New 21st Century Chemistry
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Topic 14 Unit 45
In-text activities
Discussion (page 3)
1 Petroleum
2
Sea water
3
Bauxite
4
Sulphur
Discussion (page 5)
The activity is designed to allow students to think about the impact of the chemical industry in
their lives.
1 a) Unless the student grows his / her own food, spins his / her own fibres or purchases items
from individuals who do, most purchased items (even those labelled as 100% natural) are
packaged in chemically processed paper, cardboard or plastics.
b) Packaging that helps maintain product quality (freshness and purity) during shopping and
storage is essential.
2
In most cases, synthetic products cost less than their natural alternatives, e.g.
• nylon vs silk;
• synthetic vitamins vs natural vitamins.
Other natural materials are produced in particular regions. These require more transport for
distribution than locally produced synthetics. Typically, synthetics are at least of comparable if
not superior quality (e.g. synthetic rubber for natural). Sometimes a blend of natural and
synthetic provides the best combination of cost and quality (e.g. clothing made by blending
cotton with synthetic fibres).
Discussion (page 8)
The chemical industry can make a positive contribution to improved quality of life and creation of
opportunities for quality employment. However, the chemical industry is often viewed by the
general public as causing more harm than good. General ignorance of the end use and value of the
industry’s products may be a possible reason. For example, petrochemicals do not reach the final
consumer. They are first sold to customer industries, undergo several transformations, and then go
into products that seem to bear no relation to the initial raw material.
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The industry is perceived as being polluting and causing significant environmental damage.
Another argument may be based on the concept of ‘environmental justice’ which implies a fair
sharing of the benefits as well as the costs of production. The developing world and low-income
areas of the developed world continue to attract a disproportionate share of toxic waste disposal
sites and new investment in toxic production facilities.
Finally, the teacher may bring out that the chemical industry in the 21st century needs to address
social, economic and environmental concerns simultaneously.
Discussion (page 12)
Possible questions:
• What reagents to use?
• How much of each reagent?
• What temperature?
• What pressure?
• Whether or not to use a catalyst?
• How long to allow the reaction to proceed?
Checkpoint (page 23)
1 a) i) The reaction is first order with respect to HBr(g).
The reaction is first order with respect to O2(g).
ii) The overall order of reaction is 2.
b) i) The reaction is second order with respect to NO(g).
The reaction is first order with respect to O2(g).
ii) The overall order of reaction is 3.
c) i) The reaction is second order with respect to NO2(g).
ii) The overall order of reaction is 2.
2
a) Rate = k[C4H9Cl(l)][OH–(aq)]
b) Rate = k[C12H22O11(aq)][H+(aq)]
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3
a) Withdraw samples from the reaction mixture at regular time intervals.
Quench and then
titrate the sample with standard sodium hydroxide solution.
b)
c) Plot a graph using the data shown below.
Time (hr)
Rate (mol dm–3 hr–1)
[CH3COOCH3(l)] (mol dm–3)
0.500
0.157
0.190
1.00
0.0920
0.125
1.50
0.0667
0.0800
2.00
0.0452
0.0500
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The plot of rate of hydrolysis against [CH3COOCH3(l)] is a straight line passing through
the origin.
Thus, the rate of hydrolysis is directly proportional to [CH3COOCH3(l)].
i.e. rate ∝ [CH3COOCH3(l)]
∴ the hydrolysis is first order with respect to CH3COOCH3(l).
d) Slope of graph = k =
(0.124 – 0.0500) mol dm–3 hr–1
(0.150 – 0.0625) mol dm–3
= 0.846 hr–1
e) Rate = k[(CH3COOCH3(l)]
where k is the rate constant.
Checkpoint (page 40)
a) Add a known volume of standard sodium thiosulphate solution and some starch indicator to
the I–(aq) ion before mixing with BrO3–(aq) ion and H+(aq) ion.
Mix all the chemicals (BrO3–(aq), H+(aq), I–(aq), S2O32–(aq) and starch indicator) and record
the time for the first appearance of a dark blue colour.
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The iodine formed from the I–(aq) ion and BrO3–(aq) ion will react with the S2O32–(aq) ion
present:
I2(aq) + 2S2O32–(aq)
2I–(aq) + S4O62–(aq)
When all the S2O32–(aq) ion has been used up, the free iodine will give a dark blue complex
with the starch.
b)
1
can be taken as the initial rate.
t
From experiments 1 and 2, the initial rate doubles when the concentration of I–(aq) ion is
doubled.
∴ order of reaction with respect to I–(aq) ion is 1.
From experiments 1 and 3, the initial rate doubles when the concentration of BrO3-(aq) ion is
doubled.
∴ order of reaction with respect to BrO3–(aq) ion is 1.
c) As compared to experiment 1, the initial concentration of I–(aq) ion has increased by 3 times,
and that of BrO3–(aq) ion has increased by 1.5 times.
∴ the initial rate should be 4.5 times of that in Experiment 1.
Time for the formation of the same amount of I2(aq) =
97 s
= 22 s
4.5
d) Conduct a series of experiments by varying the concentration of H+(aq) ion, but keeping the
concentrations of I–(aq) ion and BrO3–(aq) ion constant.
Measure the time required for the formation of the same amount of iodine.
OR
Conduct an experiment using the following initial concentrations:
[I–(aq)] = 0.0020 mol dm–3
[BrO3–(aq)] = 0.0080 mol dm–3
[H+(aq)] = 0.040 mol dm–3
Measure the time required for the formation of the same amount of iodine.
If the time is the same as that in Experiment 1, then the order of reaction with respect to H+(aq)
ion is 0.
If the time is
1
of that in Experiment 1, then the order of reaction with respect to H+(aq) ion
2
is 1.
If the time is
1
of that in Experiment 1, then the order of reaction with respect to H+(aq) ion
4
is 2.
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Checkpoint (page 42)
1 a) The progress of the reaction can be followed by monitoring the pressure of the gases in a
closed reaction vessel
because the number of moles of gas doubles when SO2Cl2(g) undergoes reaction.
b) i) The rate equation for the reaction can be expressed as:
Rate = k[SO2Cl2(g)]x
From experiments 1 and 2,
5.8 x 10–5 = k(0.20)x ......(1)
1.16 x 10–4 = k(0.40)x ......(2)
Dividing (1) by (2),
5.8 x 10–5
=
1.16 x 10–4
0.20
0.40
x
∴
x =1
so, the rate equation for the reaction is
Rate = k[SO2Cl2(g)]
ii) From experiment 1,
[SO2Cl2(g)] = 0.20 mol dm–3
Rate = 5.8 x 10–5 mol dm–3 s–1
Substituting these values in the rate equation,
5.8 x 10–5 mol dm–3 s–1 = k(0.20 mol dm–3)
k = 2.9 x 10–4 s–1
iii) [SO2Cl2(g)] =
0.60 mol
2.0 dm3
= 0.30 mol dm–3
Initial rate of formation of SO2(g) = (2.9 x 10–4 s–1) x 0.30 mol dm–3
= 8.7 x 10–5 mol dm–3 s–1
∴ amount of SO2(g) formed during the first 0.50 s
= 0.50 s x 8.7 x 10–5 mol dm–3 s–1 x 2.0 dm3
= 8.7 x 10–5 mol
2
a) The rate equation for the reaction can be expressed as:
Rate = k[NO(g)]x[O2(g)]y
i) From experiments 1 and 2,
0.7 x 10–4 = k(1.0 x 10–2)x(1.0 x 10–2)y ......(1)
2.8 x 10–4 = k(2.0 x 10–2)x(1.0 x 10–2)y ......(2)
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Dividing (1) by (2),
0.7 x 10–4
=
2.8 x 10–4
1.0 x 10–2
2.0 x 10–2
x
x =2
From experiments 1 and 3,
0.7 x 10–4 = k(1.0 x 10–2)x(1.0 x 10–2)y ......(1)
1.4 x 10–4 = k(1.0 x 10–2)x(2.0 x 10–2)y ......(3)
Dividing (1) by (3),
0.7 x 10–4
=
1.4 x 10–4
1.0 x 10–2
2.0 x 10–2
y
y =1
∴ the reaction is second order with respect to NO(g) and first order with respect to
O2(g).
ii) From experiments 3 and 4, the concentration of NO(g) triples while that of O2(g)
remains the same.
As the reaction is second order with respect to NO(g),
thus the initial rate of formation of NO2(g) in experiment 4 should be 9 times of that in
experiment 3.
∴ initial rate of formation of NO2(g) in experiment 4
= 9 x 1.4 x 10–4 mol dm–3 s–1
= 1.26 x 10–3 mol dm–3 s–1
b) Rate = k[NO(g)]2[O2(g)]
where k is the rate constant.
c) From experiment 2,
[NO(g)] = 2.0 x 10–2 mol dm–3
[O2(g)] = 1.0 x 10–2 mol dm–3
Rate = 2.8 x 10–4 mol dm–3 s–1
Substituting these values in the rate equation,
2.8 x 10–4 mol dm–3 s–1 = k(2.0 x 10–2 mol dm–3)2(1.0 x 10–2 mol dm–3)
k = 70 dm6 mol–2 s–1
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Unit-end exercises (pages 46 – 53)
Answers for the HKALE questions are not provided.
1 • Basic inorganic chemicals and fertilizers
• Petrochemicals and polymers
• Dyestuffs, paints and pigments
• Pharmaceuticals
• Specialities
2
Plastic household products are made from petrochemicals.
3
a) In a batch process, feedstocks are put into a reactor and allowed to react. When the reaction
is completed, the product is separated from the reaction mixture. The process is then
repeated, batch by batch, until the required amount of product is manufactured.
In a continuous process, feedstocks are fed in one end of the reactor and product is
withdrawn at the other end of the reactor in a continuous flow.
b) i) Continuous process
ii) Batch process
iii) Continuous process
4
a) The index / power of the concentration of X in the rate equation.
b) Rate = k[X][Y]2 where k is the rate constant.
c) 3
d) 8
5
a) Curve I
b) Curve III
6
a) So that the only variable is the concentration of sucrose.
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b)
c) Plot a graph using the data shown below.
Time (hr)
Rate (mol dm–3 hr–1)
[C12H22O11(aq)] (mol dm–3)
1
1.7 x 10–3
0.0080
4
8.3 x 10
–4
0.0044
8
4.2 x 10–4
0.0018
12
1.8 x 10–4
0.00070
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The plot of rate of hydrolysis against the concentration of sucrose is a straight line passing
through the origin.
Thus, the rate of hydrolysis is directly proportional to the concentration of sucrose,
i.e. rate ∝ [C12H22O11(aq)]
∴ the hydrolysis is first order with respect to C12H22O11(aq).
d) Slope =
(17.0 – 6.0) x 10–4 mol dm–3 hr–1
(0.0080 – 0.0028) mol dm–3
= 0.21 hr–1
∴ the rate constant is 0.21 hr–1.
e) Rate = k[C12H22O11(aq)]
where k is the rate constant.
7
a)
b) The rate of reaction doubles when the concentration of OH–(aq) ions is doubled.
Thus, the reaction is first order with respect to OH–(aq) ions.
c) Rate = k[ester][OH–(aq)]
where k is the rate constant.
d) From experiment 2,
[ester] = 0.0010 mol dm–3
[OH–(aq)] = 0.40 mol dm–3
Rate = 6.9 x 10–5 mol dm–3 min–1
Substituting these values in the rate equation,
6.9 x 10–5 mol dm–3 min–1 = k(0.0010 mol dm–3)(0.40 mol dm–3)
k = 0.17 dm3 mol–1 min–1
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8
The rate equation for the reaction can be expressed as:
Rate = k[RX(l)]x[OH–(aq)]y
a) From experiments 1 and 3,
1.5 x 10–6 = k(0.1)x(0.1)y ......(1)
3.0 x 10–6 = k(0.2)x(0.1)y ......(3)
Dividing (1) by (3),
1.5 x 10–6
3.0 x 10–6
=
0.1
0.2
x
x=1
∴ the reaction is first order with respect to haloalkane.
b) From experiments 1 and 2,
1.5 x 10–6 = k(0.1)x(0.1)y ......(1)
3.0 x 10–6 = k(0.1)x(0.2)y ......(2)
Dividing (1) by (2),
1.5 x 10–6
3.0 x 10–6
=
0.1
0.2
y
y=1
∴ the reaction is first order with respect to the hydroxide ion.
The rate equation for the reaction is rate = k[RX(l)][OH–(aq)] where k is the rate constant.
c) From experiment 1,
[RX(l)] = 0.1 mol dm–3
[OH–(aq)] = 0.1 mol dm–3
Rate = 1.5 x 10–6 mol dm–3 s–1
Substituting these values in the rate equation,
1.5 x 10–6 mol dm–3 s–1 = k(0.1 mol dm–3)(0.1 mol dm–3)
k = 1.5 x 10–4 dm3 mol–1 s–1
d) Rate = 1.5 x 10–4 dm3 mol–1 s–1 (0.2 mol dm–3)(0.2 mol dm–3)
= 6.0 x 10–6 mol dm–3 s–1
e) i) The number of particles per unit volume increases.
There is a greater chance for collision and the number of effective collisions increases.
Hence the initial reaction rate increases.
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ii) The particles have more energy and collide more often.
A large portion of the particles have energy equal to or greater than the activation
energy and thus can react upon collision.
Hence the initial reaction rate increases.
9
a) The rate equation for the reaction can be expressed as:
Rate = k[CH3COCH3(aq)]x[Br2(aq)]y[H+(aq)]z
From the first and second experiments,
3.8 x 10–5 = k(0.20)x(0.050)y(0.050)z ......(1)
3.8 x 10–5 = k(0.20)x(0.10)y(0.050)z ......(2)
Dividing (1) by (2),
3.8 x 10–5
3.8 x 10–5
=
y
0.050
0.10
y=0
From the first and third experiments,
3.8 x 10–5 = k(0.20)x(0.050)0(0.050)z ......(1)
7.6 x 10–5 = k(0.20)x(0.050)0(0.10)z ......(3)
Dividing (1) by (3),
3.8 x 10–5
7.6 x 10–5
=
z
0.050
0.10
z =1
From the third and fourth experiments,
7.6 x 10–5 = k(0.20)x(0.050)0(0.10) ......(3)
1.7 x 10–4 = k(0.30)x(0.050)0(0.15) ......(4)
Dividing (3) by (4),
7.6 x 10–5
1.7 x 10–4
=
0.20
0.30
x
0.10
0.15
x=1
∴ the rate equation for the reaction is
Rate = k[CH3COCH3(aq)][H+(aq)]
where k is the rate constant.
b) From the first experiment,
Rate = 3.8 x 10–5 mol dm–3 s–1 = k(0.20 mol dm–3)(0.050 mol dm–3)
k = 3.8 x 10–3 dm3 mol–1 s–1
c) i) Graph A
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ii) Graph B
10 a) From experiments 1 and 2,
1.6 x 10–5 = k(3.0 x 10–2)2(4.0 x 10–2) ......(1)
Rate2 = k(6.0 x 10–2)2(4.0 x 10–2) ......(2)
Dividing (1) by (2),
1.6 x 10–5
Rate2
=
3.0 x 10–2
6.0 x 10–2
2
Rate2 = 6.4 x 10–5 mol dm–3 s–1
From experiments 1 and 3,
1.6 x 10–5 = k(3.0 x 10–2)2(4.0 x 10–2) ......(1)
6.4 x 10–5 = k(3.0 x 10–2)2[B] ......(3)
Dividing (1) by (3),
1.6 x 10–5
6.4 x 10–5
=
4.0 x 10–2
[B]
[B] = 1.6 x 10–1 mol dm–3
From experiments 1 and 4,
1.6 x 10–5 = k(3.0 x 10–2)2(4.0 x 10–2) ......(1)
1.6 x 10–5 = k[A]2(16.0 x 10–2) ......(4)
Dividing (1) by (4),
1.6 x 10–5
1.6 x 10–5
=
3.0 x 10–2
[A]
2
4.0 x 10–2
16.0 x 10–2
[A] = 1.5 x 10–2 mol dm–3
Experiment
no.
Initial concentration
of A (mol dm–3)
Initial concentration
of B (mol dm–3)
Initial rate of reaction
(mol dm–3 s–1)
1
3.0 x 10–2
4.0 x 10–2
1.6 x 10–5
2
6.0 x 10–2
4.0 x 10–2
6.4 x 10–5
3
3.0 x 10–2
1.6 x 10–1
6.4 x 10–5
4
1.5 x 10–2
16.0 x 10–2
1.6 x 10–5
b) From the first experiment,
Rate = 1.6 x 10–5 mol dm–3 s–1 = k(3.0 x 10–2 mol dm–3)2(4.0 x 10–2 mol dm–3)
k = 0.44 dm6 mol–2 s–1
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11 a) When sodium thiosulphate solution reacts with dilute hydrochloric acid, a yellow
precipitate of sulphur forms. This changes the light transmittance of the reaction mixture.
To follow the progress of the reaction, make a cross on a piece of paper.
Put the conical flask containing some sodium thiosulphate solution on top of the paper.
Add dilute hydrochloric acid to the flask.
Record the time t for the solution mixture to become opaque, i.e. when the cross can no
longer be seen from above.
b) To keep the total volume of all reaction mixtures constant.
Thus, the initial concentration of thiosulphate ion in each reaction mixture is directly
proportional to the volume of sodium thiosulphate solution used.
c) i) When the opaque stage is reached, only a small amount of sulphur has formed and the
reaction has only proceeded a small part of the way to completion. The average rate
from start to the opaque stage can be taken to be approximately equal to the initial rate.
Initial rate ∝
1
t
ii) Plot a graph using the data shown below.
1
Relative initial rate ( ) (s–1)
t
Volume of Na2S2O3(aq) (cm3)
5.88 x 10–3
5.0
12.0 x 10–3
10.0
17.9 x 10
–3
15.0
23.8 x 10–3
20.0
30.3 x 10–3
25.0
1
y
30.0
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The plot of relative initial rate against volume of Na2S2O3(aq) is a straight line passing
through the origin.
Thus, the initial rate is directly proportional to the initial concentration of S2O32–(aq)
ion.
i.e. rate ∝ [S2O32– (aq)]
∴ the reaction is first order with respect to thiosulphate ion.
iii) When the volume of Na2S2O3(aq) is 30.0 cm3, the relative initial rate of reaction is
36.2 x 10–3 s–1, i.e. t = 28 s.
12 a) A small amount of sodium thiosulphate solution and starch solution were included in each
reaction mixture. The thiosulphate ion would turn any iodine formed back to iodide ion.
2S2O32–(aq) + I2(aq)
S4O62–(aq) + 2I–(aq)
By measuring the time taken for a dark blue colour to appear, we knew how long it took to
use up all the thiosulphate ion, and thus the time for a small, fixed amount of iodine to
form.
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b) The initial concentration of iodide ion in each reaction mixture is directly proportional to
the volume of potassium iodide solution used.
c) i) As the reaction had only proceeded a small part of the way to completion, we could
take initial rate ∝
1
.
t
ii) Plot a graph using the data shown below.
1
Relative initial rate ( ) (s–1)
t
Volume of KI(aq) (cm3)
0.010
5.0
0.021
10.0
0.032
15.0
0.043
20.0
The plot of relative initial rate against the volume of KI(aq) is a straight line passing
through the origin.
Thus, the initial rate of the reaction is directly proportional to the initial concentration
of iodide ion.
i.e. rate ∝ [I–(aq)]
∴ the reaction is first order with respect to iodide ion.
iii) In experiment 1, it took 100 s to produce iodine to react with all the thiosulphate ion
added.
2S2O32–(aq) + I2(aq)
S4O62–(aq) + 2I–(aq)
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Number of moles of S2O32– ion added = 0.010 mol dm–3 x
5.0
dm3
1 000
= 5.0 x 10–5 mol
Number of moles of iodine formed in 100 s =
5.0 x 10–5
mol
2
= 2.5 x 10–5 mol
2.5  10 -5 mol
Initial rate of formation of iodine =
46.0
dm 3  100 s
1 000
= 5.4 x 10–6 mol dm–3 s–1
Number of moles of I–(aq) ion used = 0.10 mol dm–3 x
5.0
dm3
1 000
= 5.0 x 10–4 mol
Initial concentration of I–(aq) ion =
5.0  10 - 4 mol
46.0
dm 3
1 000
= 1.1 x 10–2 mol dm–3
Rate = k[I–(aq)]
5.4 x 10–6 mol dm–3 s–1 = k(1.1 x 10–2 mol dm–3)
k = 4.9 x 10–4 s–1
∴ the rate constant is 4.9 x 10–4 s–1.
d) Prepare a series of reaction mixtures by varying the volume of H2O2(aq) or H2SO4(aq) used,
but keeping the volume of other reagents and the total volume constant. (Add water to keep
the volume of each reaction mixture constant).
Measure the time taken for a blue colour to appear for each reaction mixture.
13 —
14 —
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15 a) i)
ii) Slope =
(0.40 – 0.58) mol dm–3
(22.0 – 10.0) min
= –0.015 mol dm–3 min–1
b) i) Rate of the reaction when concentrations of both S2O82–(aq) ion and H3AsO3(aq) are
both 0.50 mol dm–3 equals to –slope of the curve obtained in (a)(ii), i.e. 0.015 mol dm–3
min–1.
Rate when concentrations of S2O82–(aq) ion and H3AsO3(aq) are both 1.0 mol dm–3:
Rate when concentrations of S2O82–(aq) ion and H3AsO3(aq) are both 0.50 mol dm–3
= 0.060 mol dm–3 min–1 : 0.015 mol dm–3 min–1
=4:1
So, when the concentrations of S2O82–(aq) ion and H3AsO3(aq) are both halved, the rate
goes down by a factor of 4.
Hence the overall order of the reaction should be 2.
ii) Any two of the following:
I Rate = k[S2O82–(aq)][ H3AsO3(aq)]
II Rate = k[S2O82–(aq)]2
III Rate = k[H3AsO3(aq)]2
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Repeat the experiment using double initial concentration of S2O82–(aq) ion, but keeping
the initial concentrations of other reagents constant.
If the initial rate doubles, then equation I is correct.
If the initial rate quadruples, then equation II is correct.
If the initial rate does not change, then equation III is correct.
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