ΔHOc = ΔHO

2 − ΔHO�

= [(–394) + 2(–286) – (–239)] kJ mol–�

= –820 + (270 x 2) kJ mol–�

= −727 kJ mol–�

b) i) Heat transferred = 200.0 x 4.�8 x 20.7 = �7 300 J

ii) Number of moles of CH3OH burnt = 0.848 g32.0 g mol–�

= 0.0265 mol

iii) Enthalpy change of combustion = −�7 300 J0.0265 mol

= −653 kJ mol−�

37

Suggested Answers

�3 a) 3C(graphite) + 2H2(g) CH3C CH(g)

b)3C(graphite) + 2H2(g) + 4O2(g)

3CO2(g) + 2H2O(l)

CH3C CH(g) + 4O2(g)∆HO

f [CH3C CH(g)]

∆HOc [CH3C CH(g)]3 x ∆HO

c [C(graphite)]+ 2 x ∆HO

c [H2(g)]

ΔH of[CH3C CH(g)] = 3 x ΔH o

c[C(graphite)] + 2 x ΔH oc[H2(g)] – ΔH o

c[CH3C CH(g)] = [3(–394) + 2(–286) – (–� 938)] kJ mol–�

= +�84 kJ mol–�

\ the enthalpy change of formation of propyne is +�84 kJ mol–�.

c) No

Carbon and hydrogen do not react to produce propyne under normal conditions.

�4 a) i) CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

ii) ∆Hr

CaO(s) + CO2(g) +CaCO3(s) + 2HCl(aq) 2HCl(aq)

∆H1 ∆H2

CaCl2(aq) + H2O(l) + CO2(g)

iii) ΔHr = ΔH� – ΔH2

b) Any two of the following:


• Density and specific heat capacity of the solution not being exactly the same as those of water.

• Experiment not carried out under standard conditions.

• Large relative error in temperature measurement.

�5 a) The enthalpy change of a reaction depends on the initial and final states of the reaction and is independent of the route by which the reaction may occur.

b) i) ∆HOr

∆HOc [CO(g)] = –283 kJ mol–1∆HO

c [CH4(g)] = –890 kJ mol–1

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

12CO(g) + 2H2O(l) + O2(g)

38


ii) ΔHr = ΔHOc [CH4(g)] – ΔHO

c [CO(g)] = [–890 – (–283)] kJ mol–�

= –607 kJ mol–�

iii) Any one of the following:

• CO2 instead of CO is formed.

• Ash is formed.

c) H2O formed in the reaction is not in the liquid state.

Less heat is evolved from the reaction because there is enthalpy change associated with the transformation of water from liquid to vapour state, which is an endothermic process.

�6 a) The pressure is � atmosphere and the temperature is 25 °C.

b) 6C(graphite) + 7H2(g) C6H�4(l)

c) Carbon does not undergo reaction with hydrogen to form hexane.

d) ΔHOf [compound] = ∑ ΔHO


ΔHOf [C6H�4(l)] = [C(graphite)] + ΔHO

c [H2(g)] − ΔHOc [C6H�4(l)]

= –394+(–286) – (–4 �63) = +3 480 kJ mol–�

�7 Number of moles of �.92 g of hydrazine = �.92 g(2 x �4.0 + 4 x�.0) g mol−�

= 0.0600 mol

The standard enthalpy of formation of hydrazine = –37.4 kJ0.0600 mol

= –623 kJ mol–�

∆HOr = –623 kJ mol–1

∆HOf [H2O(l)] = –286 kJ mol–1∆HO

f [N2H4(l)]

+

∆HOf [N2(g)] = 0 kJ mol–1

+

∆HOf [O2(g)] = 0 kJ mol–1

N2H4(l) + O2(g) N2(g) + 2H2O(g)

N2(g) + 2H2(g) + O2(g)

ΔHOr = ∑ ΔHO


ΔHOr = ΔHO

f [H2O(l)] – ΔHOf [N2H4(l)]

–623 kJ mol–� = –286 kJ mol–� – ΔHOf [N2H4(l)]

ΔHOf [N2H4(l)] = (–623 + 286) kJ mol–�

= –337 kJ mol–�

39

Suggested Answers

�8 a) The standard enthalpy change of formation of a substance, ΔHOf , is the enthalpy change when one mole

of the substance is formed from its elements in their standard states.


b) ΔHOr = ∑ ΔHO


= ΔHOf [Ca(OH)2(s)] + ΔHO

f [C2H2(g)] – ΔHOf [CaC2(s)] – 2 x ΔHO

f [H2O(l)] = [(–987) + (+227) – (–60) –2(–286)] kJ = –�28 kJ

�9 a) ΔHOr = ∑ ΔHO


ΔHOr = 2 x ΔHO

f [MgO(s)] + 4 x ΔHOf [NO2(g)] + ΔHO

f [O2(g)] – 2 x ΔHOf [Mg(NO3)2(s)]

ΔHOf [Mg(NO3)2(s)] = �

2 (2 x ΔHO

f [MgO(s)] + 4 x ΔHOf [NO2(g)] + ΔHO

f [O2(g)] – ΔHOr )

= �2

[2(–602) + 4 (+39.9) + 0 – (+538)] kJ mol–�

= –79� kJ mol–�

b) As ΔHOf [Mg(NO3)2(s)] < 0, Mg(NO3)2(s) is more stable than its constituent elements under standard

conditions.

20 a) i) Number of moles of propanone = 2.90 g58.0 g mol–�

= 0.0500 mol

ii) Heat transferred = 200.0 g x 4.�8 J g–� K–� x (78.4 – 20.2)K = 48700 J = 48.7 kJ

iii) Enthalpy change of combustion of propanone = –48.7 kJ0.0500 mol

= –974 kJ mol–�

b) i) Any one of the following:


• Incomplete combustion of propanone.

ii) Difference: Less exothermic / ΔH less negative / less heat is evolved.

Justification: Heat is absorbed in the transformation of liquid water to water vapour.

c) i) The standard enthalpy change of formation of a substance, ΔHOf , is the enthalpy change when one

mole of the substance is formed from its elements in their standard states.


40


ii) ∆HOf

∆HOc [CH3COOH(l)]2 x ∆HO

c [C(graphite)]

+

∆HOf [N2(g)] = 0 kJ mol–1

+

2 x ∆HOc [H2(g)]

2C(sraphite) + 2H2(g) + 3O2(g) CH3COOH(l) + 2O2(g)

2CO2(g) + 2H2O(g)



ΔHOf [CH3COOH(l)] = 2 x ΔHO


c [CH3COOH(l)] = [2(–394) + 2(–286) – (–870)] kJ mol–�

= –490 kJ mol–�

2� a) i) The reaction also forms CO2(g).

ii) Hess’s Law states that the enthalpy change of a reaction depends on the initial and final states of the reaction and is independent of the route by which the reaction may occur.

iii) The standard enthalpy change of combustion of a substance, ΔHOc , is enthalpy change when one

mole of the substance is completely burnt in oxygen under standard conditions.

b) ΔHOr = ∑ ΔHO


= 2 x ΔHOf [Fe(l)] + 3 x ΔHO

f [CO2(g)] – ΔHOf [Fe2O3(s)] – 3 x ΔHO

f [CO2(g)] = [2(+�4) + 3(–394) – (–822) –3(–��)] kJ mol–�

= +� kJ mol–�

c) i) C(graphite) + O2(g) CO2(g)

ii) The standard enthalpy change of formation of CO2(g) and the standard enthalpy change of combustion of carbon refer to the same thermochemical equation, and thus their values are the same.

22 a) Respiration.

b) i) Heat released = �00.0 g x 4.�8 J g–� K–� x (4�.0 – 23.7)K = 7 230 J = 7.23 kJ

ii) Number of moles of glucose burnt = 0.83� g(6 x �2.0 + �2 x �.0 + 6 x �6.0) g mol–�

= 0.00462 mol

iii) Enthalpy change of combustion of glucose = –7.23 kJ0.00462 mol

= –�560 kJ mol–�

c) ΔHOc = ∑ ΔHO


= 6 x ΔHOf [CO2(g)] + 6 x ΔHO

f [H2O(l)] – ΔHOf [C6H�2O6(s)]

= [6(–396) + 6(–286) – (–� 250)] kJ mol–�

= –2 830 kJ mol–�

4�

Suggested Answers

d) Any two of the following:

• Incomplete combustion occurs.



• Heat capacities of the apparatus are not taken into account.

• Loss of water via evaporation is possible.

23 Answers for the HKALE question are not provided.

pages 84–93Topic Exercise

� D

2 C Look at the target equation, we need � mole of CO(g) as the product. The second equation has � mole of CO(g), but it is on the reactant side. So, reverse the second equation. And combine the equations as shown below.

C(graphite) + O2(g) CO2(g) ΔH = −394 kJ mol−�

CO2(g) CO(g) + �2

O2(g) ΔH = +283 kJ mol−�

C(graphite) + �2

O2(g) CO(g) ΔH = [(−394) + (+283)] kJ mol−� = −�� kJ mol−�

\ the enthalpy change for the reaction is −�� kJ mol−�.

3 B ΔHOr = ∑ ΔHO


= ΔHOf [HCOOH(l)] − ΔHO

f [CO2(g)] = [(−409) − (−394)] kJ mol−�

= −�5 kJ mol−�

\ the enthalpy change for the reaction is −�5 kJ mol−�.



= ΔHOf [AgCl(s)] + ΔHO

f [ClO2(g)] − ΔHOf [AgClO3(s)]

ΔH[ClO2(g)] = ΔHOr − ΔHO

f [AgCl(s)] + ΔHOf [AgClO3(s)]

= [(+5) − (−�27) + (−30)] kJ mol−�

= +�02 kJ mol−�

\ the standard enthalpy change of formation of ClO2(g) is +�02 kJ mol−�.

5 A

42


6 A

7 A

8 D Graphite is more stable than diamond.

The conversion of diamond to graphite is an exothermic process.

9 C

�0 B Both HCl(aq) and HNO3(aq) are strong acids. The standard enthalpy change of neutralization between a strong acid and a strong alkali is −57.� kJ mol−�.

�� Answers for the HKCEE question are not provided.

�2 a) An enthalpy change is the heat released or taken in during any change in a system, provided the system is kept at constant pressure.

b) Hess’s Law states that the enthalpy change of a reaction depends on the initial and final states of the reaction and is independent of the route by which the reaction may occur.

c) The standard enthalpy change of formation of a substance, ΔHOf , is the enthalpy change when one


d) i) The standard state of sodium is solid with the symbol Na(s). By definition, ΔHOf [Na(s)] = 0 kJ mol–�.

There is enthalpy change associated with the process Na(s) Na(l), which is endothermic.

ii) ΔHOr = ∑ ΔHO


= 4 x ΔHOf [NaCl(s)] + ΔHO

f [Ti(s)] – ΔHOf [TiCl4(g)] – 4 x ΔHO

f [Na(l)] = [4(–4��) + 0 – (–720) – 4(+3)] kJ mol–�

= –936 kJ mol–�

iii) In the reaction, sodium acts as a reducing agent to reduce titanium(IV) chloride to titanium metal.

�3 Answers for the HKASLE question are not provided.

�4 a) i) Heat produced = 250.00 g x 4.�8 x (3�.5 – 2�.0) J = �� 000 J

ii) Number of moles of ethanol burnt = (63.2� – 62.47) g (2 x �2 + 6 x �.0 + �6.0) g mol–�

= 0.74 g46.0 g mol–�

= 0.0�60 mol

iii) Enthalpy change of combustion of ethanol = –�� 000 J0.0�60 mol

= –688 000 J mol–�

= –688 kJ mol–�

43

Suggested Answers

b) i) Any three of the following:

• Heat loss from beaker.

• Incomplete combustion of the ethanol.

• Not all the heat from the flame is used to heat up the beaker and the water.

• The distance between the flame and the beaker too large.

• Without draught excluder.

• Heat capacity of the beaker not taken into account.

• Evaporation of the ethanol/ water.

ii) ΔHOf [compound] = ∑ ΔHO


ΔHOf [C2H5OH(l)] = 2 x ΔHO


c [C2H5OH(l)] = [2(–394) – 3(–286) – (–� 368)] kJ mol–�

= � 422 kJ mol–�

�5 a) Heat change = �07 g x 4.�0 J g–� K–� x (3�.6 – 25.6) K = 2 630 J

Number of moles of SiCl4 hydrolyzed = 6.8� g�70.� g mol–�

= 0.0400 mol

Enthalpy change when � mole of SiCl4 undergo hydrolysis = 2 630 J0.0400 mol

= 65 750 kJ mol–�

= 65.8 kJ mol–�

b) • Record the temperature of water at half-minute intervals for 2 �2

minutes. Measure the time using a stop watch.

• Add SiCl4(l) to the water at the 3rd minute.

• Replace the lid, stir the mixture and record the temperature of the mixture for an additional 6 minutes.

• Plot a graph of temperature against time.

• Join the points before the addition of SiCl4(l) using a straight line and extrapolate to the 3rd minute, the time to add SiCl4(l). Label the temperature at the 3rd minute as Ti.

• Join the point after the addition of SiCl4(l) using a straight line and extrapolate back to the 3rd minute. Label the temperature at the 3rd minute as Tf.

• The separation of the lines at the 3rd minute corresponds to the maximum temperature change for the process.

\ Maximum temperature change in the course of reaction ΔT = Tf – Ti

44


�6 a) i) Heat produced = 50.0 g x 4.�8 J g–� K–� x (68.0 – 22.0)K = 9 6�0 J

ii) Number of moles of hexane burnt = 0.320 g86.0 g mol–�

= 0.00372 mol

ΔHc[C6H�4(l)] = –9 6�0 J0.00372 mol

= –2 580 kJ mol–�

iii) Any two of the following:




• Heat capacities of the calorimeter, the thermometer and the stirrer are not taken into account.

• Loss of liquid fuel/water via evaporation is possible.

iv) Error in reading the temperature from the thermometer is less than the effect of not taking heat loss into account.

b)

6CO2(g) + 7H2O(l)

∆Hr

C6H12(l) + H2(g) + O2(g)192 C6H14(l) + O2(g)19

2

∆Hc[C6H12(l)]+

∆Hc[H2(g)]

∆Hc[C6H14(l)]

ΔHr = ΔHr[C6H�2(l)] + ΔHr[H2(g)] – ΔHr[C6H�4(l)] = [(–4 003) + (–286) – (–4 �63)] kJ = –�26 kJ

�7 Answers for the HKDSE question are not provided.

�8 a) i) Measure the mass of octane and burn it.

Measure the temperature rise of a known volume of water.

Heat produced = mass of water x heat capacity of water x temperature rise

Scale up to � mole of octane.

ii) Any two of the following:




• Heat capacities of the calorimeter, the thermometer and the stirrer are not taken into account.


45

Suggested Answers

b) i) ΔH� = ΔHf[C8H�8(l)]

ΔH2 = enthalpy change of combustion of 8 moles of carbon or enthalpy change of formation of 8 moles of carbon dioxide

ΔH3 = enthalpy change of combustion of 9 moles of hydrogen or enthalpy change of formation of 9 moles of water

ΔH4 = ΔHc[C8H�8(l)]



ΔH�= ΔH4 – (ΔH2+ ΔH3) = [–5 470 – (–5 7�8)] kJ mol–�

= +248 kJ mol–�

�9 a) i) The standard enthalpy change of combustion of a substance, ΔHOc , is enthalpy change when one

mole of the substance is completely burnt in oxygen under standard conditions.



= 4 x ΔHOf [CO2(g)] + 5 × ΔHO

f [H2O(l)] – ΔHOf [CH3CH2CH2CH2OH(l)]

= [4(–394) + 5(–286) – (–327)] kJ mol–�

= –2 680 kJ mol–�

b) i) In the alkanol homologous series, there is a difference of one –CH2– unit between two neighbouring members.

As the length of the carbon chain increases, more energy has to be consumed to break one extra C–C bond and two extra C–H bonds.

ii) Any two of the following:




• Heat capacities of the apparatus are not taken into account.


• The thermometer is not precise enough.

20 a) i) Heat released = 50.0 g x 4.�8 J g–� K–� x (32.0 – �9.2) K = 2 680 J

ii) Number of moles of magnesium used = 0.486 g24.3 g mol–�

= 0.0200 mol

iii) Enthalpy change of reaction = –2 680 J0.0200 mol

= �34 kJ mol–�

46


b) i) The standard enthalpy change of formation of a substance, ΔHOf , is the enthalpy change when one


The temperature and pressure under the standard conditions are 25 °C and � atmosphere

ii)

Ca(s) + 2HCl(aq) + C(s) + O2(g) CaCO3(s) + 2HCl(aq)32

CaCl2(aq) + H2(g) + C(s) + O2(g)32

CaCl2(aq) + H2O(l) + C(s) + O2(g) CaCl2(aq) + H2O(l) + CO2(g)

∆Hf

–168

–286

–394

–54

ΔHOf [CaCO3(s)] = [–�68 – 286 – 394 –(–54)] kJ mol–�

= –794 kJ mol–�

2� a) Shorten the time for the heat loss to the surroundings during the reaction.

(The use of finely powdered magnesium carbonate increases the rate of reaction and shortens the time needed for completing the reaction, and thus shortens the time for the heat loss to the surroundings during the reaction and the temperature measured could be more precise.)

b) i) Number of moles of magnesium carbonate reacted = 3.5 g84.0 g mol–�

= 0.04�7 mol

ii) Excess HNO3 ensure all magnesium carbonate reacts.

c) i) Heat transferred = 50 g x 4.�8 J g–� K–� x (29.7 – 2�.0) K = �820 J

ii) ΔH = –� 820 J0.04�7 mol

= –43.6 kJ mol–�

d) Pipette: more accurate

Measuring cylinder: Faster / easier / more convenient

e) i) The calculated enthalpy change is less negative/ less exothermic.

Temperature rise lower than expected due to heat loss.

ii) The calculated enthalpy change is less negative / less exothermic.

Less magnesium carbonate undergoes reaction so that the temperature rise is lower.

47

Suggested Answers

f) i)

Maximum temperature change = (29.7 – 20.5) = 9.2 °C

ii) This method performs cooling correction to compensate the heat loss.

(As there is heat loss to the surroundings when the system releases heat, therefore it is not possible to measure the accurate temperature rise at that moment. Extending the time for measuring temperature enables the use of extrapolation method to perform cooling correction, in order to compensate the heat loss and to deduce the maximum temperature change without heat loss to the surroundings.)

Documents

Suggested Teaching Schemeintranet.stmgss.edu.hk/~ted/ccy/new_edition/Book3Canswer.pdf · • Standard enthalpy changes of combustion of some common substances • Practice 34.7 34