Energetics SQ Ans

Structured Questions Answers

1 (a) (i) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) [1]

(ii) The pressure increases due to the formation of hydrogen gas. [2]

(iii) Heat change at constant volume is equal to change in internal energy. [1]

(b) (i) The pressure remains constant. [1]

(ii) The hydrogen gas produced from the reaction has to push back the air in the atmosphere,

thus doing work and consuming energy. [2]

(c) Less heat will be released. [1]

2 (a) An exothermic reaction is a reaction that gives out heat. [1]

(b) The reaction mixture becomes hot [1] as heat is released from the reacting system to the

surroundings. [1]

(c) Both reactions will give out more or less the same amount of heat. [1] Although the number of

moles of H+(aq) ions of 2 M HCl is doubled than that of 1 M HCl, the number of moles of

OH(aq) ions available for the reaction is the same. [1] As a result, the number of moles of

H+(aq) ions of both solutions used for neutralization is the same. [1]

3 (a) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) [1]

(b) Burning methane is an exothermic reaction. [1]

(c) During the course of the combustion reaction, four CH bonds and two OO bonds are broken.

[1] These bond-breaking processes are endothermic. [1] To complete the reaction, two CO

bonds and four OH bonds are formed [1] and these bond-forming processes are exothermic. [1]

The formation of two CO bonds and four OH bonds give out more energy than that is needed

for the breaking of four CH bonds and two OO bonds. [1] Therefore, the combustion of

methane is an overall exothermic reaction. [1]

4 (a) Bond-forming processes: 4 OH bonds [2]

Bond-breaking processes: 2 HH bonds [2] and 1 OO bond [2]

(b)

[3]

(c) No pollutants are produced. [1] This reaction is highly exothermic, thus a large amount of energy

is released for the propulsion of space shuttles. [1]

4H + 2O

Ent

halp

y

2H (g) + O (g) 2 2

2H2O(l)

Reaction coordinate

reactants H = ve

products

5 (a) Since the reaction is exothermic, [1] the enthalpy change is negative. [1]

(b) The reaction is not very rapid as the effect of hand-warmer can last for several hours. [1]

(c) No. [1] The rate of rusting is too slow. [1] He needs some catalyst to speed up the rate of the

exothermic reaction. [1]

(d) (i) This type of hand-warmer can be reused. [1]

(ii) During the generation of heat, the chemical energy is released from the reaction mixture as

heat. [1] During the recharging by heating the warmer, heat energy is stored as chemical

energy prepared to be used next time. [1]

6 (a) The reaction is exothermic. [1]

(b) It is readily oxidized, as it has only 2 electrons in its valence shell. [2]

(c) It acts as a catalyst. [1]

(d) Mg(s) + 2H2O(l) Mg(OH)2(aq) + H2(g) [2]

(e) (i) Fe2+(aq) + Mg(s) Fe(s) + Mg2+(aq) [2]

(ii) It is a redox/displacement reaction. [1]

7 (a) 16H2S(g) + 8SO2(g) 16H2O(l) + 3S8(s) [2]

(b) It is exothermic. [1]

(c) HS bonds and S=O bonds are broken during the reaction. [2]

(d) HO bonds and SS bonds are formed during the reaction. [2]

(e)

[3]

8 (a) The above reaction is exothermic [1] because it is a combustion reaction. [1]

(b)

[3] c. [1]

32H + 16O + 24S

Ent

halp

y

16H S(g) + 8SO (g) 2 2

(c) The reverse reaction is endothermi

CH (g) + 2O (g) 4 2

CO2(g) + 2H2O(l)

Reaction coordinate

Ent

halp

y

H is negative

Reaction coordinate

16H2O(l) + 3S8(s)

reactants H = −ve

products

9 overcome the intermolecular forces (hydrogen bonds)

(b) ded to break the ionic bond between ammonium nitrate. [1]

(a) Endothermic. [1] Energy is needed to

between water molecules. [1]

Endothermic. [1] Energy is nee

(c) Endothermic. [1] Energy is needed to break the covalent bond between hydrogen atoms. [1]

10 ) In an endothermic reaction, bond-forming processes give out less energy than that is required in

(b) olving ammonium nitrate in water is an endothermic process. [1]

[3]

(a

the bond-breaking processes. [1] Then, heat is taken in from the surroundings. [1] The enthalpies

of products are relatively higher than that of reactants. [1] Thus, the above statement is incorrect.

[1]

Diss

NH4+(aq) + NO3

−(aq)

11 (a) Any TWO of the following:

Fossil fuels/hydroelectric power/nuclear power/wind power/power from biomass [2]

increased by 10.0C. [1] (b) (i) The temperature may be

(ii) It is impossible to produce the temperature rise of 20.0C as Mg is limited. [1]

(c) (i) Number of moles of Ca = 1molg1.40

g0.4 = 0.1 mol [1]

tely burned w kJ of energy,

∴

(ii)

[3]

Since 1 mole of Ca comple ill release 800.0

the amount of energy released = 800.0 kJ mol1 × 0.1 mol = 80.0 kJ [1]

12 (a) (i) When the pac t er bags are broken. [1] Then ammonium nitrate and

water inside the pack are allowed to mix. [1] The dissolving of ammonium nitrate in water is

(ii)

(b) Amm ride [1]

(c) sorb more heat from the

k is squeezed, the wo inn

an endothermic process which absorbs heat from the surroundings. Thus, it gives a cooling

effect. [1]

NH4NO3(s) + aq NH4+(aq) + NO3

(aq) H soln = +ve [2]

onium chlo

Use the pack with greater amount of reactants which can ab

ΔH = −80.0 kJ

Reacti ate on coordin

Ent

halp

y Ca(s) + Cl2(g)

CaCl2(s)

ΔH = +ve

Reaction coordinate

Ent

halp

y

NH4NO3(aq) +

surroundings. [1]

13 (a) (g) + 2O2(g) CO2(g) + 2H2O(l) [1]

For butane, C H (g) +

For methane, CH4

4 102

13O (g) 4CO (g) + 5H O(l) [1]

O2

(b) e

2 2 2

For hydrogen, 2H2(g) + (g) 2H2O(l) [1]

Heat change when burning 1.0 g m thane

= 890.0 kJ mol1 × 1molg)40.10.12(

g0.1 = 55.625 kJ

ed is 55.625 kJ. [1]

Heat change when burning 1.0 g butane

Hence, the heat releas

= 2877.0 kJ mol1 × 1molg)10.140.12( 0 = 49.6 kJ

d is 49.6 kJ [1]

Heat change when burning 1.0 g hydrogen

g0.1

Hence, the heat release

= 286.0 kJ mol1 × 1molg)20.1(

g0.1 = 143.0 kJ

(c) Hydrogen is the best rocket fuel as it can release the largest amount of energy when burning the

Hence, the heat released is 143.0 kJ. [1]

same mass of fuel. [2]

14 ing 1.0 g CH3OH = (a) Energy released by burn

1

1

molg)0.1640.10.12(

molkJ0.843

= 26.34 kJ g [1] 1

Energy released by burning 1.0 g C2H5OH = 1

1

molg)0.1660.120.12(

molkJ1371

= 29.80 kJ g [1] 1

Energy released by burning 1.0 g C3H7OH = 1

1

molg)0.1680.130.12(

molkJ2010

= 33.50 kJ g [1] 1

Hence, C3H7OH releases the largest amount of energy per gram of substance burnt. [1]

(b) st appropriate for using as a fuel as it can release the largest amount of energy C3H7OH is the mo

when burning the same mass of the alkanols. [2]

15 (a) ers) [1]

) The packs become hot when calcium chloride and water inside the packs are allowed to mix. [1]

ermic process which releases a large

(c)

Calcium chloride (accept any other possible answ

(b

The dissolving of calcium chloride in water is an exoth

amount of heat and then gives a heating effect. [1]

CaCl2(s) + aq Ca2+(aq) + 2Cl(aq) H soln = x kJ mol1 [2]

16 (a) 1.0 g cm3 = 100.0 g Mass of the reaction mixture = (50.0 + 50.0) cm3 ×

Heat released = m × c × ΔT

= 100.0 g × 4.2 J g1 K1 × 13.0 K

= 5460 J [1]

Number of moles of HCl used = 3dm0.50

× 2.01000

mol dm3 = 0.1 mol

= Number of moles of NaOH used 3 × 2.0 mol dmdm1000

0.50l

ed = 0.1 m

Heat released per mole of H O formed =

3 = 0.1 mo

∴ number of moles of H2O form ol [1]

2mol1.0

J5460 = 54 600 J mol1

1

∴ 54.6 kJ m

(b) Mass of the reaction mixture = (100.0 + 100.0) cm3 00.0 g [1]

= 54.6 kJ mol

the enthalpy change of neutralization is ol1. [1]

× 1.0 g cm3 = 2

Number of moles of HCl used = 3dm0.100

× 2.0 mol dm 3 = 0.1000

2 mol

= Number of moles of NaOH used 3 × 2.0 mol dmdm1000

0.100l

ed = 0.2 mol [1]

2 ol

The gh the volumes of two solutions are doubled, the

number of and OH(aq) ions reacting are also doubled. [1] Twice as much heat is

3 = 0.2 mo

∴ number of moles of H2O form

Heat released of 0.2 mole of H O = 54.6 kJ m 1 × 0.2 mol = 10.92 kJ


10 920 J = 200.0 g × 4.2 J g1 K1 ×ΔT

ΔT = 13.0 K [1]

temperature rise is 13.0C. [1] Althou

moles of H+(aq)

given out, but this is used to heat up twice the volume of the solution. [1] For this reason, both

experiments have the same rise in temperature.

17 (a) the acid completely is 20.0 cm3. [1]

(ii) The temperature rises by 18.0C (38.0C 20.0C). [1]

(c) change when one mole of water is

an alkali under standard conditions. [2]

Number of moles of HCl used =

(i) The volume of NaOH needed to neutralize

(b) HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) [2]

Standard enthalpy change of neutralization is the enthalpy

formed from neutralization between an acid and

(d) Mass of the reaction mixture = (20.0 + 20.0) cm3 × 1.0 g cm3 = 40.0 g


= 40.0 g × 4.2 J g1 K1 × [(273 + 38) (273 + 20)] K

= 3024 J [1]

3dm0.20

× 3.0 mol dm1000

ol

=

3 = 0.06 m

Number of moles of NaOH used 3 × 3.0 mol dmdm1000

0.20l 3 = 0.06 mo

∴ number of moles of H2O form ol [1] ed = 0.06 m

Heat released per mole of H2O formed = mol06.0

J3024

= 50 400 J mol1

∴ the enthalpy change of neutralization is 50.4 kJ mol . [1]

(e) It was due to the heat loss to the environment in the course of reaction. [1] In addition, the

that of water. [1]

= 50.4 kJ mol 1 1

specific heat capacity of the reaction mixture was not the same as

(f) Replace the expanded polystyrene cup with a vacuum flask. [1] The specific heat capacity of the

reaction mixture should be pre-determined. [1]

18 (a) 2]

) Mass of the reaction mixture = 20 000 cm3 × 1.0 g cm3 = 20 000 g [1]

4.5) (273 + 20.0)] K

(c) 1 = 180.0 g mol1

les of glucose =

C6H12O6(aq) 2CH3CH2OH(aq) + 2CO2(g) [

(b

Minimum amount of heat generated

= m × c × ΔT

= 20 000 g × 4.2 J g1 K1 × [(273 + 3

= 1218 kJ [1]

Molar mass of glucose = 12.0 × 6 + 1.0 × 12 + 16.0 × 6 g mol

Number of mo1molg0.180

g0.600 = 3.33 mol [1]

se fermented =Heat released per mole of gluco mol33.3

kJ1218= 365.8 kJ mol1

1. [1]

∴ the enthalpy change for the fermentation of glucose is 365.8 kJ mol

19 (a)

) KOH(aq) + HCl(aq) KCl(aq) + H2O(l) [1]

heat loss to the surroundings. [1]/The heat will be given out very

= m × c × ΔT

Number of m

The reaction is exothermic. [1]

(b

(c) The purpose is to minimize the

quickly. [1]

(d) Mass of the reaction mixture = (50.0 + 50.0) cm3 × 1.0 g cm3 = 100.0 g [1]

Heat released

= 100.0 g × 4.2 J g1 K1 × [(273 + 23.1) (273 + 19.6)] K

= 1470 J [1]

oles of HCl used = 3dm0.50

× 0.5 mol dm1000

=

3 = 0.025 mol

Number of moles of NaOH used 3 × 0.5 mol dmdm1000

0.50l

ed = 0.025 m

Heat released per m

3 = 0.025 mo

∴ number of moles of H2O form ol [1]

ole of H2O formed = mol025.0

J1470

= 58 800 J mol1

1

∴ the enthalpy change of reaction is 58.8 kJ mol . [1]

= 58.8 kJ mol 1

(e) The experiment was carried out in a glass beaker whi amount of heat loss to the

olystyrene cup with a lid instead of the glass beaker. [1]

(f) re rise would be

of NaOH reacted was

ch caused a large

surroundings. [1]

Surrounding the glass beaker with the insulating material e.g. cotton wool. [1]/

Use an expanded p

The temperature change would be more or less the same, and the temperatu

around 3.5C. [1] Although the amount of HCl was increased, the amount

limited. [1] The number of moles of water formed would be the same. Therefore, the temperature

change would be the same. [1]

20 (a) smaller that the theoretical value as some heat energy from the

surroundings is absorbed during the reaction. [1] Therefore, lower temperature drop of the

(ii)

ounded by cotton wool. [1]

(b)

Heat released = m ×

= 50. 1 K1 × [(273 + 25.4) (273 + 21.1)] K

Number of mo

∴ the entha the reaction =

(i) The experimental value is

solution will be observed. [1]

The experiment should be carried out in an expanded polystyrene cup with a lid. [1] The

polystyrene cup should be surr

Mass of water = density × volume

= 1.0 g cm3 × 50.0 cm3

= 50.0 g [1]

c × T

0 g × 4.2 J g

= 903.0 J [1]

les of AgNO3 = 0.1 mol dm3 × 0.050 dm3 = 0.005 mol

lpy change of mol005.0

J0.903 = 180 600 J mol1

]

= 180.6 kJ mol 1 [1

21 (a) Heat transferred to water

= mcΔT [1]

[(273 + 59.0) (273 + 19.0)] K

]

f

= 250.0 g × 4.2 J g1 K1 ×

= 42 000 J [1

Number of moles o ethanol burnt = 1molg0.46

g)1.2194.221(

= 0.05 mol

nt = Heat released per mole of ethanol burmol05.0

J00042 = l1

1 ]

(b) There is heat loss to the surroundings. [1]

(c)

eter. [1]/

y of the container into consideration. [1]/

840 000 J mo

∴ the enthalpy change of combustion of ethanol is 840.0 kJ mol . [1

Any TWO of the following:

Perform the experiment in a bomb calorim

Take the specific heat capacit

Cover the beaker with a lid. [1]/

Surround the beaker with mineral wool. [1]

22 (a) ) + 2H2O(l) [1]

) Mass of the reaction mixture = (40.0 + 40.0) cm3 × 1.0 g cm3 = 80.0 g [1]

Number of mo ol

Number of mo )2 used = 0.6 mol dm3 × 0.040 dm3 = 0.024 mol

2HNO3(aq) + Ba(OH)2(aq) Ba(NO3)2(aq

(b


= 80.0 g × 4.2 J g1 K1 × [(273 + 22.0) (273 + 18.0) K

= 1344 J [1]

les of HNO3 used = 0.6 mol dm3 × 0.040 dm3 = 0.024 m

les of Ba(OH

∴ number of moles of H2O = 0.024 mol [1]

Heat released per mole of H O formed = 2mol02.0 4

1 ]

(c) 2HNO3(aq) + Ba(OH)2(aq) Ba(NO3)2(aq) + 2H2O(l) Hneut = 56.0 kJ mol1 [2]

(d) ] There was heat loss to

meter were

J1344 = 56 000 J mol1

∴ the standard enthalpy change of neutralization = 56.0 kJ mol . [1

The calculated value would be less negative than the theoretical value. [1

the surroundings. [1] The specific heat capacities of the calorimeter and the thermo

not taken into account. [1]

23 (a) Cl(aq) + H2O(l) [1]

) Mass of the reaction mixture = (30.0 + 50.0) cm3 × 1.0 g cm3 = 80.0 g [1]

Number of mo

Number of mo 0.50 mol dm3 × 0.05 dm3 = 0.025 mol

NaOH(aq) + HCl(aq) Na

(b


= 80.0 g × 4.2 J g1 K1 × [(24.4 + 273) (20.3 + 273)] K

= 1377.6 J [1]

les of HCl = 1.0 mol dm3 × 0.03 dm3 = 0.03 mol

les of NaOH =

Since mole ratio of HCl : NaOH = 1 : 1

∴ number of moles of H2O formed = 0.025 mol [1]

Heat released per mole of H O formed =2 mol025.0

J6.1377

= 55 104 J mol1 [1]

kJ mol1. [1]

(The negative sign is added to denote that the change is exothermic.)

(c) maller.

lization. [1]

d for complete

∴ the enthalpy change of neutralization between HCl and NaOH is 55.1

Compared to the theoretical value of 57.1 kJ mol1, the calculated value is 2.0 kJ mol1 s

[1] It is due to the heat loss to the surroundings in the course of neutra

(d) The standard enthalpy change of neutralization between ethanoic acid and sodium hydroxide is

less negative. [1] Since ethanoic acid is a weak acid, some energy has to be supplie

ionization of ethanoic acid. [1]

24 (a) 2(g) [2]

) The reaction is exothermic as the temperature rose after the reaction. [2]

e magnesium is more reactive than zinc, the reaction

leases more energy. [2]

Zn(s) + 2HCl(aq) ZnCl2(aq) + H

(b

(c) The temperature rise will be greater. [1] Sinc

between magnesium and dilute hydrochloric acid is more vigorous and re

25 (a)

polystyrene cup should be surrounded by cotton wool. [1]

Number of m .040 dm3 = 0.024 mol

Heat released O formed =

The experiment should be carried out in an expanded polystyrene cup with a lid. [1] The

(b) Mass of the reaction mixture = (30.0 + 40.0) cm3 × 1.0 g cm3 = 70.0 g [1]

Heat released = m × c × T

= 70.0 g × 4.2 J g1 K1 × 4.3 K

= 1264.2 J [1]

oles of NaOH = 0.6 mol dm3 × 0

per mole of H2mol024.0

J2.1264 = 52 675 J mol1

= 52.7 k 1

(c) The temperature would decrease, as no further exothermic neutralization occurs. [1]

∴ the enthalpy change of neutralization J mol . [1]

26

) Molar mass of methanal = 12.0 + 1.0 × 2 + 16.0 g mol1 = 30.0 g mol1

(a) CH2O(g) + O2(g) CO2(g) + H2O(l) ΔH c = 517.5 kJ mol1 [2]

(b

Number of moles of 2.0 g methanal = 1molg0.30 ]

g0.2 = 0.0667 mol [1

methanal = 5 1 67 mol

= 34.5 kJ [1]

Number of moles of 2.0 g ethanal =

∴ heat released when burning 2.0 g 17.5 kJ mol × 0.06

Molar mass of ethanal = 12.0 × 2 + 1.0 × 4 + 16.0 g mol1 = 44.0 g mol1

1molg0.44

g0.2ol [1]

1 455 mol

= 48.4 kJ [1]

Number of moles of 2.0 g methanol =

= 0.0455 m

∴ heat released when burning 2.0 g ethanal = 1063.5 kJ mol × 0.0

Molar mass of methanol = 12.0 + 1.0 × 4 + 16.0 g mol1 = 32.0 g mol1

1molg0.32

g = 0.06

0.225 mol [1]

ethanol = 71 1 25 mol

= 44.7 kJ [1]

Number of moles of 2.0 g ethanol =

∴ heat released when burning 2.0 g m 5.0 kJ mol × 0.06

Molar mass of ethanol = 12.0 × 2+ 1.0 × 6 + 16.0 g mol1 = 46.0 g mol1

1molg0.46

g0.2ol [1]

1 435 mol

= 59.6 kJ [1]

(c) Ethanol is the best for use as fuel in motor cars as it can release the largest amount of energy for

= 0.0435 m

∴ heat released when burning 2.0 g ethanol = 1371.0 kJ mol × 0.0

Therefore, the organic compound is ethanol. [1]

the same mass of fuel carried. [2]

27 3 2 2(a) CH CH CH OH(l) +

2

9O (g) 2 3CO2(g) + 4H2O(l) [1]

(b) Placing cotton wool around the beaker./Covering the beak

(c) Heat transferred to water = m × × ΔT

er with a lid. [1]

c

Number of moles of prop

= 500.0 cm3 × 1.0 g cm3 × 4.2 J g1 K1 × 46.0 K

= 96 600 J [1]

an-1-ol burnt = 1molg)0.1680.130.12(

g88.2

.048 mol

Heat released per mole of propan-1-ol burnt =

= 0

mol04.0 8

J60096 = 2012.5 kJ mol1

an-1-ol is 1 ]

(No mark if negative sign is omitted.)

(d)

incomplete combustion of propane-1-ol. [1]

rried out under standard conditions, the enthalpy change cannot

∴ the enthalpy change of combustion of prop 2012.5 kJ mol . [1

(i) They were carbon soot. [1]

(ii) They were formed because of the

(e) Since the experiment was not ca

be called as ‘standard enthalpy change’. [1]

28 (a) 2(g) + H2O(l) [1]

) The temperature of the solution rises. [1]

0) cm3 × 1.0 g cm3 = 30.0 g [1]

Number of moles o 0.02 mol

Number of moles o .0 mol dm3 × 0.01 dm3 = 0.01 mol

2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + CO

(b

Gas bubbles are evolved. [1]

(c) Mass of the reaction mixture = (20.0 + 10.


= 30.0 g × 4.2 J g1 K1 × 15.0 K

= 1890 J [1]

f HCl = 1.0 mol dm3 × 0.02 dm3 =

f Na2CO3 = 1

∴ number of moles of H2O formed = 0.01 mol [1]

Heat released per mole of H O formed = 2mol01.0

J1890 = 189.0 kJ mol1

189.0 k 1

(No mark if negative sign is omitted.)

nergy. [1]

n mixture to the surroundings. [1]

∴ the enthalpy change of the reaction is J mol . [1]

(d) The carbon dioxide released brings away some of the heat e

There is heat loss from the reactio

(Accept any other possible reasons.)

29 (a) ings. [1]

) Heat transferred to water

[(58.0 +273) (25.0 + 273)] K

There was no heat loss to the surround

(b

= m × c × ΔT

= 650.0 g × 4.2 J g1 K1 ×

= 90 090 J [1]

Heat released per mole of propanol burnt = propanolofmolesofNumber

J09090

2 010 000 J = propanolofmolesofNumber

J09090

Number of moles of propanol = 0.0448 mol [1]

Molar mas = 60.0 g mol1 [1]

ol1 = 2.69 g [1]

(c)

s of propanol = 12.0 × 3 + 1.0 × 8 + 16.0 g mol1

∴ the actual mass of propanol used to heat the water = 0.0448 mol × 60.0 g m

Percentage purity of propanol = %100g69.2 = 53.8 % [1]

g0.5

The low percentage purity of propanol showed that there were many impurities in the sample of

propanol. [1] The impurities may also have standard enthalpy changes of combustion. Therefore,

the temperature rise of water may not be contributed by the combustion of propanol only. [1]

30 ) Neutralization. [1] The standard enthalpy change of neutralization is the enthalpy change when

(b) at loss to the surroundings. [1] 3 × 1.0 g cm3 = 90.0 g [1]

J g1 K1 × [(50.0 + 273) (25.0 + 273)] K

H2SO4(aq) + Na2SO4(aq) + 2H2O(l)

(a

one mole of water is formed from neutralization between an acid and an alkali under standard

conditions. [2]

There was no he

(c) Mass of the reaction mixture = (40.0 + 50.0) cm


= 90.0 g × 4.2

= 9450 J [1]

2NaOH(aq)

Number of moles of sulphuric acid = 3dm0.40

× 01000

.55 mol dm3

ol

Number of moles of sodium hydrox

= 0.022 m

ide = 3dm0.50

1000 × 1.2 mol dm3

ol

∴ number of moles of water formed = 0.022 mol × 2 = 0.044 mol [1]

= 0.06 m

Heat released per mole of water formed = mol044.0

J9450 = 214 772.7 J mol1

∴ the standard enthalpy change of the reaction is 214.77 kJ mol . [1] 1

31 (a) mole of the substance

(b)

Standard enthalpy change of combustion is the enthalpy change when one

is completely burnt in oxygen under standard conditions. [2]

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) [1]

(c)

[3]

By applying Hess’s Law,

ΔH + ΔH 1 = ΔH 2 + ΔH 3 [1]

∴ ΔH = 6 × ΔH c [C(s)] + 6 × ΔH c [H2(g)] ΔH f [C6H12O6(s)]

= 6 × (394.0) + 6 × (286.0) (1286) kJ mol1

= 2794 kJ mol1 [1]

The standard enthalpy change of combustion of glucose is 2794 kJ mol1.

Molar mass of glucose = 12.0 × 6 + 1.0 × 12 + 16.0 × 6 g mol1 = 180.0 g mol1

Number of moles of 1.0 g glucose = 1molg0.180

g0.1 = 0.00556 mol [1]

∴ the standard enthalpy change of combustion of 1.0 g glucose

= 2794 kJ mol1 × 0.00556 mol

= 15.53 kJ [1]

32 (a) CH3COCH3(l) + 4O2(g) 3CO2(g) + 3H2O(l) [1]

ΔH

(b)

[3]


ΔH + ΔH 1 = ΔH 2 + ΔH 3 [1]

∴ ΔH = 3 × ΔH f [CO2(g)] + 3 × ΔH f [H2O(l)] ΔH f [CH3COCH3(l)]

= 3 × (395.0) + 3 × (286.0) (250.0) kJ mol1

= 1793 kJ mol1

The standard enthalpy change of combustion of CH3COCH3(l) is 1793 kJ mol1. [1]

33 (a) Reaction I involves breaking N≡N bond and OO bond and forming 2 NO bonds. [1] Since the

N≡N bond and OO bond are very strong covalent bonds, a large amount of energy is needed to

break these bonds. [1] The bond-forming process gives out less energy than that is needed for the

bond-breaking process. [1] Therefore, reaction I is endothermic in nature.

(b)

ΔH 1

CH3COCH3(l) + 4O2(g) 3CO2(g) + 3H2O(l)

3C(s) + 3H2(g)

ΔH 2 ΔH 3

ΔH

+2

9O2(g) + 3O2(g)

+2

3O2(g)

ΔH 1

+9O2(g) +6O2(g

)ΔH 2

6CO (g) + 6H O(l) C H O (s) + 6O (g) 2 26 12 6 2

6C(s) + 6H2(g)

ΔH 3

+3O2(g

)

[3] By applying Hess’s Law,

3 [1]

(55.0) kJ mol1

The standard enthalpy change of formation of N2O5(g) is +11.0 kJ mol1. [1]

(c)

ΔH = ΔH 1 + ΔH 2 + ΔH

∴ ΔH = 180.0 + (114.0) +

= 11.0 kJ mol1

It is used for the preparation of explosives. [1]

34 ) The standard enthalpy change of combustion of a substance is the heat given out when one mole

4 10

(a

of that substance under its standard state is completely burned in oxygen. [1] Thus, the correct

equation should be C H (g) + 2

13O (g) 4CO (g) + 5H O(l). [1]

His suggestion is correct. Duri s, heat will be lo

2 2 2

(b) ng the combustion proces st to the surroundings.

(c) 00.0 cm3 × 1.0 g cm3 = 500.0 g

J g1 K1 40 K = 84.0 kJ [1]

Molar mass of b ol1

[1] If the beaker is not covered with a lid, the temperature rise of water will be less than the

actual value. [1]

Mass of water = 5

Energy released = m × c × ΔT

= 500.0 g 4.2

utane = 12.0 × 4 + 1.0 × 10 g mol1 = 58.0 g m

Number of moles of butane = 1molg0.58

g4.1 = 0.0707 mol [1]

∴ the standard enthalpy change of combustion of butane = mol0707.0

kJ84.0

= 1188 kJ mol [1]

1

35 (a)

2

1Cl2(g) + O2(g) ClO2(g) ΔH f = +102.0 kJ mol1 [2]

(b)

ΔH N2(g) +

2

5O (g) 2

O2(g) ΔH 2

ΔH 1

ΔH 3 2NO(g) +

2

1O2(g)

N O (g) 2 5

O (g) 2

2NO (g) 2

[3]


ΔH + ΔH 1 = ΔH 2 + ΔH 3 [1]

∴ ΔH = 2 × ΔH f [AgCl(s)] + 2 × ΔH f [ClO2(g)] 2 × ΔH f [AgClO3(s)]

= 2 × (127.0) + 2 × (+102.0) 2 × (30.0) kJ mol1

= +10.0 kJ mol1

The standard enthalpy change of the reaction is +10.0 kJ mol1. [1]

36 (a) C8H18(l) +

2

25O2(g) 8CO2(g) + 9H2O(l) [1]

ΔH

(b)

[3]


ΔH + ΔH 1 = ΔH 2 + ΔH 3 [1]

∴ ΔH = 8 × ΔH f [CO2(g)] + 9 × ΔH f [H2O(l)] ΔH f [C8H18(l)]

= 8 × (394.0) + 9 × (286.0) (250.0) kJ mol1

= 5476 kJ mol1

The standard enthalpy change of combustion of C8H18(l) is 5476 kJ mol1. [1]

(c) Molar mass of C8H18(l) = 12.0 × 8 + 1.0 × 18 g mol1 = 114.0 g mol1 [1]

Number of moles of C8H18(l) = 1molg0.114

g0.200 = 1.754 mol [1]

∴ the enthalpy change involved in the combustion of 0.2 kg of C8H18(l)

= 5476 kJ mol1 × 1.754 mol = 9607 kJ. [1]

37 (a) SO2(g) +

2

1O2(g) SO3(g) [1]

(b)

ΔH

ΔH 1

+2

25O2(g)

+8O2(g

)ΔH 2

8CO2(g) + 9H2O(l) C8H18(l) + 2

25O2(g)

8C(s) + 9H2(g)

ΔH 3

+2

9O2(g)

ΔH 1

+3O2(g) ΔH 2

2AgCl(s) + 2ClO (g) + 2AgClO (s) + Cl (g) 23 2

2Ag(s) + 2Cl2(g)

ΔH 3

+3O2(g

)

[3]


ΔH = ΔH f [SO3(g)] ΔH f [SO2(g)] [1]

]

= 396.0 (297.0) kJ mol1

= 99.0 kJ mol1

∴ the standard enthalpy change of reaction is 99.0 kJ mol1. [1]

(c)

[3]


ΔH = ΔH f [H2SO4(l)] ΔH f [SO3(g)] ΔH f [H2O(l)] [1

= 814.0 (396.0) (286.0) kJ mol1

= 132.0 kJ mol1

∴ the standard enthalpy change of reaction is 132.0 kJ mol1. [1]

38 (a) Fe2O3(s) +

2

3C(s) 2Fe(s) +

2

3CO2(g) [1]

(b)

[3]


ΔH = 2

3 × ΔH 1

2

1× ΔH 2 [1]

Fe2O3(s) + 2

3C(s)

2Fe(s) + 2

3C(s) +

2

3O2(g)

2Fe(s) + 2

3CO2(g)

ΔH

2

3 × ΔH 1 2

1 × ΔH 2

S(s) + 2

3O2(g)

SO3(g) SO2(g) + 2

1O2(g) ΔH

ΔH f [SO ΔH 2(g)] [SO (g)] f 3

ΔH H SO (l)

S(s) + H2(g) + 2O2(g)

2 4SO (g) + H O(l)3 2

ΔH f [HΔH (g)] + ΔH [SO 2SO4(l)] f 3 f

= 2

3 × (393.5)

2

1× (1648) kJ mol1

= +233.8 kJ mol1 [1]

(c) The reaction in (a) is an endothermic reaction. [1]

39 (a)

[3]


ΔH = ΔH f [Al2O3(s)] ΔH f [Fe2O3(s)] [1]

= 1676 (824.0) kJ mol1

= 852.0 kJ mol1

The standard enthalpy change of the thermite reaction is 852.0 kJ mol1. [1]

(b) When 1 mol of Fe2O3 is reacted, 1 mol of Al2O3 and 2 mol of Fe are produced. [1]

Specific heat capacity of the product mixture

= 1 mol × 79.0 J mol1 K1 + 2 mol × 25.1 J mol1 K1 = 129.2 J K1

∴ 129.2 J of energy is required to raise the product mixture by 1 K. [1]

(c) Initial temperature = 298.0 K

Total heat released by the reaction = 852.0 kJ mol1 × 1 mol = 852.0 kJ = 852 000 kJ

The rise in temperature of the product mixture

= 1KJ0.129

J000 852 = 6605 K [1]

∴ the final temperature = 298.0 + 6605 K = 6903 K [1]

(d) The heat released by the thermite reaction is capable of heating its products far above the melting

point of iron. Since iron is a major component of steel, steel can be welded. [1]

40 (a) CaCO3(s) CaO(s) + CO2(g) [1]

(b)

[3]


CaCO3(s)

CaCl2(aq) + H2O(l) +

+2HCl(aq)

CaO(s) + CO2(g)

+2HCl(aq) ΔH 1 ΔH 2

ΔH

2Al(s) + Fe2O3(s)

2Al(s) + 2Fe(s) + 2

3O2(g)

AlΔH

O (s) + 2Fe(s) 2 3

ΔH f [Fe2O3(s)] ΔH [Alf O (s)] 2 3

ΔH = ΔH 1 ΔH 2 [1]

= 364.0 (192.0) kJ mol1

= 172.0 kJ mol1

The standard enthalpy change of decomposition of calcium carbonate is 172.0 kJ mol1. [1]

41 (a) The standard enthalpy change of formation of a substance is the enthalpy change when one mole

of it is formed from its elements in their standard states under standard conditions. [2]

(b) CS2(l) + 3O2(g) CO2(g) + 2SO2(g) [2]

(c)

[3]


ΔH + ΔH 3 = ΔH 1 + ΔH 2 [1]

∴ ΔH = ΔH 1 + ΔH 2 ΔH 3

ΔH = ΔH f [CO2(g)] + 2 × ΔH f [SO2(g)] ΔH c [CS2(l)]

= 395.0 + 2 × (297.0) (1076) kJ mol1

= 87.0 kJ mol1

The standard enthalpy change of formation of carbon disulphide is +87.0 kJ mol1. [1]

42 (a) The standard enthalpy change of formation of a substance is the enthalpy change when one mole

of it is formed from its elements in their standard states under standard conditions. [2]

(b) The overall enthalpy change of a chemical reaction is the same, regardless of the route by which

the reaction takes place. [2]

(c) (i)

[3]

(ii) By applying Hess’s Law,

ΔH + ΔH 3 = ΔH 1 + ΔH 2 [1]

∴ ΔH = ΔH 1 + ΔH 2 ΔH 3

ΔH = 2 × (395.0) + 2 × (286.0) (1409) kJ mol1

ΔH 2C(s) + 2H2(g)

+O2(g)

2ΔH 1

+3O2(g)

2CO2(g) + 2H2O(l)

C2H4(g)

+2O2(g) 2ΔH 2 ΔH 3

ΔH C(s) + 2S(s)

+2O2(g)

ΔH 2 ΔH 1 ΔH 3

+O2(g) +3O2(g)

CS (l) 2

CO (g) + 2SO (g) 2 2

= 47.0 kJ mol1

The standard enthalpy change of formation of ethylene is +47.0 kJ mol1. [1]

43 (a) Standard enthalpy change of combustion of a substance is the enthalpy change when one mole of

it is completely burnt in oxygen under standard conditions. [2]

(b) C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(l) [1]

(c) Heat transferred to water

= 550.0 g × 4.2 J g1 K1 × [(273 + 41.2) (273 + 25.5)] K

= 36 267 J [1]

Number of moles of sucrose burnt = 1molg)110.16220.1120.12(

g2.2

= 0.00643 mol [1]

Heat released per mole of sucrose burnt = mol00643.0

J26736 = 5640 kJ mol1

∴ the standard enthalpy change of combustion of sucrose is 5640 kJ mol1. [1]

(d) 12C(s) + 11H2(g) + 2

11O2(g) C12H22O11(s) [1]

(e)

[3]


ΔH f + ΔH c [C12H22O11(s)] = 12 × ΔH f [CO2(g)] + 11 × ΔH f [H2O(l)] [1]

ΔH f = 12(394.0) + 11(286.0) (5640) kJ mol1

= 2234 kJ mol1 [2]

44 (a) C6H4(OH)2(aq) + H2O2(aq) C6H4O2(aq) + 2H2O(l) [1]


(b)

[3]

C6H4(OH)2(aq) +

H O ( )

ΔH r

ΔH 1

C6H4O2(aq) + H2(g) + H2O(l) + 2

1O2(g)

C6H4O2(aq) + 2H2O(l)

2

1ΔH 2

ΔH 3

ΔH 12C(s) + 11H2(g) +

2

11 O2(g)

+12O2(g)

f C H O (s) 12 22 11

12ΔH f [CO2(g)]

+12O (g) 2 11ΔH [H O(l)] f 2

ΔH [C

12CO2(g) + 11H2O(l)

c H O (s)] 12 22 11

ΔH = ΔH + r 12

1ΔH + 2 ΔH 3 [1]

= 177.0 + 2

1(190.0) + (286.0) kJ mol1

∴ y change of reaction is 204.0 kJ mol1. [1]

(c) rogen peroxide. [1] It is an

= 204.0 kJ mol1

the standard enthalp

The origin of hot spray is the reaction between hydroquinone and hyd

exothermic reaction. [1] The heat generated from the reaction is used to boil the liquid or water

inside the abdomen of the beetle. [1] Then the hot vapour is released in the form of a hot spray.

[1]

45 ) Standard enthalpy change of formation of a substance is the enthalpy change when one mole of it

(b) lf. [1]

negative

(ii)

[3]


ol1

∴ ge for the reaction is 0.33 kJ mol1. [1]

(a

is formed from its elements in their standard states under standard conditions. [2]

This is because no heat changes are involved when an element is formed from itse

(c) (i) The reactions are exothermic because the standard enthalpy changes of reaction are

in value. [2]

ΔH 2 = ΔH r + ΔH 1 [1]

ΔH r = ΔH 2 ΔH 1

= 296.83 (297.16) kJ m

= 0.33 kJ mol1

the enthalpy chan

ΔH r

46 (a)

[3]


Ag(s) + 2

1Cl2(g) ΔH 3

ΔH 2 ΔH 1

Ag+(aq) + Cl(aq)

AgCl(s)

ΔH r e +e

ΔH 2 ΔH 1

S (rhombic) S (monoclinic)

SO2(g)

+O2(g) +O (g) 2

ΔH 3 = ΔH 1 + ΔH 2 + ΔH r [1]

(167.15) kJ mol1

The standard enthalpy change of reaction is 65.48 kJ mol1. [1]

(b)

50.0 + 50.0) cm3 × 1.0 g cm3 = 100.0 g

J g1 K1 × 2.8 K

Number of m 3 used = 0.1 mol dm3 × 0.05 dm3 = 0.005 mol

∴ ΔH r = 127.07 105.56

= 65.48 kJ mol1

(i) AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq) [1]

(ii) The reaction is exothermic. [1]

(iii) Mass of the reaction mixture = (


= 100.0 g × 4.2

= 1176 J [1]

oles of AgNO

Number of moles of HCl used = 0.1 mol dm3 × 0.05 dm3 = 0.005 mol

∴ Number of moles of AgCl formed = 0.005 mol [1]

Heat given out per mole of AgCl formed = mol005.0

J1176 = 235.2 kJ mol1

∴ the standard enthalpy change of reaction is 235.2 kJ mol . [1]

(iv)

(c) mation reactions cannot be easily or safely carried out in a

1

A calorimeter [1]

This is because some of the for

calorimeter [1] and some reactions involve the formation of side products. [1] Take the formation

of carbon monoxide as an example. When carbon and oxygen react, a mixture of carbon

monoxide and carbon dioxide may be formed. [1] Hence, the heat change measured is not only

due to the formation of carbon monoxide. [1]

47 (a)

[3]


1

s)] ΔH f [Fe2O3(s)] 1)

(b) From the r n is exothermic. [1]

ΔH r + ΔH 1 = ΔH 2 [1]

∴ ΔH r = ΔH 2 ΔH

= ΔH f [Al2O3(

= 1669 kJ mol1 (822.0 kJ mol

= 847.0 kJ mol1 [1]

esult above, the reactio

(c)

ΔH r

ΔH 1

+2

O2(g) 3

Fe2O3(s) + 2Al(s) 2Fe(s) + Al2O3(s)

2Fe(s) + 2Al(s)

ΔH 2

+2

23 O (g)

[3]

48 ) C6H14(g) + (a

2

19 O2(g) 6CO2(g) + 7H2O(l) [1]

[3]


ΔH 2 + ΔH 3 [1]

+ 7 × ΔH f [H2O(l)] ΔH f [C6H14(g)] [1]

The standard enthalpy change of combustion of hexane is 4199 kJ mol1. [1]

(b)

ΔH c [C6H14(g)] + ΔH 1 =

∴ ΔH c [C6H14(g)] = 6 × ΔH f [CO2(g)]

= 6 × (394.0) + 7 × (286.0) (167.0) kJ mol1

= 4199 kJ mol1

49 ) O2(g) + 2Cl2(g) 2OCl2(g) [2]

[3]


H 2 = ΔH 3 [1]

ΔH ΔH f [H2O(g)]

The standard enthalp 2(g) is 103.4 kJ mol1. [1]

(a

(b)

ΔH f [OCl2(g)] + ΔH 1 + Δ

∴ ΔH f [OCl2(g)] = 2 × ΔH f [HCl(g)]

= 2 × (92.3) (46.0) (242.0) kJ mol1

= 103.4 kJ mol1

y change of formation of OCl

ΔH 1

ΔH 2 Δ f [OCl2(g)] H

+2

12O (g) +O2(g)

OCl2(g) + H2O(g) O2(g) + 2HCl(g)

Cl2(s) + 2H (g)

ΔH 3

+2

1O2(g)

+6O2(g

)

ΔH c [C6H14(g)]

ΔH 2 ΔH 1

+2

19 O2(g) +

2

7 O2(g)

C6H14(g) +

19 O2(g)

6CO2(g) +

6C(s) + 7H2(g)

ΔH 3

Reaction coordinate

Fe O (s) + 2Al(s) 2 3

Ent

halp

y

2Fe(s) + Al2O3(s)

1 ΔH = 847.0 kJ molr

50 ) (i) N2(g) + 2H2(g) N2H4(l) [1]

[3]


f [N2H4(l)] + ΔH3

[1]

(b) (i)

[3]


3 [1]

ΔH f [N2H4(l)] 2 × ΔH f [H2O2(l)]

∴ the st nge of reaction is 818.2 kJ mol1. [1]

(a

(ii)

2 × ΔH1 + 2 × ΔH2 = ΔH

∴ ΔH f [N2H4(l)] = 2ΔH1 + 2ΔH2 ΔH3

N2H4(l) + 2H2O2(l) N2(g) + 4H2O(g) [1]

(ii)

ΔH r + ΔH 1 + ΔH 2 = ΔH

∴ ΔH r = 4 × ΔH f [H2O(g)]

= 4 × (285.8) (50.6) 2 × (187.8) kJ mol1

= 818.2 kJ mol1

andard enthalpy cha

51 ) C(s) + 2H2(g) CH4(g) ΔH f = 74.0 kJ mol1 [2]

[3]


(a

(b)

ΔH r + ΔH 1 = ΔH 2 [1]

CH4(g) C(g) + 4H(g)

C(s) + 2H2(g)

ΔH r

ΔH 1 ΔH 2

N2(g) + 2H2(g) ΔH f [N2H4(l)]

+O2(g) +2O2(g) +3O2(g)

N2H4(l)

2NO2(g) + 2H2O(g)

ΔHH2 2 × Δ2 × ΔH 3

1

+2O2(g

)

ΔH 1

N2H4(l) + 2H2O2(l) N2(g) + 4H2O(g)

N2(g) + 4H2(g)

+2O2(g) ΔH 2

ΔH 3

ΔH r

∴ ΔH r = ΔH 2 ΔH 1

18.0) (74.0) kJ mol1

∴ the standard enthalpy change of reaction is 1663 kJ mol1. [1]

= 717.0 + 4 × (2

= 1663 kJ mol1

52 ) 2N2H4 + N2O4 3N2 + 4H2O [1]

ΔH f (reactants) [1]

(c) Heat l1 ×

(a

(b) By applying Hess’s Law,

ΔH = ΔH f (products)

= 3(0) + 4(241.8) 2(50.6) 9.16 kJ mol1

= 1077.56 kJ mol1 [1]

change = 1077.56 kJ mo2

mol08.0 [1]

= 43.1 kJ

∴ the heat released is 43.1 kJ. [1]

53 ) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) [1] (a

(b) Number of moles of propane required = 1molkJ2220

kJ0.350 = 0.158 mol [1]

Mass of propane required = 0.158 mol × (12.0 × 3 + 1.0 × 8) g mol = 6.95 g [1]

(c)

(g) C3H8(g) [1]

ΔH c (products) [1]

(iii) The propane is exothermic. [1] Because the enthalpy change is

1

Flammable [1]

(d) (i) 3C(s) + 4H2

(ii) By applying Hess’s Law,

ΔH = ΔH c (reactants)

= 3(394.0) + 4(285.0) (2220) kJ mol1

= 102 kJ mol1 [1]

formation process of

negative in sign which indicates heat is released. [1]

54 CH3OH(l) + (a) 2O

2

3(g) CO2(g) + 2H2O(l) [2]

s Law,

ΔH f (reactants) [1]

(c) Entha ethanol

(b) By applying Hess’

ΔH = ΔH f (products)

= 393.5 + 2(241.8) (238.7) kJ mol1

= 638.4 kJ mol1 [1]

lpy change of combustion of 50.0 g liquid m

= 638.4 kJ mol1 × 1molg)0.1640.10.12(

g0.50 [1]

= 997.5 kJ [1]

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Energetics SQ Ans