Hydrogen Peroxide and iodine ions
Hydrogen peroxide and iodide ions
Aim: To determine the rate equation for the reaction between Hydrogen peroxide and iodide ions, and to investigate the effects of a catalyst and temperatures on the reaction and to derive the activation enthalpy.
Background knowledge:
1Hydrogen peroxide reacts with iodide ions producing iodine, when in an aqueous acid solution.
H2O2 (aq) + 2I- (aq) + 2H3O+(aq) I2 (aq) + 2H2O(l)
To detect iodine you can look at the color shown by the addition of starch solution.
Iodine also reacts with sodium thiosulphate solution, which acts as a delaying agent. Adding sodium thiosulphate to the reaction mixture will allow the iodine produced initially to react with the thiosulphate ions, which reduce the iodine to iodide ions, leaving the starch unchanged.
2S2O32-(aq) + I2 (aq) S4O62- (aq) + 2I- (aq)
Once all the thiosulphate ions have reacted with the iodine, the iodine will then go on to react with the starch, producing a blue-black complex.
2The reason for the blue-black complex is due to the formation of polyiodide chains during the reaction between starch and iodine. The amylose in starch forms helices with which the iodine molecules align, causing a transfer of charge. This charge transfer corresponds to the absorption spectrum, in which the blue-black colour is the complementary colour. The details of this reaction are not fully known yet. The strength and deepness of the colour is dependent on the amount of amylose present.
3The rate of the reaction can be explained by the collision theory, which shows that the rate can be altered by4: concentrations, pressure, temperature, intensity of radiation, particle size, surface area and a catalyst. In this experiment I will be looking at concentrations, temperatures and the affect of a catalyst. The collision theory also states a reaction will only take place if three conditions are met:
1. Reactant particles collide with each other
2. The reactant particles must collide with the correct orientation.
3. The collision must provide enough energy to overcome the activation energy.
2This is due to the fact, if particles collide but are not orientated correctly the molecules will just bounce of each other, this is often due to charge of the molecules which causes repulsion if the orientation is incorrect.
If particles collide with the correct orientation, they must collide with a minimum energy otherwise they will just bounce of each other. The activation energy is used to break some of the original bonds, which is essential for a reaction to occur. The activation energy is this minimum energy and can be shown on an energy profile. We can see from the below graphs how in an endothermic reaction the energy of the reactants is lower than the energy of the products, this is why it absorbs heat. Whereas an exothermic reaction will give out heat as the reactants have a higher energy than the products.
2The activation energy can be marked on the Maxwell-Boltzmann distribution curve. The Maxwell-Boltzmann distribution refers directly to gases, however the principles can be taken and applied to liquid reactions also.
5The area underneath the curve represents the different particles and their level of energy. Only the particles that have energy higher than the activation energy will undergo a reaction. We can see that the kinetic energy of a molecule can never be zero, but theoretically can be ever increasing, as there is no defined maximum energy value.
You can then change the shape of the curve or move the activation energy in order to increase the number of collisions that overcome the activation enthalpy. Changing certain variables can do this, which is what will be done in this experiment.
6 7 The first variable I will be varying is concentration of the different reactants; I will be using 5 different concentrations for each reactant.
Increasing the concentration of a reactant means that there are a higher proportion of particles per unit volume. This then causes the rate of reaction to increase because particles have a higher chance of colliding with sufficient kinetic energy to cause a reaction.
8By varying the concentration of each reactant it allows me to calculate the order of the reaction, by investigating their effects. By finding the order I can then calculate the rate constant and rate equation.
The chemical kinetics of this reaction can be used to explain the mechanisms and rates of the reaction. I will plot a graph of the effects of the concentrations and this will enable me to determine the order. The rate equation shows how the reaction rate depends on the concentration of each substance.
9The rate equation of a reaction is:
Rate = k [A]a[B]b
K is considered the rate constant, and this can be found once the experiment has been conducted. The rate constant is 10‘A coefficient of proportionality relating the rate of a chemical reaction at a given temperature to the concentration of reactant (in a unimolecular reaction) or to the product of the concentrations of reactants’
The rate constant will change if the temperature is changed or a catalyst is added. The rate is measured in mol dm-3 s-1. The a and b are the order of the reaction in respect to A and B respectively. These orders show how the concentrations of A and B affect the reaction.
The order of a reaction is found in respect to a reactant and is defined as ‘the power to which its concentration term in the rate equation is raised. The overall order of a reaction is all of the separate orders added together.
Overall order = a + b.
The order of a reaction can only be found experimentally, it is not possible to deduce the order just by looking at the equation.
The order of a reaction can be determined graphically or by looking directly at how the concentration affects the rate.
If the order of a reactant is 0, this means that the concentration of the particular reactant does not affect the rate of reaction. Therefore this would not appear in the rate equation, as anything to the power of 0 is 1.
A graph of a zero order reaction would be a horizontal straight line if the initial rate was plotted against concentration.
If you plotted concentration against time the graph would be a decreasing straight line.
A first order reaction is identified when doubling the concentration of a reactant doubles the reaction rate. This means that the rate of reaction is proportional to the concentration of the reactant.
Rate [A]
Which can also be written as:
Rate = k [A]
A graph of the initial rate against concentration will be a straight line.
A second order reaction is identified when doubling the concentration of a reactant quadruples the reaction rate. This means that the rate of reaction is proportional to the square of the concentration of the reactant.
Rate [A]2
Which can also be written as:
Rate = k [A]2
A graph of initial rate against concentration would be a curve.
A graph of initial rate against concentration2 would be a straight line.
A reaction mechanism shows the steps in which a reaction breaks or makes bonds, all chemical changes occur due to the making and breaking of bonds. The rate equation also allows you to establish the rate-determining step. This is the slowest step, which controls the overall rate of a reaction, because the reaction cannot react any faster than the slowest step. This then allows you to determine the reaction mechanism.
11The reaction mechanism of this particular reaction is as followed.
The first step is a slow reaction between the iodide ions and the hydrogen peroxide.
H2O2 (aq) + I- (aq) IO- (aq) + H2O (l)
The second step reaches equilibrium quickly.
IO- (aq) + H3O+(aq) ⇌ HIO (aq) + H2O (l)
The final step is when the Iodine is formed
HIO (aq) + H3O (aq) + I- (aq) I2 (aq) + 2H2O (l)
The slowest step is step 1; this means this is the rate-determining step.
The second variable I will investigate is the effects of temperature, by investigating 5 different temperatures. An approximation is that if you increase the temperature by 10 °C you will double the rate of the reaction. The collision theory demonstrates how at higher temperatures the rate of a reaction will be faster, because each particle has a higher kinetic energy and therefore moves at a faster rate. This will increase the likelihood of a collision and the proportion that will have enough energy to overcome the activation enthalpy. I will be able to calculate the activation enthalpy, which is ‘the minimum kinetic energy required by a pair of colliding particles before the reaction will occur’. By increasing the temperature, this then changes the shape of the Maxwell-Boltzmann distribution curve. Meaning there are more particles that will collide with enough energy to overcome the activation energy.
I will be using the Arrhenius equation to find the activation enthalpy, the energy needed in order for a reaction to occur, by looking at how rate equations vary with respect to temperature.
The Arrhenius equation is:
Which can also be written in the form:
The different variables in this equation are:
T – Temperature, in kelvin
R – Gas constant (8.314J K-1 mol-1.), Derives from the equation pV=nRT.
Ea – Activation energy, the minimum energy needed for a reaction.
A – Arrhenius constant
The Arrhenius equation allows us to see the effect of a change of temperature on the rate constant and the rate of reaction.
The final variable I will be investigating is the effect of a catalyst, I will be using Ammonium Molybdate, which is ‘a substance which speeds up a reaction, but remains chemically unchanged at the end of the reaction’. A catalyst provides an alternative route for the reaction with a lower activation enthalpy, this will mean that when particles collide a higher proportion will have enough kinetic energy to overcome the lower activation enthalpy provided by the catalyst. This can be seen on an energy profile.
By adding a catalyst we are lowering the activation energy needed, this can be shown on the Maxwell-Boltzmann distribution curve. We can see how the activation energy has moved to the left, meaning a higher number of particles have enough energy to overcome it.
The catalyst I will be using is Ammonium Molybdate, which is a transition metal. A transition metal can be defined as ‘one which forms one or more stable ions which have incompletely filled d orbitals.’ And can be referred to as the d block elements. Transition metals are commonly used as catalysts due to their capability to change oxidation state, or because they can adsorb substances onto their surface and activate them in the process.
Catalysts can be either homogeneous or heterogeneous. Homogeneous catalysts are in the same physical state as the reactants, whereas heterogeneous catalysts are in a different state. In this experiment my catalyst with my heterogeneous as Ammonium Molybdate is a solid.
When the reactant is adsorbed on the surface of the catalyst the bonds within the reactant are weakened. The bonds break and the reactant atoms bond with the surface of the catalyst. New bonds form between the atoms and the products are released and diffuse away.
12Below is a diagram using Ethene, showing hydrogenation of the C=C.
Step 1 – the reactants adsorbing onto the catalyst surface and the H has become dissociated.
Step 2 – Products begin to form with new bonds being made.
Step 3 – The product is formed and diffuses away.
Risk Assessment1314
Chemical
Hazards
Key Risks
Hydrogen peroxide
0.25 mol dm-3
Low hazard at concentrations lower than 1.5 mol dm-3.
Contact with eyes
Sulphuric Acid
0.5 mol dm-3
Irritant
Contact with eyes/skin.
Solid potassium iodide/
Irritant
Contact with eyes
Starch
Low Hazard
Solid Sodium Thiosulphate
Mild Irritant
Contact with skin/eyes
Ammonium Molybdate
Irritant
Contact with eyes
Iodine (Produced)
Low hazard at concentrations less that 1 mol dm-3
Contact with skin and eyes.
Ways to minimize the potential chance of accidents:
Safety glasses + lab coat.
Keep workbench tidy and organized.
Clean up all spills immediately with plenty of water to dilute the chemicals.
Ensure the laboratory is well ventilated by opening windows.
Do not eat or drink in the laboratory.
If contact with skin occurs wash thoroughly with water
If contact with the eye gently flood the eye for at least 10 minutes with water.
As the concentrations of my chemicals are very low the risks are minimal.
Apparatus:
1cm3 graduated pipettes
25cm3 bulb pipette
250cm3 volumetric flask
50cm3 volumetric flask
Burette
Beakers
100cm3 conical flasks
Stop watch
Boiling tubes
Test tubes
Electronic Balance (3 decimal place and 4 decimal place)
Spatula
Glass Rod
Funnel
Water Baths at varying temperatures
Colourimeter
Cuvette
Distilled water
White tile and black marker pen
Dropping pipette
Method:
Preliminary experiment
A preliminary experiment must be conducted to obtain optimum volumes and concentrations to be used. This was to ensure the experiment would be long enough to record, but would not take too long to produce the blue-black complex. This is in order to produce a sufficient number of results from which my conclusions will be drawn. A ‘trial and error’ approach was used to come to the conclusion of which volumes and concentrations to use.
The concentrations I decided on were:
Hydrogen peroxide 0.25 mol dm-3
Potassium iodide 0.5 mol dm-3
Sulphuric acid 0.5 dm-3
Sodium thiosulphate 0.01 mol dm-3
Making up Potassium Iodine solution:
1. To work out the mass of potassium iodine to be dissolved:
a.
2. Weigh out 20.75g into a small beaker using a 3 decimal place measuring balance.
3. Add distilled water and dissolve the potassium iodine stirring with a glass rod.
4. Transfer solution into a 250cm3 volumetric flask using a funnel.
5. Wash the beaker, glass rod and funnel with distilled water into the volumetric flask.
6. Add distilled water up to the calibration mark on the volumetric flask, ensuring the bottom of the meniscus is on the calibration mark, and read from eye-level to ensure an accurate reading.
7. Insert the stopper into the volumetric flask and invert several times, to thoroughly mix the solution.
Making up Sodium Thiosulphate solution:
1. To work out the mass of Sodium Thiosulphate to be dissolved:
a.
2. Weigh out 0.6205g into a small beaker using a 4 decimal place measuring balance.
3. Add distilled water and dissolve the sodium thiosulphate stirring with a glass rod.
4. Transfer solution into a 250cm3 volumetric flask using a funnel.
5. Wash the beaker, glass rod and funnel with distilled water into the volumetric flask.
6. Add distilled water up to the calibration mark on the volumetric flask, ensuring the bottom of the meniscus is on the calibration mark, and read from eye-level to ensure an accurate reading.
7. Insert the stopper into the volumetric flask and invert several times, to thoroughly mix the solution, this is to ensure the concentration is the same throughout.
Making up Hydrogen Peroxide solution:
1. To convert the number of volumes into mol dm-3
a. Starting with a 20 volume solution
a.i. 20/12 = 1.667 mol dm-3
2. To find the volume of original solution to dilute:
a.
3. Measure 75 cm3 of 20 volumes Hydrogen Peroxide using a burette
4. Transfer into a 250cm3 volumetric flask.
5. Fill the volumetric flask up to the calibration mark with distilled water, ensuring the meniscus is on the calibration mark, reading at eye-level to ensure an accurate reading.
6. Insert the stopper and invert several times to thoroughly mix the solution; this is to ensure the concentration is the same throughout.
Diluting Sulphuric Acid solution
1. Work out the volume of solution needed to be diluted
a.
2. Measure 125 cm3 of 1.0 mol dm-3 Sulphuric Acid solution using a burette.
3. Add the Sulphuric Acid into a 250cm3 volumetric flask, using a funnel.
4. Fill the volumetric flask up to the calibration mark with distilled water, ensuring the meniscus is on the calibration mark, reading at eye-level to ensure an accurate reading.
5. Insert the stopper and invert several times to thoroughly mix the solution; this is to ensure the concentration is the same throughout.
Diluting any concentration to the desired concentration
Throughout the experiment it will be necessary to dilute solutions down to the desired concentration
1. Using the equation below
2. The volume calculated will be measured and distilled water will be added to make up the necessary volume.
The basic clock reaction – Basic Experiment
1. Add 25cm3 of the desired concentration of Sulphuric Acid into a conical flask, using a bulb pipette.
2. Using a graduated pipette add to the 100cm3 conical flask (ensure you use a clean graduated pipette for each chemical to avoid contamination), I will ensure the meniscus is read at eye level to ensure an accurate reading:
a. 1cm3 of the 0.01 mol dm-3 sodium thiosulphate
b. 1cm3 of the desired concentration of potassium iodide
c. 1cm3 of starch solution (ensuring you use a freshly made batch each day, as it has a short shelf life, due to it undergoing photochemical reactions forming unwanted products)
3. Using a clean graduated pipette add 1cm3 of the desired concentration of Hydrogen Peroxide solution into a test tube, A.
4. Add test tube A solution into the conical flask and start the stopwatch immediately.
5. Swirl the conical flask as evenly as possible over a white tile with a black cross drawn on.
6. Stop the stopwatch as soon as you can no longer see the black cross.
7. Record results.
Varying the Hydrogen Peroxide, Potassium Iodide and Sulphuric Acid concentrations
1. Hydrogen Peroxide (Experiment 1)
a. Make up 5 different concentrations via dilutions.
a.i. The concentrations I will use will be 0.5 mol dm-3, 0.4 mol dm-3, 0.3 mol dm-3, 0.25 mol dm-3, 0.15 mol dm-3, 0.1 mol dm-3
This will allow me to see the affect of doubling the concentration.
a.ii. To make up these concentrations I will dilute the previously made 0.5 mol dm-3 Hydrogen Peroxide solution for each desired concentration, using the previously mentioned equation.
a.iii.
a.iv. I will measure the value calculated using a graduated pipette and transfer it into a 50cm3 volumetric flask and add the required amount of water to fill the volumetric flask up to the pipette.
b. Carry out the basic experiment (the basic iodine clock reaction) for each concentration using:
b.i. 0.5 mol dm-3 Sulphuric Acid
b.ii. 0.5 mol dm-3 Potassium Iodide
b.iii. And varying the concentration of hydrogen peroxide for each reaction.
2. Potassium Iodide (Experiment 2)
a. Make up 5 different concentrations via dilutions.
a.i. The concentrations I will use will be 0.5 mol dm-3, 0.3 mol dm-3, 0.25 mol dm-3, 0.15 mol dm-3, and 0.1 mol dm-3. This will allow me to see the affect of doubling the concentration.
a.ii. To make up these concentrations I will dilute the previously made 0.5 mol dm-3 Potassium Iodide solution for each desired concentration, using the previously mentioned equation.
i. I will measure the value calculated above using a graduated pipette and transfer it into a 50cm3 volumetric flask and add the required amount of water to fill the volumetric flask up to the pipette.
b. Carry out basic experiment (the basic iodine clock reaction) for each concentration using:
i. 0.5 mol dm-3 Sulphuric Acid ii. 0.25 mol dm-3 Hydrogen Peroxide iii. And varying the concentration of potassium iodide for each reaction.
3. Sulphuric Acid (Experiment 3)
a. Make up 5 different concentrations via dilutions.
a.i. The concentrations I will use are 1.0 mol dm-3, 0.5 mol dm-3, 0.25 mol dm-3, 0.3 mol dm-3 and 0.15 mol dm-3
This will allow me to see the affect of doubling the concentration.
a.ii. To make up these concentrations I will dilute the previously made 0.5 mol dm-3 Sulphuric Acid solution for each desired concentration, using the previously mentioned equation.
a.iii. I will measure the value calculated above using a graduated pipette and transfer it into a 250cm3 volumetric flask and add the required amount of water to fill the volumetric flask up to the pipette.
b. Carry out the basic experiment (the basic iodine clock reaction) for each concentration using:
b.i. Varying concentrations of Sulphuric Acid for each reaction
b.ii. 0.25 mol dm-3 Hydrogen Peroxide
b.iii. 0.5 mol dm-3 Potassium Iodide
Varying the temperature of the reaction (Experiment 4)
1. I will be using 60 °C, 50°C, 40°C, 30°C, 10°C and room temperature.
2. Add 25cm3 of the 0.5 mol dm-3 of Sulphuric Acid into a conical flask., using a bulb pipette.
3. Using a graduated pipette add to the boiling tube, A (ensure you use a clean graduated pipette for each chemical to avoid contamination):
a. 1cm3 of the 0.01 mol dm-3 sodium thiosulphate
b. 1cm3 of the 0.1 mol dm-3 Potassium Iodide
c. 1cm3 of starch solution (ensuring you use a freshly made batch each day)
4. Using a clean graduated pipette add 1cm3 of 0.1 mol dm-3 Hydrogen Peroxide solution into a test tube, B.
5. Place both A and B into a water bath for 5 minutes. Measure the temperature of the solutions ensuring to clean the thermometer for each to avoid contamination. Record the temperature as the water bath temperature is likely not to be 100% correct as to what is labeled on the bath.
6. Add test tube B solution into the boiling tube and start the stopwatch immediately.
7. Quickly add a bung and invert the boiling tube to ensure the Hydrogen Peroxide is evenly distributed, take out the bung once inverted once.
8. Swirl the boiling tube as evenly as possible keeping it in the water bath.
9. Stop the stopwatch as soon as a blue-black colour appears.
10. Record results.
Varying the temperature of the reaction with a catalyst – Experiment 5
1. Repeat experiment4 adding 0.02 g of Ammonium Molybdate into boiling tube A during stage 1.
Using a colourimeter to monitor the rate of reaction - Basic Colourimeter Experiment
1. Set the colourimeter filter to 440, as it has the complementary colour to that of the substance being formed. To find this, I carried out the experiment whilst varying the filter until the best one was decided.
2. Fill a cuvette with distilled water, ensure when picking up the cuvette you only hold the cloudy side, and placing it with the arrow facing towards you. Press the calibration button; this will set the absorbance for zero.
3. Add 25cm3 of the desired concentration of Sulphuric Acid into a boiling tube, A, using a volumetric pipette.
a. Using a graduated pipette add to the boiling tube, 1cm3 of the desired concentration of potassium iodide
4. Using a clean graduated pipette add 1cm3 of the desired concentration of Hydrogen Peroxide solution into a test tube, B.
5. Add test tube B solution into the boiling tube, A and start the stopwatch immediately.
6. Quickly use a dropping pipette fill the cuvette, and place in the colorimeter, ensuring you only hold it on the cloudy sides, the arrow is facing you when placed in the colourimeter, and the outside of the cuvette is dry.
7. Measure the absorbance every 15 seconds, until the maximum absorbance is reached or you reach 25 minutes.
Varying the concentrations of sulphuric acid, hydrogen peroxide and potassium iodide using a colourimeter
Varying Hydrogen Peroxide (Experiment 7)
1. I will be using 0.25 mol dm-3 and 0.1 mol dm-3
a. To get these concentrations I will dilute using the previously mentioned equation and method.
2. Carry out the Basic colourimeter experiment using:
a. 0.5 mol dm-3 potassium iodide
b. 0.5 mol dm-3 sulphuric acid
Varying Potassium Iodide (Experiment 8)
1. I will be using 0.25 mol dm-3 and 0.5 mol dm-3
a. To get these concentrations I will dilute using the previously mentioned equation and method.
2. Carry out the Basic colourimeter experiment using:
a. 0.25 mol dm-3 hydrogen peroxide
b. 0.5 mol dm-3 sulphuric acid
Varying Sulphuric Acid (Experiment 9)
1. I will be using 0.25 mol dm-3 and 0.5 mol dm-3
a. To get these concentrations I will dilute using the previously mentioned equation and method.
2. Carry out the Basic colourimeter experiment using:
a. 0.5 mol dm-3 potassium iodide
b. 0.25 mol dm-3 hydrogen peroxide
Results
Table 1
Varying the concentration of Hydrogen peroxide
Concentration mol dm-3
Time to change (s)
Rate (1/time)
0.50
13
0.077
0.40
14
0.071
0.30
17
0.059
0.25
26
0.038
0.15
31
0.032
0.10
46
0.022
Table 2
Varying the concentration of Potassium Iodide
Concentration mol dm-3
Time to change (s)
Rate (1/time)
0.50
26
0.038
0.30
45
0.022
0.25
53
0.019
0.15
93
0.011
0.10
111
0.009
Table 3
Varying the concentration of Sulphuric Acid
Concentration mol dm-3
Time to change (s)
Rate (1/time)
0.50
14
0.071
0.30
26
0.038
0.25
54
0.019
0.15
90
0.011
0.10
104
0.010
Table 4
Affects of a change in temperature
Temperature °C
Time to change (s)
59
12
49
32
38
47
28
85
21
109
11
247
Table 5 – Affects of a change in temperature Arrhenius calculations
Temp (°C)
Time (s)
Rate (1/time)
Ln Rate
Temp (K)
1/Temp (K)
59
12
0.083
-2.48
332
0.003012048
49
32
0.031
-3.47
322
0.00310559
38
47
0.021
-3.85
311
0.003215434
28
85
0.012
-4.44
301
0.003322259
21
109
0.009
-4.69
294
0.003401361
11
247
0.004
-5.51
284
0.003521127
Table 6 – Affects of a change in temperature catalyzed Arrhenius calculations
Temp (°C)
Time (s)
Rate (1/Time)
Ln Rate
Temp (K)
1/Temp (K)
60
3
0.333
-1.10
333
0.003003003
48
6
0.167
-1.79
321
0.003115265
41
8
0.125
-2.08
314
0.003184713
30
8
0.125
-2.08
303
0.00330033
20
12
0.083
-2.48
293
0.003412969
11
29
0.034
-3.37
284
0.003521127 Without catalyst 20°C = 149 s
Colorimeter results
Using: filter 440.
0.25 mol dm-3 Hydrogen Peroxide
0.5 mol dm-3 Potassium Iodide
0.5 mol dm-3 Sulphuric Acid
Table 7
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.08
495
1.52
30
0.17
510
1.51
45
0.25
525
1.56
60
0.33
540
1.57
75
0.42
555
1.57
90
0.47
570
1.60
105
0.54
585
1.61
120
0.59
600
1.64
135
0.64
615
1.67
150
0.70
630
1.69
165
0.84
645
1.69
180
0.82
660
1.72
195
0.84
675
1.74
210
0.88
690
1.74
225
0.98
705
1.79
240
0.97
720
1.77
255
1.01
735
1.80
270
1.10
750
1.81
285
1.13
765
1.81
300
1.13
780
1.86
315
1.19
795
1.91
330
1.22
810
1.87
345
1.30
825
1.94
360
1.29
840
1.91
375
1.32
855
1.93
390
1.38
870
1.91
405
1.37
885
1.95
420
1.39
900
1.96
435
1.42
915
1.96
450
1.45
930
1.99
465
1.50
945
2.00
480
1.52
960
2.00
Using: 440 Filter
0.25 mol dm-3 Hydrogen Peroxide
0.25 mol dm-3 Potassium Iodide
0.5 mol dm-3 Sulphuric Acid
Table 8
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.07
765
1.09
30
0.10
780
1.14
45
0.15
795
1.12
60
0.20
810
1.12
75
0.19
825
1.13
90
0.25
840
1.15
105
0.28
855
1.18
120
0.30
870
1.20
135
0.34
885
1.21
150
0.36
900
1.22
165
0.42
915
1.21
180
0.44
930
1.23
195
0.48
945
1.26
210
0.49
960
1.27
225
0.51
975
1.28
240
0.54
990
1.29
255
0.57
1005
1.29
270
0.59
1020
1.31
285
0.59
1035
1.30
300
0.63
1050
1.32
315
0.64
1065
1.29
330
0.67
1080
1.31
345
0.68
1095
1.34
360
0.71
1110
1.34
375
0.73
1125
1.36
390
0.74
1140
1.35
405
0.80
1155
1.36
420
0.82
1170
1.36
435
0.83
1185
1.34
450
0.83
1200
1.32
465
0.84
1215
1.34
480
0.84
1230
1.34
495
0.88
1245
1.37
510
0.87
1260
1.36
525
0.90
1275
1.36
540
0.91
1290
1.39
555
0.95
1305
1.39
570
0.94
1320
1.38
585
0.95
1335
1.38
600
0.95
1350
1.37
615
0.99
1365
1.39
630
0.99
1380
1.43
645
1.00
1395
1.41
660
1.00
1410
1.42
675
1.02
1425
1.42
690
1.05
1440
1.45
705
1.05
1455
1.43
720
1.07
1470
1.44
735
1.08
1485
1.44
750
1.08
1500
1.44
Using: 440 Filter
0.25 mol dm-3 Hydrogen Peroxide
0.01 mol dm-3 Potassium Iodide
0.5 mol dm-3 Sulphuric Acid
Table 9
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.12
765
0.28
30
0.12
780
0.28
45
0.13
795
0.29
60
0.13
810
0.29
75
0.14
825
0.29
90
0.13
840
0.30
105
0.15
855
0.30
120
0.15
870
0.30
135
0.16
885
0.30
150
0.16
900
0.30
165
0.17
915
0.30
180
0.17
930
0.31
195
0.18
945
0.31
210
0.18
960
0.31
225
0.18
975
0.32
240
0.19
990
0.31
255
0.19
1005
0.32
270
0.19
1020
0.32
285
0.19
1035
0.31
300
0.19
1050
0.32
315
0.20
1065
0.33
330
0.20
1080
0.33
345
0.21
1095
0.33
360
0.21
1110
0.34
375
0.21
1125
0.34
390
0.22
1140
0.34
405
0.21
1155
0.34
420
0.22
1170
0.34
435
0.22
1185
0.34
450
0.23
1200
0.34
465
0.23
1215
0.35
480
0.23
1230
0.35
495
0.24
1245
0.35
510
0.24
1260
0.35
525
0.24
1275
0.35
540
0.24
1290
0.35
555
0.24
1305
0.35
570
0.25
1320
0.36
585
0.25
1335
0.36
600
0.25
1350
0.37
615
0.26
1365
0.37
630
0.26
1380
0.37
645
0.26
1395
0.36
660
0.26
1410
0.36
675
0.26
1425
0.37
690
0.27
1440
0.37
705
0.27
1455
0.37
720
0.28
1470
0.37
735
0.28
1485
0.37
750
0.28
1500
0.37
Using: 440 Filter
0.25 mol dm-3 Hydrogen Peroxide
0.5 mol dm-3 Potassium Iodide
0.25 mol dm-3 Sulphuric Acid
Table 10
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.08
660
1.40
30
0.19
675
1.43
45
0.20
690
1.46
60
0.26
705
1.48
75
0.31
720
1.45
90
0.35
735
1.51
105
0.41
750
1.52
120
0.46
765
1.54
135
0.50
780
1.58
150
0.50
795
1.64
165
0.54
810
1.57
180
0.60
825
1.58
195
0.64
840
2.59
210
0.68
855
1.63
225
0.68
870
1.65
240
0.74
885
1.68
255
0.76
900
1.67
270
0.78
915
1.72
285
0.86
930
1.70
300
0.86
945
1.73
315
0.89
960
1.73
330
0.94
975
1.75
345
0.99
990
1.74
360
0.98
1005
1.76
375
0.99
1020
1.79
390
1.02
1035
1.83
405
1.05
1050
1.83
420
1.09
1065
1.82
435
1.10
1080
1.80
450
1.10
1095
1.85
465
1.16
1110
1.85
480
1.17
1125
1.83
495
1.19
1140
1.85
510
1.22
1155
1.86
525
1.25
1170
1.86
540
1.27
1185
1.86
555
1.29
1200
1.87
570
1.33
1215
1.92
585
1.32
1230
1.93
600
1.34
1245
1.95
615
1.36
1260
1.99
630
1.38
1275
2.00
645
1.38
1290
2.00
Using: 440 Filter
0.1 mol dm-3 Hydrogen Peroxide
0.5 mol dm-3 Potassium Iodide
0.5 mol dm-3 Sulphuric Acid
Table 11
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.08
765
0.90
30
0.13
780
0.90
45
0.19
795
0.93
60
0.19
810
0.89
75
0.26
825
0.91
90
0.28
840
0.92
105
0.31
855
0.92
120
0.34
870
0.92
135
0.33
885
0.95
150
0.36
900
0.96
165
0.38
915
0.95
180
0.41
930
0.95
195
0.44
945
0.96
210
0.46
960
0.98
225
0.48
975
0.97
240
0.49
990
0.97
255
0.52
1005
0.99
270
0.54
1020
0.95
285
0.53
1035
0.97
300
0.54
1050
0.99
315
0.59
1065
1.00
330
0.63
1080
1.02
345
0.65
1095
1.02
360
0.63
1110
1.02
375
0.64
1125
1.01
390
0.66
1140
1.02
405
0.66
1155
1.02
420
0.70
1170
1.03
435
0.69
1185
1.03
450
0.71
1200
1.01
465
0.72
1215
1.05
480
0.74
1230
1.05
495
0.76
1245
1.06
510
0.76
1260
1.05
525
0.76
1275
1.08
540
0.78
1290
1.07
555
0.80
1305
1.07
570
0.83
1320
1.08
585
0.83
1335
1.07
600
0.84
1350
1.07
615
0.82
1365
1.05
630
0.86
1380
1.06
645
0.84
1395
1.06
660
0.85
1410
1.06
675
0.86
1425
1.05
690
0.89
1440
1.07
705
0.85
1455
1.08
720
0.86
1470
1.08
735
0.88
1485
1.08
750
0.90
1500
1.08
Analysis
By graphing my results I can find the order of each reactant. Graph 1 shows how the concentration of Hydrogen peroxide affects the rate of reaction. This graph shows a first order reaction due to the straight line; by looking at the raw data I can confirm this by looking the values for the 0.5 mol dm-3 and 0.25 mol dm-3, I can see that the time taken for the colour to change is doubles when the concentration is halved, at 0.5 mol dm-3 the time is 13 seconds and at 0.25 mol dm-3 the time is 26 seconds. This shows the rate of reaction is proportional to the concentration of the reactant.
The reason for the decrease in time as the concentration increases is due to there being more molecules available to react, meaning there will be a higher proportion of reactions that overcome the activation enthalpy, thus speeding up the reaction.
Looking at both graphs 2 and 3, showing how the concentration of Potassium Iodide and sulphuric acid, respectively affect the rate of reaction. I can see that both are first order.
The overall order of this reaction is then 3, as you can add the individual orders of each reactant. We can see that the rate is proportional to all three reactants.
Therefore I can work the rate equation out to be:
Rate = k [A]a[B]b[C]c
Rate = k [H2O2][I-][H+]
According to the Nuffield book of Data I can see that there are two possible rate equations for this particular experiment, and the affect that the H+ ions have on the rate is differing, and is dependent on the concentrations of H+, it can be either zero order or first order. In my experiment we can see how it is a first order reaction with respect to sulphuric acid.
I first found the rate of reaction in mol s-1 by dividing the iodine concentration by the time for the reaction to occur. The concentration of iodine produced can be found by looking at the equation
2S2O32-(aq) + I2 (aq) S4O62- (aq) + 2I- (aq)
This shows us that for every 1 mole of iodine there are two moles of S2O32- . In this experiment 0.01 mol dm-3 was used, therefore the concentration of iodine produced was 0.005 mol dm-3.
To calculate the rate when hydrogen peroxide concentration is 0.5 mol dm-3, I divided the iodine concentration by the time for the colour to be produced:
=0.005 x 13 = 3.75 x 10-4
I then did this for the rest of the concentrations for all three reactants:
Varying the concentration of Hydrogen peroxide
Concentration mol dm-3
Time to change (s)
Rate
0.50
13
3.85E-04
0.40
14
3.57E-04
0.30
17
2.94E-04
0.25
26
1.92E-04
0.15
31
1.61E-04
0.10
46
1.09E-04
Varying the concentration of Potassium Iodide
Concentration mol dm-3
Time to change (s)
Rate
0.50
26
1.92E-04
0.30
45
1.11E-04
0.25
53
9.43E-05
0.15
93
5.38E-05
0.10
111
4.50E-05
Varying the concentration of Sulphuric Acid
Concentration mol dm-3
Time to change (s)
Rate
0.50
14
3.57E-04
0.30
26
1.92E-04
0.25
54
9.26E-05
0.15
90
5.56E-05
0.10
104
4.81E-05
To calculate the rate constant the equation needs to be rearranged
In theory the rate constant should remain the exact same throughout the above experiments as the concentration does not affect it, however due to experimental error there will be slight fluctuations in it’s value. Calculating K:
I then calculated the rate constant for all concentrations to see how widely it fluctuated.
Varying the concentration of Hydrogen peroxide
Concentration mol dm-3
Rate
Rate Constant
0.50
3.85E-04
3.08E-03
0.40
3.57E-04
3.57E-03
0.30
2.94E-04
3.92E-03
0.25
1.92E-04
3.08E-03
0.15
1.61E-04
4.30E-03
0.10
1.09E-04
4.35E-03
Varying the concentration of Potassium Iodide
Concentration mol dm-3
Rate
Rate Constant
0.50
1.92E-04
3.08E-03
0.30
1.11E-04
2.96E-03
0.25
9.43E-05
3.02E-03
0.15
5.38E-05
2.87E-03
0.10
4.50E-05
3.60E-03
Varying the concentration of Sulphuric Acid
Concentration mol dm-3
Rate
Rate Constant
0.50
3.57E-04
5.71E-03
0.30
1.92E-04
5.13E-03
0.25
9.26E-05
2.96E-03
0.15
5.56E-05
2.96E-03
0.10
4.81E-05
3.85E-03
From the above tables we can see the rate constant fluctuated between 2.87 x 10-3 and 5.71 x 10-3
To calculate the units of the rate constant:
Looking at Graph 4, which shows the affects of a temperature change, I can derive the activation enthalpy by using Arrhenius equation.
To simplify Arrhenius I can then take the natural logarithms, which would imply that a graph of Ln K (rate) against 1/temp would give a straight line.
As rate is proportional to K we can see that a graph of Ln (rate) against 1/T would also give a straight line.
The time taken for an initial reaction is inversely proportional to K rate.
Therefore a graph of Ln (1/Time) against (1/Temperature in Kelvin) will give a straight line with a gradient of . To calculate the activation enthalpy in Joules we can multiply the gradient by R (the Universal gas constant, 8.314J K-1 mol-1.) This is then divided by 1000 to get KJ mol-1.
I converted the temperature into Kelvin by adding 273 to the temperature in centigrade.
Temp (°C)
Time (s)
Rate (1/time)
Ln Rate
Temp (K)
1/Temp (K)
59
12
0.083
-2.48
332
0.003012048
49
32
0.031
-3.47
322
0.00310559
38
47
0.021
-3.85
311
0.003215434
28
85
0.012
-4.44
301
0.003322259
21
109
0.009
-4.69
294
0.003401361
11
247
0.004
-5.51
284
0.003521127 On a graph I plotted the Ln Rate against 1/time in K.
My results:
Gradient = Change in X/ Change in Y = -5941.090194
Activation energy in J = -5941.090194 *-8.314 = 49394.22387
Activation energy in KJ = 49.294
According to Nuffield Book of Data15 the activation enthalpy for this reaction should come out to be: 43.5 KJ mol-1.
Therefore I can see that my results are very close to the published result, and reasons for the slight variations are likely to be down to human and precision errors.
During experiment 5, I investigated the effects of a catalyst whilst varying the temperature. This allowed me to work out the activation enthalpy whilst using Ammonium Molybdate.
Temp (°C)
Time (s)
Rate (1/Time)
Ln Rate
Temp (K)
1/Temp (K)
60
3
0.333
-1.10
333
0.003003003
48
6
0.167
-1.79
321
0.003115265
41
8
0.125
-2.08
314
0.003184713
30
8
0.125
-2.08
303
0.00330033
20
12
0.083
-2.48
293
0.003412969
11
29
0.034
-3.37
284
0.003521127
My results:
Gradient = Change in X/ Change in Y = -4381.192653
Activation energy in J = -4381.192653*-8.314 = 36425.23572
Activation energy in KJ = 36.425
The activation enthalpy of the catalyzed reaction is about 13KJ mol-1 lower than that of the non-catalyzed. Therefore we can see how a catalyst provides a route with a lower alternative energy, thus speeding up the reaction.
The final part of the investigation involved looking at using a colourimeter to monitor the rate of the reaction. This allows us to see the gradual increase of the concentration of Iodine produced. We can see how the lower concentration of all three reactants causes the % absorbance at any given time. It is clear how this backs up my data found in the previous experiments. As the concentration increases the amount of light which is absorbed as it passes through the solution will decrease, as the solution gets darker.
If we look at the colourimetery graph for the concentration of potassium iodide, we can choose a certain level of absorbance and then divide 1 by the time taken to get that absorbance to calculate the rate for each equation. For example:
At 1.00% absorbance:
For 0.5 mol dm-3 concentration it took 255 = 1/255 = 0.002922
For 0.25 mol dm-3 concentration it took 645 = 1/645 = 0.00155
We can see that approximately as the concentration doubles so does the rate of reaction, therefore backing up the previous found order. The rate is not quite due to experimental errors that may have altered the rate of reaction.
I then did this for both the hydrogen peroxide concentration and sulphuric acid concentration:
Hydrogen Peroxide:
At 74% absorbance
0.25 mol dm-3: rate = 0.001852
0.1 mol dm-3: rate = 0.003704
Sulphuric Acid:
At 74% absorbance
0.5 mol dm-3: rate = 0.002778
0.25 mol dm-3: rate = 0.004444
With the above calculations it its clear the reactants are first order.
Evaluation
Below is a table showing the percentage error of the equipment for measurements used during my investigation. The percentage error is calculated by:
Equipment
Precision
Percentage error calculations
Percentage error (%)
250 cm3 Volumetric Flask
±0.30 cm3
0.3/250 x 100
0.12
50 cm3 Volumetric Flask
±0.12 cm3
0.12/50 x 100
0.24
3 Decimal place balance
±0.0005g
0.0005/20.75 x 100
0.002409639
4 Decimal place balance
±0.00005g
0.0005/0.6205 x 100
0.080580177
Burette
±0.10 cm3
0.1/75 x 100
0.133333333
0.1/125 x 100
0.08
25cm3 bulb pipette
±0.06 cm3
0.06/25 x 100
0.24
1 cm3 graduated pipette
±0.01 cm3
0.01/1 x 100
1
Colourimeter
±0.005 A
0.005/0.07 x 100
7.142857143
0.005/2 x 100
0.25
Range of colourimeter percentage error is between 0.25% - 7.14%
Stop clock
±0.5 secs
All times to the nearest second
Thermometer
±0.5°C
0.5/60 x 100
0.833333333
0.5/11 x 100
4.545454545
Range of thermometer percentage error is between 0.83% - 4.55%
As shown above the majority if the percentage errors are minimal, however we can see the largest error was that of the colourimeter which came to 7.14%. Unfortunately there is little that can be done to prevent this, as the reading is so small. I ensured the same colourimeter was used each time, to eliminate as much error as possible.
In replacement of the thermometer, a digital thermometer could be used to gain a more accurate reading of the temperature, as the precision error for it is ±0.05°C. Other than this the equipment used was well chosen as it gave the smallest amount of error possible.
The investigation was accurate as most of the percentage errors are very low and mostly insignificant. Other errors are likely to be that the room temperature varied between 19 and 23 degrees C, throughout the experiment on a day-to-day basis, a factor out of my control, therefore this would have affected the rate constant. To reduce this error and improve the experiment, each experiment could be done in a thermostatically controlled water bath.
After looking at the published data of the activation enthalpy for a non catalysed, I can see my experiment is reliable as the value I got was only 5.794 KJ mol-1 off the actual activation enthalpy, however in order to be able to fully justify my findings I would need to repeat each experiment numerous times, a minimum of three, to then allow me to calculate an average and to disregard any anomalous data. As each experiment was only conducted once, it could be that all the data is anomalous.
By using two different methods at looking at how concentration affects the rate of reaction, my results are more reliable as both methods back each other up. To improve this aspect of the investigation I would use more and a wider variety of concentrations of each reactant, during the colourimetery, as only two concentrations of each were used. The main reason for this was the time allocated to this aspect of the investigation, as each run of colourimetery took around 25 minutes.
To improve and investigate further into this experiment I would create a calibrate curve through the colourimeter by making up solutions of the coloured substance of known concentration, then measuring the absorbance of each, ensuring to use the same conditions as the experiment will be done as. The graph of absorbance against concentration will give your calibration curve. This would then allow me to see how much iodine was produced at each stage of the reaction. Allowing me to closely monitor the rate of reaction.
A limitation of my experiment would be the catalysed experiment; due to the fact the reaction occurred so rapidly. The human reaction time is only accurate to 0.5 seconds, and in some cases the reaction took only 3 seconds, meaning the percentage error is 16.7%. To look into this particular catalyst further, I would dissolve the catalyst and dilute it to lower the concentration; this may give a longer time period before the blue-black complex forms. You would also be able to investigate if the concentration of catalyst changed the rate of reaction. Another potential way to improve this would be to investigate how different catalysts may affect the reaction and to what extent do they lower the activation enthalpy. A catalyst that could be tried would be ammonium iron sulphate, using the iron (III) ions to catalyse the reaction.
Another limitation of the experiment that would have reduced the accuracy would be the fact the blue-black complex forms gradually, therefore it subject able as to when to stop the stopwatch. In order to reduce this error a black cross was marked and the experiment was stopped once the black cross could no loner be seen. However whilst conducting the temperature experiments the reaction was done in test tubes and no cross could be used, to minimise the error the stop clock was stopped immediately at the first sign of the blue-black complex.
The final main limitation would be that the conical flask was swirled in order to mix the two solutions. Although best efforts were made to ensure the solutions were swirled evenly in each experiment, it is difficult to control. An improvement would be to use a magnetic stirrer, set at the same speed to ensure the same kinetic energy throughout the solutions, ensuring this did not affect the reaction.
As found in the Nuffield Book of Data, I can see that the order of the reaction with respect to H+ ions is dependent on the concentration. To further my investigation it would be interesting to look further into how the concentration of sulphuric acid may cause the order to be either 1 or 0 with respect to the H+ ions. I could then look at the point at which it changed from being zero order to first order. It would then be interesting to see how this affected the activation enthalpy of the experiment.
I can conclude my results are accurate and reliable, due to the fact the equipment was chosen with low precision errors and any errors given were too small to have a large impact. This can be backed up by the published data found in the Nuffield Book of Data.
Aim: To determine the rate equation for the reaction between Hydrogen peroxide and iodide ions, and to investigate the effects of a catalyst and temperatures on the reaction and to derive the activation enthalpy.
Background knowledge:
1Hydrogen peroxide reacts with iodide ions producing iodine, when in an aqueous acid solution.
H2O2 (aq) + 2I- (aq) + 2H3O+(aq) I2 (aq) + 2H2O(l)
To detect iodine you can look at the color shown by the addition of starch solution.
Iodine also reacts with sodium thiosulphate solution, which acts as a delaying agent. Adding sodium thiosulphate to the reaction mixture will allow the iodine produced initially to react with the thiosulphate ions, which reduce the iodine to iodide ions, leaving the starch unchanged.
2S2O32-(aq) + I2 (aq) S4O62- (aq) + 2I- (aq)
Once all the thiosulphate ions have reacted with the iodine, the iodine will then go on to react with the starch, producing a blue-black complex.
2The reason for the blue-black complex is due to the formation of polyiodide chains during the reaction between starch and iodine. The amylose in starch forms helices with which the iodine molecules align, causing a transfer of charge. This charge transfer corresponds to the absorption spectrum, in which the blue-black colour is the complementary colour. The details of this reaction are not fully known yet. The strength and deepness of the colour is dependent on the amount of amylose present.
3The rate of the reaction can be explained by the collision theory, which shows that the rate can be altered by4: concentrations, pressure, temperature, intensity of radiation, particle size, surface area and a catalyst. In this experiment I will be looking at concentrations, temperatures and the affect of a catalyst. The collision theory also states a reaction will only take place if three conditions are met:
1. Reactant particles collide with each other
2. The reactant particles must collide with the correct orientation.
3. The collision must provide enough energy to overcome the activation energy.
2This is due to the fact, if particles collide but are not orientated correctly the molecules will just bounce of each other, this is often due to charge of the molecules which causes repulsion if the orientation is incorrect.
If particles collide with the correct orientation, they must collide with a minimum energy otherwise they will just bounce of each other. The activation energy is used to break some of the original bonds, which is essential for a reaction to occur. The activation energy is this minimum energy and can be shown on an energy profile. We can see from the below graphs how in an endothermic reaction the energy of the reactants is lower than the energy of the products, this is why it absorbs heat. Whereas an exothermic reaction will give out heat as the reactants have a higher energy than the products.
2The activation energy can be marked on the Maxwell-Boltzmann distribution curve. The Maxwell-Boltzmann distribution refers directly to gases, however the principles can be taken and applied to liquid reactions also.
5The area underneath the curve represents the different particles and their level of energy. Only the particles that have energy higher than the activation energy will undergo a reaction. We can see that the kinetic energy of a molecule can never be zero, but theoretically can be ever increasing, as there is no defined maximum energy value.
You can then change the shape of the curve or move the activation energy in order to increase the number of collisions that overcome the activation enthalpy. Changing certain variables can do this, which is what will be done in this experiment.
6 7 The first variable I will be varying is concentration of the different reactants; I will be using 5 different concentrations for each reactant.
Increasing the concentration of a reactant means that there are a higher proportion of particles per unit volume. This then causes the rate of reaction to increase because particles have a higher chance of colliding with sufficient kinetic energy to cause a reaction.
8By varying the concentration of each reactant it allows me to calculate the order of the reaction, by investigating their effects. By finding the order I can then calculate the rate constant and rate equation.
The chemical kinetics of this reaction can be used to explain the mechanisms and rates of the reaction. I will plot a graph of the effects of the concentrations and this will enable me to determine the order. The rate equation shows how the reaction rate depends on the concentration of each substance.
9The rate equation of a reaction is:
Rate = k [A]a[B]b
K is considered the rate constant, and this can be found once the experiment has been conducted. The rate constant is 10‘A coefficient of proportionality relating the rate of a chemical reaction at a given temperature to the concentration of reactant (in a unimolecular reaction) or to the product of the concentrations of reactants’
The rate constant will change if the temperature is changed or a catalyst is added. The rate is measured in mol dm-3 s-1. The a and b are the order of the reaction in respect to A and B respectively. These orders show how the concentrations of A and B affect the reaction.
The order of a reaction is found in respect to a reactant and is defined as ‘the power to which its concentration term in the rate equation is raised. The overall order of a reaction is all of the separate orders added together.
Overall order = a + b.
The order of a reaction can only be found experimentally, it is not possible to deduce the order just by looking at the equation.
The order of a reaction can be determined graphically or by looking directly at how the concentration affects the rate.
If the order of a reactant is 0, this means that the concentration of the particular reactant does not affect the rate of reaction. Therefore this would not appear in the rate equation, as anything to the power of 0 is 1.
A graph of a zero order reaction would be a horizontal straight line if the initial rate was plotted against concentration.
If you plotted concentration against time the graph would be a decreasing straight line.
A first order reaction is identified when doubling the concentration of a reactant doubles the reaction rate. This means that the rate of reaction is proportional to the concentration of the reactant.
Rate [A]
Which can also be written as:
Rate = k [A]
A graph of the initial rate against concentration will be a straight line.
A second order reaction is identified when doubling the concentration of a reactant quadruples the reaction rate. This means that the rate of reaction is proportional to the square of the concentration of the reactant.
Rate [A]2
Which can also be written as:
Rate = k [A]2
A graph of initial rate against concentration would be a curve.
A graph of initial rate against concentration2 would be a straight line.
A reaction mechanism shows the steps in which a reaction breaks or makes bonds, all chemical changes occur due to the making and breaking of bonds. The rate equation also allows you to establish the rate-determining step. This is the slowest step, which controls the overall rate of a reaction, because the reaction cannot react any faster than the slowest step. This then allows you to determine the reaction mechanism.
11The reaction mechanism of this particular reaction is as followed.
The first step is a slow reaction between the iodide ions and the hydrogen peroxide.
H2O2 (aq) + I- (aq) IO- (aq) + H2O (l)
The second step reaches equilibrium quickly.
IO- (aq) + H3O+(aq) ⇌ HIO (aq) + H2O (l)
The final step is when the Iodine is formed
HIO (aq) + H3O (aq) + I- (aq) I2 (aq) + 2H2O (l)
The slowest step is step 1; this means this is the rate-determining step.
The second variable I will investigate is the effects of temperature, by investigating 5 different temperatures. An approximation is that if you increase the temperature by 10 °C you will double the rate of the reaction. The collision theory demonstrates how at higher temperatures the rate of a reaction will be faster, because each particle has a higher kinetic energy and therefore moves at a faster rate. This will increase the likelihood of a collision and the proportion that will have enough energy to overcome the activation enthalpy. I will be able to calculate the activation enthalpy, which is ‘the minimum kinetic energy required by a pair of colliding particles before the reaction will occur’. By increasing the temperature, this then changes the shape of the Maxwell-Boltzmann distribution curve. Meaning there are more particles that will collide with enough energy to overcome the activation energy.
I will be using the Arrhenius equation to find the activation enthalpy, the energy needed in order for a reaction to occur, by looking at how rate equations vary with respect to temperature.
The Arrhenius equation is:
Which can also be written in the form:
The different variables in this equation are:
T – Temperature, in kelvin
R – Gas constant (8.314J K-1 mol-1.), Derives from the equation pV=nRT.
Ea – Activation energy, the minimum energy needed for a reaction.
A – Arrhenius constant
The Arrhenius equation allows us to see the effect of a change of temperature on the rate constant and the rate of reaction.
The final variable I will be investigating is the effect of a catalyst, I will be using Ammonium Molybdate, which is ‘a substance which speeds up a reaction, but remains chemically unchanged at the end of the reaction’. A catalyst provides an alternative route for the reaction with a lower activation enthalpy, this will mean that when particles collide a higher proportion will have enough kinetic energy to overcome the lower activation enthalpy provided by the catalyst. This can be seen on an energy profile.
By adding a catalyst we are lowering the activation energy needed, this can be shown on the Maxwell-Boltzmann distribution curve. We can see how the activation energy has moved to the left, meaning a higher number of particles have enough energy to overcome it.
The catalyst I will be using is Ammonium Molybdate, which is a transition metal. A transition metal can be defined as ‘one which forms one or more stable ions which have incompletely filled d orbitals.’ And can be referred to as the d block elements. Transition metals are commonly used as catalysts due to their capability to change oxidation state, or because they can adsorb substances onto their surface and activate them in the process.
Catalysts can be either homogeneous or heterogeneous. Homogeneous catalysts are in the same physical state as the reactants, whereas heterogeneous catalysts are in a different state. In this experiment my catalyst with my heterogeneous as Ammonium Molybdate is a solid.
When the reactant is adsorbed on the surface of the catalyst the bonds within the reactant are weakened. The bonds break and the reactant atoms bond with the surface of the catalyst. New bonds form between the atoms and the products are released and diffuse away.
12Below is a diagram using Ethene, showing hydrogenation of the C=C.
Step 1 – the reactants adsorbing onto the catalyst surface and the H has become dissociated.
Step 2 – Products begin to form with new bonds being made.
Step 3 – The product is formed and diffuses away.
Risk Assessment1314
Chemical
Hazards
Key Risks
Hydrogen peroxide
0.25 mol dm-3
Low hazard at concentrations lower than 1.5 mol dm-3.
Contact with eyes
Sulphuric Acid
0.5 mol dm-3
Irritant
Contact with eyes/skin.
Solid potassium iodide/
Irritant
Contact with eyes
Starch
Low Hazard
Solid Sodium Thiosulphate
Mild Irritant
Contact with skin/eyes
Ammonium Molybdate
Irritant
Contact with eyes
Iodine (Produced)
Low hazard at concentrations less that 1 mol dm-3
Contact with skin and eyes.
Ways to minimize the potential chance of accidents:
Safety glasses + lab coat.
Keep workbench tidy and organized.
Clean up all spills immediately with plenty of water to dilute the chemicals.
Ensure the laboratory is well ventilated by opening windows.
Do not eat or drink in the laboratory.
If contact with skin occurs wash thoroughly with water
If contact with the eye gently flood the eye for at least 10 minutes with water.
As the concentrations of my chemicals are very low the risks are minimal.
Apparatus:
1cm3 graduated pipettes
25cm3 bulb pipette
250cm3 volumetric flask
50cm3 volumetric flask
Burette
Beakers
100cm3 conical flasks
Stop watch
Boiling tubes
Test tubes
Electronic Balance (3 decimal place and 4 decimal place)
Spatula
Glass Rod
Funnel
Water Baths at varying temperatures
Colourimeter
Cuvette
Distilled water
White tile and black marker pen
Dropping pipette
Method:
Preliminary experiment
A preliminary experiment must be conducted to obtain optimum volumes and concentrations to be used. This was to ensure the experiment would be long enough to record, but would not take too long to produce the blue-black complex. This is in order to produce a sufficient number of results from which my conclusions will be drawn. A ‘trial and error’ approach was used to come to the conclusion of which volumes and concentrations to use.
The concentrations I decided on were:
Hydrogen peroxide 0.25 mol dm-3
Potassium iodide 0.5 mol dm-3
Sulphuric acid 0.5 dm-3
Sodium thiosulphate 0.01 mol dm-3
Making up Potassium Iodine solution:
1. To work out the mass of potassium iodine to be dissolved:
a.
2. Weigh out 20.75g into a small beaker using a 3 decimal place measuring balance.
3. Add distilled water and dissolve the potassium iodine stirring with a glass rod.
4. Transfer solution into a 250cm3 volumetric flask using a funnel.
5. Wash the beaker, glass rod and funnel with distilled water into the volumetric flask.
6. Add distilled water up to the calibration mark on the volumetric flask, ensuring the bottom of the meniscus is on the calibration mark, and read from eye-level to ensure an accurate reading.
7. Insert the stopper into the volumetric flask and invert several times, to thoroughly mix the solution.
Making up Sodium Thiosulphate solution:
1. To work out the mass of Sodium Thiosulphate to be dissolved:
a.
2. Weigh out 0.6205g into a small beaker using a 4 decimal place measuring balance.
3. Add distilled water and dissolve the sodium thiosulphate stirring with a glass rod.
4. Transfer solution into a 250cm3 volumetric flask using a funnel.
5. Wash the beaker, glass rod and funnel with distilled water into the volumetric flask.
6. Add distilled water up to the calibration mark on the volumetric flask, ensuring the bottom of the meniscus is on the calibration mark, and read from eye-level to ensure an accurate reading.
7. Insert the stopper into the volumetric flask and invert several times, to thoroughly mix the solution, this is to ensure the concentration is the same throughout.
Making up Hydrogen Peroxide solution:
1. To convert the number of volumes into mol dm-3
a. Starting with a 20 volume solution
a.i. 20/12 = 1.667 mol dm-3
2. To find the volume of original solution to dilute:
a.
3. Measure 75 cm3 of 20 volumes Hydrogen Peroxide using a burette
4. Transfer into a 250cm3 volumetric flask.
5. Fill the volumetric flask up to the calibration mark with distilled water, ensuring the meniscus is on the calibration mark, reading at eye-level to ensure an accurate reading.
6. Insert the stopper and invert several times to thoroughly mix the solution; this is to ensure the concentration is the same throughout.
Diluting Sulphuric Acid solution
1. Work out the volume of solution needed to be diluted
a.
2. Measure 125 cm3 of 1.0 mol dm-3 Sulphuric Acid solution using a burette.
3. Add the Sulphuric Acid into a 250cm3 volumetric flask, using a funnel.
4. Fill the volumetric flask up to the calibration mark with distilled water, ensuring the meniscus is on the calibration mark, reading at eye-level to ensure an accurate reading.
5. Insert the stopper and invert several times to thoroughly mix the solution; this is to ensure the concentration is the same throughout.
Diluting any concentration to the desired concentration
Throughout the experiment it will be necessary to dilute solutions down to the desired concentration
1. Using the equation below
2. The volume calculated will be measured and distilled water will be added to make up the necessary volume.
The basic clock reaction – Basic Experiment
1. Add 25cm3 of the desired concentration of Sulphuric Acid into a conical flask, using a bulb pipette.
2. Using a graduated pipette add to the 100cm3 conical flask (ensure you use a clean graduated pipette for each chemical to avoid contamination), I will ensure the meniscus is read at eye level to ensure an accurate reading:
a. 1cm3 of the 0.01 mol dm-3 sodium thiosulphate
b. 1cm3 of the desired concentration of potassium iodide
c. 1cm3 of starch solution (ensuring you use a freshly made batch each day, as it has a short shelf life, due to it undergoing photochemical reactions forming unwanted products)
3. Using a clean graduated pipette add 1cm3 of the desired concentration of Hydrogen Peroxide solution into a test tube, A.
4. Add test tube A solution into the conical flask and start the stopwatch immediately.
5. Swirl the conical flask as evenly as possible over a white tile with a black cross drawn on.
6. Stop the stopwatch as soon as you can no longer see the black cross.
7. Record results.
Varying the Hydrogen Peroxide, Potassium Iodide and Sulphuric Acid concentrations
1. Hydrogen Peroxide (Experiment 1)
a. Make up 5 different concentrations via dilutions.
a.i. The concentrations I will use will be 0.5 mol dm-3, 0.4 mol dm-3, 0.3 mol dm-3, 0.25 mol dm-3, 0.15 mol dm-3, 0.1 mol dm-3
This will allow me to see the affect of doubling the concentration.
a.ii. To make up these concentrations I will dilute the previously made 0.5 mol dm-3 Hydrogen Peroxide solution for each desired concentration, using the previously mentioned equation.
a.iii.
a.iv. I will measure the value calculated using a graduated pipette and transfer it into a 50cm3 volumetric flask and add the required amount of water to fill the volumetric flask up to the pipette.
b. Carry out the basic experiment (the basic iodine clock reaction) for each concentration using:
b.i. 0.5 mol dm-3 Sulphuric Acid
b.ii. 0.5 mol dm-3 Potassium Iodide
b.iii. And varying the concentration of hydrogen peroxide for each reaction.
2. Potassium Iodide (Experiment 2)
a. Make up 5 different concentrations via dilutions.
a.i. The concentrations I will use will be 0.5 mol dm-3, 0.3 mol dm-3, 0.25 mol dm-3, 0.15 mol dm-3, and 0.1 mol dm-3. This will allow me to see the affect of doubling the concentration.
a.ii. To make up these concentrations I will dilute the previously made 0.5 mol dm-3 Potassium Iodide solution for each desired concentration, using the previously mentioned equation.
i. I will measure the value calculated above using a graduated pipette and transfer it into a 50cm3 volumetric flask and add the required amount of water to fill the volumetric flask up to the pipette.
b. Carry out basic experiment (the basic iodine clock reaction) for each concentration using:
i. 0.5 mol dm-3 Sulphuric Acid ii. 0.25 mol dm-3 Hydrogen Peroxide iii. And varying the concentration of potassium iodide for each reaction.
3. Sulphuric Acid (Experiment 3)
a. Make up 5 different concentrations via dilutions.
a.i. The concentrations I will use are 1.0 mol dm-3, 0.5 mol dm-3, 0.25 mol dm-3, 0.3 mol dm-3 and 0.15 mol dm-3
This will allow me to see the affect of doubling the concentration.
a.ii. To make up these concentrations I will dilute the previously made 0.5 mol dm-3 Sulphuric Acid solution for each desired concentration, using the previously mentioned equation.
a.iii. I will measure the value calculated above using a graduated pipette and transfer it into a 250cm3 volumetric flask and add the required amount of water to fill the volumetric flask up to the pipette.
b. Carry out the basic experiment (the basic iodine clock reaction) for each concentration using:
b.i. Varying concentrations of Sulphuric Acid for each reaction
b.ii. 0.25 mol dm-3 Hydrogen Peroxide
b.iii. 0.5 mol dm-3 Potassium Iodide
Varying the temperature of the reaction (Experiment 4)
1. I will be using 60 °C, 50°C, 40°C, 30°C, 10°C and room temperature.
2. Add 25cm3 of the 0.5 mol dm-3 of Sulphuric Acid into a conical flask., using a bulb pipette.
3. Using a graduated pipette add to the boiling tube, A (ensure you use a clean graduated pipette for each chemical to avoid contamination):
a. 1cm3 of the 0.01 mol dm-3 sodium thiosulphate
b. 1cm3 of the 0.1 mol dm-3 Potassium Iodide
c. 1cm3 of starch solution (ensuring you use a freshly made batch each day)
4. Using a clean graduated pipette add 1cm3 of 0.1 mol dm-3 Hydrogen Peroxide solution into a test tube, B.
5. Place both A and B into a water bath for 5 minutes. Measure the temperature of the solutions ensuring to clean the thermometer for each to avoid contamination. Record the temperature as the water bath temperature is likely not to be 100% correct as to what is labeled on the bath.
6. Add test tube B solution into the boiling tube and start the stopwatch immediately.
7. Quickly add a bung and invert the boiling tube to ensure the Hydrogen Peroxide is evenly distributed, take out the bung once inverted once.
8. Swirl the boiling tube as evenly as possible keeping it in the water bath.
9. Stop the stopwatch as soon as a blue-black colour appears.
10. Record results.
Varying the temperature of the reaction with a catalyst – Experiment 5
1. Repeat experiment4 adding 0.02 g of Ammonium Molybdate into boiling tube A during stage 1.
Using a colourimeter to monitor the rate of reaction - Basic Colourimeter Experiment
1. Set the colourimeter filter to 440, as it has the complementary colour to that of the substance being formed. To find this, I carried out the experiment whilst varying the filter until the best one was decided.
2. Fill a cuvette with distilled water, ensure when picking up the cuvette you only hold the cloudy side, and placing it with the arrow facing towards you. Press the calibration button; this will set the absorbance for zero.
3. Add 25cm3 of the desired concentration of Sulphuric Acid into a boiling tube, A, using a volumetric pipette.
a. Using a graduated pipette add to the boiling tube, 1cm3 of the desired concentration of potassium iodide
4. Using a clean graduated pipette add 1cm3 of the desired concentration of Hydrogen Peroxide solution into a test tube, B.
5. Add test tube B solution into the boiling tube, A and start the stopwatch immediately.
6. Quickly use a dropping pipette fill the cuvette, and place in the colorimeter, ensuring you only hold it on the cloudy sides, the arrow is facing you when placed in the colourimeter, and the outside of the cuvette is dry.
7. Measure the absorbance every 15 seconds, until the maximum absorbance is reached or you reach 25 minutes.
Varying the concentrations of sulphuric acid, hydrogen peroxide and potassium iodide using a colourimeter
Varying Hydrogen Peroxide (Experiment 7)
1. I will be using 0.25 mol dm-3 and 0.1 mol dm-3
a. To get these concentrations I will dilute using the previously mentioned equation and method.
2. Carry out the Basic colourimeter experiment using:
a. 0.5 mol dm-3 potassium iodide
b. 0.5 mol dm-3 sulphuric acid
Varying Potassium Iodide (Experiment 8)
1. I will be using 0.25 mol dm-3 and 0.5 mol dm-3
a. To get these concentrations I will dilute using the previously mentioned equation and method.
2. Carry out the Basic colourimeter experiment using:
a. 0.25 mol dm-3 hydrogen peroxide
b. 0.5 mol dm-3 sulphuric acid
Varying Sulphuric Acid (Experiment 9)
1. I will be using 0.25 mol dm-3 and 0.5 mol dm-3
a. To get these concentrations I will dilute using the previously mentioned equation and method.
2. Carry out the Basic colourimeter experiment using:
a. 0.5 mol dm-3 potassium iodide
b. 0.25 mol dm-3 hydrogen peroxide
Results
Table 1
Varying the concentration of Hydrogen peroxide
Concentration mol dm-3
Time to change (s)
Rate (1/time)
0.50
13
0.077
0.40
14
0.071
0.30
17
0.059
0.25
26
0.038
0.15
31
0.032
0.10
46
0.022
Table 2
Varying the concentration of Potassium Iodide
Concentration mol dm-3
Time to change (s)
Rate (1/time)
0.50
26
0.038
0.30
45
0.022
0.25
53
0.019
0.15
93
0.011
0.10
111
0.009
Table 3
Varying the concentration of Sulphuric Acid
Concentration mol dm-3
Time to change (s)
Rate (1/time)
0.50
14
0.071
0.30
26
0.038
0.25
54
0.019
0.15
90
0.011
0.10
104
0.010
Table 4
Affects of a change in temperature
Temperature °C
Time to change (s)
59
12
49
32
38
47
28
85
21
109
11
247
Table 5 – Affects of a change in temperature Arrhenius calculations
Temp (°C)
Time (s)
Rate (1/time)
Ln Rate
Temp (K)
1/Temp (K)
59
12
0.083
-2.48
332
0.003012048
49
32
0.031
-3.47
322
0.00310559
38
47
0.021
-3.85
311
0.003215434
28
85
0.012
-4.44
301
0.003322259
21
109
0.009
-4.69
294
0.003401361
11
247
0.004
-5.51
284
0.003521127
Table 6 – Affects of a change in temperature catalyzed Arrhenius calculations
Temp (°C)
Time (s)
Rate (1/Time)
Ln Rate
Temp (K)
1/Temp (K)
60
3
0.333
-1.10
333
0.003003003
48
6
0.167
-1.79
321
0.003115265
41
8
0.125
-2.08
314
0.003184713
30
8
0.125
-2.08
303
0.00330033
20
12
0.083
-2.48
293
0.003412969
11
29
0.034
-3.37
284
0.003521127 Without catalyst 20°C = 149 s
Colorimeter results
Using: filter 440.
0.25 mol dm-3 Hydrogen Peroxide
0.5 mol dm-3 Potassium Iodide
0.5 mol dm-3 Sulphuric Acid
Table 7
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.08
495
1.52
30
0.17
510
1.51
45
0.25
525
1.56
60
0.33
540
1.57
75
0.42
555
1.57
90
0.47
570
1.60
105
0.54
585
1.61
120
0.59
600
1.64
135
0.64
615
1.67
150
0.70
630
1.69
165
0.84
645
1.69
180
0.82
660
1.72
195
0.84
675
1.74
210
0.88
690
1.74
225
0.98
705
1.79
240
0.97
720
1.77
255
1.01
735
1.80
270
1.10
750
1.81
285
1.13
765
1.81
300
1.13
780
1.86
315
1.19
795
1.91
330
1.22
810
1.87
345
1.30
825
1.94
360
1.29
840
1.91
375
1.32
855
1.93
390
1.38
870
1.91
405
1.37
885
1.95
420
1.39
900
1.96
435
1.42
915
1.96
450
1.45
930
1.99
465
1.50
945
2.00
480
1.52
960
2.00
Using: 440 Filter
0.25 mol dm-3 Hydrogen Peroxide
0.25 mol dm-3 Potassium Iodide
0.5 mol dm-3 Sulphuric Acid
Table 8
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.07
765
1.09
30
0.10
780
1.14
45
0.15
795
1.12
60
0.20
810
1.12
75
0.19
825
1.13
90
0.25
840
1.15
105
0.28
855
1.18
120
0.30
870
1.20
135
0.34
885
1.21
150
0.36
900
1.22
165
0.42
915
1.21
180
0.44
930
1.23
195
0.48
945
1.26
210
0.49
960
1.27
225
0.51
975
1.28
240
0.54
990
1.29
255
0.57
1005
1.29
270
0.59
1020
1.31
285
0.59
1035
1.30
300
0.63
1050
1.32
315
0.64
1065
1.29
330
0.67
1080
1.31
345
0.68
1095
1.34
360
0.71
1110
1.34
375
0.73
1125
1.36
390
0.74
1140
1.35
405
0.80
1155
1.36
420
0.82
1170
1.36
435
0.83
1185
1.34
450
0.83
1200
1.32
465
0.84
1215
1.34
480
0.84
1230
1.34
495
0.88
1245
1.37
510
0.87
1260
1.36
525
0.90
1275
1.36
540
0.91
1290
1.39
555
0.95
1305
1.39
570
0.94
1320
1.38
585
0.95
1335
1.38
600
0.95
1350
1.37
615
0.99
1365
1.39
630
0.99
1380
1.43
645
1.00
1395
1.41
660
1.00
1410
1.42
675
1.02
1425
1.42
690
1.05
1440
1.45
705
1.05
1455
1.43
720
1.07
1470
1.44
735
1.08
1485
1.44
750
1.08
1500
1.44
Using: 440 Filter
0.25 mol dm-3 Hydrogen Peroxide
0.01 mol dm-3 Potassium Iodide
0.5 mol dm-3 Sulphuric Acid
Table 9
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.12
765
0.28
30
0.12
780
0.28
45
0.13
795
0.29
60
0.13
810
0.29
75
0.14
825
0.29
90
0.13
840
0.30
105
0.15
855
0.30
120
0.15
870
0.30
135
0.16
885
0.30
150
0.16
900
0.30
165
0.17
915
0.30
180
0.17
930
0.31
195
0.18
945
0.31
210
0.18
960
0.31
225
0.18
975
0.32
240
0.19
990
0.31
255
0.19
1005
0.32
270
0.19
1020
0.32
285
0.19
1035
0.31
300
0.19
1050
0.32
315
0.20
1065
0.33
330
0.20
1080
0.33
345
0.21
1095
0.33
360
0.21
1110
0.34
375
0.21
1125
0.34
390
0.22
1140
0.34
405
0.21
1155
0.34
420
0.22
1170
0.34
435
0.22
1185
0.34
450
0.23
1200
0.34
465
0.23
1215
0.35
480
0.23
1230
0.35
495
0.24
1245
0.35
510
0.24
1260
0.35
525
0.24
1275
0.35
540
0.24
1290
0.35
555
0.24
1305
0.35
570
0.25
1320
0.36
585
0.25
1335
0.36
600
0.25
1350
0.37
615
0.26
1365
0.37
630
0.26
1380
0.37
645
0.26
1395
0.36
660
0.26
1410
0.36
675
0.26
1425
0.37
690
0.27
1440
0.37
705
0.27
1455
0.37
720
0.28
1470
0.37
735
0.28
1485
0.37
750
0.28
1500
0.37
Using: 440 Filter
0.25 mol dm-3 Hydrogen Peroxide
0.5 mol dm-3 Potassium Iodide
0.25 mol dm-3 Sulphuric Acid
Table 10
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.08
660
1.40
30
0.19
675
1.43
45
0.20
690
1.46
60
0.26
705
1.48
75
0.31
720
1.45
90
0.35
735
1.51
105
0.41
750
1.52
120
0.46
765
1.54
135
0.50
780
1.58
150
0.50
795
1.64
165
0.54
810
1.57
180
0.60
825
1.58
195
0.64
840
2.59
210
0.68
855
1.63
225
0.68
870
1.65
240
0.74
885
1.68
255
0.76
900
1.67
270
0.78
915
1.72
285
0.86
930
1.70
300
0.86
945
1.73
315
0.89
960
1.73
330
0.94
975
1.75
345
0.99
990
1.74
360
0.98
1005
1.76
375
0.99
1020
1.79
390
1.02
1035
1.83
405
1.05
1050
1.83
420
1.09
1065
1.82
435
1.10
1080
1.80
450
1.10
1095
1.85
465
1.16
1110
1.85
480
1.17
1125
1.83
495
1.19
1140
1.85
510
1.22
1155
1.86
525
1.25
1170
1.86
540
1.27
1185
1.86
555
1.29
1200
1.87
570
1.33
1215
1.92
585
1.32
1230
1.93
600
1.34
1245
1.95
615
1.36
1260
1.99
630
1.38
1275
2.00
645
1.38
1290
2.00
Using: 440 Filter
0.1 mol dm-3 Hydrogen Peroxide
0.5 mol dm-3 Potassium Iodide
0.5 mol dm-3 Sulphuric Acid
Table 11
Time (s)
Absorbance
Time (s)
Absorbance
(absorbance units) (absorbance units)
15
0.08
765
0.90
30
0.13
780
0.90
45
0.19
795
0.93
60
0.19
810
0.89
75
0.26
825
0.91
90
0.28
840
0.92
105
0.31
855
0.92
120
0.34
870
0.92
135
0.33
885
0.95
150
0.36
900
0.96
165
0.38
915
0.95
180
0.41
930
0.95
195
0.44
945
0.96
210
0.46
960
0.98
225
0.48
975
0.97
240
0.49
990
0.97
255
0.52
1005
0.99
270
0.54
1020
0.95
285
0.53
1035
0.97
300
0.54
1050
0.99
315
0.59
1065
1.00
330
0.63
1080
1.02
345
0.65
1095
1.02
360
0.63
1110
1.02
375
0.64
1125
1.01
390
0.66
1140
1.02
405
0.66
1155
1.02
420
0.70
1170
1.03
435
0.69
1185
1.03
450
0.71
1200
1.01
465
0.72
1215
1.05
480
0.74
1230
1.05
495
0.76
1245
1.06
510
0.76
1260
1.05
525
0.76
1275
1.08
540
0.78
1290
1.07
555
0.80
1305
1.07
570
0.83
1320
1.08
585
0.83
1335
1.07
600
0.84
1350
1.07
615
0.82
1365
1.05
630
0.86
1380
1.06
645
0.84
1395
1.06
660
0.85
1410
1.06
675
0.86
1425
1.05
690
0.89
1440
1.07
705
0.85
1455
1.08
720
0.86
1470
1.08
735
0.88
1485
1.08
750
0.90
1500
1.08
Analysis
By graphing my results I can find the order of each reactant. Graph 1 shows how the concentration of Hydrogen peroxide affects the rate of reaction. This graph shows a first order reaction due to the straight line; by looking at the raw data I can confirm this by looking the values for the 0.5 mol dm-3 and 0.25 mol dm-3, I can see that the time taken for the colour to change is doubles when the concentration is halved, at 0.5 mol dm-3 the time is 13 seconds and at 0.25 mol dm-3 the time is 26 seconds. This shows the rate of reaction is proportional to the concentration of the reactant.
The reason for the decrease in time as the concentration increases is due to there being more molecules available to react, meaning there will be a higher proportion of reactions that overcome the activation enthalpy, thus speeding up the reaction.
Looking at both graphs 2 and 3, showing how the concentration of Potassium Iodide and sulphuric acid, respectively affect the rate of reaction. I can see that both are first order.
The overall order of this reaction is then 3, as you can add the individual orders of each reactant. We can see that the rate is proportional to all three reactants.
Therefore I can work the rate equation out to be:
Rate = k [A]a[B]b[C]c
Rate = k [H2O2][I-][H+]
According to the Nuffield book of Data I can see that there are two possible rate equations for this particular experiment, and the affect that the H+ ions have on the rate is differing, and is dependent on the concentrations of H+, it can be either zero order or first order. In my experiment we can see how it is a first order reaction with respect to sulphuric acid.
I first found the rate of reaction in mol s-1 by dividing the iodine concentration by the time for the reaction to occur. The concentration of iodine produced can be found by looking at the equation
2S2O32-(aq) + I2 (aq) S4O62- (aq) + 2I- (aq)
This shows us that for every 1 mole of iodine there are two moles of S2O32- . In this experiment 0.01 mol dm-3 was used, therefore the concentration of iodine produced was 0.005 mol dm-3.
To calculate the rate when hydrogen peroxide concentration is 0.5 mol dm-3, I divided the iodine concentration by the time for the colour to be produced:
=0.005 x 13 = 3.75 x 10-4
I then did this for the rest of the concentrations for all three reactants:
Varying the concentration of Hydrogen peroxide
Concentration mol dm-3
Time to change (s)
Rate
0.50
13
3.85E-04
0.40
14
3.57E-04
0.30
17
2.94E-04
0.25
26
1.92E-04
0.15
31
1.61E-04
0.10
46
1.09E-04
Varying the concentration of Potassium Iodide
Concentration mol dm-3
Time to change (s)
Rate
0.50
26
1.92E-04
0.30
45
1.11E-04
0.25
53
9.43E-05
0.15
93
5.38E-05
0.10
111
4.50E-05
Varying the concentration of Sulphuric Acid
Concentration mol dm-3
Time to change (s)
Rate
0.50
14
3.57E-04
0.30
26
1.92E-04
0.25
54
9.26E-05
0.15
90
5.56E-05
0.10
104
4.81E-05
To calculate the rate constant the equation needs to be rearranged
In theory the rate constant should remain the exact same throughout the above experiments as the concentration does not affect it, however due to experimental error there will be slight fluctuations in it’s value. Calculating K:
I then calculated the rate constant for all concentrations to see how widely it fluctuated.
Varying the concentration of Hydrogen peroxide
Concentration mol dm-3
Rate
Rate Constant
0.50
3.85E-04
3.08E-03
0.40
3.57E-04
3.57E-03
0.30
2.94E-04
3.92E-03
0.25
1.92E-04
3.08E-03
0.15
1.61E-04
4.30E-03
0.10
1.09E-04
4.35E-03
Varying the concentration of Potassium Iodide
Concentration mol dm-3
Rate
Rate Constant
0.50
1.92E-04
3.08E-03
0.30
1.11E-04
2.96E-03
0.25
9.43E-05
3.02E-03
0.15
5.38E-05
2.87E-03
0.10
4.50E-05
3.60E-03
Varying the concentration of Sulphuric Acid
Concentration mol dm-3
Rate
Rate Constant
0.50
3.57E-04
5.71E-03
0.30
1.92E-04
5.13E-03
0.25
9.26E-05
2.96E-03
0.15
5.56E-05
2.96E-03
0.10
4.81E-05
3.85E-03
From the above tables we can see the rate constant fluctuated between 2.87 x 10-3 and 5.71 x 10-3
To calculate the units of the rate constant:
Looking at Graph 4, which shows the affects of a temperature change, I can derive the activation enthalpy by using Arrhenius equation.
To simplify Arrhenius I can then take the natural logarithms, which would imply that a graph of Ln K (rate) against 1/temp would give a straight line.
As rate is proportional to K we can see that a graph of Ln (rate) against 1/T would also give a straight line.
The time taken for an initial reaction is inversely proportional to K rate.
Therefore a graph of Ln (1/Time) against (1/Temperature in Kelvin) will give a straight line with a gradient of . To calculate the activation enthalpy in Joules we can multiply the gradient by R (the Universal gas constant, 8.314J K-1 mol-1.) This is then divided by 1000 to get KJ mol-1.
I converted the temperature into Kelvin by adding 273 to the temperature in centigrade.
Temp (°C)
Time (s)
Rate (1/time)
Ln Rate
Temp (K)
1/Temp (K)
59
12
0.083
-2.48
332
0.003012048
49
32
0.031
-3.47
322
0.00310559
38
47
0.021
-3.85
311
0.003215434
28
85
0.012
-4.44
301
0.003322259
21
109
0.009
-4.69
294
0.003401361
11
247
0.004
-5.51
284
0.003521127 On a graph I plotted the Ln Rate against 1/time in K.
My results:
Gradient = Change in X/ Change in Y = -5941.090194
Activation energy in J = -5941.090194 *-8.314 = 49394.22387
Activation energy in KJ = 49.294
According to Nuffield Book of Data15 the activation enthalpy for this reaction should come out to be: 43.5 KJ mol-1.
Therefore I can see that my results are very close to the published result, and reasons for the slight variations are likely to be down to human and precision errors.
During experiment 5, I investigated the effects of a catalyst whilst varying the temperature. This allowed me to work out the activation enthalpy whilst using Ammonium Molybdate.
Temp (°C)
Time (s)
Rate (1/Time)
Ln Rate
Temp (K)
1/Temp (K)
60
3
0.333
-1.10
333
0.003003003
48
6
0.167
-1.79
321
0.003115265
41
8
0.125
-2.08
314
0.003184713
30
8
0.125
-2.08
303
0.00330033
20
12
0.083
-2.48
293
0.003412969
11
29
0.034
-3.37
284
0.003521127
My results:
Gradient = Change in X/ Change in Y = -4381.192653
Activation energy in J = -4381.192653*-8.314 = 36425.23572
Activation energy in KJ = 36.425
The activation enthalpy of the catalyzed reaction is about 13KJ mol-1 lower than that of the non-catalyzed. Therefore we can see how a catalyst provides a route with a lower alternative energy, thus speeding up the reaction.
The final part of the investigation involved looking at using a colourimeter to monitor the rate of the reaction. This allows us to see the gradual increase of the concentration of Iodine produced. We can see how the lower concentration of all three reactants causes the % absorbance at any given time. It is clear how this backs up my data found in the previous experiments. As the concentration increases the amount of light which is absorbed as it passes through the solution will decrease, as the solution gets darker.
If we look at the colourimetery graph for the concentration of potassium iodide, we can choose a certain level of absorbance and then divide 1 by the time taken to get that absorbance to calculate the rate for each equation. For example:
At 1.00% absorbance:
For 0.5 mol dm-3 concentration it took 255 = 1/255 = 0.002922
For 0.25 mol dm-3 concentration it took 645 = 1/645 = 0.00155
We can see that approximately as the concentration doubles so does the rate of reaction, therefore backing up the previous found order. The rate is not quite due to experimental errors that may have altered the rate of reaction.
I then did this for both the hydrogen peroxide concentration and sulphuric acid concentration:
Hydrogen Peroxide:
At 74% absorbance
0.25 mol dm-3: rate = 0.001852
0.1 mol dm-3: rate = 0.003704
Sulphuric Acid:
At 74% absorbance
0.5 mol dm-3: rate = 0.002778
0.25 mol dm-3: rate = 0.004444
With the above calculations it its clear the reactants are first order.
Evaluation
Below is a table showing the percentage error of the equipment for measurements used during my investigation. The percentage error is calculated by:
Equipment
Precision
Percentage error calculations
Percentage error (%)
250 cm3 Volumetric Flask
±0.30 cm3
0.3/250 x 100
0.12
50 cm3 Volumetric Flask
±0.12 cm3
0.12/50 x 100
0.24
3 Decimal place balance
±0.0005g
0.0005/20.75 x 100
0.002409639
4 Decimal place balance
±0.00005g
0.0005/0.6205 x 100
0.080580177
Burette
±0.10 cm3
0.1/75 x 100
0.133333333
0.1/125 x 100
0.08
25cm3 bulb pipette
±0.06 cm3
0.06/25 x 100
0.24
1 cm3 graduated pipette
±0.01 cm3
0.01/1 x 100
1
Colourimeter
±0.005 A
0.005/0.07 x 100
7.142857143
0.005/2 x 100
0.25
Range of colourimeter percentage error is between 0.25% - 7.14%
Stop clock
±0.5 secs
All times to the nearest second
Thermometer
±0.5°C
0.5/60 x 100
0.833333333
0.5/11 x 100
4.545454545
Range of thermometer percentage error is between 0.83% - 4.55%
As shown above the majority if the percentage errors are minimal, however we can see the largest error was that of the colourimeter which came to 7.14%. Unfortunately there is little that can be done to prevent this, as the reading is so small. I ensured the same colourimeter was used each time, to eliminate as much error as possible.
In replacement of the thermometer, a digital thermometer could be used to gain a more accurate reading of the temperature, as the precision error for it is ±0.05°C. Other than this the equipment used was well chosen as it gave the smallest amount of error possible.
The investigation was accurate as most of the percentage errors are very low and mostly insignificant. Other errors are likely to be that the room temperature varied between 19 and 23 degrees C, throughout the experiment on a day-to-day basis, a factor out of my control, therefore this would have affected the rate constant. To reduce this error and improve the experiment, each experiment could be done in a thermostatically controlled water bath.
After looking at the published data of the activation enthalpy for a non catalysed, I can see my experiment is reliable as the value I got was only 5.794 KJ mol-1 off the actual activation enthalpy, however in order to be able to fully justify my findings I would need to repeat each experiment numerous times, a minimum of three, to then allow me to calculate an average and to disregard any anomalous data. As each experiment was only conducted once, it could be that all the data is anomalous.
By using two different methods at looking at how concentration affects the rate of reaction, my results are more reliable as both methods back each other up. To improve this aspect of the investigation I would use more and a wider variety of concentrations of each reactant, during the colourimetery, as only two concentrations of each were used. The main reason for this was the time allocated to this aspect of the investigation, as each run of colourimetery took around 25 minutes.
To improve and investigate further into this experiment I would create a calibrate curve through the colourimeter by making up solutions of the coloured substance of known concentration, then measuring the absorbance of each, ensuring to use the same conditions as the experiment will be done as. The graph of absorbance against concentration will give your calibration curve. This would then allow me to see how much iodine was produced at each stage of the reaction. Allowing me to closely monitor the rate of reaction.
A limitation of my experiment would be the catalysed experiment; due to the fact the reaction occurred so rapidly. The human reaction time is only accurate to 0.5 seconds, and in some cases the reaction took only 3 seconds, meaning the percentage error is 16.7%. To look into this particular catalyst further, I would dissolve the catalyst and dilute it to lower the concentration; this may give a longer time period before the blue-black complex forms. You would also be able to investigate if the concentration of catalyst changed the rate of reaction. Another potential way to improve this would be to investigate how different catalysts may affect the reaction and to what extent do they lower the activation enthalpy. A catalyst that could be tried would be ammonium iron sulphate, using the iron (III) ions to catalyse the reaction.
Another limitation of the experiment that would have reduced the accuracy would be the fact the blue-black complex forms gradually, therefore it subject able as to when to stop the stopwatch. In order to reduce this error a black cross was marked and the experiment was stopped once the black cross could no loner be seen. However whilst conducting the temperature experiments the reaction was done in test tubes and no cross could be used, to minimise the error the stop clock was stopped immediately at the first sign of the blue-black complex.
The final main limitation would be that the conical flask was swirled in order to mix the two solutions. Although best efforts were made to ensure the solutions were swirled evenly in each experiment, it is difficult to control. An improvement would be to use a magnetic stirrer, set at the same speed to ensure the same kinetic energy throughout the solutions, ensuring this did not affect the reaction.
As found in the Nuffield Book of Data, I can see that the order of the reaction with respect to H+ ions is dependent on the concentration. To further my investigation it would be interesting to look further into how the concentration of sulphuric acid may cause the order to be either 1 or 0 with respect to the H+ ions. I could then look at the point at which it changed from being zero order to first order. It would then be interesting to see how this affected the activation enthalpy of the experiment.
I can conclude my results are accurate and reliable, due to the fact the equipment was chosen with low precision errors and any errors given were too small to have a large impact. This can be backed up by the published data found in the Nuffield Book of Data.
You May Also Find These Documents Helpful
-
4) What is the original colour of iodine, and what colour does it turn in presence of starch and glycogen?…
- 2549 Words
- 11 Pages
Good Essays -
What does the color change in #3 indicate? Water and IKI was in jar and the iodine caused the color change of the water.…
- 354 Words
- 2 Pages
Satisfactory Essays -
1. No, the Starch did not. Because once we added Iodine to the Beaker the color did not change.…
- 288 Words
- 2 Pages
Satisfactory Essays -
9) Then add one dropperful of starch solution. The blue color of the starch-iodine complex should appear.…
- 501 Words
- 3 Pages
Powerful Essays -
If the amylase fails to digest the starch, I would expect the solution inside the intestine (Dialysis tubing) to change color to blue-black due to the chemical action that will occur when the Lugol’s regent (I2Kl) permeates through the intestine (Dialysis tubing).…
- 656 Words
- 3 Pages
Good Essays -
Introduction: I know prior to doing this experiment that iodine mixed with starch creates a dark color and that most objects, organic and inorganic, naturally experience isotonic reactions.…
- 893 Words
- 3 Pages
Better Essays -
Note: the color of the medium immediately surrounding the colonies after addition of iodine on PDA plate.…
- 961 Words
- 7 Pages
Better Essays -
Then you titrate the iodine remaining with sodium thiosulfate solution using starch indicator. Because you know how much iodine was produced from iodate(V) and iodide and you have found by titration how much is left over you can calculate how much iodine must have reacted with the ascorbic acid.…
- 4606 Words
- 19 Pages
Powerful Essays -
The iodine test was used to test for starch and glycogen in a particular solution. If the result turned out to be neither a blue-black or reddish-brown colour, it would be considered a negative result. If it turned blue-black, starch was…
- 1158 Words
- 5 Pages
Good Essays -
Therefore there are proteins present because of the significant purple colour exposed from the reaction of the Biuret solution…
- 1105 Words
- 5 Pages
Powerful Essays -
Baking soda did not react with water and baking powder did. One of our pure substances produced a characteristic color (blue-back) when it reacted with the iodine-alcohol solution. Starch is an organic compound you have probably encountered in your biology course. Identify which of the pure substances is…
- 699 Words
- 3 Pages
Good Essays -
3. Iodine will change from a brown colour to dark blue/black colour when it comes into contact with which chemical substance?…
- 1286 Words
- 7 Pages
Good Essays -
This experiment was done to determine the effects of the nature of the reactants, concentration, temperature, surface area and catalyst on the rate of chemical reactions. The nature of the reactants implies a difference if the reactants are aqueous or organic, acidic or basic or if they occur in the same phase or not. Acid-base reactions, formation of salts, and exchange of ions are fast reactions while reactions in which large molecules are formed or broken apart are usually slow. Generally, the rate of reaction is directly proportional to the concentration of the reactants, as well as to the temperature. Greater surface areas and addition of catalysts also increase the rate of chemical reactions.…
- 1408 Words
- 6 Pages
Powerful Essays -
In the Iodine test for starch, lonely two of the substance reacted with the iodine. The potato juice and the starch solution turn to a dark blackish color when the iodine was added shows that there are coiled carbohydrates that are with the juice and starch solution. As for the solutions and substance that did not react with the iodine that does not meant that there are not carbohydrates with in the solutions only that the carbohydrates are not in a coiled structure, since iodine will not react with carbohydrates unless they are coiled…
- 1210 Words
- 5 Pages
Better Essays -
The purpose of this experiment was to determine the reaction order of the oxidation of iodine…
- 2648 Words
- 24 Pages
Good Essays