innova junior college 2012 jc2 h2 chemistry practical ... - ASKnLearn

INNOVA JUNIOR COLLEGE
2012 JC2 H2 CHEMISTRY PRACTICAL
REACTION KINETICS
I PRINCIPLES OF REACTION KINETICS
RATE EQUATION
For a reaction: aA + bB
The rate equation takes the form:
products
Rate = k [A] m [B] n
The powers m and n are the orders of reaction with respect to reactant A and reactant B
respectively. The overall order is m + n.
m and n can only be determined experimentally.
The rate constant, k, is only constant at a particular temperature.
For elementary and non-elementary reactions, please refer to Reaction Kinetics lecture notes.
The rate of reaction can be varied by changing the concentrations of reactants, changing the
temperature and adding a catalyst.
II METHODS TO DETERMINE ORDER OF REACTION
OVERVIEW
1

(A) Continuous method with sampling (ie. involve titration)
Reactants of known concentrations are mixed and the time noted.
Reaction samples are withdrawn regularly at fixed time intervals.
Reaction samples are stopped by either dilution or addition of a ‘quenching’ reagent (i.e one
which can stop the reaction chemically).
Samples are titrated with a standard solution to determine the reactant/product concentration
at each sample time.
Rate of reaction can be determined with the change in concentration of reactant/product
with time.
(B) Continuous method without sampling
Reactants of known concentrations are mixed and the time noted.
Reaction mixture is continuously monitored with time using a convenient physical property
(must be related to concentration).
Rate of reaction can be determined with the change in concentration of reactant/product
with time.
(C) Discontinuous method (ie. initial rate method)
For reactions that are accompanied by prominent visual changes, the rate of the reaction
may be studied by “clock reaction” (ie. by measuring the time taken for a prescribed visual
change to occur).
Rate of reaction is measured by:
(I) Colourless reactants form coloured products such as precipitate/coloured solution
(II) Coloured reactants form colourless products
III DATA ANALYSIS : CONCENTRATION TIME GRAPHS
A concentration-time graph shows either how the concentration of a product increases with time,
or how the concentration of a reactant decreases with time. The gradient (slope) of a
concentration-time graph at any point measures the rate of reaction at that time.
Method 1 : Gas Collection method
One method to monitor the changes in concentration over time is the gas collection method. The
apparatus shown below can be used to study the rate of reaction between marble chips (calcium
carbonate) and dilute hydrochloric acid.
CaCO 3 (s) + 2HCl (aq)
CaCl 2 (aq) + CO 2 (g) + H 2 O (l)
2

Set up 1
100 cm 3
conical flask
100 cm 3
measuring
cylinder
CaCO 3
1000 cm 3
beaker
50 cm 3 of
HCl
OR Set up 2
The reaction is allowed to proceed and the total volume of carbon dioxide gas collected is
recorded. The CaCO 3 used is in excess and the reaction slows down as the hydrochloric acid gets
less concentrated. The total volume of gas collected when the reaction stops = V final . This final
volume is proportional to the concentration and number of moles of dilute hydrochloric acid used.
At time t after the start of timing, the volume of gas collected is V t . This varies with the amount of
acid that has reacted by that time. As time passes, the volume of gas collected V t increases as the
volume of gas produced at time t is proportional to the number of moles of acid used up.
(V final – V t ) is proportional to the concentration of dilute hydrochloric acid unreacted. As time
passes, the concentration of dilute hydrochloric acid decreases. Hence plotting (V final – V t ) against
time gives a concentration-time graph for the experiment.
(V final – V t ) ∝ [HCl] unreacted
3

Example of results obtained:
Time/min 1 2 3 4 5 6 7 8 9
Volume of gas 1.7 3.4 4.9 6.6 8.1 9.5 11.2 12.2 12.2
collected, V t /cm 3 (V final )
V final – V t /cm 3 10.5 8.8 7.3 5.6 4.1 2.7 1.0 0 0
[HCl] / mol dm -3 Time / min
(V final – V t ) / cm 3 Time / min
Tangent
at t=0
Tangent
at t=0
Analysis of results:
1) Find the initial rate of the reaction by constructing a tangent to the curve at time = 0s to obtain
the gradient.
2) Vary the concentration of the dilute hydrochloric acid used and plot graphs of (V final – V t )
against time on the same axis.
3) Find the order of reaction with respect to hydrochloric acid by comparing the initial rate of
reaction measured to the concentration of hydrochloric acid used.
V final – V t / cm 3
tangents
[HCl] = 1 mol dm -3
[HCl] = 2 mol dm -3
Time / min
4

Possible Errors and Improvements:
In identifying suggestion for improvements, you must take note that the
suggested improvement must address the error highlighted.
Error 1 (for both set ups): Difficult to stopper the conical flask when starting the stop watch
without loss of CO 2 gas, thus leading to a smaller volume of gas collected.
Improvement 1: Use a pressure-equalizing funnel to introduce the hydrochloric acid.
(containing HCl)
CaCO 3
Error 2 (for set up 1): The volume of gas was not measured at constant pressure. The volume of
a gas is greatly affected by changes in pressure due to compression of air.
Improvement 2: Use a well-greased syringe (refer to set up 2) to collect the gas so that gas is
collected at constant pressure.
Error 3 (for set up 1): Measuring cylinder is used for the measurement of hydrochloric acid. The
volume of HCl measured is inaccurate since the measuring cylinder can only measure to the
nearest 0.5 cm 3 .
Improvement 3: Use a more precise apparatus, eg. burette to measure out 50 cm 3 of hydrochloric
acid. The volume of HCl measured is more accurate since the burette can measure to the nearest
0.05 cm 3 .
5

Error 4 (for set up 1): Initially the gas that was produced will cause the displacement of water
which is present in the tubing. It will thus give a lower volume of gas being collected.
Improvement 4: Use an inverted burette for gas collection by connecting the tubing to the burette
tip. Gas is collected by pushing the liquid level down in the burette.
CaCO 3 in
dilute HCl
OR
Use a frictionless gas syringe (refer to set up 2) to collect the gas that was produced, so that the
gas need not be collected over water.
Error 6 (for set up 1): As CO 2 gas is slightly soluble in water, the volume of gas collected is less
than the actual volume of CO 2 evolved.
Improvement 6: Saturate the water with CO 2 gas before starting gas collection.
OR
Use a frictionless gas syringe (refer to set up 2) to collect the CO 2 gas that was produced, so that
the gas need not be collected over water.
Other examples of reactions involving gas collection:
(1) Metals reacting with acid to form hydrogen gas such as:
Mg(s) + H 2 SO 4 (aq) MgSO 4 (aq) + H 2 (g)
(2) Decomposition of hydrogen peroxide which is catalysed by manganese(IV) oxide
2H 2 O 2 (aq)
2H 2 O(l) + O 2 (g)
6

Method 2 : Mass loss of reaction mixture
The rate of a reaction that produces a gas can also be measured by following the mass loss as the
gas is formed and escapes from the reaction flask. When calcium carbonate reacts with dilute
hydrochloric acid, carbon dioxide is evolved. While the reaction is proceeding, the mass of the
reaction flask decreases as gas escapes. The rate of the reaction is followed by observing the loss
in mass (which is the mass of gas evolved).
CaCO 3 (s) + 2HCl (aq)
CaCl 2 (aq) + CO 2 (g) + H 2 O (l)
The method is suitable for reactions producing carbon dioxide or oxygen but not very accurate for
reactions producing hydrogen gas (as the mass loss is too low for it to be accurate).
Analysis of results:
1) A graph of mass of reaction flask (g) against time is plotted. This is the mass loss curve.
2) Find the initial rate of reaction by measuring the tangent at t = 0s. The reaction rate is
expressed as the rate of loss in mass from the flask in e.g. g/min based on the initial gradient
(see graph below).
3) Vary the concentration of the dilute hydrochloric acid used and plot graphs of mass of reaction
flask against time.
4) Find the order of reaction with respect to hydrochloric acid by comparing the initial rate of
reaction measured to the concentration of hydrochloric acid.
7

Method 3 : Clock Reaction
A clock reaction is a reaction that is set up to produce a sudden observable colour change after a
certain time when it has produced a fixed amount of one reactant/product.
Rate of reaction is measured by:
Colourless reactants form coloured
products such as precipitate/coloured
solution
Example:
Na 2 S 2 O 3 (aq) + 2HCl (aq)
S (s) + SO 2 (g) + 2NaCl (aq)+ H 2 O (l)
yellow ppt
Rate ∝
1
time
Coloured reactants form colourless
products
Example:
CH 2 =CH 2 + Br 2
reddish
brown
Rate ∝
V coloured reactant
time
CH 2 BrCH 2 Br
V Br 2
e.g Rate ∝
time
when total volume of solution is kept
constant.
Eg 1. Reaction between sodium thiosulfate and hydrochloric acid
This expt. will be done
in RK1.
Na 2 S 2 O 3 (aq) + 2HCl (aq) S (s) + SO 2 (g) + 2NaCl (aq) + H 2 O (l)
The reaction of sodium thiosulfate (VI) with dilute hydrochloric acid produces a yellow precipitate
of sulfur.
8

The rate equation for the reaction is in the form:
Rate = k [Na 2 S 2 O 3 ] x [HCl] y
Where k is the rate constant
x is the order of reaction with respect to sodium thiosulfate (VI)
y is the order of reaction with respect to hydrochloric acid.
A simple and effective procedure is to measure the time taken for enough sulfur to form to obscure
a cross beneath the conical flask containing the reaction mixture. A stop watch is used to measure
the time taken. It is reasonable to assume the same amount of sulfur is needed to obscure the
cross at each time when different concentration of sodium thiosulfate (VI) or acid is used.
The rate of the reaction can be determined by calculating the amount of sulfur produced in the
time recorded. This is given by
Rate ∝
amount of sulfur produced
time
Amount of sulfur produced is assumed to be the same in each reaction so
Rate ∝
1
time
The order of reaction for each reactant is determined by varying the concentration of each reactant
in turn, keeping the others constant. The concentration of each reactant is varied by using various
volumes of sodium thiosulfate (VI) and hydrochloric acid. Water is added to keep the total
volume of solution constant so that the volume of each reagent is proportional to its
concentration.
9

Analysis of results:
Graphs of
1
time
The reason is that
(y axis) against volume of reactant that is varied (x axis) are plotted.
1
time
is proportional to the rate of reaction and the volume of reactant is
proportional to its concentration, since the total volume is constant.
Concentration of reactant ∝
volume of reactant
total volume of solution
Hence by plotting
1
time
against concentration of reactant (x axis).
(y axis) against volume of reactant is the same as plotting Rate (y axis)
For a zero order reaction,
Rate
[reactant] /
mol dm -3
Vol. of
reactant
/cm3
For a first order reaction,
Rate
[reactant] /
mol dm -3
Vol. of
reactant
/cm3
10

For a second order reaction,
Rate
[reactant] /
mol dm -3
Vol. of
reactant
/cm3
This expt. is
similar to
Eg 2. Reaction between hydrogen peroxide and acidified solution of potassium iodide
RK2.
When hydrogen peroxide is added to an acidified solution of potassium iodide, the iodide ions are
slowly oxidized to form iodine.
2I – (aq) + 2H + (aq) + H 2 O 2 (aq)
colourless
I 2 (aq) + 2H 2 O(l)
brown
Small quantities of starch and sodium thiosulfate (VI) are also present. As the iodine is produced, it
reacts with thiosulfate (VI) and is converted back to iodide ions:
I 2 (aq) + 2S 2 O 3
2–
2I – (aq) + S 4 O 6
2–
brown colourless colourless colourless
When the iodine produced in the reaction is in excess of the sodium thiosulfate (VI), a blue colour
suddenly appears as iodine reacts with starch.
As long as excess thiosulfate (VI) ions are present in the solution, no free iodine can accumulate
because it is immediately turned into iodide ions which are colourless.
The thiosulfate (VI) ions are the limiting reagent. So once all the thiosulfate (VI) ions are
consumed, iodine starts to form in the solution. Aqueous iodine is brown (or pale yellow,
depending on its concentration) in colour. If starch is added to the solution then a more dramatic
blue solution is formed by the complex of starch–iodine. The colour change is sharp, and the time
elapsed to this point is determined simply by the use of a stop watch.
11

The time taken for the blue colour to appear with a constant quantity of sodium thiosulfate (VI)
present depends on the rate of formation of iodine.
Rate of reaction ∝ 1 t
where t is the time from the addition of the peroxide solution to the appearance of the blue colour.
The volume of hydrogen peroxide and potassium iodide are varied, keeping the volume of
thiosulfate (VI) and acid constant. Water is also added keeping the total volume of solution
constant so that the volume of each reagent is proportional to its concentration.
Analysis of results: Method 1
When the mixture is added to the beaker, the stop watch is started and when the blue colour
appears, the time is recorded as t. The product (V H2 O 2
x t) is recorded when the volume of H 2 O 2 is
varied and the rest of the reactants volumes are kept constant. The product (V KI x t) is recorded
when the volume of KI is varied and the rest of the reactants volumes are kept constant. Both the
products are constant when the volumes of KI and H 2 O 2 are varied. Thus, it is first order with
respect to hydrogen peroxide and first order with respect to potassium iodide.
Hence the rate equation is
Rate = k [H 2 O 2 ] [KI]
The reason the products are constant is because:
For a first order reaction,
Rate = k [reactant]
i.e. Rate ∝ concentration of reactant
Since volume of reactant ∝ concentration of reactant and rate∝ 1
time ,
Rate ∝ concentration of reactant
≡
1
∝ volume of reactant
time
∴ volume of reactant = k (
1
time )
∴ volume of reactant x time = k
where k is a constant
12

FOR YOUR KNOWLEDGE…
For a second order reaction,
Rate = k[reactant] 2
ie. Rate α (concentration of reactant) 2
≡
1
time
α (volume of reactant)2
∴ (volume of reactant) 2 = k (
1
time )
∴ (volume of reactant) 2 x time = k
For a zero order reaction,
Rate = k
∴
1
time = k
In addition, the concentration of iodide ion in the solution remains constant at its original value.
Even though the peroxide–iodide reaction tends to consume iodide ions, the thiosulfate (VI) ion
immediately returns the iodine to the iodide ion form:
As long as there are excess thiosulfate (VI) ions present, there is no change in the concentration
of the iodide ion. This means that the rate of the reaction will change only because the
concentration of the KI and H 2 O 2 changes.
Analysis of results: Method 2
From the relationship volume of reactant x time = k,
⇒
1
time
Taking logs on both sides of the equation:
log (
= volume of reactant x k
1
) = log (volume of reactant) + log k
time
13

Plotting log (
1
) against log (volume of reactant) gives a straight line. The gradient of the graph
time
gives the order of reaction with respect to the reactant.
Eg 3. Decolourisation of coloured reactants [Extracted from RJC Prelim 2009]
Ethene reacts with bromine in tetrachloromethane to form 1,2−dibromoethane as shown by the
equation:
CH 2 =CH 2 + Br 2 CH 2 BrCH 2 Br
To find out the orders of reaction with respect to ethene and bromine, ethene and bromine were
first dissolved separately in tetrachloromethane. Various volumes of these solutions and
tetrachloromethane were mixed and the time taken for the colour of bromine to disappear was
recorded. The results are shown in the table below:
Experiment
Volume of ethene
solution/ cm 3
Volume of bromine
solution/ cm 3
Volume of
tetrachloromethane/
cm 3
Time taken for
colour of bromine to
disappear/ s
1 20 20 0 15
2 12 20 8 25
3 20 10 10 15
4 40 20 20 t 4
Discuss (i) to (iii) with reference to experiments 1 to 3.
14

(i) Explain why varying volumes of tetrachloromethane were used.
To keep the total volume constant so that the concentration of ethene or bromine used is
directly proportional to the volume used.
(ii) State the relationship between the rate of reaction and
• time taken for the colour of bromine to disappear
• volume of bromine used
Rate of reaction is inversely proportional to the time taken for the colour of bromine to
disappear and directly proportional to the volume of bromine used.
V Br 2
Rate ∝
time
when total volume of solution is kept constant
(iii) Deduce the order of reaction with respect to ethene and show that the order of reaction with
respect to bromine is 1.
V Br 2
Rate ∝
time
Experiment
Volume of
ethene
solution/ cm 3
Volume of
bromine solution/
cm 3
Volume of
tetrachloromethane/
cm 3
Time taken for
colour of
bromine to
disappear/ s
1 20 20 0 15
2 12 20 8 25
3 20 10 10 15
Rate of
reaction
20
15 = 1.33
20
25 = 0.8
10
15 = 0.667
Comparing experiments 1 and 2,
When [ethene] used is decreased by 0.6 times, keeping [Br 2 ] constant, rate also
decrease by 0.6 times.
Hence, order with respect to ethane = 1
Comparing experiment 1 and 3,
When [Br 2 ] used is doubled, keeping [ethene] constant, rate is also doubled.
Hence, order with respect to bromine = 1
(iv) Suggest a value for t 4 , time taken for the colour of bromine to disappear in experiment 4.
15

Experiment
Volume of
ethene solution/
cm 3
Volume of
Volume of bromine
solution/ cm 3 tetrachloromethane/
cm 3
Time taken for
colour of
bromine to
disappear/ s
4 40 20 20 t 4
The total volume of solution in Experiment 4 is 80 cm 3 which is double that of other
experiments. Halved all the volumes in experiment 4 will result in same volumes as that of
experiment 3.
Volume of
Volume of
Volume of bromine
Experiment ethene solution/
cm 3
solution/ cm 3 tetrachloromethane/
cm 3
3 40÷2 = 20 20÷2 = 10 20÷2 = 10
Hence the concentration of ethene and bromine are the same for Experiment 3 and 4. The
rate of reaction is the same. Hence the time taken for colour of bromine to disappear is the
same. Therefore, t 4 = 15 s.
Method 4 : Titrimetric (Quenching) Method
For the reaction between propanone and iodine:
CH 3 COCH 3 (aq) + I 2 (aq)
H +
CH 3 COCH 2 I (aq) + HI (aq)
The rate equation would be
Rate = k [CH 3
COCH 3
] m [I 2
] n +
[ H ] p
where m, n and p are the orders of the reaction with respect to propanone, iodine and acid
respectively.
The rate of reaction may be studied by determining the amount of unreacted iodine in the reaction
mixture at different times.
Dilute sulfuric acid is used as the catalyst. Sodium hydrogencarbonate solution is used to quench
(stop) the reaction instantly by neutralizing the acid.
The unreacted iodine was then titrated with the diluted sodium thiosulfate (VI) using starch as
indicator.
2S 2 O 3 2– (aq) + I 2 (aq)
S 4 O 6 2– (aq) + 2I – (aq)
16

Experimental Procedure:
(1) Preparation of reaction mixture (2) Withdrawal of samples for analysis
Reactants of known volumes
and concentrations
Stop-watch
started at the
point of mixing
At regular time
intervals (eg, 5 min),
samples (eg, 10.0 cm 3 )
are withdrawn from the
reaction mixture using
a pipette
The reaction vessel should be kept in a
constant-temperature water bath
(3) Quenching of reaction in sample (4) Analysis of sample
Addition of ‘quenching’ reagent (NaHCO 3 )
to remove acid
catalyst
Titration with a standard solution to
determine the concentrations of I 2 at different
time.
Standard thiosulfate solution
with known concentration
Withdrawn sample is placed in conical flask.
The exact time at which the reaction is
“Quenched” sample
quenched is noted.
IV EFFECT OF TEMPERATURE ON REACTION RATE
Aim of expt: To determine a value for the activation energy of a reaction.
Background of reaction:
Peroxodisulfate ions, S 2 O 2- 8 , oxidises I - ions as follows:
S 2 O 2- 8 (aq) + 2I - (aq) I 2 (aq) + 2SO 2- 4 (aq) ……… Reaction 1
The reaction is first order with respect to both the concentration of peroxodisulfate ions and the
concentration of iodide ions.
17

The iodine produced in Reaction 1 reacts immediately with thiosulfate ions, S 2 O 3 2- as shown below.
Thus, the concentration of iodine is kept very low.
2S 2 O 3 2- (aq) + I 2 (aq) 2I - (aq) + S 4 O 6 2- (aq) …………. Reaction 2
Once all the thiosulfate ions have reacted, the concentration of iodine rapidly increases and due to
the presence of starch in the reaction mixture, the dark blue colouration of an iodine-starch
complex is formed.
The amount of iodine required to react with the thiosulfate ions can be deduced from the fixed
volume of S 2 O 2- 3 used in each experiment. By measuring the time it takes for this fixed amount of
iodine to be produced, it is possible to determine values for the rate of Reaction 1 at different
temperatures.
Method:
Six experiments are conducted at different temperatures. For all the experiments, the
concentrations of I - , S 2 O 2- 3 and S 2 O 2- 8 are kept constant. In each experiment, the time taken, t, for
the dark blue colour to appear is measured.
Procedure:
For each experiment, solution 1 is prepared as follows.
To a dry 100 cm 3 beaker, add 20 cm 3 I - , 10 cm 3 S 2 O 3 2- and 5 cm 3 starch.
(i) Experiment 1
1. Pour 20 cm 3 2-
of S 2 O 8 solution rapidly, in a steady stream into the beaker containing
solution 1. Start the stopwatch when about half the S 2 O 2- 8 solution has been added.
2. Sir the mixture using the thermometer.
3. Note the initial temperature, T i , of the reaction mixture.
4. Stop the stopwatch when the dark blue colour first appears.
5. Note the final temperature, T f , of the reaction mixture.
6. Note the time elapsed on the stopwatch, t, to the nearest second.
(ii) Experiment 2
1. Prepare solution 1 as described above.
2. ** Half-fill a 250 cm 3 beaker with ice/salt water to make a cooling bath.
3. Place the 100 cm 3 beaker containing solution 1, into the cooling bath and carefully stir
solution 1 until its temperature is between 5 o C and 10 o C.
4. Remove the beaker from the cooling bath and, using the method described in Experiment
1, determine the time taken, t, for the solution to turn dark blue.
18

Experiment 2 gives the slowest experiment.
(iii) Experiment 3
1. Prepare solution 1 as described above.
2. ** Half-fill a 250 cm 3 beaker with hot water to make a heating bath.
3. Place the 100 cm 3 beaker containing solution 1, into the heating bath and carefully stir
solution 1 until its temperature is between 50 o C and 60 o C.
4. Remove the beaker from the heating bath and, using the method described in Experiment
1, determine the time taken, t, for the solution to turn dark blue.
Experiment 3 gives the fastest experiment.
(iv) Experiments 4 to 6
Select three other suitable reaction temperatures, between the temperatures of Experiments
2 and 3, for use in three additional experiments. In each case, use the heating bath or the
cooling bath to bring solution 1 to an appropriate temperature and perform these
experiments.
Analysis of result:
Sample results:
Temp, T/K 280 303 309 312 318 324
Time, t/s 270 204 138 115 75 55
ln (1/t) -5.60 -5.32 -4.93 -4.75 -4.32 -4.01
1/T / K -1 3.57 x 10 -3 3.30 x 10 -3 3.24 x 10 -3 3.21 x 10 -3 3.14 x 10 -3 3.09 x 10 -3
Graphical determination of activation energy
The value for activation energy, E a , can be determined by performing a series of experiments,
where the reaction mixture is kept constant, but different reaction temperatures are used, and the
same end-point (appearance of the dark blue colour) is times. The activation energy for the
reaction can be obtained using the Arrhenius Equation.
k =
By plotting a graph of ln (1/t) against 1/T, a straight line of best-fit will be obtained. The gradient of
the line is –E a /R, where R is the ideal gas constant.
19

ln (1/t)
1/T
x
x
x
x
x
x
Suppose the magnitude of the slope is -6.36 x 10 3 K.
Thus, -E a /R = -6.36 x 10 3 K
Hence, -E a = -6.36 x 10 3 K x 8.81 J K -1 mol -1
E a = 52.9 kJ mol -1
20