PROBLEMS FOR 1.4
Unless stated otherwise, you should assume that the volumes of all gases are measured at
the same temperature and pressure.
1
10cm3 of hydrogen are burnt in oxygen to form water.
a Write a balanced equation, including state symbols, for this reaction.
H2(g) + ½O2 (g)  H2O(l)
(or multiples of this)
b What volume of oxygen is needed to burn the hydrogen completely?
5cm3
2
100cm3 of propane gas (C3H8) are burnt in oxygen to form carbon dioxide and water.
a Write a balanced equation, including state symbols, for this reaction.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
b What volume of oxygen is needed?
500cm3
c What volume of carbon dioxide is formed?
300cm3
3
10cm3 of a gaseous hydrocarbon reacts completely with 40cm3 of oxygen to produce
30cm3 of carbon dioxide.
a How many moles of carbon dioxide must have been formed from one mole of the
hydrocarbon?
3 moles
b How many carbon atoms must there be in the formula of the hydrocarbon?
3 carbon atoms
c How many moles of oxygen were used in burning one mole of the hydrocarbon?
4 moles
d How many moles of water must have been formed in burning one mole of the
hydrocarbon?
2 moles (4O2 used up; 3 CO2 formed so 1 mole O2 left, forming 2 moles H2O)
e What is the formula of the hydrocarbon?
C3H4
4
18g of pentane (C5H12) are completely burnt in a car engine to form carbon dioxide
and water.
a How many moles of pentane are burnt?
Moles = mass/Mr = 18/72 = 0.25 moles
b How many moles of oxygen are needed to burn all the pentane?
C5H12 + 8O2  5CO2 + 6H2O
So moles O2 = 0.25 x 8 = 2 moles
c What volume of oxygen is needed assuming one mole of gas occupies 24dm3.
Volume = 2 x 24 = 48dm3
d What volume of air is needed (assume air contains 20% oxygen by volume)?
48 = 20% x volume
Volume = 48/0.2 = 240dm3
e What volume of carbon dioxide is formed?
Moles CO2 = 5 x 0.25 = 1.25
Volume = 1.25 x 24 = 30dm3
5
A hydrocarbon contains 85.7% C and 14.3% H by mass. Its relative molecular
mass is 28.
a Find its empirical formula.
Carbon
Hydrogen
85.7/12
14.3/1
7.14
14.3
Ratio: 1
2
Empirical formula = CH2
b Suggest a molecular formula for this compound.
Mr(CH2) = 14 so formula of compound with Mr 28 = C2H4
6
A hydrocarbon contains 92.3% carbon. Its relative molecular mass is 78. Find
a its empirical formula.
b its molecular formula.
Carbon
Hydrogen
92.3/12
7.7/1
7.7
7.7
Ratio: 1
:
1
Empirical formula = CH
Mr(CH) = 13
78/13 = 6 therefore molecular formula = C6H6
7
There are two different compounds with the molecular formula C4H10. Draw skeletal
formulae for these and give their systematic names.
8
9
10
Write balanced equations for each of the following reactions.
a Pentane, C5H12, burns in a plentiful supply of air to form carbon dioxide and water.
C5H12(l) + 8O2(g)  5CO2(g) + 6H2O(l)
b Pentane burns in a limited supply of air to form carbon monoxide and water.
C5H12(l) + 5.5O2(g)  5CO(g) + 6H2O(l)
(or multiples)
Write the equations, including state symbols, to which the following enthalpy changes
apply.
a standard enthalpy change of formation of ethanol, C2H5OH
2C(s) + 3H2(g) + ½O2(g)  C2H5OH(l)
(in this case multiples NOT allowed as it
is a standard enthalpy change of formation of one mole)
b standard enthalpy change of combustion of ethanol
C2H6O(l) + 3O2(g)  2CO2(g) + 3H2O(l)
c standard enthalpy change of formation of butane, C4H10
4C(s) + 5H2(g)  C4H10(g)
d standard enthalpy change of combustion of butane
C4H10(g) + 6.5O2(g)  4CO2(g) + 5H2O(l)
e standard enthalpy change of formation of glucose, C6H12O6
6C(s) + 6H2(g) + 3O2(g)  C6H12O6(s)
f standard enthalpy change of combustion of glucose.
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
Look at the following hydrocarbons:
a Draw a skeletal formula for each of the hydrocarbons A to E.
b Which hydrocarbon(s) would you expect to be solid at 298 K? D
c Which hydrocarbon would you expect to be the most volatile? A
d Which hydrocarbon does not fit the general formula CnH2n+2 for alkanes? B
e Which two hydrocarbons are isomers? C + E
f Give an example of a hydrocarbon that might have been formed by cracking
another of the hydrocarbons shown here. (Identify both alkanes.) Write a balanced
equation for this process.
More than one example; should be D as the reactant and could be A, C or E as
products.
g For which hydrocarbon(s) are there no isomeric compounds? A
11
Ethene reacts with bromine to form 1,2-dibromoethane according to the following
equation:
H2C CH2(g) + Br2(g)
BrH2C
CH2Br(g)
Use bond enthalpies to calculate a value for the enthalpy change of the reaction. The
bond enthalpy of Br
Br is +193 kJ mol-1 and that of C
Br is +290 kJ mol-1.
12
Look at the values for energy density of the different fuels in the Table 1.
Fuel
Formula
Standard enthalpy
change of
combustion, H c /kJ
Relative
molecular
mass
mol-1
hexane
C6H14(l)
-4163
86
methane
CH4(g)
-890
16
methanol
CH3OH(l)
-726
32
carbon
C(s)
-393
12
hydrogen
H2(g)
-286
2
Table 1 – Energy densities of some important fuels.
Energy density
(energy transferred on
burning 1kg of
fuel)/kJ kg-1
-48 400
-55 600
-22 700
-32 800
-143 000
a
On the basis of energy density, which is the best fuel in the table? What are the
practical difficulties involved in using this particular fuel?
Hydrogen. Problems with storage, transport, modification of cars, production.
b
Compare hydrogen and hexane. Explain why hydrogen has the higher energy density,
even though it has the lower (least negative) enthalpy change of combustion.
Mr of hydrogen is only 2, hexane is 86. So a kilogram of hydrogen contains many more
moles than a kilogram of hexane.
c
Here are some data for octane, C8H18, and decane, C10H22, both of which are
components of petrol:
octane, C8H18
decane, C10H22
H c /kJ mol-1
Relative molecular mass
-5470
-6778
114
142
Use the data to calculate the energy density for each of these compounds.
Compare your two answers. How do they compare with the values of energy density given
for the fuels in Table 1?
Octane: moles in 1 kg = 1000/114 = 8.77 Energy density = -5470 x 8.77 = -47982 kJ kg-1
Decane: moles in 1 kg = 1000/142 = 7.04 Energy density = -6778 x 7.04 = -47732 kJ kg-1
Both are higher than methanol, carbon and hydrogen, lower than methane and similar to
hexane.
13
For the purposes of this assignment, assume that petrol is pure heptane, C7H16.
a
Write an equation for the complete combustion of heptane vapour.
C7H16(g) + 11O2(g)  7CO2(g) + 8H2O(l)
How many moles of oxygen are needed for the combustion of 1 mole of heptane?
11
How many moles of heptane are there in 1g?
Mr heptane = 100
moles = 1/100 = 0.01 moles
What mass of oxygen is needed to burn 1g of heptane (Ar: O, 16)?
Moles O2 = 11 x 0.01 = 0.11
mass = 0.11 x 32 = 3.52g
What mass of air is needed to burn 1g of heptane? (Assume that air is 22% oxygen
and 78% nitrogen by mass.)
3.52 = 22% x mass of air
mass of air = 3.52/.22 = 16g
What volume of air (measured at 25oC and 1 atmosphere pressure) is needed to burn
1g of heptane? (Assume that 1 mole of oxygen has a volume of 24 dm3 under these
conditions and that air is 21% oxygen and 79% nitrogen by volume.)
Moles O2 = 0.11
Volume O2 = 0.11 x 24 = 2.64 dm3
Volume of air = 2.64 / 0.21 = 12.6 dm3
b
c
d
e
f
14
a
Why is it important that catalytic converters start working at as low a temperature as
possible?
The lower the temperature that the catalyst starts to act, the quicker the exhaust
gases are controlled after the engine is switched on (as it will take less time for the
catalyst to heat to a lower temperature).
b
What is meant by a catalyst poison?
A substance which significantly reduces the effect of the catalyst.
c
Why is the catalyst used in the form of a fine powder?
To increase surface area.
d
Suggest a reason why the catalytic converter has eventually to be replaced.
Poisoning of catalyst; the catalyst also wears away as a result of engine vibrations and
the movement of hot gases.
e
Catalytic converters convert the pollutants CO, CxHy and NOx into harmless gases. This
is still only a partial solution to the emissions problem. Why?
Large amounts of CO2 are still released, which contributes to greenhouse effect.
f
Why is it possible to fit oxidation catalytic converters to existing cars, but not threeway converters?
Three-way converters need special fuel injection systems and oxygen sensors (you
don’t need to know this!)
15
The petrol tank of a typical car holds about 45 litres of petrol (approximately 10
gallons).
a
Calculate the amount of energy that is released by burning 45 litres of petrol.
Use the following information to help you.
 Assume that petrol is octane (C8H18).
 The standard enthalpy change of combustion of C8H18 is -5500 kJ mol-1.
 The density of octane is 0.70g cm-3.
 1 litre is 1000cm3.
Mass petrol = density x volume = 0.70 gcm-3 x 45000 cm3 = 31500 g
Moles petrol = mass/Mr = 31500 / 114 = 276
Energy released = energy per mole x number of moles = -5500 x 276 = -1518000kJ
b
Calculate the mass and volume (at 20oC and 1 atmosphere pressure) of hydrogen
needed to provide the same amount of energy as 45 litres of octane.
 The standard enthalpy change of combustion of H2 is -286 kJ mol-1.

1 mole of a gas at 20oC and atmospheric pressure has a volume of about 24
litres.
(You can remind yourself about calculations involving gases by reading Chemical Ideas 1.4)
Number of moles needed = energy released / energy per mole
= -1518000 / -286 = 5308 moles
Mass = moles x Mr = 5307 x 2 = 10615g (previous answer was rounded up)
Volume = moles x 24 = 5308 x 24 = 127392 dm3
Hydrogen engines are efficient. In motorway driving conditions a hydrogen engine can be
over 20% more efficient than a petrol engine. In city 'stop-go' driving conditions the
hydrogen engine is about 50% more efficient than a petrol engine.
c
Taking efficiencies into account, what mass and volume of hydrogen are needed to
give the same mileage as 45 litres of petrol in
i motorway driving conditions?
ii city driving conditions?
i) mass needed 20% less; 80% x 10615 = 8492 g
volume needed = 80% x 127392 = 101913 dm3
ii) mass needed 50% less; 50% x 10615 = 5307 g
volume needed = 50% x 127392 = 63696 dm3
15
The potential energy diagrams A-D
represent four different
reactions. All the diagrams are
drawn to the same scale.
Which of the potential energy
diagrams, A – D:
(a)
(b)
(c)
(d)
(e)
(f)
represent exothermic
reactions? B, C
represent endothermic
reactions? A, D
shows the largest activation
enthalpy? D
shows the smallest activation enthalpy? B
represents the most exothermic reaction? B
Represents the most endothermic reaction? A
16 An alcohol burner is used to heat 200 cm3 of water in a copper can. The temperature of
the water increases from 18oC to 22oC and 0,002 mol of alcohol is burned. Calculate the
molar enthalpy of combustion of the alcohol. The specific heat capacity of water is 4.17 J g-1
deg-1
Energy released = mass x temp rise x S.H.C = 200 x 4 x 4.17 = 3336 J = 3.336 kJ
Energy per mole = 3.336 / 0.002 = 1668 kJmol-1
17 Look at the enthalpy cycle below:
H1
CH4 (g) + 2O2
CO2 (g) + 2 H2O
(g)
(l)


C (s) + 2 H2 (g) + 2 O2 (g)
(a)
What enthalpy change is represented by H1 ?
Enthalpy change of combustion of methane
(b) What enthalpy change is represented by H2 ?
Enthalpy change of formation of methane
(c) What are the combined enthalpy changes represented by H3 ?
Enthalpy change of formation of CO2 + 2 x enthalpy change formation H2O
(d) Write an equation that links H1, H2 and H3.
H3 = H1 + H2
or
H1 = H3 - H2
(e) Use your answers to parts (a) to (d) to help you calculate a value for the standard
enthalpy change of combustion of methane. You will also need to make use of the
following standard enthalpy changes:
Hf (CH4)
Hf (CO2)
Hf (H2O)
=
=
=
- 75 kJ mol-1
- 393 kJ mol-1
- 286 kJ mol-1
H1 =(-393 + (2x-286)) - (-75)
= -104 kJmol-1