IJSO – Training Material
Phase 3 – Practical Training
Experiment 1
Investigating the Effects of Concentration and Temperature on the Rate of a
Chemical Reaction
Objective:
To investigate the concentration and temperature effects on the rate of
a chemical reaction
Introduction
You have learnt in the classroom session that the rate of a chemical reaction
depends on the concentration of the reactants as well as the reaction temperature.
We are going to investigate this chemical phenomenon. To facilitate our
investigation, we need to choose a chemical reaction that exhibits some easily
observable physical changes so that we can monitor the progress of the reaction
conveniently. In this regard we will study the reaction between sodium thiosulphate
and diluted hydrochloric acid.
S2O32-(aq) + 2 H+(aq)

SO2(g) + S(s)
+ H2O(l)
Sodium thiosulphate reacts with diluted acid to give sulphur dioxide, sulphur and
water. Both sodium thiosulphate and diluted hydrochloric acid are colorless solution.
Sulphur dioxide is a very soluble gas and dissolves completely in the aqueous solution.
The sulphur formed, however, is not soluble and will exist in the mixture as white or
yellow precipitate (or colloidal). It makes the reaction mixture becomes opaque as
the reaction occurs. Therefore, we can study the reaction rate by monitoring the
opaqueness of the reaction. This can be easily done by measuring the time taken for
forming a certain amount of precipitate.
In the experiment, you will carry out the reaction by mixing the two reactants
into a small beaker, and put the beaker on top of a piece of white paper which has a
mark on it. Before the reaction starts you can see the mark clearly from the top of
the beaker through the solution. However, as the reaction proceeds, sulphur
precipitate forms and makes the solution become more opaque, and so eventually the
mark is completely masked. The time taken for the mark to become totally
disappeared indicates how fast the reaction has occurred.
1
Chemicals:
0.15 M sodium thiosulphate solution
1 M hydrochloric acid
Apparatus:
100 mL beakers
100 mL measuring cylinder
Thermometer (alcohol, up to 110oC)
Stopwatch
Part A: To study the concentration effect on the rate of a chemical reaction.
In this part of the experiment you will study the concentrate effect on the rate of
the above reaction by varying the concentration of the sodium thiosulphate solution.
Procedure:
1. Take a piece of white paper and make a clear mark on it.
2.
3.
4.
5.
6.
7.
Put a dry clean beaker on top of the mark.
Measure 60 mL of 0.15 M sodium thiosulphate solution and put it into the
beaker.
Quickly add 10 mL of 1 M HCl into the beaker in a single portion and start the
stopwatch immediately.
Stir the mixture gently with a glass rod for a few seconds.
Look at the mark from vertically above the beaker through its content, and record
the time taken for the mark to become totally disappeared.
Repeat steps 1-6 by substituting the 60 mL 0.15 M sodium thiosulphate solution
with:
Solution
Vol. of 0.15M Na2S2O3
solution
Vol. of water
Total volume
A
45 mL
15 mL
60 mL
B
30 mL
30 mL
60 mL
C
15 mL
45 mL
60 mL
Data
1. Time for the disappearance of the cross
Volume of
Na2S2O3(aq)
(mL)
60
Volume of
water
(mL)
0
Volume of
HCl(aq)
(mL)
10
Initial concentration of
Na2S2O3(aq) in the
reaction mixture/M
0.1500M
Time for the disappearance
of the cross/s
Trial 1 Trial 2 Average
16.6
17.4
17.00
45
15
10
0.1125M
24.7
25.0
24.85
30
30
10
0.0750M
36.6
37.1
36.85
15
45
10
0.0375M
83.4
85.0
84.20
2
Data Analysis
1. Calculate the initial concentration of sodium thiosulphate in the reaction
mixtures.
Run 1
Initial concentration of Na2S2O3
0.15  (60 / 1000)
=
M
60 / 1000
= 0.1500 M
Run 2
Initial concentration of Na2S2O3
0.15  (45 / 1000)
=
M
60 / 1000
= 0.1125 M
Run 3
Initial concentration of Na2S2O3
0.15  (30 / 1000)
=
M
60 / 1000
= 0.0750 M
Run 4
Initial concentration of Na2S2O3
0.15  (15 / 1000)
=
M
60 / 1000
= 0.0375 M
3
2.
Plot the time required for the disappearance of the cross against the initial
concentration of sodium thiosulphate on a graph paper. Label your graph
properly.
Time required (s)
Time required for the disappearance of the cross against
initial concentration of sodium thiosulphate
90
80
70
60
50
40
30
20
10
0
0
0.04
0.08
0.12
0.16
0.2
Ini t i a l c o nc e nt r i o o f s o d i um t hi o s ul p ha t e ( M )
3.
How is the rate of reaction affected by the concentration of the reactant?
The rate of reaction increases when the concentration of sodium thiosulphate
increases.
4
Part B: To study the temperature effect on the rate of a chemical reaction.
In this part of the experiment you will study the temperature effect on the rate of
the about reaction. You will carry out the reaction at different temperature and
compare the results.
Temperature: 0oC (ice-water bath)
Room temperature (refer to the data taken in Part A)
50oC (by warming the Na2S2O3 solution with a 50oC water bath)
Procedure
Reaction at 0oC
1.
2.
3.
4.
5.
6.
7.
8.
Measure 60 mL of 0.15 M sodium thiosulphate solution and put it into a clean,
dry 100 mL beaker.
Cool the solution in an ice-water bath for 15 minutes.
Measure the temperature of the solution.
Carry out the reaction of the cooled Na2S2O3 solution with 1 M HCl by following
the steps 1-6 described in part A.
Measure 60 mL of 0.15 M sodium thiosulphate solution and put it into another
dry, clean 100 mL beaker.
Warm the solution in a 50oC water bath for 15 minutes.
Measure the temperature of the warmed solution.
Carry out the reaction of the warmed Na2S2O3 solution with 1 M HCl by
following the steps 1-6 described in part A.
Data
Time for the disappearance of the cross
Initial concentration of
Na2S2O3(aq) in the reaction
moisture/M
0.15M
0.15M
0.15M
Temperature of the reaction
mixture/oC
51.5
23.8
7.8
Time for the disappearance of the
cross/s
Trial 1
Trial 2
Average
5.9
5.7
5.80
16.6
17.4
17.00
46.0
45.3
45.65
Data Analysis
1. From the data in the above table, how is the rate of reaction affected by
temperature?
The rate of reaction increases when the temperature of reaction increases.
5
Experiment 2
Investigating the effect of surface area (particle size of a solid reactant) on the
rate of a chemical reaction.
Objective:
To investigate the effect of surface area on the rate of a chemical
reaction
Introduction
If a chemical reaction involves one or more solid reactants, the particle sizes of
the solid reactants will affect the rate of the reaction. It is because surface area
increases as the particles become smaller. Increasing the reactants’ surface area
allows the reactants approach to each other more frequently, and so the reaction rates
are often enhanced. A daily example is the burning of charcoal. Powdered
charcoal burns more fiercely than large lumps of charcoal. It is because powdered
charcoal has a much larger surface for reacting with oxygen.
Zinc metal reacts with diluted hydrochloric acid to give zinc chloride and
hydrogen gas. The rate of hydrogen gas formation allows us to observe the reaction
rate conveniently. In this experiment we will compare the reaction rates of diluted
hydrochloric acid with zinc powder and zinc granules.
Chemicals:
1M hydrochloric acid
Zinc powder
Zinc granules
Apparatus:
test tubes, spatula
Procedure:
1. Measure about 5 mL of 1 M hydrochloric acid into a test tube and add some zinc
2.
3.
granules. Record the observation.
Measure about 5 mL of 1 M hydrochloric acid into another clean test tube and
add zinc powder. The amount of zinc powder added should be approximately
equal to the amount of zinc granules added in step 1. Record the observation.
Compare the rate of these two reactions.
6
Data Analysis
1.
Which of the two reactions occurs faster?
give a faster reaction rate?
Did zinc granules or zinc powder
The reaction with zinc powder occurs faster, so zinc powder give a faster
reaction rate.
2.
How does the particle size of solid reactant affect the reaction rate?
When the particle size of solid reactant smaller, its surface area increases, and
allows reactants approach to each other more frequently. Thus, the reaction rate
will be enhanced.
7
Experiment 3
Determination of the Rate Equation for a Chemical Reaction
Objective:
To determine the rate equation for the reaction of iodine and propanone
in acidic medium
Introduction
From experiment 1 you should have realized that the rate of a chemical equation
depends on its reactants’ concentration. The relationship between the rate of a
chemical reaction and its reactants’ concentration can be expressed into a rate
equation. For example, propanone reacts with iodine in acidic medium as shown
below:
CH3COOCH3(aq) + I2(aq)  CH3COCH2I(aq) + H+(aq) + I-(aq)
The rate of the reaction may depend on the concentration of propanone
(CH3COCH3), iodine (I2) and acid (H+). The rate equation of the reaction can be
expressed in the form of:
Rate = k[CH3COCH3]a[I2]b[H+]c
where [CH3COCH3], [I2] and [H+] are the concentration of propanone, iodine
and acid, respectively. k is the rate constant which is a quantity that depends on
reaction temperature. a, b and c are the orders of reaction with respect to propanone,
iodine and acid, respectively. Reaction orders (a, b and c) are quantities that must be
determined experimentally and cannot be deduced from the chemical equation. The
reaction order with respect to a reactant is not necessarily equal to the reactant’s
coefficient present in the balanced chemical equation.
In this experiment you are going to determine the values of k, a, b, and c for the
reaction of propanone with iodine. You will do so by running the reaction several
times and varying the concentration of the reactants one at a time.
Chemicals:
1 M hydrochloric acid
1 M propanone
0.002 M iodine solution (prepared by dissolving 0.5 g of I2 and 3.3 g of
KI in 1 L of deionized water)
8
Apparatus:
measuring cylinders, conical flasks
Procedure:
1. Use measuring cylinders to measure the appropriate amounts of hydrochloric
acid, propanone solution and deionized water into dry conical flasks, according
to the table shown below.
2. Add appropriate amounts of iodine solution into the conical flasks. Start the
stop watch.
3. Swirl the flasks gently.
4. Measure the time taken for the colour of iodine disappears completely.
Data
Volume of 1 M HCl(aq) (mL)
Volume of 1 M CH3COCH3(aq) (mL)
Volume of deionised water (mL)
Volume of 0.002M I2(aq) (mL)
 [I2] in the time taken (mol dm-3)
Time for the disappearance of the
Trial 1
colour (s)
Trial 2
Average
Initial rate of reaction (mol dm-3 s-1)
Run 1
20
8
0
4
2.5  10-4
52.5
49.5
51.00
4.90  10-4
Run 2
10
8
10
4
2.5  10-4
102.8
102.7
102.75
2.43  10-4
Run 3
20
4
4
4
2.5  10-4
109.0
104.5
106.75
2.34  10-4
Run 4
20
8
2
2
1.25  10-4
40.0
38.5
39.25
3.18  10-4
Data Analysis
1.
Calculate the initial concentration of I2 in the reaction mixture and hence the
change in iodine concentration in the time taken.
Run 1
Initial concentration of I2
0.002  (4 / 1000)
=
M
32 / 1000
= 2.5  10-4 M
Run 2
Initial concentration of I2
0.002  (4 / 1000)
=
M
32 / 1000
= 2.5  10-4 M
9
Run 3
Initial concentration of I2
0.002  (4 / 1000)
=
M
32 / 1000
= 2.5  10-4 M
Run 4
Initial concentration of I2
0.002  (2 / 1000)
=
M
32 / 1000
= 1.25  10-4 M
2.
Calculate the rate of consumption of I2, this is the initial rate of reaction in
terms of I2.
Run 1
Initial rate of reaction in term of I2
2.5  10-4
=
51
= 4.90  10-6 mol dm-3 s-1
Run 2
Initial rate of reaction in term of I2
2.5  10-4
=
102.75
= 2.43  10-6 mol dm-3 s-1
Run 3
Initial rate of reaction in term of I2
2.5  10-4
=
106.75
= 2.34  10-6 mol dm-3 s-1
10
Run 4
Initial rate of reaction in term of I2
2.5  10-4
=
39.25
= 3.18  10-6 mol dm-3 s-1
3.
Determine the order of reaction with respect to each reactant.
Rate = k[CH3COCH3] a[I2] b[H+] c
Compare Run 1 and Run 2,
[I2] & [CH3COCH3] are the same in these runs by using the same volume, and
they have the same k value.
 The rate equation can be writted as Rate = k[H+]a
20
32
)/
4.90  10
1000 1000

10
32
2.43  10 6
(1 
)/
1000 1000
2.16 = [2] c
log 2.16 = c log 2
c = log 2.16 / log 2
= 1.11
~1
 The order of H+ is 1
6
(1 
c
Compare Run 1 and Run 3,
[I2] & [H+] are the same in these runs by using the same volume, and they have
the same k value.
 The rate equation can be writted as Rate = k[CH3COCH3]a
8
32
)/
4.90  10
1000 1000

6
4
32
2.34  10
(1 
)/
1000 1000
2.09 = [2] a
log 2.09 = a log 2
a = log 2.09 / log 2
= 1.06
~1
 The order of CH3COCH3 is 1
6
(1 
a
11
Compare Run 1 and Run 4,
[CH3COCH3] & [H+] are the same in these runs by using the same volume, and
they have the same k value.
 The rate equation can be writted as Rate = k[I2]b
4
32
)/
4.90  10
1000 1000

8
32
3.18  10 6
(0.002 
)/
1000 1000
1.54 = [2] a
log 1.54 = c log 2
b = log 1.54 / log 2
= 0.62
<1
 The order of I2 is 0
6
4.
(0.002 
b
Estimate the rate constant (with unit) for the reaction using the data in Run 1.
Rate = k[CH3COCH3] 1[I2] 0[H+] 1
Rate = k[CH3COCH3][ H+]
4.90  10-6 = k [(1  0.008/0.032][ (1  0.020/0.032]
k = 3.136  10-5 mol-1 dm-3 s-1
5.
Write the rate equation for the reaction.
Rate = 3.136  10-5 [CH3COCH3] 1 [H+] 1
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