Iodine Clock Reaction
- Pages: 23
- Word count: 5664
- Category: College Example
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1. Investigate the effect of temperature on the rate of reaction.
For this aim 3 sets of results will be obtained by timing how long it takes for the colour change to occur in different temperatures determined by the use of an electric water bath. With these results, the effect of temperature on the rate of reaction will be investigated.
2. Determine the activation enthalpy with and without the catalyst ammonium molybdate(VI) and with different protic acids and use this to compare the effectiveness.
To find out which catalyst is most effective, this aim will be carried out as an iodine clock reaction. The goal of this aim is to find out what catalyst is best to make this reaction occur at the fastest rate.
3. Determine the effects of the presence of ethanol on the rate equation.
It is known that ethanol effects hydrogen peroxide and therefore it has an effect on the rate equation. This aim will find out the effect of ethanol by carrying out the iodine clock reaction with and without ethanol present and the results will be compared to draw a conclusion.
4. Investigate the order of reaction with respect to hydrogen peroxide, iodide and acid.
The goal of this aim is to find out what order of reaction each chemical produced by making a graph for each of the reactants and analysing the line of best fit.
5. Investigate the rate equation, rate constant and possible mechanism for this reaction.
Investigating the possible mechanism will depict why the reaction occurs as it does. This aim is theory based and it will inform as to how the chemicals react and how this forms the products. By finding the rate equation, the dependence of rate on the concentration can be worked out. The rate constant will help to work out the rate equation.
6. Determine how the presence of a catalyst changes the mechanism of the Harcourt-Essen reaction and whether the mechanism is effected by monoprotic, diprotic or tripotic acids.
Catalysts speed up chemical reactions and so this theory based aim will find out how this happens and which catalysts is the most efficient at increasing the rate of reaction.
7. To find out how the concentration of iodide ions and peroxodisulphate (VI) ions affects the rate of reaction.
Potassium iodide reacts with hydrogen peroxide to form iodine, in this aim the concentration of iodide and peroxodisulphate ions will be measured to see how this effects the rate of reaction.
The equation for this experiment is as follows:
H2O2(l) + 2I-(l) + 2H+ ( I2(s) + 2H2O(l)
2I- ( I2 + 2e-
H202 + 2H+ + 2e- ( 2H2O
In this reaction the iodide ions are oxidised by the hydrogen peroxide and turned into iodine, the hydrogen peroxide gains electrons and is turned into water.
The iodine clock reaction occurs in two parts the first of which is the reactions between iodine and sodium thiosulfate, which produces iodide ions and a tetrathionate anion. The blue-black colour of the starch-iodide complex shows up when all of the thiosulphate ions have reacted with the iodine. The reaction is shown below:
I2(s) + 2S2O32-(l) ( 2I- + S4O6(l)-
The iodide ions being reacting with the starch once the thiosulphate ions have been exhausted, this produces the starch-I5- complex that gives the blue-black colour that is recorded to give the amount of hydrogen peroxide that has reacted and the time that the reaction took. The equation is shown below:
I2 + starch ( starch-I5- + I-
The specifics of the reaction between the starch and iodine are not completely known but it is thought that the coils of amylose are where the iodine fits. The iodine transfers charge to the starch, which combined with the spacing between the energy levels in the complex that is formed, corresponds to the absorption spectrum thus the blue-black colour is given off.
The second equation is as follows:
2KMnO4-(L) + 5H2O2(L) + 6H+( 2KMn2+(aq) + 5O2(l) + 8H2O(l)
In this equation potassium permanganate is the reagent as it brings about the colour change. Potassium permanganate is an inorganic compound which is a salt because it has k+ and MnO4- ions. The oxidizing ability of this compound is very strong and it dissolves in water to produce very dark purple solution. Potassium permanganate was formerly known as Condy’s crystals because the solution in water would evaporate to give prismatic crystals of a purple/black colour that glistened.
How the Rate of Reaction Varies:
According to the collision theory, every reaction has a minimum energy requirement for the molecules to begin reacting. Activation energy is the minimum energy required for the reaction to begin, the energy is required to break bonds and form new ones so that products can form. If two molecules collide with energy less than the activation energy then the molecules will not break any bond, which means no new bonds can form and therefore no reaction has taken place. Temperature affects the activation energy because by heating the reaction, molecules gain kinetic energy thus lowering the activation energy and therefore the reaction can take place faster because the kinetic theory states that molecules move faster. Faster moving molecules are more likely to collide with more energy than the activation energy and therefore thus the reaction takes place faster. Activation energy holds great importance in any reaction and so knowing the proportions of reacting particles that have enough energy to reaction is very helpful. All particles have numerous energies that can be shown with gases for which the Maxwell-Boltzmann Distribution shows all the energies of a gas. The Maxwell-Boltzmann Distribution is a graph where the number of particles is plotted against energy as shown below:
The number of molecules is shown by the area under the graph, this graph shows the kinetic energy to never be equal to zero. As there is no defined maximum value this graph could potentially continue onto infinity and it also shows that are not many molecules with a high energy.
Increasing the concentration of the reactants can make the reaction occur faster because for a reaction to take place particles must collide with each other. Increasing the concentration increases the frequency of collisions and thus more successful collisions take place and so the overall reaction happens faster. This is because when the concentration increases there are more reactant particles present to collide. This idea can be displayed with the following diagram:
In this diagram each particle has an equal chance of being hit this means that the chances of a red particle colliding with the solid blue particle is 25%. If another blue particle is added, (represented by the dashed particle), then the concentration has increased and the chances of a blue particle being hit is now 50% which is a 50% increase.
The greater the surface area of a reactant, the faster the reaction will take place. This is because a greater surface area means more of the reactant is available to react in a certain amount of time for example the powder form of a molecule will react faster than a solid lump of that molecule because in the powder more molecules are available to collide and therefore to react. Below is a visual representation of the powder example:
In the above illustration shows that when the red reactant is a powder more particles are available to react which means that the reaction happens faster and the rate of reaction increases.
Pressure only effects reactions with gases, it has no effect on solids or liquids. When the pressure of a reaction is increased the rate of reaction also increases. Increasing the pressure means compressing the gaseous reactant into a smaller volume, this increases the likelihood of collisions and as the collision theory states; collisions are necessary for a reaction to happen. Pressure uses the same concept as concentration, increases the concentration of a reaction means there are more particles in a given space and increasing the pressure means there are more particles in a smaller amount of space. The below equation demonstrates the relationship between pressure and the rate of reaction:
pV = nRT
p = pressure,
V = volume
n = number of moles
R = molar gas constant
T = Temperature
In this equation RT is a constant therefore it can be represented by the letter k and the entire equation can be rewritten to read:
p = k(n/v)
(n/v) is the same as the concentration and so the equation can also be written as p = kC however because the concentration is not always known it is not the best form for the equation. This equation shows a proportionality so the final equation would be as below:
p ( k(n/v)
A catalyst speeds up a reaction by providing an alternative pathway that has a lower activation enthalpy and it remains chemically unchanged at the end of the reaction. By providing a pathway that requires a lower activation enthalpy means that there will be more particles that have enough energy to react and start breaking bonds.
The change in activation enthalpy would have the following impact on the Maxwell – Boltzmann diagram:
The effect it would have on an enthalpy profile diagram would be as follows:
Catalysts can either adsorb or form intermediate compounds. Adsorption is commonly used car exhausts to make the fumes safe and create less pollution. This is when molecules attach to a surface using very weak forces such as instantaneous dipole – induced dipole. The bonds between the reactants weaken and break and then new bonds are formed between the reactants that form products. The products then diffuse away from the surface of the catalyst.
Enzymes work by forming an intermediate that is known as an enzyme-substrate complex. The reactants combine with the catalyst to create an intermediate compound, which is very unstable and reacts almost straight away. When the intermediate compound reacts it forms the final product and the catalyst remains chemically unchanged.
Orders of Reaction:
The rate of reaction is the rate of change of a reaction as time increases which is why rate laws are differential equations. Zero order reactions occur rarely, most are either first order or second order. For zero order laws the rate is independent of concentration.
The graph below is of a zero order reaction. In this graph the rate of reaction remains constant throughout the reaction which means that the gradient is also constant which produces the straight diagonal line that is visible. The graph also shows that the half life decreases as the concentration decreases as there is a smaller each time.
The zero order graph shows that the rate of reaction is unaffected by the reactant and therefore if a reactant is zero order it is not needed in the rate equation as it has no impact on the rate of reaction.
First order graphs show varying rate of reactions throughout the reaction which then means the gradients is also constantly changing. This produces a curve like the one in the graph below and this graph also shows that the half-life decreases as concentration decreases.
The first order graph shows that the rate of reaction is affected by the reactant and therefore first order reactants are included in the rate equation is the reactant has a direct affect on the rate of reaction.
Second order graphs show that the rate of reaction varies throughout the reaction just like first order graphs however for second order graphs the constantly changing gradient produces and exponential curve as shown below. In this graph also the half-life decreases as the concentration decreases.
The relationship between the rate of reaction and the concentration of the reactant is exponential which means that if the concentration doubles then the rate of reaction quadruples. The rate of reaction is directly proportional to the concentration squared.
I have chosen to carry out the reaction using colorimetry because I am measuring a sudden colour change and a colorimeter will give an accurate reading of how much the colour has changed by. Colorimeters are very accurate pieces of equipment and can detect very small changes in light absorbency which is much more accurate than trying to see the colour change by sight.
I am going to calibrate the hydrogen peroxide because hydrogen peroxide is known to degrade in the bottle into oxygen and water, which means the concentration written on the bottle is not the true value. Calibrating the hydrogen peroxide will ensure that the correct concentration of hydrogen peroxide is used.
I have decided to make the starch solution because this will be more accurate than using starch solution from a bottle which may be contaminated from previous use. By making my own solution I will easily be able to change the volume or concentration of the starch solution depending on the results of my preliminary experiment, this will ensure that I obtain results that are as accurate as possible.
The Arrhenius equation is used to show how the rate constant changes when the temperature changes or when the activation energy is changed by adding a catalyst.
The Arrhenius equation is shown below:
k = Ae-(EA/RT)
k is the rate constant
EA is the activation energy
T is the temperature (in kelvins)
R is the gas constant (8.31JK-1mol-1)
A is apporoximately constant, and is taken as constant over small temperature ranges. e is a mathematical entity.
The equation above can be rewritten to read:
Ln k = c – EA (1/t) R
When this equation is plotted on a graph, it produces a straight line graph which can be used to work out the activation energy. In this graph the gradient is represented by (EA/R) because this corresponds to ‘m’ in the general equation for a line – y = mx + c.
Background of Chemicals:
Hydrogen peroxide has the chemical formula H2O2 and it as a pH of 4.5 this is the simplest compound with a single oxygen-oxygen bond and is therefore known as peroxide. The physical appearance of hydrogen peroxide is that it is a clear liquid that appears colorless in dilute solutions and it is a slightly thicker consistency than water. Hydrogen peroxide is a highly reactive oxygen species and this property makes it ideal to use in bleaches or other cleaning agents, it can also be used as a propellant in rocketry. Hydrogen peroxide is a by-product of oxidative metabolism in organisms and most living species have enzymes known as catalyse peroxidases, which are used in the decomposition of low concentrations of hydrogen peroxide into water and oxygen.
The formula for hydrogen peroxide is below:
When a bottle of hydrogen peroxide is left in the presence of light it undergoes decomposition, for this reason hydrogen peroxide is kept in a brown bottle in a cool place. Due to this nature of hydrogen peroxide, it will be titrated with potassium permanganate to find out it’s true concentration and then it will be diluted to the required amount.
The products that hydrogen peroxide give off when it decomposes are water and oxygen which are shown in the reaction below:
2H2O2 ( 2H2O + O2
Hydrogen peroxide decomposes at a rate which is related to the temperature and the pH of the chemicals within the reaction. Acid can be added to the solution of hydrogen peroxide to stabilize it because hydrogen peroxide is incompatible with many substances that have the ability of to catalyse the reaction, transition metals and their compounds inclusive.
In this reaction the iodide ions come from the inorganic compound potassium iodide which has the chemical formula KI . Potassium iodide is a white salt that is less hygroscopic than sodium iodide therefore it is easier to use. Potassium iodide is commercially used for medicinal purposes such as protecting the thyroid for radioactive iodine, it is also used in radiation detectors and Geiger counters.
The structural formula for potassium iodide is shown below:
Positive hydrogen ions are a proton that hydrates readily they cannot however exist freely in solution. They are formed when atomic hydrogen loses an electron and forms a positive ion and they are found in all aqueous solutions of acids. Hydrogen ions are formed when an electron is removed from a hydrogen atom, which leaves a single proton because hydrogen only has one electron. Hydrogen ions are very reactive and thus can only exist in an environment that is virtually particle free such as a vacuum or in its gaseous state.
Sodium thiosulphate is usually found in its pentahydrate state in which it is a colourless crystalline compound. The formula for sodium thiosulphate is Na2S2O3.5H2O and in this experiment it will be used in this state. This compound can be used to tan leather for chemical manufacture or it can fix film in photography.
The structural formula for the pentahydrate form of sodium thiosulphate is shown below:
Potassium permanganate had the chemical formula KMnO4 and it is an inorganic chemical compound. This compound is also known as permanganate of potash or Condy’s crystals and it is a salt made up of K+ and KMnO4- ions. Potassium permanganate has a strong oxidizing ability. This compound is dissolvable in water and gives off a strong purple color and upon evaporation it leaves prismatic dark purple crystals. This property allows us to determine the concentration of hydrogen peroxide because the colorless hydrogen peroxide turns slightly pink which is the endpoint for this reaction.
The structure for KMnO4 is below:
Hydrogen chloride is a monoprotic acid because it only has one proton that it is able to donate. This solution is hydrogen chloride dissolved in water, which is very corrosive and therefore must be diluted before use. Hydrochloric acid is naturally occurring in gastric acid.
Strategy and Justification:
Calculating The Rate:
The rate can be calculated by measuring various things such as volume or pressure or an analysis of results can tell us what the rate is.
By measuring the volume of the product that is produced, the rate constant can be calculated. This can be done with such methods as an inverted burette or a gas syringe in a reaction that produces a gas such as hydrogen peroxide and catalase. The oxygen that is produced in the boiling tube is passed through some rubber tubing; this then displaced the water in the inverted burette or pushes the stopper in a gas syringe. This enables us to be able to measure the product that is formed. The rate would be worked out by drawing a graph for each experiment in which the initial rate of reaction would act as the gradient of volume of product produced against time graph. The value obtained for the initial rate of reaction would be plotted against the concentration producing a graph, which would tell us the order of reaction. However for the Harcourt-Essen reaction, no gas is produced so the volume cannot be measured.
A varying pressure is detectable when two molecules react together to give off a gaseous product. A graph can be plotted by comparing a reaction that has taken place under normal room pressure and a reaction that has taken place under compression, for example placing a bung on a test tube would alter the pressure. The pressure would be plotted against time and the gradient for the graph would give the initial rate of reaction. This data can then be plotted on another graph to find the overall order of reaction.
Titration is often used for chemical analysis to determine the concentration of a known reactant. A substance is put into a burette and is slowly poured into a conical flask, which allows for the exact amount of a reactant to be determined before the endpoint is reached. Small amounts of the substance in burette must be taken out at intervals to ensure that the reaction does not go past the endpoint. The measurements that are made using this method are the concentrations of the reactant at the moment the sample was taken out. By plotting concentration against time, a graph can be drawn.
Colorimetery can be used to calculated the rate. For this type of experiment, choosing the correct size and color of the filter is of high importance because the substance has to absorb the same wavelength of light which is being transmitted by the colorimeter. This is known as the complementary color for the color of the substance because they are the same. During the experiment, the percentage abosorbancy is measured at certain intervals usually at about 10 seconds. This data can then be plotted on a graph against times that has the units of mol s-1. This can also produce a calibration curve for which absorbance is plotted against time.
In this method, the conductance of a solution is measured. If the solution is aqueous then the ions in the solution can conduct electricity while the reaction is taking place. By measuring the electrical conductivity, the rate of reaction can be measured. To produce a graph from this data, the conductivity of the substance would be plotted against concentration in which the initial curve gives the initial rate of reaction.
Clock Reaction method:
In a clock reaction small amounts of solutions are reacted together by adding them to a conical flask from a burette, which produces intermediate substances and then the final product. An indicator is used so that the color of the final product can be observed. Other substances are also added to block the production of the reactant through means of a chemical reaction. This works because the reactant and the indicator to react until the additional substance is used up, once this had been used up the indicator and reactant form a colored complex. The most commonly used example of this is the iodine clock reaction in which sodium thiosulphate is added to the reaction producing iodine. Not until all the thiosulphate ions have been used up in the reaction will the iodine build up.
In this experiment the clock reaction method will be used to calculate the rate of reaction, this is because the Harcourt-Essen reaction does not produce a gaseous product and therefore the first two methods are not applicable. The colorimetric analysis will also be used to determine the effect of concentration on the rate of reaction because colorimeters are very accurate pieces of equipment that can detect tiny changes in absorbance so therefore they will able to detect the absorbency of the different blue colors that are observed. The conductimetric method can not be used because an aqueous solution is not produced and therefore no electrical conduction can be detected.
• 3 burettes, burettes are highly accurate measuring apparatus and therefore they are appropriate to use for this experiment because exact measurements of solutions are required. • 2 boiling tubes, boiling tubes are much larger than test tubes and therefore when heated, the solutions will heat more evenly and there is less chance of the contents spitting out. Boiling tubes are also larger which makes it easier for the various solutions to mix properly. • A thermometer to measure the temperatures of the boiling tubes to keep the test reliable. • A colorimeter, colorimeters are able to measure the slightest color change and therefore they are being used in this experiment because color changes of various intensities will take place. • Cuvettes to pour the mixture into to place into the colorimeter. • Stopwatch to time how long it takes for the color change to take place. • Test tube rack to hold the boiling tubes in.
• A stirring rod to stir the mixture and make sure that it is evenly spread.
• Conical flask to hold the hydrogen peroxide in.
• Weighing scale to measure the mass of the leftover potassium permanganate.
• Hydrogen peroxide – 0.3 moldm-3
• Sulfuric acid – 0.3 moldm-3
• Sodium thiosulfate – 0.005 moldm-3
• Potassium iodide – 0.02 moldm-3
• Starch solution
• Ammonium molydbate (catalyst)
• Hydrochloric acid (monoprotic acid)
• Sulphuric acid (diprotic acid)
• Phosphoric acid
• Potassium permanganate – 0.1 moldm-3
Kinetics of the Harcourt Essen reaction
REACTION: H2O2 + 2I– + 2H+ → I2 + 2H2O
1. Investigate the effect of temperature on the rate of reaction. Equipment:
• 150cm3 Hydrogen peroxide
• 100cm3 Sodium thiosulphate
• 150cm3 Sulphuric acid
• 2 boiling tubes
• Test tube rack
• pH strips
• 3 Burettes
1. Set up 3 burettes with hydrogen peroxide, sodium thiosulphate and sulphuric acid 2. Place 150cm3 of 0.300moldm-3 of hydrogen peroxide, 100cm3 of 0.00500moldm-3 of sodium thiosulphate, 150cm3 of 0.300moldm-3 of sulphuric acid and water into a boiling tube and label this A. Using a boiling tube will ensure that all the components are mixed and heated equally. 3. The boiling tube will then be placed in a test tube rack. 4. Measure and record the temperature each time. The temperature must be kept constant in every experiment to keep the experiment reliable. 5. Next, pipette 2cm3 of potassium iodide into a different boiling tube and label it B. 6. Pour boiling tube B into the other A and immediately start the timing from the time the solution is added. 7. Stir the mixture.
8. Stop the time when the first blue colour appears and record the time in the table of results. 9. After the reaction finishes, check the pH and record it down to ensure that the pH is kept constant in each experiment. This ensures a reliable test. 10. Repeat the reaction three times. This helps to reduce anomalies and gives an accurate set of results, with the experiment being a fair test. If there are is an anomalous result, repeat the reading. 11. Next, plot a graph for the time against the concentration for each set of (averaged) results. A tangent will then be drawn on the first part of the graph. The gradient of this tangent will give the initial rate of the reaction. Doing this for each set of results will give initial rates for all of the experiments. 12. These would then be plotted against concentration, allowing the order of reaction to be found.
2. Determine the activation enthalpy with and without the catalyst ammonium molybdate(VI) and with different protic acids and use this to compare the effectiveness. Equipment list:
• 3 Burettes
• 150cm3 Hydrogen peroxide
• Starch solution
• Sodium thiosulphate
• 150cm3 sulphuric acid
• 2 boiling tubes
• Stop watch
• pH strips
1. Set up 3 burettes
2. Place 150cm3 of 0.300cm-3 solutions of hydrogen peroxide, starch solution, sodium thiosulphate, 150cm3 of 0.300moldm-3 sulphuric acid and water into a boiling tube and label this A. The only concentrations that change are the volumes of hydrogen peroxide and water. The total volume stays the same. 3. Measure and record the temperature of the boiling tube. The temperature must be kept constant in every experiment to keep the test reliable. 4. Next, pipette 150cm3 of 0.0200moldm-3 of potassium iodide into another boiling tube label this B. Doing this will stop the reaction taking place before it is required. 5. Calibrate and reset the colorimeter using a cuvette filled with distilled water. This will keep the test reliable because it will make sure the colorimeter has not been tampered with. 6. Pour the solution from boiling tube B into boiling tube A and immediately start timing from the second the solution is added. 7. Quickly transfer some of the solution from the boiling tube to the cuvette and place this into the colorimeter 8. Every ten seconds measure the percentage absorbency of the solution, and record this in a table. 9. Stop after 120 seconds.
10. After the reaction finishes, check the pH and record it down to ensure that the pH is kept constant in each experiment, this makes the test reliable. 11. Repeat the reaction twice for each catalyst. This helps to reduce anomalies and gives an accurate set of results. If there are any anomalous results, repeat the reading. 12. Next, plot a graph for the time against the % absorbency for each set of (averaged) results. A tangent will then be drawn on the first part of the graph. The gradient of this tangent will give the initial rate of the reaction. Doing this for each set of results will give initial rates for all of the experiments. 13. These would then be plotted against concentration, allowing the order of reaction to be found.
3. Determine the effects of the presence of ethanol on the rate equation. Method: Titration
2MnO4-(aq) + 5H2O2(aq) + 6H+(aq) ( 2Mn2+(aq) + 5O2(g) + 8H2O(l)
1. Dissolve 8.3g of potassium permanganate into 200g of water in a beaker. 2. Fill a beaker with water and then add 30g of hydrogen peroxide and 10g of concentrated sulphuric acid. 3. Using a burrette, add potassium permanganate into the acidified hydrogen peroxide and stop when the purple color is persistent. 4. Once the purple color is persistent, take out the remaining potassium permanganate and weigh how much is left over. 5. Use the calculation described in ‘How techniques Work’ to find out the concentration of the hydrogen peroxide. 6. Carry out the iodine clock reaction mentioned in the above method to carry out the experiment. 7. Repeat the experiment 3 times with ethanol and 3 times without ethanol. This keeps the experiment reliable and reduces the chances of anomalous results.
4. Investigate the order of reaction with respect to hydrogen peroxide, iodide and acid. Equipment:
• Starch powder
• 3 burettes
• 50cm3 Sulphuric acid
• 50cm3 sodium thiosulfate
• 50cm3 Potassium iodide
• Conical flask
• Stop watch
1. Prepare a starch solution using 20g of starch powder and 100 cm3 of water, once the starch powder has dissolved boil the solution. 2. Set up 3 burettes with 50 cm3 of sulfuric acid, one with 50 cm3 of sodium thiosulfate and the third with 50 cm3 of potassium iodide. 3. Measure out 10.0 cm3 of sulfuric acid 6.0 cm3 of sodium thiosulfate, 25.0 cm3 of potassium iodide 1.0 cm3 of starch solution and 4.0 cm3 of water into a small beaker. 4. Pour 10.0 cm3 of hydrogen peroxide into a conical flask. 5. Pour the contents of the small beaker into the conical flask and immediately start the stopwatch. 6. Stop the timer when a sudden color change is observed and record the time taken. 7. Repeat the above steps 4 times and each time dilute the solution being investigated following a ten-fold serial dilution. 8. Plot a graph for concentration against time, draw a line of best fit and determine the order of reaction.
2. Investigate the rate equation, rate constant and possible mechanism for this reaction. Method:
The rate equation can be worked out with the Arrhenius equation and the rate constant and mechanism can be found by plotting a graph.
3. Determine how the presence of a catalyst changes the mechanism of the Harcourt-Essen reaction and whether the mechanism is effected by monoprotic, diprotic or tripotic acids. Equipment:
• 3 burettes
• 50cm3 sulphuric acid
• 50cm3 sodium thiosulphate
• 50cm3 potassium iodide
• 1cm3 starch solution
• Conical flask
• Stop watch
1. Set up 3 burettes with 50 cm3 of sulfuric acid, one with 50 cm3 of sodium thiosulfate and the third with 50 cm3 of potassium iodide. 2. Measure out 10.0 cm3 of sulfuric acid 6.0 cm3 of sodium thiosulfate, 25.0 cm3 of potassium iodide, 1.0 cm3 of starch solution and 4.0 cm3 of water into a small beaker. 3. Add 10.0 cm3 the solution being investigated into the small beaker. 4. Pour 10.0 cm3 of hydrogen peroxide into a conical flask. 5. Pour the contents of the small beaker into the conical flask and immediately start the stopwatch. 6. Stop the timer when a sudden color change is observed and record the time taken. – Repeat the above steps for all the different catalysts and protic acids being investigated. – Work out the mechanism for each experiment.
5. To find out how the concentration of iodide ions and peroxodisulphate (VI) ions affects the rate of reaction. Equipment:
• 3 burettes
• 50cm3 Sulphuric acid
• 50cm3 Sodium thiosulphate
• 50cm3 Potassium iodide
• 1cm3 Starch solution
• Stop watch
1. Set up 3 burettes with 50 cm3 of sulfuric acid, one with 50 cm3 of sodium thiosulfate and the third with 50 cm3 of potassium iodide. 2. Measure out 10.0 cm3 of sulfuric acid 6.0 cm3 of sodium thiosulfate, 25.0 cm3 of potassium iodide, 1.0 cm3 of starch solution and 4.0 cm3 of water into a small beaker. 3. Pour 10.0 cm3 of hydrogen peroxide into a conical flask. 4. Pour the contents of the small beaker into the conical flask and immediately start the stopwatch. 5. Stop the timer when a sudden color change is observed and record the time taken. 6. Repeat the above steps 4 times and serial dilute the solution being investigated following a ten-fold dilution each time. 7. Plot a graph for concentration against time and work out the rate of reaction.