Download 9. Solution Guide to Supplementary Exercises

Topic
10
Rate of Reaction
Part A Unit-based exercise
Unit 36 An introduction to rate of reaction
Fill in the blanks
1
concentration; time
2
instantaneous
3
a) the gaseous product / carbon dioxide evolved
b) loss in mass
4
a) colorimeter
b) standard alkali
c) the gaseous product / carbon dioxide evolved
1
time to reach the opaque stage
5
transmittance;
6
hydroxide; titrimetric
True or false
7
T In the oxidation of methanoic acid by bromine, the intensity of the orange colour of bromine decreases
as the reaction proceeds.
When we shine light upon the reaction mixture, the absorbance of the reaction mixture is directly
proportional to the colour intensity of the reaction mixture and the concentration of the bromine in the
reaction mixture.
Hence the progress of the reaction can be followed by a colorimeter.
8
T
9
F
10 F
11 T During the decomposition of hydrogen peroxide solution, oxygen gas is formed.
2H2O2(aq)
2H2O(l) + O2(g)
If the reaction vessel is a closed system, the pressure inside the vessel will increase. We can follow the
progress of the reaction by measuring the pressure inside the vessel with a pressure sensor connected to
a data-logger interface and a computer.
1
12 F
13 F
14 F
15 T The rate of a reaction decreases with time. At the start, there are plenty of reactant particles per unit
volume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.
the concentration of reactants falls. So, the reaction slows down.
Multiple choice questions
16 C
17 D Option D — A limestone status damage caused by acid rain may take years.
18 D Option D — The reaction between two solutions has the greatest reaction rate.
19 B
20 C
0.105 mol dm–3
20 s
= 5.25 x 10–3 mol dm–3 s–1
21 C Average reaction rate =
22 B
$PODFOUSBUJPONPMENm
m
m
NPMEN
m
T
5JNFT
(0.165 – 0.09) mol dm–3
(60 – 0) s
= 1.25 x 10–3 mol dm–3 s–1
Instantaneous rate of reaction at 30 s =
23 C
24 A
2
25 B
26 C
0.30
mol
2
= 0.15 mol
27 D Number of moles of Fe2O3 consumed in 10.0 seconds =
0.15 mol
10.0 s
= 0.015 mol s–1
Rate of consumption of Fe2O3 =
∴ the rate of consumption of Fe2O3 is 1.5 x 10–2 mol s–1.
28 A
1
2
d[N2O5(g)]
dt
1 d[NO2(g)]
=
4
dt
d[O2(g)]
=
dt
29 B Rate = –
d[N2O5(g)]
dt
2 d[NO2(g)]
=
4
dt
2
=
(3.8 x 10–5 mol dm–3 s–1)
4
= 1.9 x 10–5 mol dm–3 s–1
Instantaneous rate of decomposition of N2O5(g) = –
d[O2(g)]
dt
1 d[NO2(g)]
=
4
dt
1
=
(3.8 x 10–5 mol dm–3 s–1)
4
= 9.50 x 10–6 mol dm–3 s–1
30 D Instantaneous rate of decomposition of O2(g) =
31 A CaCO3(s) and HCl(aq) react according to the following equation:
CaCO3(s) + 2HCl(aq)
CaCl2(aq) + H2O(l) + CO2(g)
Number of moles of HCl consumed in 10.0 minutes = 0.200 mol min–1 x 10.0 min
= 2.00 mol
According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl.
2.00
mol
2
= 1.00 mol
i.e. number of moles of CaCO3 consumed =
Molar mass of CaCO3 = (40.1 + 12.0 + 3 x 16.0) g mol–1
= 100.1 g mol–1
3
Mass of CaCO3 consumed = number of moles of CaCO3 x molar mass of CaCO3
= 1.00 mol x 100.1 g mol–1
= 100.1 g
Mass of CaCO3 remaining = (145.0 – 100.1) g
= 44.9 g
∴ there are 44.9 g of CaCO3(s) remaining.
32 C The sealed flask was a closed system. Therefore the mass of the flask plus its contents remained the
same.
33 A
34 D
35 A
36 B In the oxidation of propanone by iodine, the intensity of the brown colour of iodine decreases as the
reaction proceeds.
When we shine light upon the reaction mixture, the absorbance of the reaction mixture is directly
proportional to the colour intensity of the reaction mixture and the concentration of the iodine in the
reaction mixture.
Hence the progress of the reaction can be followed by a colorimeter.
37 D
38 D
39 A
40 A The rate of a reaction decreases with time. At the start, there are plenty of reactant particles per unit
volume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.
the concentration of reactants fall. So, the reaction slows down.
41 D
42 B
43 C (1) Following the progress of a reaction using physical property may require the use of expensive
instructments, e.g. colorimeter.
44 C
45 B
46 A
4
47 B
48 B
49 A (1) The intensity of the yellow-brown colour of Br2(aq) decreases as the reaction proceeds.
Hence the progress of reaction can be followed by using a colorimeter.
(2) The reaction mixture becomes turbid as the sulphur forms.
A colorimeter can be used to follow the change in light level passing through the reaction mixture
as the turbidity increases.
(3) The reaction does NOT involve any colour change.
Hence a colorimeter CANNOT be used to follow the progress of the reaction.
50 B (1) During the acid catalyzed hydrolysis of methyl methanoate, the concentration of ethanoic acid
(CH3COOH) in the reaction mixture increases as the reaction proceeds.
Hence the progress of the reaction can be followed by titration with an alkali.
(3) During the decomposition of a dicarboxylic acid, the concentration of the acid in the reaction
mixture decreases as the reaction proceeds.
Hence the progress of the reaction can be followed by titration with an alkali.
51 D
52 A
53 B Carbon dioxide gas is produced in the reaction between sodium carbonate and dilute hydrochloric acid.
As the carbon dioxide gas escapes, the reaction mixture gets lighter as the reaction proceeds.
54 B In the reaction between iodine and propanone, the intensity of the brown colour of iodine decreases
as the reaction proceeds.
I2(aq) + CH3COCH3(aq)
brown
colourless
CH3COCH2I(aq) + HI(aq)
colourless
Hence the progress of the reaction can be followed by using a colorimeter.
55 C
56 A
5
Unit 37 Factors affecting the rate of a reaction
Fill in the blanks
1
a) surface area
b) concentration
c) temperature
2
faster; larger
3
catalyst; chemical
4
a) right orientation
b) activation energy
5
iron
6
platinum; vanadium(V) oxide
7
Enzymes
8
fermentation
True or false
9
F
10 F
The volume of a liquid would NOT affect the rate of a reaction.
The surface area of marble chips is smaller than that of powdered marble of the same mass.
The reaction between dilute hydrochloric acid and marble chips is slower than that between the acid and
powdered marble.
11 F
During the reaction between magnesium and an acid, magnesium would react with hydrogen ions in
the acid.
Hydrochloric acid is a strong acid while ethanoic acid is a weak acid. Hence 2 mol dm–3 hydrochloric acid
has a higher concentration of hydrogen ions than 2 mol dm–3 ethanoic acid does.
Thus the reaction between magnesium and 2 mol dm–3 hydrochloric acid is faster than that between
magnesium and 2 mol dm–3 ethanoic acid.
12 T
13 T
14 F
The physical state of a catalyst may be the same as those of the reactants.
For example, the hydrolysis of ethyl ethanoate can be catalyzed by dilute hydrochloric acid.
15 T Manganese(IV) oxide can catalyze the decomposition of hydrogen peroxide.
6
16 F
17 F
18 T
19 F
20 T
Multiple choice questions
21 B Option A — The antacid tablets in Experiments I and II have the same surface area. So, they CANNOT
be used for comparison.
Option B — The only difference between Experiments I and IV is the surface area of the antacid
tablets.
Option C — The antacid tablets in Experiments II and III have the same surface area. So, they CANNOT
be used for comparison.
Option D — The concentration and temperature of the acids in Experiments III and IV are different. So,
they CANNOT be used for comparison.
22 C Option A — The concentration and temperature of the acids in Experiments I and II are different. So,
they CANNOT be used for comparison.
Option B — The only difference between Experiments I and IV is the surface area of the antacid
tablets. So, they CANNOT be used for comparison.
Option C — The only difference between Experiments II and III is the concentration of the acids.
Option D — The surface area of the tablets, concentration and temperature of the acids in Experiments
III and IV are different. So, they CANNOT be used for comparison.
23 C
24 B
25 A
26 B The hydrochloric acid was in excess in each case. Hence the amount of calcium carbonate limited the
amount of carbon dioxide produced, i.e. equal mass of gas was produced in both cases. Thus the total
loss in mass of contents of the reaction flask was the same in both cases.
Powdered marble had a greater surface area than marble lumps of the same mass.
The rate of the reaction increased when the surface area of marble was increased.
27 D The initial rate of reaction would decrease when hydrochloric acid of a lower concentration was used.
7
28 D During the reaction between marble and an acid, marble would react with hydrogen ions in the acid.
Hydrochloric acid is a strong acid while ethanoic acid is a weak acid.
–3
Therefore 1 mol dm
ethanoic acid does.
–3
hydrochloric acid has a higher concentration of hydrogen ions than 1 mol dm
Hence the initial rate of reaction would decrease when ethanoic acid was used.
29 A The initial rate of the reaction would increase with an increase in temperature.
30 C
31 D
32 A
Reaction
mixture
Number of moles of Zn
mass
=
molar mass
2 mol dm–3 x
1
5 g
65.4 g mol–1
2
Number of moles of H2SO4
= molarity of solution x
volume of solution
= 0.076 mol
Reaction between Zn and H2SO4
100
dm3
1 000
Zn(s) +
0.076 mol
H2SO4(aq)
0.2 mol
ZnSO4(aq) + H2(g)
100
dm3
1 000
Zn(s) +
0.076 mol
H2SO4(aq)
0.1 mol
ZnSO4(aq) + H2(g)
= 0.2 mol
1 mol dm–3 x
= 0.1 mol
According to the equation, 1 mole of Zn reacts with 1 mole of H2SO4 to produce 1 mole of H2. During
the reaction, 0.076 mole of Zn reacted with 0.076 mole of H2SO4. Therefore the amount of Zn limited
the amount of H2 produced (0.076 mole of H2 produced in both cases).
Option A — The hydrogen produced would escape. Thus the mass of each reaction mixture would
decrease.
As the same amount of gas was produced in both reactions, the total mass lost of both
reaction mixtures would be the same.
Options B, C and D — The rate of reaction between zinc and 2 mol dm–3 sulphuric acid was higher
than that between zinc and 1 mol dm–3 sulphuric acid.
Hence the reaction between zinc and 2 mol dm–3 sulphuric acid took less time
to complete.
33 C As calcium carbonate disappeared in both experiments after reaction, it could be deduced that calcium
carbonate was the limiting reactant, i.e. its amount limited the amount of carbon dioxide gas formed.
The carbon dioxide formed would escape. Thus the masses of both reaction mixtures would decrease.
Curve Y showed a smaller loss in mass, and thus the mass of carbon dioxide gas formed was smaller.
It could be deduced the mass of calcium carbonate used was smaller (i.e. Option C).
Furthermore, the tangent to curve Y was less steep than that to curve X, i.e. the rate of the reaction
represented by curve Y was lower.
This agreed with the change stated in Option C. The rate of the reaction would decrease when a lump
of calcium carbonate was used instead of calcium carbonate powder.
8
34 C Curve II showed a greater volume of oxygen formed. Thus the number of moles of hydrogen peroxide
decomposed was greater, i.e. some hydrogen peroxide solution was added (Option C).
Furthermore, the tangent to curve II was less steep than that to curve I, i.e. the rate of decomposition
represented by curve II was lower.
This agreed with the change stated in Option C. Adding some 0.1 mol dm–3 hydrogen peroxide
solution to the original 1 mol dm–3 hydrogen peroxide solution would dilute the original hydrogen
peroxide solution. The rate of decomposition would decrease.
35 D
Reaction
Number of moles of HCl
= molarity of solution x
volume of solution
Number of moles of CaCO3
mass
=
molar mass
–3
1 mol dm
1
8 g
= 0.08 mol
100.1 g mol–1
100
x
dm3
1 000
= 0.1 mol
2 mol dm–3 x
2
100
dm3
1 000
= 0.2 mol
Reaction between CaCO3 and HCl
CaCO3(s) + 2HCl(aq)
0.08 mol
0.1 mol
CaCl2(aq) + H2O(l)
+ CO2(g)
CaCO3(s) + 2HCl(aq)
0.08 mol
0.2 mol
CaCl2(aq) + H2O(l)
+ CO2(g)
According to the equation, 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of CaCl2.
Option A — In Reaction 1, 0.1 mole of HCl reacted with 0.05 mole of CaCO3 to give 0.05 mole of
CaCl2.
Thus CaCO3 was in excess, i.e. Option A was INCORRECT.
Option B — In Reaction 2, 0.08 mole of CaCO3 reacted with 0.16 mole of HCl to give 0.08 mole of
CaCl2.
Thus HCl was in excess, i.e. Option B was INCORRECT.
Option C — Different amounts of CaCl 2 , and hence calcium chloride solutions of different
concentration, were produced in the two reactions.
Option D — The initial rate of the reaction between CaCO3 and a less concentrated HCl(aq) (Reaction 1)
is smaller than that between CaCO3 and a more concentrated HCl(aq) (Reaction 2).
36 B HCl(aq) and CH3COOH(aq) react with calcium carbonate according to the following equations:
Beaker A
CaCO3(s) + 2HCl(aq)
CaCl2(aq) + H2O(l) + CO2(g)
number of moles of HCl = 1 mol dm–3 x
Beaker B
CaCO3(s) + 2CH3COOH(aq)
100
dm3 = 0.1 mol
1 000
(CH3COO)2Ca(aq) + H2O(l) + CO2(g)
number of moles of CH3COOH = 1 mol dm–3 x
100
dm3 = 0.1 mol
1 000
Option A — During the reaction between CaCO3 and an acid, CaCO3 reacts with hydrogen ions in the
acid.
HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
1 mol dm –3 HCl(aq) has a higher concentration of hydrogen ions than 1 mol dm –3
CH3COOH(aq) does.
Hence the initial rate of reaction in Beaker A is higher than that in Beaker B.
9
Option B — As equal numbers of moles of HCl and CH3COOH are used, the same amount of marble
chips would be consumed in both cases.
Hence equal masses of marble chips remain in the two beakers after reaction.
Options C and D — As excess marble chips are used, HCl(aq) and CH3COOH(aq) are the limiting
reagents. Their amounts limit the amount of carbon dioxide gas produced.
0.1 mole of HCl and 0.1 mole of CH3COOH would produce the same amount of
gas.
37 D
Experiment
Number of moles of Zn
mass
=
molar mass
1
5 g
–1
65.4 g mol
2
= 0.076 mol
Number of moles of acid
= molarity of solution x
volume of solution
number of moles of HCl
100
dm3
= 1 mol dm–3 x
1 000
= 0.1 mol
number of moles of H2SO4
100
= 1 mol dm–3 x
dm3
1 000
= 0.1 mol
Reaction between Zn and acid
Zn(s) +
0.076 mol
2HCl(aq)
0.1 mol
ZnCl2(aq) + H2(g)
Zn(s) +
0.076 mol
H2SO4(aq)
0.1 mol
ZnSO4(aq) + H2(g)
Options A and B — According to the first equation, 1 mole of Zn reacts with 2 moles of HCl to
produce 1 mole of H2.
In Experiment 1, 0.1 mole of HCl reacts with 0.05 mole of Zn to produce 0.05
mole of H2.
According to the second equation, 1 mole of Zn reacts with 1 mole of H2SO4.
In Experiment 2, 0.076 mole of Zn reacts with 0.076 mole of H2SO4 to produce 0.076
mole of H2.
Thus the volume of gas produced in Experiment 1 is smaller than that produced in
Experiment 2, i.e. Options A and B are INCORRECT.
Option D — During the reaction between Zn and an acid, Zn reacts with hydrogen ions in the acid.
HCl(aq) is a monobasic acid while H2SO4(aq) is a dibasic acid.
1 mol dm –3 HCl(aq) has a lower concentration of hydrogen ions than 1 mol dm –3
H2SO4(aq) does.
Hence the initial rate of reaction in Experiment 1 is lower than that in Experiment 2.
Thus the curves in Option D show the results of the experiments.
38 A The concentration of the brown colour iodine in the reaction mixture decreased as the reaction
proceeded. The reaction mixture became lighter in colour gradually. Thus the reaction mixture absorbed
less and less light and so the absorbance went down.
Options B and C — The concentration of I2 in Sample 1 was the same as that in Sample 2.
Hence the initial absorbance of Sample 1 was the same as that of Sample 2, i.e.
Options B and C are INCORRECT.
10
Option D — The concentration of propanone in Sample 1 was lower than that in Sample 2.
The rate of reaction for Sample 1 was lower and thus the tangent to its curve would be
less steep.
Thus the curves in Option D show the results.
39 C The same amount of hydrogen is produced in each case as the same mass of zinc granules is used.
Different temperatures affect only the reaction rates.
40 D
41 C
42 B
43 B
44 A
45 D
46 B
47 B
48 B
49 D Option D — Manganese(IV) oxide can act as a catalyst in the decomposition of hydrogen peroxide.
50 A
51 B
52 D
53 C
54 B
Reaction
Number of moles of Mg
mass
=
molar mass
2 mol dm–3 x
1
1 g
24.3 g mol–1
2
Number of moles of HCl
= molarity of solution x
volume of solution
= 0.04 mol
50
dm3
1 000
Mg(s) + 2HCl(aq)
0.04 mol
0.1 mol
MgCl2(aq) + H2(g)
50
dm3
1 000
Mg(s) + 2HCl(aq)
0.04 mol
0.2 mol
MgCl2(aq) + H2(g)
= 0.1 mol
4 mol dm–3 x
= 0.2 mol
Reaction between Mg and acid
According to the equation, 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2.
In Reaction 1, 0.04 mole of Mg reacted with 0.08 mole of HCl to give 0.04 mole of H2.
11
Thus HCl was in excess.
In Reaction 2, 0.04 mole of Mg reacted wtih 0.08 mole of HCl to give 0.04 mole of H2.
Thus HCl(aq) was in excess.
(1) Magnesium was the limiting reactant in both reactions. Thus magnesium disappeared in both cases
after reaction.
(2) The same amount of hydrogen (0.04 mole) was produced in both reactions.
(3) The initial rate of Reaction 2 was greater as a more concentrated hydrochloric acid was used.
55 A
56 A
57 B
58 D
59 A The rate of a reaction decreases with time. At the start, there are plenty of reactant particles per unit
volume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.
the concentration of reactants fall. So, the reaction slows down.
60 D When dilute sulphuric acid reacts with calcium carbonate, insoluble calcium sulphate forms. The calcium
sulphate covers the surface of calcium carbonate and prevents further reaction.
Nitric acid is a monobasic acid while sulphuric acid is a dibasic acid.
1 mol dm–3 nitric acid has a lower concentration of hydrogen ions than 1 mol dm–3 sulphuric acid
does.
61 A
62 D In most cases, the rate of a reaction increases when the temperature is increased.
63 C A catalyst CANNOT change the amount of product formed in a reaction.
Unit 38 Gas volume calculations
Fill in the blanks
12
1
volumes; numbers
2
molar
3
molar volume
True or false
4
F
5
T
6
F
7
F
8
T
9
F
At room temperature and pressure, water is a liquid while carbon dioxide is a gas. Therefore one mole
of carbon dioxide occupies a much greater volume than one mole of water.
At room temperature and pressure, one mole of any gas occupies 24 dm3. However, the mass of one
mole of any gas is not the same.
10 T
Multiple choice questions
Volume of Cl2 (at room temperature and pressure)
Molar volume of the gas (at room temerature and pressure)
960 cm3
=
3
–1
24 000 cm mol
= 0.0400 mol
11 B Number of moles of chlorine =
Volume of SO2 (at room temperature and pressure)
Molar volume of the gas (at room temerature and pressure)
3
19.2 dm
=
3
–1
24.0 dm mol
= 0.800 mol
12 D Number of moles of sulphur dioxide =
Number of sulphur dioxide molecules = Number of moles of SO2 x L
= 0.800 mol x 6.02 x 1023 mol–1
= 4.82 x 1023
Volume of NO2 (at room temperature and pressure)
Molar volume of the gas (at room temerature and pressure)
3
14.4 dm
=
24.0 dm3 mol–1
= 0.600 mol
13 A Number of moles of NO2 present =
Molar mass of NO2 = (14.0 + 2 x 16.0) g mol–1
= 46.0 g mol–1
Mass of NO2 = Number of moles of NO2 x Molar mass of NO2
= 0.600 mol x 46.0 g mol–1
= 27.6 g
13
14 A
15 B
Gas
Number of moles of gas present
H2
2.0 g
= 1.0 mol
2.0 g mol–1
O2
16.0 g
= 0.500 mol
32.0 g mol–1
NH3
28.0 dm3
= 1.17 mol
24.0 dm3 mol–1
SO2
0.80 mol
16.0 g of oxygen contain the smallest number of moles of molecules, i.e. the smallest number of
molecules.
16 D
Gas
Ar
N2
HCl
Number of moles of atoms
in one mole of gas
Number of moles of
atoms present
= 1.20 mol
1
1.20 mol
48.0 dm3
= 2.00 mol
24.0 dm3 mol–1
2
2 x 2.00 mol = 4.00 mol
= 1.32 mol
2
2 x 1.32 mol = 2.64 mol
48.0 dm
= 2.00 mol
24.0 dm3 mol–1
3
3 x 2.00 mol = 6.00 mol
Number of moles of gas present
48.0 g
40.0 g mol–1
48.0 g
36.5 g mol–1
3
CO2
48.0 dm3 of carbon dioxide contain the greatest number of moles of atoms, i.e. the greatest number
of atoms.
17 B Number of moles of hydrogen / sulphur dioxide
Volume of gas (at room temperature and pressure)
=
Molar volume of the gas (at room temerature and pressure)
3
1 dm
=
24.0 dm3 mol–1
1
=
mol
24
Number of hydrogen atoms = 2 x Number of moles of H2 molecules x Avogadro constant
1
= 2 x
x L
24
= x
Number of atoms in 1 dm3 of sulphur dioxide
= 3 x Number of moles of SO2 molecules x Avogadro constant
1
= 3 x
x L
24
3
=
x
2
14
18 C Molar mass of CO2 = (12.0 + 2 x 16.0) g mol–1
= 44.0 g mol
Molar volume of CO2 (under the experimental condition)
Molar mass of CO2
=
Density of CO2 (under the experimental condition)
44.0 g mol–1
=
1.76 g dm–3
= 25.0 dm3 mol–1
19 D Option A — One mole of oxygen gas (O2) has a mass of 32.0 g.
Option B — One mole of oxygen gas contains 6.02 x 1023 molecules, i.e. 2 x 6.02 x 1023 atoms.
Option C — Water is a liquid at room temperature and pressure. It occupies a much smaller volume
than a gas.
Option D — Molar mass of helium is 4.0 g mol–1. One mole of oxygen gas and one mole of helium
gas occupy the same volume at room temperature and pressure.
20 A
Option
Gas
Molar mass of gas
Mass
of gas
present
Number of moles of
gas present
Number of moles of
gas present
Nitrogen
—
—
—
9.60 dm
11.2 g
11.2 g
= 0.400 mol
–1
28.0 g mol
0.400 mol x 24.0 dm3 mol–1
3
= 9.60 dm
12.0 dm
A
–1
Carbon (12.0 + 16.0) g mol
–1
monoxide = 28.0 g mol
3
3
Chlorine
—
—
—
Sulphur
dioxide
—
—
2.71 x 10
= 0.450 mol
6.02 x 1023 mol–1
0.450 mol x 24.0 dm3 mol–1
3
= 10.8 dm
Chlorine
—
—
0.300 mol
3
–1
0.300 mol x 24.0 dm mol
3
= 7.20 dm
Sulphur
dioxide
—
—
—
8.40 dm
22.4 g
22.4 g
= 0.800 mol
28.0 g mol–1
0.800 mol x 24.0 dm3 mol–1
= 19.2 dm3
—
1 mol
24.0 dm
B
23
C
–1
Nitrogen
2 x 14.0 g mol
–1
= 28.0 g mol
D
Carbon
monoxide
21 D
—
?X2
+
?Y2
3
200 cm
600 cm3
1 molecule 3 molecules
i.e. X2 + 3Y2
3
3
?Z
400 cm3
2 molecules
2Z
∴ The molecular formula of Z is XY3.
15
22 A 2Ca(NO3)2(s)
0.0500 mol
2CaO(s) + 4NO2(g) + O2(g)
? cm3
? cm3
According to the equation, 2 moles of Ca(NO3)2 produce 4 moles of NO2 and 1 mole of O2 on
decomposition.
5 x 0.0500
∴ Number of moles of gas obtained =
mol
2
= 0.125 mol
Volume of gas obtained (at room temperature and pressure)
= Number of moles of gas x Molar volume of the gas (at room temperature and pressure)
= 0.125 mol x 24.0 dm3 mol–1
= 3.00 dm3
23 B 2LiOH(s) + CO2(g)
7.17 g
? dm3
Li2CO3(s) + H2O(l)
Molar mass of LiOH = (6.9 + 1.0 + 16.0) g mol–1
= 24.0 g mol–1
Mass
Molar mass
7.17 g
=
–1
24.0 g mol
= 0.299 mol
Number of moles of LiOH =
According to the equation, 2 moles of LiOH can absorb 1 mole of CO2.
0.299
mol
2
= 0.150 mol
∴ Number of moles of CO2 absorbed =
Volume of CO2 absorbed (at room temperature and pressure)
= Number of moles of CO2 x Molar volume of the gas (at room temperature and pressure)
= 0.150 mol x 24.0 dm3 mol–1
= 3.60 dm3
24 D Fe2O3(s) + 3CO(g)
? dm3
2Fe(s) + 3CO2(g)
672 g
Molar mass of Fe = 55.8 g mol–1
Mass
Molar mass
672 g
=
–1
55.8 g mol
= 12.0 mol
Number of moles of Fe =
According to the equation, 3 moles of CO are required to produce 2 moles of Fe.
3
x 12.0 mol
2
= 18.0 mol
∴ Number of moles of CO required =
16
Volume of CO required (at room temperature and pressure)
= Number of moles of CO x Molar volume of the gas (at room temperature and pressure)
= 18.0 mol x 24.0 dm3 mol–1
= 432 dm3
25 A
26 B
27 A 8NH3(g) + 3Cl2(g)
28.8 dm3 ? dm3
6NH4Cl(s) + N2(g)
Volume of NH3 (at room temperature and pressure)
Molar volume of the gas (at room temerature and pressure)
3
28.8 dm
=
24.0 dm3 mol–1
= 1.20 mol
Number of moles of NH3 =
According to the equation, 8 moles of NH3 require 3 moles of Cl2 for complete reaction.
3
x 1.20 mol
8
= 0.450 mol
∴ Number of moles of Cl2 required =
Volume of Cl2 required (at room temperature and pressure)
= Number of moles of Cl2 x Molar volume of the gas (at room temperature and pressure)
= 0.450 mol x 24.0 dm3 mol–1
= 10.8 dm3
28 D
29 B
30 B
31 D
32 C
33 B
34 B
35 C
36 C
37 D
38 B (1) One mole of chlorine gas has a mass of 71.0 g.
(2) At room temperature and pressure, one mole of any gas occupies a volume of 24.0 dm3.
(3) One mole of chlorine gas contains 6.02 x 1023 chlorine molecules, i.e. 2 x 6.02 x 1023 chlorine
atoms.
39 A (1) Avogadro’s Law states that equal volumes of gases at the same temperature and pressure contain
equal numbers of particles. The particles are molecules in this case.
17
40 A The molar mass of CO2 is 44.0 g mol–1. 88.0 g of CO2 contain 2 moles of CO2 molecules.
(1) There are 2 moles of carbon atoms in 88.0 g of CO2.
3
(3) 88.0 g of CO2 have a volume of 48.0 dm , at room temperature and pressure.
41 A
42 B (1) At room temperature and pressure, oxygen is a gas while sulphur is a solid. The mass of 20 cm3 of
sulphur is much greater than that of 10 cm3 of oxygen.
(2) Suppose the mass of one mole of oxygen atoms is x g, then the mass of one mole of sulphur
atoms is 2x g.
2 g
Number of moles of atoms in 2 g of sulphur =
–1
2x g mol
1
=
mol
x
1 g
Number of moles of atoms in 1 g of oxygen =
x g mol–1
1
=
mol
x
∴ 2 g of sulphur and 1 g of oxygen contain the same number of moles of atoms, i.e. the same
number of atoms.
(3) At room temperature and pressure, oxygen is a gas while sulphur is a solid. Therefore one mole of
oxygen atoms occupies a much greater volume than one mole of sulphur atoms.
43 C One mole of sulphur dioxide and one mole of oxygen contain the same number of molecules, not
atoms.
44 D
45 C Avogadro’s Law states that equal volumes of gases at the same temperature and pressure contain equal
numbers of particles. Under the same conditions, 10 cm3 of chlorine and 10 cm3 of hydrogen contain
equal numbers of molecules.
46 B At room temperature and pressure, oxygen is a gas while carbon is a solid. Therefore oxygen occupies
a much greater volume than carbon.
47 C 2 moles of hydrogen react with 1 mole of oxygen to give 2 moles of water. At room temperature and
pressure, 0.50 mole of hydrogen (i.e. 12 dm3) reacts with 0.25 mole of oxygen (i.e. 6 dm3) to give
0.50 mole of water. Water is a liquid at room temperature and pressure. Therefore 0.50 mole of water
occupies a volume much less than 6 dm3.
48 C
49 D
50 C
18
Part B
Topic-based exercise
Multiple choice questions
1
C
2
D Average reaction rate =
3
D
4
C Rate = –
1
4
d[NH3(g)]
dt
= –
1
5
d[O2(g)]
dt
2.0 g
3.0 min
–1
= 0.67 g min
d[NH3(g)]
dt
d[O
4
2(g)]
=
dt
5
4
=
(1.80 mol dm–3 s–1)
5
= 1.44 mol dm–3 s–1
Rate of consumption of NH3(g) = –
5
A Rate = –
=
1
6
1
5
d[O2(g)]
dt
d[H2O(g)]
dt
d[H2O(g)]
dt
d[O2(g)]
6
= –
dt
5
6
–3 –1
=
(1.80 mol dm s )
5
= 2.16 mol dm–3 s–1
Rate of formation of H2O(g) =
6
C
7
B
8
A
9
D Option D — The intensity of the orange colour of Cr2O72– ions increases as the reaction proceeds.
19
10 C Consider a reaction in which a gas is produced. If the reaction vessel is a closed system, the pressure
inside the vessel will increase. We can follow the progress of the reaction by measuring the pressure
inside the vessel with a pressure sensor connected to a data-logger interface and a computer.
Option C — NO gas is produced in the reaction between copper(II) oxide and dilute nitric acid. Hence
the progress of the reaction CANNOT be followed by using a data-logger with a pressure
sensor.
11 A
12 A
13 C
14 D
15 C
16 B
17 C
18 B
19 B
20 D
21 B The reaction represented by curve Y took more time to complete, i.e. the rate of this reaction was
lower.
The changes is pressure were the same in both experiments. It could be deduced that equal volumes
of gas were produced.
Option B — Calcium carbonate powder had a greater surface area than calcium carbonate lumps of
the same mass.
The rate of the reaction decreases when the surface area of calcium carbonate is
decreased, i.e. when using calcium carbonate lumps instead of calcium carbonate powder.
Option D — Calcium carbonate was the limiting reactant as it disappeared in both experiments after
reaction.
Using a lower mass of calcium carbonate would decrease the volume of gas formed, i.e.
decrease the pressure. This Option D is INCORRECT.
22 D The tangent to curve II was less steep than that to curve I, i.e. the rate of the reaction represented by
curve II was lower.
However, a greater volume of hydrogen was formed in this reaction.
Option A — Increasing the temperature would increase the rate of the reaction.
20
Option B — Using magnesium powder would increase the rate of the reaction.
Option C — Using a more concentrated sulphuric acid would increase the rate of the reaction.
Option D — Using a less concentrated sulphuric acid would decrease the rate of the reaction.
However, increasing the number of moles of H2SO4 (from 0.1 mole to 0.14 mole) would
increase the volume of hydrogen formed.
23 A The concentration of MnO4– ions in the reaction mixture decreased as the reaction proceeded. The
reaction mixture became lighter in colour gradually. Thus the reaction mixture absorbed less and less
light and so the absorbance went down.
Option A — The concentration of MnO4–(aq) ion in reaction mixture 1 was higher than that in reaction
mixture 2.
So, the rate of reaction for reaction mixture 1 was also higher and thus the tangent to its
curve would be steeper.
Thus the curves in Option A show the results.
Options B, C and D — The concentration of MnO4– ions in reaction mixture 1 was higher than that in
reaction mixture 2.
Hence the initial absorbance of reaction mixture 1 was higher than that of
reaction mixture 2.
∴ Options B, C and D are INCORRECT.
24 B Options A and C — At the start, the tangent to the curve for Experiment 1 was less steep than that
for Experiment 2, i.e. the initial rate of reaction of Experiment 1 was lower than
that of Experiment 2.
Thus Options A and C are INCORRECT.
Option B — The curve for Experiment 2 went flat earlier than that for Experiment 1, i.e. the rate of
reaction for Experiment 2 fell to zero earlier.
Hence the graph shown in Option B is correct.
25 A
26 C
27 C
28 C Molar mass of SO2 = 64.0 g mol–1
Molar volume of SO2 (under the experimental condition)
Molar mass of SO2
=
Density of SO2 (under the experimental condition)
64.0 g mol–1
=
2.46 g dm–3
= 26.0 dm3 mol–1
29 B
30 D
21
31 A
32 D
33 D
34 D CaC2(s) + H2O(l)
? g
CaO(s) + C2H2(g)
1 440 cm3
Volume of C2H2 (at room temperature and pressure)
Molar volume of the gas (at room temerature and pressure)
1 440 cm3
=
24 000 cm3 mol–1
= 0.0600 mol
Number of moles of C2H2 =
According to the equation, 1 mole of CaC2 produces 1 mole of C2H2.
∴ Number of moles of CaC2 reacted = 0.0600 mol
Molar mass of CaC2 = 64.1 g mol–1
Mass of CaC2 reacted = Number of moles of CaC2 x Molar mass of CaC2
= 0.0600 mol x 64.1 g mol–1
= 3.85 g
35 C
36 D Fe + 2HCl
or Fe + 2H+
FeCl2 + H2
Fe2+ + H2
Moles or iron = 4.25 x 10–3 mol
Mass of iron = 4.25 x 10–3 x 55.8
= 0.237 g
0.237 x 100 / 0.263 = 90.1%
37 B
38 B
2H2(g) + O2(g)
1 200 cm3 1 200 cm3
2H2O(l)
? g
Volume of H2 / O2 (at room temperature and pressure)
Molar volume of the gas (at room temerature and pressure)
1 200 cm3
=
24 000 cm3 mol–1
= 0.0500 mol
Number of moles of H2 / O2 =
According to the equation, 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O. O2 is in
excess in this case. The amount of H2 limits the amount of H2O produced.
Number of moles of H2O produced = 0.0500 mol
Molar mass of H2O = 18.0 g mol–1
22
Mass of H2O produced = Number of moles of H2O x Molar mass of H2O
= 0.0500 mol x 18.0 g mol
= 0.900 g
39 A
40 D
41 B
42 A
43 C
44 A
45 C
46 B (1) A catalyst would NOT change the amount of product formed in a reaction.
(3) The physical state of a catalyst may NOT be the same as those of the reactants.
For example, the solid magnese(IV) oxide acts as a catalyst in the decomposition of hydrogen
peroxide solution.
47 B
48 B
49 B
50 B
Beaker
Number of moles of Mg
mass
=
molar mass
Number of moles of acid
= molarity of solution x
volume of solution
Reaction between Mg and acid
Mg(s) +
0.074 mol
Number of moles of HCl
100
= 1 mol dm–3 x
dm3
1 000
= 0.1 mol
1
1.8 g
–1
24.3 g mol
2
2HCl(aq)
0.1 mol
MgCl2(aq) + H2(g)
According to the equation, 1 mole
of Mg reacts with 2 moles of HCl
to produce 1 mole of H2.
Hence 0.1 mole of HCl reacted
with 0.05 mole of Mg to give 0.05
mole of H2.
Thus Mg was in excess.
= 0.074 mol
Mg(s) +
0.074 mol
Number of moles of H2SO4
100
= 1 mol dm–3 x
dm3
1 000
= 0.1 mol
H2SO4(aq)
0.1 mol
MgSO4(aq) + H2(g)
According to the equation, 1 mole
of Mg reacts with 1 mole of H2SO4
to produce 1 mole of H2.
Hence 0.074 mole of Mg reacted
with 0.074 mole of H2SO4 to give
0.074 mole of H2.
Thus H2SO4(aq) was in excess.
23
(1) The hydrogen produced would escape. Thus the mass of each reaction mixture would decrease.
Different amounts of hydrogen gas were produced in the two cases.
(2) HCl(aq) is a monobasic acid while H2SO4(aq) is a dibasic acid.
Therefore the concentration of hdyrogen ions in 1 mol dm–3 HCl(aq) was lower than that in 1 mol
dm–3 H2SO4(aq).
Hence the initial rate of Reaction 1 would be smaller than that in Reaction 2.
(3) H2SO4(aq) was in excess in Reaction 2.
51 A In the oxidation of oxalate ions by permanganate ions, the intensity of the purple colour of
permanganate ions decreases as the reaction proceeds.
When we shine light upon the reaction mixture, the absorbance of the reaction mixture is directly
proportional to the colour intensity of the reaction mixture and the concentration of the permanganate
ions in the reaction mixture.
Hence the progress of the reaction can be followed by a colorimeter.
52 B
53 A
54 C Hydrochloric acid is a monobasic acid while sulphuric acid is a dibasic acid.
Therefore 1 mol dm–3 hydrochloric acid has a lower concentration of hydrogen ions than 1 mol dm–3
sulphuric acid does.
Hence the reaction between 100 cm3 of 1 mol dm–3 hydrochloric acid and zinc granules is slower than
that between 100 cm3 of 1 mol dm–3 sulphuric acid and zinc granules.
55 D
56 D A catalyst would NOT undergo any permanent chemical changes at the end of a reaction.
57 B At room temperature and pressure, carbon dioxide is a gas while water is a liquid. Therefore one mole
of carbon dioxide occupies a much greater volume than one mole of water.
58 C
24
Short questions
59 a) Determine the concentration of hydroxide ions remaining in the reaction mixture by titration with standard
acid
(1)
because the concentration of hydroxide ions in the reaction mixture decreases as the reaction proceeds.
(1)
b) Any one of the following:
• By using a colorimeter
(1)
because the intensity of the yellow-brown colour of bromine decreases as the reaction proceeds. (1)
• By measuring the amount of carbon dioxide gas collected over water saturated with carbon dioxide
(1)
because carbon dioxide is formed as the reaction proceeds.
(1)
c) By measuring the pressure of the reaction mixture in a closed vessel
(1)
because 4 moles of gas react to give 3 moles of gas and the pressure decreases as the reaction
proceeds.
(1)
d) By measuring the time to reach an opaque stage
(1)
because the light transmittance of the reaction mixture changes as the yellow precipitate sulphur is
formed.
(1)
60 • Increase in temperature
(1)
• Increase in the surface area of carbon
(1)
• Increase in the concentration of oxygen
(1)
• Use of a catalyst
(1)
61 A reaction does not occur if the reactant particles collide with each other in a wrong orientation.
(1)
A reaction does not occur if the colliding reactant particles have insufficient energy (less than the activation
energy).
(1)
62 a) The number of moles per unit volume increases.
The chance of collision increases and the number of effective collisions increases.
b) The molecules have more energy and collide more often.
A larger portion of the molecules have energy equal to or greater than the activation energy.
(1)
(1)
(1)
(1)
c) A catalyst is a substance which alters the rate of a reaction
(1)
without itself undergoing any permanent chemical changes.
(1)
25
Structured questions
63 a)
HBTTZSJOHF
EJMVUFIZESPDIMPSJDBDJE
NBHOFTJVNDBSCPOBUFMVNQT
(1 mark for correct set-up; 1 mark for correct labels; 0 mark if the set-up is not workable)
(2)
b) Volume of acid
(1)
Concentration of acid
(1)
Temperature
(1)
c) The hydrochloric acid was consumed as the reaction proceeded.
(1)
Hence the concentration of the acid in the reaction mixture dropped during the course of the reaction.
(1)
d) i) D
(0.5)
ii) A
(0.5)
e)
Rate of reaction
Acid used
3
lower
–3
100 cm of 2 mol dm
3
–3
200 cm of 1 mol dm
3
HCl(aq)
✔
✔
(0.5)
HCl(aq)
–3
200 cm of 0.5 mol dm
26
higher
HCl(aq)
–3
100 cm of 0.5 mol dm
3
same
HCl(aq)
✔
✔
(0.5)
(0.5)
(0.5)
64 a)
.BTTPGCFBLFSBOEDPOUFOUTH
m
H
m
NJO
5JNFNJO
(2 marks for correctly plotted points; 1 mark for correctly drawn curve of best fit)
(3)
b) Loss in mass of contents of beaker for the time period 2.0 min to 6.0 min
= (193.9 – 187.3) g
= 6.60 g
= mass of NO(g) formed
mass of NO(g) formed
time
6.60 g
=
(6.0 – 2.0) min
= 1.65 g min–1
Average rate of formation of NO(g) =
(1)
(1)
c) Mass of NO(g) formed in the first 3 minutes = (200.0 – 191.9) g
= 8.10 g
8.10 g
30.0 g mol–1
= 0.270 mol
Number of moles of NO(g) formed =
(1)
According to the equation, 3 moles of Cu reacts with HNO3(aq) to give 2 moles of NO(g).
3
x 0.270 mol
2
= 0.405 mol
∴ number of moles of Cu consumed =
(1)
27
Mass of Cu consumed = 0.405 mol x 63.5 g mol–1
= 25.7 g
(1)
(185.0 – 191.6) g
(7.5 – 3.0) min
= 1.47 g min–1
d) Instantaneous rate of formation of NO(g) = –
(1)
(1)
e) i) Increase
(1)
There is an increase in the total surface area for the contact of reactants.
ii) Decrease
(1)
(1)
0.3 mol dm–3 nitric acid has a smaller concentration of NO3–(aq) ions than 0.5 mol dm–3 nitric acid.
(1)
iii) Increase
(1)
Reactant particles have more energy and collide more often.
(1)
A larger portion of the particles have energy equal to or greater than the activation energy.
(1)
65 a) 2H2O2(aq)
2H2O(l) + O2(g)
(1)
b) The rates of decomposition of hydrogen peroxide solution was A > B > C.
At A, the concentration of hydrogen peroxide solution was high. So, the rate of decomposition was
high.
(1)
At B, the concentration of hydrogen peroxide solution was lower than that at A. So, the rate of
decomposition was lower.
(1)
c)
At C, the rate of decomposition was zero as all the hydrogen peroxide was used up.
(1)
(168.235 – 168.143) g
4.0 min
= 0.023 g min–1
(1)
(1)
d) i) (168.235 – 168.123) g = 0.112 g
(1)
0.112 g
–1
32.0 g mol
= 0.00350 mol
ii) Number of moles of O2 given off =
(1)
According to the equation, 2 moles of H2O2 decompose to give 1 mole of O2.
i.e. number of moles of H2O2 = 2 x 0.00350 mol
= 0.00700 mol
Original molarity of hydrogen peroxide solution =
(1)
0.00700 mol
50.0
dm3
1 000
(
)
= 0.140 mol dm
28
–3
(1)
e) i) As a catalyst. / To increase the rate of decomposition.
ii) The shape of the curve would be the same.
The rate of decomposition is independent of the amount of catalyst added.
(1)
(1)
(1)
f) Any one of the following:
• As the change in the mass is very small in this experiment, the use of a data-logger can give more
accurate results.
(1)
• The experimental results in the form of graph could be obtained immediately.
(1)
g) Any one of the following:
• Measure the volume of gas given off.
(1)
• Measure the pressure of the mixture in a closed reaction vessel.
(1)
(76 – 46) cm3
(40 – 20) s
= 1.5 cm3 s–1
66 a) Average rate of formation of H2 =
b) i) The rate of reaction decreases as the reaction proceeds.
(1)
(1)
(1)
ii) As the reaction proceeds, the reactants are consumed gradually, i.e. the concentration decreases. (1)
The chance of collision decreases and the number of effective collisions decreases.
c) i) The time taken for the reaction to complete would be longer.
(1)
(1)
Using zinc granules decreases the contact surface area between zinc and the acid.
(1)
There is a smaller chance for collision and the number of effective collisions decreases.
(1)
So, the rate of the reaction decreases.
ii) The total volume of hydrogen evolved is the same.
As the acid is in excess, the amount of zinc used limits the amount of gas evolved.
d) • Use hydrochloric acid of different concentrations.
(1)
(1)
(1)
• Use the same mass of zinc powder.
(1)
• Use the same volume of hydrochloric acid.
(1)
• Keep the temperature the same.
(1)
29
67 a) CaCO3(s) + 2HCl(aq)
CaCl2(aq) + H2O(l) + CO2(g)
(1)
b) Add a suitable amount of calcium carbonate to the acid until the acid is saturated with dissolved carbon
dioxide.
(1)
c)
7PMVNFPGHBTDN
DVSWF9
5JNFT
d) At the start, there are plenty of reactant particles per unit volume.
(1)
(1)
There is a great chance for collision and the number of effective collisions is also great. Hence the rate
of reaction is high.
(1)
As the reactant particles are consumed gradually, the concentration of reactants fall. So, the reaction
slows down.
(1)
The reaction stops when one of the reactant is used up.
1.00 g
100.1 g mol–1
= 0.00999 mol
(1)
e) Number of moles of CaCO3 =
Number of moles of HCl = 0.100 mol dm–3 x
(1)
100.0
3
dm
1 000
= 0.0100 mol
(1)
According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CO2. Thus
CaCO3 was in excess. The amount of HCl limited the amount of CO2 produced.
0.0100
mol
2
= 0.00500 mol
i.e. number of moles of CO2 produced =
Volume of CO2 produced = 0.00500 mol x 24.0 dm3 mol–1
= 0.120 dm3
= 120 cm3
(1)
(1)
f) i) Hydrochloric acid is a strong acid and it almost completely dissociates in water. A weak acid only
partially dissociates in water.
(1)
Therefore hydrochloric acid has a higher concentration of hydrogen ions than a weak monobasic acid
of the same concentration.
(1)
Hence the rate of reaction between calcium carbonate and hydrochloric acid would be higher than
that between calcium carbonate and the weak acid.
(1)
30
ii)
7PMVNFPGHBTDN
DVSWF9
5JNFT
(1 mark for showing the slower production of carbon dioxide, i.e. curve less steep; 1 mark for showing
the same final volume as that of curve X)
(2)
g) When sulphuric acid reacted with calcium carbonate, insoluble calcium sulphate formed.
The calcium sulphate covered the surface of calcium carbonate and prevented further reaction.
68 a) i) Colorimeter
(1)
(1)
(1)
ii) The intensity of the brown colour due to iodine decreases as the reaction proceeds.
(1)
The colorimeter measures the amount of light absorbed by the reaction mixture when a beam of light
passes through the reaction mixture, i.e. the absorbance.
(1)
The absorbance is directly proportional to the colour intensity of the reaction mixture and the
concentration of iodine in the reaction mixture.
(1)
"CTPSCBODF
b)
TBNQMF
TBNQMF
5JNF
(1 mark for correct labels of the axes; 1 mark for showing the same absorbance for both samples; 1
mark for showing the curve for sample 1 being less steep)
(3)
c) Add sodium carbonate / hydrogencarbonate
to neutralize the H+(aq) ions.
(1)
(1)
d) Any one of the following:
• Use a pH meter
to measure the concentration of H+(aq) ions.
(1)
(1)
• Measure the electrical conductivity
(1)
as ions are formed in the reaction.
(1)
31
69
$PODFOUSBUJPOPG.O0mBR
JPOTNPMENm
Ym
Ym
Ym
Ym
m
Y m
m
Y H
m
Y
m
T
5JNFT
a) The intensity of the purple colour of permanganate ions decreases as the reaction proceeds.
(1)
The colorimeter measures the amount of light absorbed by the reaction mixture when a beam of light
passes through the reaction mixture, i.e. the absorbance.
(1)
The absorbance is directly proportional to the colour intensity of the reaction mixture and the concentration
of permanganate ions in the reaction mixture.
(1)
b) Prepare a calibration curve by plotting the absorbance of permanganate ions of known concentration
against the concentration of permanganate ions.
(1)
From the calibration curve, read off the concentration of permanganate ions according to the absorbance
recorded.
(1)
c) The concentration of permanganate ions was the only factor which affected the reaction rate.
(1)
d) i) Instantaneous rate of consumption of MnO4–(aq) ions
(1.0 x 10–4 – 6.3 x 10–4) mol dm–3
(366 – 335) s
–5
= 1.7 x 10 mol dm–3 s–1
= –
(1)
(1)
ii) Instantaneous rate of consumption of C2O42–(aq) ions
5
(1.7 x 10–5 mol dm–3 s–1)
2
= 4.3 x 10–5 mol dm–3 s–1
=
32
(1)
e) Any one of the following:
• Measuring the loss of CO2(g) by weighing
(1)
• Measuring the volume of CO2(g) produced by gas collection
(1)
• Titrating the H+(aq) ions with a standard alkali
(1)
• Measuring the pH change of the reaction mixture
(1)
70 a) When sodium thiosulphate solution reacts with dilute sulphuric acid, a yellow precipitate of sulphur forms.
This changes the light transmittance of the reaction mixture.
(1)
To follow the progress of the reaction, mark a cross on a piece of paper.
(1)
Put the conical flask containing some sodium thiosulphate solution on top of the paper. Add dilute
sulphuric acid to the flask.
(1)
Record the time t for the solution mixture to become opaque, i.e. when the cross can no longer be seen
from above.
(1)
b) The average rate of reaction from the start to the opaque stage
∝
1
time to reach the opaque stage (t)
(1)
c) i)
5JNFUT
5FNQFSBUVSFž$
(1 mark for correctly drawn curve of best fit; 2 marks for correctly plotted points)
ii) 64 – 68 s
d) Change the concentration of the reactants.
(2)
(1)
(1)
33
71 a)
DPUUPOXPPM
EJMVUFIZESPDIMPSJD
BDJE
DBMDJVNDBSCPOBUF
XXX.XX
FMFDUSPOJD
CBMBODF
(1 mark for correct set-up; 2 marks for 4 correct labels; 0 mark if the set-up is not workable)
(3)
b) The reaction rate was higher at X than at Y. This was because the reaction got slower as it proceeded.
(1)
c) • The loss in mass of the contents of the reaction flask for sample A is greater than that for sample B,
i.e. sample A produces more carbon dioxide gas. Therefore sample A has a higher purity of calcium
carbonate than sample B.
(1)
• The initial rate for sample B is faster than that for sample A. It is because sample B has a smaller
particle size.
(1)
d) Mass of CO2 given off = 3.70 g
CaCO3(s) + 2HCl(aq)
? g
CaCl2(aq) + CO2(g) + H2O(l)
3.70 g
3.70 g
44.0 g mol–1
= 0.0841 mol
Number of moles of CO2 =
(1)
According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl to give 1 mole of CO2.
i.e. number of moles of CaCO3 in sample A = 0.0841 mol
(1)
Mass of CaCO3 in sample A = 0.0841 mol x 100.1 g mol–1
= 8.42 g
8.42 g
10.0 g
= 84.2%
∴ percentage by mass of CaCO3 in sample A =
x 100%
(1)
72 a) A gas was evolved in the reaction.
(1)
b)
TVDUJPOGMBTL
QSFTTVSF
TFOTPS
NBHOFTJVN
SJCCPO
EJMVUF
TVMQIVSJDBDJE
EBUBMPHHFS
JOUFSGBDF
DPNQVUFS
(1 mark for correct set-up; 3 marks for 6 correct labels; 0 mark if the set-up is not workable)
34
(4)
2.00 g
24.3 g mol–1
= 0.0823 mol
c) Number of moles of Mg present =
–3
Number of moles of H2SO4 present = 1.00 mol dm x
(1)
100.0
dm3
1 000
= 0.100 mol
Mg(s) + H2SO4(aq)
(1)
MgSO4(aq) + H2(g)
According to the equation, 1 mole of Mg reacts with 1 mole of H2SO4. During the reaction, 0.0823 mole
of Mg reacted with 0.0823 mol of H2SO4. Therefore H2SO4 was in excess.
(1)
d) As the reaction proceeded, sulphuric acid was consumed gradually.
At A, the concentration of sulphuric acid was high. So, the rate of reaction was high.
(1)
At B, the concentration of sulphuric acid was lower than that at A. So, the rate of reaction was lower.
(1)
At C, the rate of reaction was zero as all the magnesium was used up.
(1)
e) i) Sulphuric acid is a dibasic acid while hydrochloric acid is a monobasic acid. Therefore 1 mol dm–3
sulphuric acid has a higher concentration of hydrogen ions than 1 mol dm–3 hydrochloric acid.
(1)
Hence the initial rate of the reaction between magnesium and hydrochloric acid was lower than that
between magnesium and sulphuric acid.
(1)
ii)
1SFTTVSF
DVSWF9
DVSWF:
5JNF
(1 mark for showing a lower rate of reaction, i.e. tangent to curve less steep; 1 mark for showing a
5
of the original value)
(2)
smaller final pressure, around
8
73 a) From the curves, 1 mole of P(g) reacts with 3 moles of Q(g) to give 2 moles of R(g).
Equation: P(g) + 3Q(g)
2R(g)
b) The time required will be longer.
(1)
(1)
In a large container, the concentrations of reactants become smaller.
(1)
The chance of collision decreases and the number of effective collisions decreases.
(1)
35
c) Colliding molecules will undergo reaction if they possess an energy equal to or greater than the activation
energy, and
(1)
collide in the right orientation
74 a) 2H2O2(aq)
(1)
2H2O(l) + O2(g)
(1)
b) As a catalyst. / Increase the rate of decomposition.
(1)
c)
7PMVNFPGPYZHFODN
5JNFT
(2 marks for correctly plotted points; 1 mark for correctly drawn curve of best fit)
(3)
d) 60 – 62 cm3
(1)
e) All hydrogen peroxide has decomposed.
(1)
f)
7PMVNFPGPYZHFODN
5JNFT
(1 mark for showing the slower production of oxygen, i.e. tangent to curve less steep; 1 mark for showing
a smaller final volume, about half that of the original curve)
(2)
36
g) Add 1 g of manganese(IV) oxide to 50 cm3 of hydrogen peroxide solution. Wait until the reaction
stops.
(1)
Filter to remove manganese(IV) oxide from the solution.
(1)
Add ‘fresh’ manganese(IV) oxide to the filtrate. No oxygen is given off. This indicates that the oxygen
does not come from manganese(IV) oxide.
(1)
Add ‘fresh’ hydrogen peroxide solution to the residue (manganese(IV) oxide). Oxygen is given off. This
indicates that the oxygen comes from hydrogen peroxide.
(1)
h) i) An enzyme is a protein-based catalyst.
(1)
ii) The enzyme was denatured / destroyed.
(1)
75 a) Burette
(1)
b) Place each test tube containing the solution(s) in a thermostat for 5 – 10 minutes.
(1)
c) i) To keep the total volume of each sample constant. Thus, the concentration of sulphuric acid in the
sample is directly proportional to the volume of acid used.
(1)
ii) To ensure that the only variable is the change in the concentration of sulphuric acid in the sample.
(1)
d) i) The average rate of reaction from the start to decolorization
∝
1
time to decolorize
(1)
3FMBUJWFBWFSBHFSBUFPGSFBDUJPOTm
ii) (1)
Relative average
Volume of
rate of reaction
Sample sulphuric acid
1
3
( ) (s–1)
(cm )
t
1
1.5
0.0105
2
2.0
0.0139
3
2.5
0.0172
4
3.0
0.0208
(1)
7PMVNFPGTVMQIVSJDBDJEDN
(1 mark for correctly plotted points; 1 mark for a straight line drawn through the points)
(2)
37
(2)
3FBDUJPOBWFSBHFSBUFPGSFBDUJPOTm
Sample
Temperature
(°C)
Relative average
rate of reaction
(
1
–1
) (s )
t
2
30
0.0139
5
40
0.0286
6
50
0.0714
(1)
5FNQFSBUVSFž$
(1 mark for correctly plotted points; 1 mark for reasonably smooth curve drawn)
(2)
iii) The rate of the reaction is directly proportional to the concentration of sulphuric acid. / The rate of
the reaction increases with the concentration of sulphuric acid.
(1)
The rate of the reaction approximately doubles for each 10 °C rise in temperature.
(1)
76 a) Any two of the following:
• Measure the loss in mass of the reaction mixture
(1)
because a gas is given off.
(1)
• Measure the electrical conductivity of the reaction mixture
(1)
because there is a decrease in the number of ions.
(1)
• Measure the pH of the reaction mixture
(1)
because hydrochloric acid is consumed.
b) Initially some carbon dioxide dissolves in the solution (until the solution is saturated).
(1)
c) i) 53.0 cm3
(1)
ii)
38
(1)
Time (s)
Volume of CO2 (Vt) (cm3)
(Vfinal – Vt) (cm3)
5
13.0
40.0
10
22.0
31.0
20
35.0
18.0
30
45.0
8.0
(0.25)
40
48.5
4.5
(0.25)
50
51.0
2.0
(0.25)
60
52.0
1.0
(0.25)
iii) The concentration of hydrochloric acid
(1)
iv)
7GJOBMm7U
DN
5JNFT
(1 mark for correctly plotted points; 1 mark for reasonably smooth curve drawn)
53.0 cm3
24 000 cm3 mol–1
= 0.00221 mol
(2)
v) Number of moles of CO2 =
(1)
According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of
CO2.
i.e. number of moles of HCl = 2 x 0.00221 mol
= 0.00442 mol
Molarity of hydrochloric acid =
(1)
0.00442 mol
50.0
dm3
1 000
(
)
= 0.0884 mol dm–3
(1)
d) The rate increases
(1)
because powdered marble has a greater surface area.
(1)
77 a)
TVDUJPOGMBTL
QSFTTVSF
TFOTPS
EBUBMPHHFS
JOUFSGBDF
MJWFS
IZESPHFO
QFSPYJEFTPMVUJPO
DPNQVUFS
(1 mark for correct set-up; 3 marks for 6 correct labels; 0 mark if the set-up is not workable)
(4)
39
b) 2H2O2(aq)
2H2O(l) + O2(g)
(1)
c) A gas (oxygen) was evolved in the reaction.
(1)
1SFTTVSF
d)
5JNF
(1 mark for showing the faster increase in pressure, i.e. tangent to curve steeper; 1 mark for showing a
greater final pressure, about twice that of the original one)
(2)
174 cm3
24 000 cm3 mol–1
= 0.00725 mol
e) Number of moles of O2 =
(1)
According to the equation, 2 moles of H2O2 decompose to produce 1 mole of O2.
i.e. number of moles of H2O2 decomposed = 2 x 0.00725 mol
= 0.0145 mol
(1)
Mass of H2O2 = 0.0145 mol x 34.0 g mol–1
= 0.493 g
0.493 g
x 100%
6.95 g
= 7.09%
(1)
0.330 mol dm–3
40
= 8.25 x 10–3 mol dm–3 min–1
(1)
Percentage by mass of H2O2 in sample =
78 a) Average rate of formation of X(g) =
b) i) Final concentration of X(g) = 0.440 mol dm–3
0.440
mol dm–3
2
= 0.220 mol dm–3
Initial concentration of X2(g) =
Number of moles of X2(g) = 0.220 mol dm–3 x 1.00 dm3
= 0.220 mol
40
(1)
ii)
$PODFOUSBUJPONPMENm
9H
9H
5JNFNJO
(2 marks for correctly plotted points; 1 mark for reasonably smooth curve drawn)
c) The molecules has more energy at a higher temperature.
A larger portion of the molecules have energy equal to or greater than the activation energy.
(3)
(1)
(1)
3 000 cm3
24 000 cm3 mol–1
= 0.125 mol
(1)
4 800 cm3
24 000 cm3 mol–1
= 0.200 mol
(1)
d) Number of moles of SO2 =
Number of moles of Cl2 =
According to the equation, 1 mole of SO2 reacts with 1 mole of Cl2 to produce 1 mole of SO2Cl2.
Therefore, Cl2 was in excess. The amount of SO2 limited the amount of SO2Cl2 formed.
Number of moles of SO2Cl2 formed = 0.125 mol
(1)
Mass of SO2Cl2 formed = 0.125 mol x 135.1 g mol–1
= 16.9 g
(1)
79 a) Any two of the following:
• A blue soution was formed.
(1)
• Malachite dissolved.
(1)
• Effervescence occurred.
(1)
b) CuCO3(s) + H2SO4(aq)
CuSO4(aq) + H2O(l) + CO2(g)
(1)
41
3
60.0 cm
3
–1
24 000 cm mol
= 0.00250 mol
c) i) Number of moles of CO2 collected =
(1)
ii) According to the equation, 1 mole of CuCO3 reacts with 1 mole of H2SO4 to produce 1 mole of
CO2.
i.e. number of moles of CuCO3 = 0.00250 mol
(1)
Mass of CuCO3 in the sample = 0.00250 mol x 123.5 g mol–1
= 0.309 g
0.309 g
x 100%
0.440 g
= 70.2%
% by mass of CuCO3 in sample =
d) Some carbon dioxide dissolved in the acid.
(1)
(1)
3
3.60 dm
24.0 dm3 mol–1
= 0.150 mol
e) Number of moles of CO2 =
Number of moles of oxygen in the oxide = 2 x 0.150 mol
= 0.300 mol
11.2 g
55.8 g mol–1
= 0.201 mol
(1)
(1)
Number of moles of M in the oxide =
∴ the chemical formula of the oxide is M2O3.
1 440 cm3
24 000 cm3 mol–1
= 0.0600 mol
= number of moles of carbon
(1)
(1)
80 a) i) Number of moles of CO2 =
Mass of carbon in 1.32 g of ester = 0.0600 mol x 12.0 g mol–1
= 0.720 g
(1)
(1)
1.08 g
18.0 g mol–1
= 0.0600 mol
ii) Number of moles of H2O =
Number of moles of hydrogen = 2 x 0.0600 mol
= 0.120 mol
Mass of hydrogen in 1.32 g of ester = 0.120 mol x 1.0 g mol–1
= 0.120 g
iii) Mass of oxygen in 1.32 g of ester = (1.32 – 0.720 – 0.120) g
= 0.480 g
42
(1)
(1)
iv) In 1.32 g of the ester, there are 0.720 g of carbon, 0.120 g of hydrogen and 0.480 g of oxygen.
Carbon
Hydrogen
Oxygen
Mass of element
in the compound
0.720 g
0.120 g
0.480 g
Number of moles
of atoms that
combine
0.720 g
= 0.0600 mol
–1
12.0 g mol
Simplest ratio of
atoms
0.0600 mol
= 2.00
0.0300 mol
0.120 g
0.480 g
= 0.120 mol
= 0.0300 mol (1)
–1
–1
1.0 g mol
16.0 g mol
0.120 mol
= 4.00
0.0300 mol
0.0300 mol
= 1.00
0.0300 mol
(1)
∴ the empirical formula of the ester is C2H4O.
8.00 g
32.0 g mol–1
= 0.250 mol
= number of moles of ester
b) Number of moles of oxygen gas =
22.0 g
0.250 mol
= 88.0 g mol–1
(1)
Molar mass of ester =
(1)
c) Let (C2H4O)n be the molecular formula of the ester.
Relative molecular mass of ester = n(2 x 12.0 + 4 x 1.0 + 16.0)
= 44n
i.e. 44n = 88.0
n= 2
∴ the molecular formula of the ester is C4H8O2.
(1)
d) Any one of the following:
O
CH3
O
CH2
CH3
(1)
CH2
CH2
CH3
(1)
O
CH3
(1)
C
O
H
C
O
CH3
CH2
C
43
81 a) Mass of dry air = 0.00119 g cm–3 x 830 cm3
= 0.988 g
(1)
b) Mass of sealed flask itself = (167.70 – 0.988) g
= 166.71 g
(1)
c) Mass of unknown gas = (168.02 – 166.71) g
= 1.31 g
(1)
3
830 cm
3
–1
24 000 cm mol
= 0.0346 mol
d) Number of moles of unknown gas =
1.31 g
0.0346 mol
= 37.9 g mol–1
(1)
Molar mass of unknown gas =
e) The calculated molar mass of the unknown gas would be lower.
(1)
(1)
The density of the dry air left in the flask was lower than that of the unknown gas.
(1)
Dry air remaining in the flask would result in a lower mass of gas in the flask.
(1)
f) Any one of the following:
• Find the mass of the empty flask.
Fill the flask with a liquid of known density (e.g. water at 25 °C). Measure the mass of the filled
flask.
(1)
Subtract to find the mass of the liquid. Use the known density and mass to calculate the volume of
the flask.
(1)
• Measure 830 cm3 of a liquid (e.g. water) in a 1 000 cm3 graduated cylinder.
(1)
Transfer the liquid into the flask to see if the liquid fills the flask completely.
(1)
82 a) There are three components of a lightstick: two chemicals that interact to release energy and a fluorescent
dye.
To activate a lightstick, bend the plastic stick so as to mix the chemicals.
(1)
The fluorescent dye accepts the energy released and converts it into light.
(1)
b) Increasing the temperature increases the rate of the reaction between the two chemicals that interact to
release energy.
(1)
More light is produced.
(1)
The reaction is faster and thus the chemicals are used up in a shorter time.
(1)
c) i) 2H2O2(aq)
44
2H2O(l) + O2(g)
(1)
ii) Any one of the following:
• Measure the loss in mass of the solution
(1)
because oxygen gas is given off and the solution gets lighter as the decomposition proceeds. (1)
• Measure the volume of gas given off
(1)
because oxygen gas is produced when hydrogen peroxide solution decomposes.
• Measure the pressure of hydrogen peroxide solution kept in a closed vessel
because oxygen gas is produced when hydrogen peroxide solution decomposes.
(1)
(1)
(1)
iii) A catalyst is a substance which alters the rate of a reaction
(1)
without itself undergoing any permanent chemical changes.
(1)
83 a) Vinegar / lemon juice / sour milk / buttermilk / yogurt / tart fruits / cream of tartar
b) NaHCO3(s)
2.00 g
+ H+(aq)
0.800 mol dm–3
25.0 cm3
Na+(aq)
+
H2O(l)
2.00 g
84.0 g mol–1
= 0.0238 mol
+
(1)
CO2(g)
? cm3
Number of moles of NaHCO3 =
Number of moles of H+ = 0.800 mol dm–3 x
(1)
25.0
dm3
1 000
= 0.0200 mol
(1)
According to the equation, 1 mole of NaHCO3 reacts with 1 mole of H+ to produce 1 mole of CO2. Thus,
NaHCO3 was in excess. The amount of H+ limited the amount of CO2 formed.
i.e. number of moles of CO2 produced = 0.0200 mol
(1)
Volume of CO2 produced = 0.0200 mol x 24.0 dm3 mol–1
= 0.480 dm3
= 480 cm3
(1)
c) NH4HCO3(s)
NH3(g) + CO2(g) + H2O(l)
(1)
d) i) Fermentation is the slow breakdown of large organic molecules into small molecules by
microorganisms
(1)
in the absence of oxygen.
ii) C6H12O6(aq)
2C2H5OH(aq) + 2CO2(g)
(1)
(1)
iii) By measuring the volume of carbon dioxide formed. / By passing the carbon dioxide gas into limewater
and measuring the light transmittance of the limewater.
(1)
45
84 Place a known volume of potassium permanganate solution and a known volume of a solution containing
oxalic acid and sulphuric acid separately in a thermostat of known temperature.
(1)
Follow the progress of the reaction by using a colorimeter:
Mix the two solutions in a cuvette. Lower the cuvette into the cell compartment of the colorimeter and start
recording for 5 minutes.
The intensity of the purple colour of permanganate ions decreases as the reaction proceeds.
(1)
The colorimeter measures the amount of light absorbed by the reaction mixture when a beam of light passes
through the reaction mixture, i.e. the absorbance.
(1)
The absorbance is directly proportional to the colour intensity of the reaction mixture and the concentration
of permanganate ions in the reaction mixture.
(1)
If the reaction is fast, the absorbance of the reaction mixture would fall rapidly.
If the reaction is slow, the absorbance of the reaction mixture would fall slowly.
(1)
Repeat the experiment at different temperatures. Compare the absorbance-time curves obtained at different
temperatures and deduce the relationship between the reaction rate and temperature.
(1)
(3 marks for organization and presentation)
85 During the hydrolysis, each molecule of ethyl ethanoate hydrolyzed produces one molecules of ethanoic
acid. As hydrochloric acid is a catalyst in the reaction, its amount remains the same. Hence the increase in
total amount of acid in the reaction mixture is a direct measure of the amount of ethyl ethanoate that has
undergone hydrolysis.
(1)
Mix 5.00 cm3 of ethyl ethanoate and 100 cm3 of dilute hydrochloric acid in a reaction flask. Start the stop
watch at the same time.
(1)
Immediately withdraw 5.00 cm3 of the reaction mixture. Run the sample into 50 cm3 of water which has
been chilled in an ice bath. Titrate with sodium hydroxide solution. Let the volume of sodium hydroxide
solution consumed be V0.
(1)
Withdraw 5.00 cm3 samples at regular time intervals and titrate with the sodium hydroxide solution. Let the
volume of sodium hydroxide solution consumed at time t be Vt.
(1)
Stopper the reaction flask. Leave for two days to allow the hydrolysis to complete.
Withdraw a 5.00 cm3 sample and titrate with the sodium hydroxide solution. Let the volume of sodium
hydroxide solution consumed be V∞.
(1)
V∞ is the volume of alkali required to neutralize the hydrochloric acid and the ethanoic acid formed upon
complete hydrolysis. Hence (V∞ – V0) is proportional to the concentration of ethyl ethanoate at the beginning
of the reaction and (V∞ – Vt) is proportional to the concentration of unreacted ethyl ethanoate at time t.
(1)
(3 marks for organization and presentation)
46
86 Weigh a certain mass of zinc carbonate (w g).
(1)
Place the zinc carbonate in a conical flask connected to a gas syringe.
(1)
Add 2 mol dm–3 hydrochloric acid to the carbonate until in excess.
(1)
Collect the carbon dioxide gas evolved using the syringe.
(1)
Measure the volume of carbon dioxide gas collected (v cm3)
(1)
Treatment of data:
ZnCO3(s) + 2HCl(aq)
ZnCl2(aq) + CO2(g) + H2O(l)
Molar volume of carbon dioxide (in cm3)
=
v
x molar mass of ZnCO3
w
(1)
(3 marks for organization and presentation)
47