suggested answers to in-text activities and...

New 21st Century Chemistry

Suggested answers to in-text activities and unit-end exercises

Topic 9 Unit 34

In-text activities

Checkpoint (page 7)

1 Amount of heat gained = 1 000 g x 4.18 J g–1 K–1 x (14.40 – 11.30) K

= 13 000 J

= 13.0 kJ

the amount of heat gained from 0 s to 200 s is 13.0 kJ.∴

2 Amount of heat used = m x c x ΔT

24 300 kJ = m x 4.18 J g–1 K–1 x (351.2 – 293.1) K

m = 100 g

the mass of water being warmed is 100 g.∴

3 a) Heat capacity of the tin block =

= 65.2 J K–1

b) Specific heat capacity of tin block =

=

= 0.217 J g–1 K–1

the heat capacity and specific heat capacity of the tin block are 65.2 J K∴ –1 and 0.217 J g–1

K–1 respectively.

Checkpoint (page 15)

1 a) Endothermic

b) Exothermic

c) Exothermic

d) Endothermic

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.

Suggested answers to in-text activities and unit-end exercises 1 © Jing Kung. All rights reserved.Topic 9 Unit 34

2 040 J31.3 K

heat capacity of tin block mass of tin block

65.2 J K–1

300 g



1 a)

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

2H2O2(l) 2 H2O + O2(g) ΔH = 2(–98.2 kJ) = –196 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 = +196 kJ

2 a) The thermochemical equation shows that the explosion reaction of 1 mole of sucrose

releases 2 192 kJ of heat.

Number of moles of sucrose exploded =

= 0.0133 mol

Heat released from 4.56 g of sucrose = 0.0133 mol x 2 192 kJ mol–1

= 29.2 kJ

29.2 kJ of heat are released when one jelly-baby is exploded.∴

b) Number of moles of SrCl2 vaporized =

= 0.0851 mol

Mass of SrCl2 vaporized = 0.0851 mol x 158.6 g mol–1

= 13.5 g

13.5 g of SrCl∴ 2 can be vaporized by exploding one jelly-baby.


1 a) i) 2C(graphite) + 3H2(g) C2H6(g) ΔHӨ f [C2H6(g)] = –84.7 kJ mol–1

ii) 2Al(s) + O 2(g) Al2O3(s) ΔHӨ f [Al2O3(s)] = –1 676 kJ mol–1


4.56 g342.0 g mol–1

29.2 kJ343 kJ mol -1

32


b)

2 a) Consider the two allotropes of carbon, graphite and diamond.

C(diamond) C(graphite) ΔHӨ r = –2 kJ mol–1

Graphite is slightly more stable than diamond. Consequently, ΔHӨ f of CO2(g) should relate to

the allotrope graphite, rather than diamond.

∴ΔHӨ r ≠ΔHӨ

f

b) ΔHӨ r ≠ΔHӨ

f

c) Sulphur exists in many allotropic forms, but the most stable form at 25 °C and 1 atm is

rhombic sulphur.

Δ∴ HӨ r ≠ΔHӨ

f


1 a) ΔHӨ r or ΔHӨ

c

b) ΔHӨ r

2 a) i) C8H18(l) + O 2(g) 8CO2 (g) + 9H2O(l) ΔHӨ[C8H18(l)] = –5 470 kJ mol–1

ii) C3H7OH (l) + O 2(g) 3CO2 (g) + 4H2O(l) ΔHӨ[C3H7OH(l)] = –2 021 kJ mol–1

b)

Discussion (page 32)

1 The molar enthalpy change of combustion of alkanes increases as the number of carbon atoms

in alkane molecule increases (there is a linear relationship). The enthalpy change of combustion


252

92


per gram decreases slightly.

2 The molar enthalpy change of combustion increases by about 655 kJ per carbon atom in an

alkane.

Molar enthalpy change of combustion of decane

= –[5 470 + 2 x 655] kJ mol–1

= –6 780 kJ mol–1

The enthalpy change of combustion per gram of decane should be slightly less than that of

octane. The exact number is difficult to predict, owing to the non-linear relationship.

3 The molar enthalpy change of combustion was easier to predict, owing to the linear relationship.


a) standard enthalpy change of combustion of ethane

b) standard enthalpy change of formation of magnesium oxide or

standard enthalpy change of combustion of magnesium

c) standard enthalpy change of formation of water or

standard enthalpy change of combustion of hydrogen

d) standard enthalpy change of neutralization between hydrofluoric acid and sodium hydroxide

solution


• Replace the polystyrene calorimeter by a vacuum flask calorimeter.

• Determine the heat capacities of the 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.


1 Amount of heat released during reaction = 250.0 g x 4.18 J g–1 K–1 x 14.2 K

= 14 800 J

Number of moles of Cu2+(aq) ions reacted = 0.200 mol dm–3 x dm 3

= 0.0500 mol

Enthalpy change for the reaction = x 3

= –888 000 J mol–1

= –888 kJ mol–1

the enthalpy change for the reaction is –888 kJ mol∴ –1.


250.01 000

–14 800 J0.0500 mol


2 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 4.18 J g–1 K–1 x 3.4 K

= 711 J

Number of moles of HCl reacted = number of moles of NaOH reacted

= 0.500 mol dm–3 x dm 3

= 0.0125 mol

0.0125 mole of HCl reacted with 0.0125 mole of KOH to produce 0.0125 mole of H2O.

Amount of heat released when 1 mole of water is formed =

= 56 900 J mol–1

= 56.9 kJ mol–1

the enthalpy change of neutralization is –56.9 kJ mol∴ –1.

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


= 543 J

Number of moles of CH3COOH reacted = number of moles of KOH reacted

= number of moles of H2O formed

= 0.0125 mol

Amount of heat released when 1 mole of water is formed =

= 43 400 J mol–1

= 43.4 kJ mol–1

the enthalpy change of neutralization is –43.4 kJ mol∴ –1.

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 polystyrene calorimeter, the thermometer and the stirrer being

ignored;

• thermometer not being precise enough; and

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

of water.


25.01 000

711 J0.0125 mol

543 J0.0125 mol


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.

3 a)

b) Maximum temperature rise: ΔT = Tf – Ti

= 80.0 °C – 26.0 °C

= 54.0 °C

c) Amount of heat released = 25.0 g x 4.18 J g–1 K–1 x 54.0 K

= 5 640 J

Number of moles of CuSO4 used = 1.00 mol dm–3 x dm 3

= 0.0250 mol

Enthalpy change of the reaction =

= –226 000 J mol–1

= –226 kJ mol–1

the enthalpy change of the reaction is –226 kJ mol∴ –1。


25.01 000

–5 640 J0.0250 mol



1 • Incomplete combustion occurs.

• Heat loss to the surroundings.

• The experiment is not carried out under standard conditions.

• Heat capacities of the metal container, the thermometer and the stirrer are ignored.

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

• The thermometer is not precise enough.

2 • Replace the metal container with a flame calorimeter to reduce heat loss.

• Oxygen is supplied to ensure complete combustion.

• Determine the heat capacities of the metal container, 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.


1 Temperature change of water = (69.0 – 26.2) °C

= 42.8 °C

Amount of heat released during combustion= 250.0 g x 4.18 J g–1 K–1 x 42.8 K

= 44 700 J

= 44.7 kJ

Number of moles of CH3CHO burnt =

= 0.0375 mol

Enthalpy change of combustion =

= –1 190 kJ mol–1

the enthalpy change of combustion of ethanal is –1 190 kJ mol∴ –1。

2 a) Amount of heat released during hydrogenation = 500.0 g x 4.18 J g–1 K–1 x 15.4 K

= 32 200 J

= 32.2 kJ

Number of moles of C6H8 used =

= 0.150 mol


1.65 g44.0 g mol–1

–44.7 kJ0.0375 mol

12.0 g80.0 g mol–1


Enthalpy change of hydrogenation =

= –215 kJ mol–1

the enthalpy change of hydrogenation of cyclohexa-1,3-diene is –215 kJ mol∴ –1。

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-1,3-diene undergoes complete hydrogenation.

• The thermometer is of high precision.

3 a) To minimize heat loss to the surroundings.

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

c) Heat gain for water = 200.0 g x 4.18 J g–1 K–1 x ΔT

Number of moles of propan-1-ol burnt =

= 0.0100 mol

Amount of heat released by propan-1-ol = 2 021 kJ mol–1 x 0.0100 mol

= 20.2 kJ

= 20 200 J

= 200.0 g x 4.18 J g–1 K–1 x ΔT

ΔT = 24.2 K

Final temperature of water = (24.2 + 21.0) °C

= 45.2°C

∴ the final temperature of the water is 45.2 °C.

STSE Connections (page 53)

1 The properties include

• non-toxic;

• environmentally friendly;

• odourless; and

• non-combustible.

2 The freezing point of a liquid is the temperature at which the liquid changes to a solid.

The reverse process of converting a solid to a liquid is called melting. The melting point of a

solid is the same temperature as the freezing point of the liquid.

3 The solid ionic compound (e.g. calcium chloride) undergoes an exothermic dissolving process.


–32.2 kJ0.150 mol

0.600 g60.0 g mol–1


Unit-end exercises (pages 57 – 65)

Answers for the HKCEE (Paper 1) and HKALE questions are not provided.

1 A possible concept map is shown below:

2 a) Endothermic

b) Exothermic

c) Endothermic

d) Exothermic

e) Endothermic

3 a) The specific heat capacity of a substance is the amount of heat required to raise the

temperature of 1 g of the substance by 1 K (or 1 °C).

b) The heat capacity of a substance is the amount of heat required to raise the temperature of

the substance by 1 K (or 1 °C).

4 a) 4C(graphite) + 5H2(g) C4H10(g)

b) C4H10(g) + O2(g) 4CO2(g) + 5H2O(l)

c) 6C(graphite) + 6H2(g) + 3O2(g) C6H12O6(s)

d) C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

5 a) i) The standard conditions are:

• at a temperature of 25 °C (298 K);

• at a pressure of 1 atmosphere (atm);

• all the substances involved are in their standard state; i.e. the most stable physical

state at 25 °C and 1 atm; and

• all the solutions involved have a concentration of 1 mol dm–3.

ii) The standard enthalpy change of formation of a substance is the enthalpy change when

one mole of the substance is formed

from its elements in their standard states.


132


b) i) H2(g) + O2(g) 5H2O(l)

ii) The two different terms refer to the same thermochemical equation with the same ΔH

value.

6 C

7 D Option D — Gold has the lowest specific heat capacity.

Hence it will have the greatest temperature change with equal amount of heat

added.

8 D Option D — The enthalpy level diagram for an exothermic reaction is shown below:

In an exothermic reaction, the total enthalpy of the products is less than that of

the reactants.

9 A When the reaction proceeds in the reverse direction, the magnitude of ΔH for the equation

remains the same, but its sign changes.

i.e. 2H2(g) + O2(g) 2H2O(l) ΔH = –572 kJ

10 C Amount of heat required = 40.0 g x 0.90 J g–1 K–1 x (32.3 – 20.0) K

= 443 J

11 D Number of moles of propane burnt =

= 1 364 mol

Amount of heat produced = 2 219 kJ mol–1 x 1 364 mol

= 3 030 000 kJ

= 3.03 x 106 kJ

12 D Maximum number of moles of water that can be converted =

= 20.2 mol

Maximum mass of water that can be converted = 20.2 mol x 18.0 g mol–1

= 364 g


12

60.0 x 1 000 g44.0 g mol–1

890 kJ44.0 kJ mol–1


13 A (1) & (2) The standard enthalpy change of combustion of a substance is the enthalpy

change when one mole of the substance is completely burnt in oxygen under

standard conditions.

Carbon may burn to form two types of oxide:

C(s) + O2(g) CO2(g)

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

Complete combustion of carbon produces carbon dioxide, so only the enthalpy

change of the first process is regarded as the standard enthalpy change of

combustion of carbon.

14 Amount of heat required = m x c x Δ T

947 J = 125 g x c x 15.7 K

c = 0.483 J g–1 K–1

∴ the specific heat capacity of stainless steel is 0.483 J g–1 K–1.

15 Number of moles of NaOH dissolved =

= 0.625 mol

Amount of heat released = 44.7 kJ mol–1 x 0.625 mol

= 27.9 kJ

∴ 27.9 kJ of heat were released.

16 Amount of heat given off = 222 g x 4.18 J g–1 K–1 x (41 – 21) K

= 18 600 J

= 18.6 kJ

Number of moles of sulphur burnt =

= 0.0623 mol

Enthalpy change of combustion of sulphur =

= –299 kJ mol–1

the enthalpy change of combustion of sulphur is –299 kJ mol∴ –1。

17 a) Ca2+(aq) + CO32–(aq) CaCO3(s)

b) i) Total volume of the reaction mixture = 50 cm3 + 50 cm3

= 100 cm3

Mass of the reaction mixture = 100 g

Heat taken in = 100 g x 4.18 J g–1 K–1 x 1.5 K

= 627 J


12

25.0 g40.0 g mol–1

2.00 g32.1 g mol–1

–18.6 kJ0.0623 mol


ii) Number of moles of CaCl2 used = 1.00 mol dm–3 x dm3

= 0.0500 mol

iii) Enthalpy change for the reaction =

= +12 500 J mol–1

= +12.5 kJ mol–1

iv) Temperature

because ΔT is so small and thus leads to relatively large percentage error.

v) Vacuum flask / two expanded polystyrene cups with a lid

c) The temperature fall would be the same.

18 Number of moles of NH4NO3 in the cold-pack =

= 0.625 mol

Amount of heat taken in = 25.8 kJ mol–1 x 0.625 mol

= 16.1 kJ

= 16 100 J

Amount of heat taken in = m x c x ΔT

16 100 J = 125 g x 4.18 J g–1 K–1 x ΔT

ΔT = 30.8 K

ΔT = (Tf – 25.0) °C

= 30.8 °C

Tf = 55.8 °C

the final temperature of the cold-pack is 55.8 °C.∴

19 Average temperature of CH3COOH(aq) and NH3(aq) =

= 18.4 °C

Temperature rise of the reaction mixture= (31 – 18.4) °C

= 12.6 °C

Total volume of the reaction mixture= 75.0 cm3 + 75.0 cm3

= 150.0 cm3

Mass of the reaction mixture = 150.0 g


= 7 900 J

= 7.90 kJ

Number of moles of CH3COOH = number of moles of NH3


501 000

+627 J0.0500 mol

50.0 g80.0 g mol–1

(18.2 + 18.6) °C2


= 2.00 mol dm-3 x dm3

= 0.150 mol

= number of moles of H2O formed

Enthalpy change of neutralization =

= –52.7 kJ mol–1

the enthalpy change of neutralization for the reaction is –52.7 kJ mol∴ –1.

20 a)

b) Amount of heat released when 5.0 g of butan-1-ol are burnt = 102 kJ

Number of moles of butan-1-ol burnt =

= 0.0676 mol

Enthalpy change of combustion of butan-1-ol =

= –1 510 kJ mol–1

c) Any answer in the range of –2 650 to –2 700 kJ mol–1

d) Any one of the following:

• Incomplete combustion of butan-1-ol.

• Heat capacities of the container, the thermometer and the stirrer are ignored.

• Experiment not carried out under standard conditions.

• Loss of butan-1-ol / water via evaporation.

• The thermometer not precise enough.

21 —

22 Amount of heat released during hydrogenation = 300.0 g x 4.18 J g–1 K–1 x 10.1 K


75.01 000

–7.90 kJ0.150 mol

5.0 g74.0 g mol–1

mol

–102 kJ0.0676 mol

mol


= 12 700 J

= 12.7 kJ

Number of moles of C6H12 used = 0.10 mol

Enthalpy change of hydrogenation =

= –127 kJ mol–1

∴ the enthalpy change of hydrogenation of hex-1-ene is –127 kJ mol–1.

23 a) The standard enthalpy change of combustion of a substance is the enthalpy change when one

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

b) C3H7OH(l) + O2(g) 3CO2(g) + 4H2O(l)

c) i) Amount of heat produced = 50.0 g x 4.18 J g–1 K–1 x 12.8 K

= 2 680 J

= 2.68 kJ

ii) Number of moles of propan-1-ol burnt =

= 1.67 x 10–3 mol

iii) Enthalpy change of combustion of propan-1-ol =

= –1 600 kJ mol–1

d) Any two of the following:

• Heat loss to the surroundings.

• Incomplete combustion of propan-1-ol.

• Heat capacities of the container and the thermometer are ignored.

• Experiment not carried out under standard conditions.

• Loss of propan-1-ol / water via evaporation.

• The thermometer not precise enough.

24 —


–12.7 kJ0.10 mol

92

0.100 g60.0 g mol–1

–2.68 kJ1.67 x 10–3 mol

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