The Kinetocs Experiment of Iodide Ions and Persulphate Ions Essay

The catalytic effect of D-block ions and the kinetics of reactions iodine clock reaction: By Stephen Parsons 6K2 Centre number: 61813 Candidate number: 8270 Table of Contents My aim and my reaction:3 Rate of reaction:4 Activation enthalpy:5 Collision theory:6 The effect of temperature on reaction rate:7 The effect of concentration on reaction rate:7 The effect of a catalyst on reaction rate:8 D-block elements:9 The effect of extra kinetic energy (from stirring etc. ):10 Where do we use D-block ion catalysts? 10 Arrhenius equation:12 Orders of reaction:10 Oxidation states of D-block ions:13

Electronic configuration and transition metals:13 Experiment:13 Method:14 Risk assessment:15 How to make my standard solutions of each of my reactants:15 Changing the concentration of potassium iodide:18 Changing the concentration of potassium peroxodisulphate:19 Changing the quantity of iron (III) sulphate:20 Changing the temperature at which different runs are done at:21 Changing the temperature of the reaction and using a catalyst:21 Results:22 Iron (III) sulphate:22 Potassium iodide:27 Potassium peroxodisulphate:31 Temperature change:35 Arrhenius equation:37 Percentage error:42

Conclusion:43 Evaluation:44 Bibliography46 Introduction: My aim and my reaction: I will be adding Potassium iodide, distilled water, Potassium peroxodisulphate (VI), Sodium thiosulphate (VI), starch and when varying the use of a catalyst I will also add Iron (III) sulphate. I will start a stop watch when I mix the appropriate concentrations of solutions together, When I react these solutions a slow reaction will occur, tri-iodide is formed and this then reacts with the sodium thiosulphate(VI) this will happen until all of the sodium thiosulphate has reacted with the tri-iodide that is produced.

The further tri-iodide that is produced is reacts with the starch in the solution and this changes the solutions colour to a bluey-black and then the stop watch is stopped and the time is recorded. My chemical reactions equations: Step 1: Potassium iodide and potassium peroxodisulphate and produces iodine 2KI(aq)+K2S2O8(aq) I2 + K4S2O8 Step 2: Sodium thiosulphate then reacts with this iodine and this reaction occurs until all of the sodium thiosulphate is reacted and this leads to step 3. Na2S2O3+I2(aq)S4O6-2 +2I- Step 3:

Iodide ions produced from step 1 that bi-pass step 2 because of the lack of sodium thiosulphate to react with. 2I-+ starchstarch bided to I- mixture (blue in colour) This signals the reactions(s) end point I will be investigating a variety of kinetics experiments, one of the variables I will change will include the use of a D-Block catalyst, I will be using Iron (III) sulphate as my catalyst. I also aim to vary the concentrations of the potassium iodide I use and also the concentration of potassium peroxodisulphate (VI) in another separate set of runs.

I will also change the temperature at which I do the reactions at as another variable. My experiments will be iodine clock reactions and I will be investigating the rate of reaction as it changes as I change variables such as the use of a catalyst, the variable of concentration changes of the reactants and the variable of temperature. From these experiments I aim to determine the orders of the reaction. I also will compare and contrast the effects of each variable between the others to determine which variable is the most drastic and effective to change and compare this to how these variables are varied in industry.

Rate of reaction: Reaction rate is the rate at which the reactants are used up or the time taken for a given amount of product to form or how much of the products are formed in a certain amount of time. This allows us to compare similar reaction with a set individual variable and see how this variable affects the reactions rate. Rate equation= change in amount of substance/concentration Time taken Reaction rate can be changed by a change in: * Concentration of reactants Temperature * Pressure * Catalyst * Surface area The concentration of a reactant will change the reaction rate in most reactions because there are more particles of the reactants in the same amount of solution this makes the number of successful collisions rise so this makes the reaction rate faster if the concentration is higher of any of the reactants. The temperature affects most reactions and increases there rates unless the reaction uses an enzyme that only is effective up to or down to a certain temperature.

This change has this effect because it increases collision rate and speed so more collisions with the correct reaction energy occur so more successful collisions occur increasing the reaction rate. The pressure is an important factor in gas reactions because the gas concentration is proportional to pressure. The use of a catalyst affect reaction rate by lowering the reaction enthalpy making the proportion of successful collisions higher.

Surface area of reactants is a key variable because a higher surface area provides a higher chance of 2 reactants colliding and producing a successful reaction. Activation enthalpy: I just talked a lot about activation enthalpy I will now explain what is meant by this. Activation enthalpy is the minimum amount of kinetic energy required by a pair of colliding particles for a reaction to occur. it is the amount of energy required for the particles of the reactants to break the bonds holding them together so that they can form the new bonds of the product. 1) This diagram is a energy profile it shows how a certain enthalpy must be achieved before a reaction can take place this enthalpy (energy) is represented by the highest point of the curve measured down to the line representing the reactants, this must be reached or higher. The activation energy is the amount of energy that is required to stretch and break the bonds of the reactants so that the constituent atoms can be re-arranged and form the products.

In this diagram the products are at a lower energy level than the reactants and this means the reaction is exothermic and releases energy as it takes place, this energy quantity can be calculated by the distance on the graph between the enthalpy of the reactants and the products in this case the enthalpy is a minus number as there is a drop this is the same for all exothermic reactions, this also encourages the same reaction of other molecules to occur as this energy released helps the next reaction achieve the reaction enthalpy.

My reaction is slightly exothermic and doesn’t need the introduction of heat to initiate or carry out the experiment. Collision theory: Collision theory is the theory of how the number of collisions between 2 reactants affects the rate of a reaction. The higher number of collisions the more likely that these collisions will produce a higher number of uccessful reactions. Gasses and liquids move more than solids so more collisions occur because liquids and gasses particles are always moving. In solutions and gasses collisions are constantly occurring not all of these collisions cause a reaction, this is because the orientation of the collision and energy involved in the collision must be enough.

Collision theory states that reactants must have enough kinetic energy when they collide and for the two particles to hit each other in the correct manor. This is the Maxwell Boltzmann energy distribution curve: Figure 1: This diagram shows how only particles with over a certain amount of energy can react. (1) Figure 2:This diagram shows how when a catalyst is used the activation energy is lowered so a higher proportion of the molecules will have high enough activation energy. (2)

The Maxwell Boltzmann distribution graph shows that in a solution or mixture whether it be liquid gas or solid not all the particles have the same amount of energy and it shows because of this often not all of the particles have the capability to react, but it also shows that as the temperature increases (energy) that a higher proportion of particles are capable of reacting, it also shows that with the use of an appropriate catalyst that the energy needed is lowered so a higher proportion of particles can react than before the use of a catalyst.

The effect of temperature on reaction rate: Temperature has an effect on the rate of a reaction, as temperature rises so does the rate of reaction. Temperature affects reaction rate dramatically because it increases the kinetic energy of the particles, this means that a higher proportion of the reactants will have enough energy to cause a successful collision and react at higher temperatures. I am going to investigate the change in reaction rate when the same reaction is performed at various temperatures. The effect of concentration on reaction rate:

Concentration of reactants also affects the reaction rate; this is due to the fact that there will be more particles of reactants in the same space at higher concentrations. This means that there will be more collisions then there would be at lower concentrations increasing the chances of successful collisions. The effect of a catalyst onreaction rate: A catalyst is a substance that increases the speed of a reaction, but it is not chemically changed by the reaction, this means there will be the same mass of catalyst at the end of the reaction as there was at the start.

A catalyst does not cause an increase in the amount of product formed, it only increases the reaction rate by providing an alternative route for the reaction which has a lower activation enthalpy. This means that more molecules can react when they collide. There are two types of catalyst homogeneous and heterogeneous. Homogeneous catalysts are in the same physical state as the reactants e. g. liquid-liquid or aqueous-aqueous etc. A heterogeneous catalyst is a catalyst that is in a different state than the reactants e. g. solid-gas.

I will be using a homogeneous catalyst of Iron (III) sulphate which will be dissolved in water so it is in an aqueous state as are the reactants. Homogeneous catalysts usually form a middle step and form what is called an intermediate, this is a unstable ion and it is only present for a very short period of time before it reacts to form the products. (Ea = activation energy) Figure 3 * This diagram works for both heterogeneous and homogeneous catalysts. * Ea1 is the activation energy leading to the formation of an intermediate complex. Ea2 is the activation energy for the change of the intermediate complex into products. * Ea3 is the activation energy of the un-catalysed reaction. (5) This diagram shows that when a catalyst is used the activation enthalpy is always lower than when the reaction is un-catalysed, it also shows how there is often 2 peaks of activation energy when the catalyst is used this is because when a catalyst is used a intermediate is formed, this intermediate is unstable so reacts very quickly to finish the reaction. D-block elements: Figure 4 (3) This picture shows a section of the periodic table and shows in turquoise the

D-block elements that I am interested in; such elements include scandium, titanium, vanadium, chromium, manganese, iron, cobalt, and nickel, copper and zinc all these elements are transition metals and D-block ions. These transition metals can act as catalysts because they have varying oxidising states this is what allows these elements to be catalysts to some reactions this is because these molecules can have varying oxidation states so they can attract and donate electrons but then also accept electrons from other particles this is what allows them to catalyse reactions just like the reaction I am producing in my coursework.

The effect of extra kinetic energy (from stirring etc. ): The introduction of extra kinetic energy also increases molecules energy and the proportion that are able to react successfully as shown by the Maxwell-boltzmann distribution graph, kinetic energy can be increased by stirring shaking etc. but it is not easy to control scientifically and this is why I am not investigating its effect. Where do we use D-block ion catalysts? We use these transition metals as catalysts in a lot of things we take for granted for example a nickel catalyst is used in catalytic converters in the exhaust pipe of cars, busses motorbikes etc. his is an important use as it converts a lot of the dangerous and poisonous gasses into less harmful products e. g. carbon monoxide reacts with nitrogen monoxide in the catalytic converter to produce carbon dioxide which isn’t poisonous and harmful to life and nitrogen which also isn’t poisonous or harmful to life. An iron catalyst is used to produce ammonia in the Haber process, this is important as ammonia is used in the production of fertilisers so this catalyst helps us grow our food!

Nickel is used to hydrogenise the double carbon-carbon bond in unsaturated vegetable oils to produce margarine a more spread able butter. Also Vanadium (V) oxide, V2O5, is used as a catalyst in the ‘Contact Process’ in the production of sulphur trioxide which is used in the manufacture of sulphuric acid. Orders of reaction: The reactants of a reaction usually are related to the rate of a reaction, so if the concentration of a reactant is changed sometimes this will cause a increase in reaction rate.

This is the rate equation below: (7) This simple equation shows the relationship between the reactants and their concentrations and the rate of their reaction. If the order of the reaction is first order with respect to a. then as the concentration of reactant a. is increased so will the rate here is a graph to express this: (8) If the order of the reaction with respect to a. is second order the rate of the reaction will increase proportionally2here is another graph to represent this: (9)

If the order with respect to reactant a. is zero order than as the concentration increases the rate stays the same as this reactant isn’t the rate determining step. Here is yet another graph to show this: (10) When you vary the concentration of reactants in an experiment you can determine the effect they have on the rate and what order they are with respect to the reaction. This can be done by plotting a graph of time against reaction progress. This is done by using the reciprocal of the reaction time 1/t against concentration.

This graph will show that if the reaction time is short the rate is fast by producing a sharper gradient on the graph but also if the reaction takes longer the rate is slower and the gradient is shallower, if the graph is straight but rising it will be first order, if it is curved and rising it will be second order and if it is horizontal it is zero order. Another method can be used this is the half-life method. A graph can be produced from this which plots the quantity of reactant remaining against time. On this graph we can measure the change that has occurred at different intervals e. if 1000cm3 of reactant is present to begin with we would measure the time between there being this much reactant being present until 500cm3 remained, this would then be repeated down from 500cm3 to 250cm3 and so on this should give equal time intervals if the reaction is first order. (11) Fractional orders exist and are commonly found in oscillating reactions but my reactions will not produce fractional orders so explanation is not relevant for my experiment. Arrhenius equation: (6) K: is the rate constant Ea is the activation enthalpy R is the gas constant 8. 31j?? mol??

T is temperature in Kelvin This is the logarithmic form of the equation. How do you do this? Workout the rate constant, do this by; changing how long the reaction took by working out its reciprocal number. Then discover the logarithm of each rate by pressing the In button on the calculator and then put the reciprocal number in this will give you the InK part of the equation. Then find out the value of the temperature remember to this in Kelvin (add 273 degrees on to standard) now plot a graph with the InK on the y-axis and then the value of temperature in Kelvin on the x-axis.

Measure the gradient you have plotted and multiply this by -8. 31 to get the activation energy value and multiply this further by 1000 to get the units to kj mol?? (6) Oxidation states of D-block ions: D-block ions have various oxidation states; this is what enables them to be catalysts. Iron(II) is a D-block ion it has (II) after iron because this is the oxidation state that it is in at this point but irons oxidation state can vary from (II) to (VI) this allows the iron to donate up to six electrons in a reaction.

As these elements have the ability to donate electrons and regain them they can act as catalysts by donating and then retrieving these free electrons. (11) Electronic configuration and transition metals: Transition metals are D-block elements this is because the final shell to fill up in these elements is the 3d shell, they also have the 4s shell but this fills before the 3d shell because it makes the atom more stable. Even though the 4s sub shell fills before the 3d shell it also empties first when ionisation occurs this is because the 4s sub shell is further from the nucleus of the atom so the attraction is weaker.

All these D-Block elements produce numerous ions because they can be ionised more than once this allows them to be catalysts in even more reactions than if they could simply just ionise once. Experiment: Equipment: * 0-110®C Thermometer * Stopwatch * Burettes (or graduated pipettes) * 10 boiling tubes * 5 test tubes * 6 100 ml beakers (for making reactants in) * 6 volumetric flasks with bungs ( for keeping reactants in and accurately making reactants) * Water bath * Kettle * Test tube rack * Glass rod (stirrer) * Ice * Marker pen * Paper towels * Freezer Weighing boat * Spatula Materials: * 100 cm? of freshly prepared starch solution * Distilled water * Potassium iodide solution 1. 00 mol dm?? (500cm? ) * Potassiumperoxodisulphate(VI)(K2S2O8) solution 0. 0400 mol dm?? (500cm? ) * Sodium thiosulphate(VI)(Na2S2O3) solution 0. 0100 mol dm?? (200cm? ) * Iron (III)sulphate solution 0. 0002 mol dm?? (100cm? ) Method: I will be changing the concentration of the potassium iodide solution for 1 set of runs. Then the concentration of the potassium peroxodisulphate on a different set of runs which I will do.

I will then do a further set of runs where I alter the temperature of each run in a series of runs. And finally I will alter the concentration of catalyst that I am adding (which is iron (III) sulphate). This is the iodine clock reaction, and I will use this to measure how the rate changes as the variables are changed. First the apparatus must be set up: the burettes must be filled with each solution, the boiling tubes need to be kept in a test tube rack or in the water bath to control temperature for the temperature experiments.

The solutions need to be measured out accurately, all the solutions can be mixed together at once except for the potassium iodide. Then the potassium iodide can be added and the stopwatch can be started. The boiling tube should then be observed until the clear colourless mixture changes to a blue-black colour and then the stopwatch must be stopped at the first sign of the blue/black complex forming and the time should be recorded. This should then be repeated with all the other mixtures e. g. different concentrations of potassium iodide. The temperature of the solution must also be recorded at the end of the experiment.

This blue/black complex forms after a certain amount of iodine is produced as it reacts with the sodium thiosulphate and when all of the sodium thiosilphate has reacted with iodine the solution will immediately turn black as so no further sodium thiosulphate can react with the iodine so it reacts with the starch and forms the blue/black complex, the sodium thiosulphate is important to distinguish a clear change in rate of the reaction, this is why the sodium thiosulphate concentration and volume is kept the same throught all the experiments.

Risk assessment: Chemical used:| Hazard:| Sodium thiosulphate| May irritate eyes low risk. | Potassium peroxodisulphate| May irritate eyes low risk. | Starch solution| May irritate eyes low risk. | Iron (III) sulphate| Harmful if consumed. | Potassium iodide| May irritate eyes low risk. | Course of action? | | Wear safety spectacles, tuck in stalls and bags, tuck in tie and shirt to avoid flames of Bunsen burners and don’t sit down while working. | the chemicals I am using are irritants at high concentrations, I will be using fairly low concentrations throughout my experiment so I will still be cautious to remove any (if any) reactants are spilt that they are cleaned up quickly and washed if spilt on skin. I will wear goggles to protect my eyes from my reactants and from other people’s reactants that could potentially be hazardous. I will keep all bags and coats well under the desk to prevent any trip hazards. How to make my standard solutions of each of my reactants: Sodium thiosulphate 0. 100 mol dm-3 100cm3: Apparatus and chemicals: * 100cm3pyrexbeaker * 100cm3 volumetric flask * Measuring scales down to 3 decimal places * Weighing boat * Spatula * Bunsen burner * Protective goggles * Distilled water * Sodium thiosulphate I will be producing a sodium thiosulphate solution at a concentration of 0. 0100 moles per deci-meter3. To do this I will require 0. 24818 grams to produce 100cm3 of 0. 0100 mol dm-3 solution. This is because sodium thiosulphate has a molecular mass of 248. 18grams per mole so if I wanted a 1 molar solution of 100cm3 I would need 24. 18 grams but I need a 0. 0100 mol dm-3 solution so I need a hundredth of this which is 0. 24818 grams so 0. 248 grams on the 3 decimal place scales. Using the scales and the weighing boat zero the scales then spatula in the exact mass of the required chemical in this case sodium thiosulphate. Then after this gently put all of the sodium thiosulphate into the 100cm3 beaker and add around about 75 cm3 of distilled water. Then using a Bunsen burner and a clamp stand gently but thoroughly heat the solution till the entirety of the solid has dissolved.

Allow this to cool to room temperature (so that the solution doesn’t take up a higher volume than it should) then pour carefully into the 100cm3 volumetric flask; fill the volumetric flask up to the line (with the meniscus on the line) with more distilled water to make the exact concentration required. Potassium iodide 1 mol dm-3 100cm3: Apparatus and chemicals: * 100cm3pyrexbeaker * 100cm3 volumetric flask * Measuring scales down to 3 decimal places * Weighing boat * Spatula * Bunsen burner * Protective goggles * Distilled water * Potassium iodide

I will be producing a potassium iodide solution at a concentration of 1. 00 moles per deci-meter3. To do this I will require 16. 6 grams to produce 100cm3 of 1. 00 mol dm-3 solution. This is because potassium iodide has a molecular mass of 166 grams per mole so if I wanted a 1 molar solution of 100cm3 I would need 16. 6 grams. So I would measure out exactly 16. 600 grams of potassium iodide on the 3 decimal place scales. Using the scales and the weighing boat zero the scales then spatula in the exact mass of the required chemical in this case potassium iodide.

Then after this gently put all of the potassium iodide into the 100cm3 beaker and add around about 75 cm3 of distilled water. Then using a Bunsen burner and a clamp stand gently but thoroughly heat the solution till the entirety of the solid has dissolved. Allow this to cool to room temperature (so that the solution doesn’t take up a higher volume than it should) then pour carefully into the 100cm3 volumetric flask; fill the volumetric flask up to the line (with the meniscus on the line) with more distilled water to make the exact concentration required. Potassium peroxodisulphate(VI) 0. 400 mol dm-3 100cm3 of solution required: Apparatus and chemicals: * 100cm3pyrexbeaker * 100cm3 volumetric flask * Measuring scales down to 3 decimal places * Weighing boat * Spatula * Bunsen burner * Protective goggles * Distilled water * Potassium peroxodisulphate I will be producing a potassium peroxodisulphatesolution at a concentration of 0. 0400 moles per deci-meter3. To do this I will require 1. 081 grams to produce 100cm3 of 0. 0400 mol dm-3 solution. This is because potassium peroxodisulphatehas a molecular mass of 270. 31 grams per mole so if I wanted a 1 molar solution of 100cm3 I would need 27. 31 grams but I need a 0. 0400 mol dm-3 solution so I need 4 hundredths of this which is 1. 081 grams so 1. 081 grams on the 3 decimal place scales. Using the scales and the weighing boat zero the scales then spatula in the exact mass of the required chemical in this case potassium peroxodisulphate. Then after this gently put all of the potassium peroxodisulphate into the 100cm3 beaker and add around about 75 cm3 of distilled water. Then using a Bunsen burner and a clamp stand gently but thoroughly heat the solution till the entirety of the solid has dissolved.

Allow this to cool to room temperature (so that the solution doesn’t take up a higher volume than it should) then pour carefully into the 100cm3 volumetric flask; fill the volumetric flask up to the line (with the meniscus on the line) with more distilled water to make the exact concentration required. Starch solution: The concentration of this doesn’t really matter so I will use a spatula of the solid starch and dissolve this in the distilled water by heating it with a Bunsen burner in a 100cm3 beaker. Iron (III) sulphate catalyst 0. 00001 mol dm-3 100cm3: Apparatus and chemicals: 100cm3pyrexbeaker * 100cm3 volumetric flask * Measuring scales down to 3 decimal places * Weighing boat * Spatula * Bunsen burner * Protective goggles * Distilled water * Iron (III) sulphate These reactants and these pieces of equipment are precise enough to provide an accurate set of results for my experiments. I will be producing an iron (III) sulphate solution at a concentration of 0. 00001 moles per deci-meter3. To do this I will require 0. 001grams to produce 100cm3 of 0. 00001mol dm-3 solution. This is because iron (III) sulphate has a molecular mass of 51. 908 grams per mole so if I wanted a 1 molar solution of 100cm3 I would need 15. 191grams. But I need a solution of o. oooo1 mols perdm-3So I would need only 1 in 100,000 of that so I would only need 0. 001 grams so I would measure this out measure out exactly on the 3 decimal place scales. Using the scales and the weighing boat zero the scales then spatula in the exact mass of the required chemical in this case iron (III) sulphate. Then after this gently put all of the iron (III) sulphate into the 100cm3 beaker and add around about 75 cm3 of distilled water.

Then using a Bunsen burner and a clamp stand gently but thoroughly heat the solution till the entirety of the solid has dissolved. Allow this to cool to room temperature (so that the solution doesn’t take up a higher volume than it should) then pour carefully into the 100cm3 volumetric flask; fill the volumetric flask up to the line (with the meniscus on the line) with more distilled water to make the exact concentration required. This solution will not be very accurate but this will not matter too much as the chemical will be used relatively against itself. Changing the concentration of potassium iodide:

I will vary the concentration of potassium iodide (one of my reactants) as part of my experiment I will do this by varying the quantity of a standard solution I have made of this reactant and evening out the volume with distilled water so if I wanted 5cm3 of 1. 0 mol per dm-3 I would use 5cm3 of my standard 1molar potassium iodide solution, but if I wanted 5cm3 of 0. 5 mol per dm-3 potassium iodide I would add 2. 5cm3 of 1. 0 mol per dm-3 potassium iodide with 2. 5cm3 of distilled water. For 5 cm3 of potassium iodide1. 0 mol dm-3=5. 0 cm3of 1. 0 mol dm-3 For 5 cm3 of potassium iodide0. mol dm-3=4. 0 cm3 of 1. 0 mol dm-3 + 1cm3 of distilled water For 5 cm3 of potassium iodide0. 6 mol dm-3=3. 0cm3 of 1. 0 mol dm-3 + 2cm3 0f distilled water For 5 cm3 of potassium iodide0. 5 mol dm-3=2. 5cm3 of 1. omol dm-3 potassium iodide +2. 5 cm3 of distilled water. For 5 cm3 of potassium iodide0. 4 mol dm-3=2. 0cm3 of 1. omol dm-3 potassium iodide +3. 0 cm3 of distilled water. For 5 cm3 of potassium iodide0. 3 mol dm-3=1. 5cm3 of 1. omol dm-3 potassium iodide +3. 5 cm3 of distilled water. For 5 cm3 of potassium iodide0. 2 mol dm-3=1. 0cm3 of 1. omol dm-3 potassium iodide +4. cm3 of distilled water. For 5 cm3 of potassium iodide0. 1 mol dm-3=0. 5cm3 of 1. omol dm-3 potassium iodide +4. 5 cm3 of distilled water. Procedure: Prepare a 1. 0 mol per dm-3 solution of potassium iodide. Add accurately using burettes the correct amount of potassium iodide and then vary the concentration by adding the correct amount of distilled water. Changing the concentration of potassium peroxodisulphate: I will use the same method to prepare these solutions as I have just described but I will be aiming to achieve different concentrations than before because of the nature of this reactant.

For 5 cm3 of potassium peroxodisulphate0. 0400mol dm-3 =5. 0 cm3of 0. 0400 mol dm-3 of potassium peroxodisulphate 0. 0400 mol dm-3 For 5 cm3 of potassium peroxodisulphate0. 0320 mol dm-3=4. 0 cm3 0f 0. 0400 mol dm-3 of potassium peroxodisulphate 0. 0400 mol dm-3 + 1. 0 cm3 of distilled water. For 5 cm3 of potassium peroxodisulphate0. 0240 mol dm-3=3. 0 cm3 0f 0. 0400 mol dm-3 of potassium peroxodisulphate 0. 0400 mol dm-3 + 2. 0 cm3 of distilled water. For 5 cm3 of potassium peroxodisulphate0. 0200 mol dm-3=2. 5 cm3 0f 0. 0400 mol dm-3 of potassium peroxodisulphate 0. 0400 mol dm-3 + 2. cm3 of distilled water. For 5 cm3 of potassium peroxodisulphate0. 0160 mol dm-3=2. 0 cm3 0f 0. 0400 mol dm-3 of potassium peroxodisulphate 0. 0400 mol dm-3 + 3. 0 cm3 of distilled water. For 5 cm3 of potassium peroxodisulphate0. 0120 mol dm-3=1. 5 cm3 0f 0. 0400 mol dm-3 of potassium peroxodisulphate 0. 0400 mol dm-3 + 3. 5 cm3 of distilled water. For 5 cm3 of potassium peroxodisulphate0. 0080 mol dm-3=1. 0 cm3 0f 0. 0400 mol dm-3 of potassium peroxodisulphate 0. 0400 mol dm-3 + 4. 0 cm3 of distilled water. For 5 cm3 of potassium peroxodisulphate0. 0040 mol dm-3=0. 5 cm3 0f 0. 400 mol dm-3 of potassium peroxodisulphate 0. 0400 mol dm-3 + 4. 5 cm3 of distilled water. Procedure: Prepare a 0. 0400 mol per dm-3 solution of potassium peroxodisulphate. Add accurately (using burettes) the correct amount of potassium peroxodisulphate and then vary the concentration by adding the correct amount of distilled water. Changing the quantity of iron (III) sulfate: I was originally going to use different concentrations of my catalyst but then after preliminary experiments I decided to dilute my catalyst from 0. 0002 mol dm-3 down to 0. 00001 mol dm-3 and to use different quantities of his catalyst rather than different concentrations. This affects the concentration of the other reactants but not enough to make a significant difference. Quantities of 0. 00001 mol dm-3 of iron (III) sulphate in solution: 1. 0cm3 0. 8cm3 0. 6cm3 0. 5cm3 0. 4cm3 0. 3cm3 0. 2cm3 0. 1cm3 Procedure: Make up the 0. 00001 mol dm-3 iron (III) sulphate solution Add to the other reactants the designated quantities. Changing the temperature at which different runs are done at: To do this I will use a water bath and 2 thermometers (1 in each boiling tubes containing the reactants) and ice cubes a kettle and water.

I will be doing 8 runs at different temperatures being: 60 degrees Celsius 50 degrees Celsius 40 degrees Celsius 30 degrees Celsius 25 degrees Celsius 20 degrees Celsius 15 degrees Celsius 10 degrees Celsius I will only mix and start the stop watch at the point when both boiling tubes mixtures are at the starting temperature and to keep them at this temperature by keeping the boiling tube in the water bath. Changing the temperature of the reaction and using a catalyst: To do this I will use a water bath and 2 thermometers (1 in each boiling tubes containing the reactants) and ice cubes a kettle and water.

The reactants will also contain 1. 0 cm3 of 0. 00001 mol dm-3 iron (III) sulphate this is more catalyst than is needed so it doesn’t need to be very accurate but I will be just to be on the safe side. I will be doing 8 runs at different temperatures being: 60 degrees Celsius 50 degrees Celsius 40 degrees Celsius 30 degrees Celsius 25 degrees Celsius 20 degrees Celsius 15 degrees Celsius 10 degrees Celsius I will only mix and start the stop watch at the point when both boiling tubes mixtures are at the starting temperature and to keep them at this temperature by keeping the boiling tube in the water bath.

This set of test will allow me to see when the catalyst actually begins to play a part when it has the correct activation enthalpy. Results: Iron (III) sulfate: These are the results are for the concentration change variable of the Iron (III) sulphate catalyst being altered between the concentrations of o. oooo1 moles per deci-meter-3 up to a concentration of 0. 00020 moles per deci-meter-3. I did 3 preliminary runs to determine the concentration of the catalyst to provide a range of results; this is because the catalyst is so effective even at very low concentrations that a higher concentration of catalyst isn’t required.

Here are the results below. Mixture:| Concentration of catalyst mol dm3| Results:| Repeats:| Preliminary 1| 0. 01000| 44. 98| | Preliminary 2| 0. 00100| 45. 71| | Preliminary 3| 0. 00010| 63. 82| | 1| 0. 00001| 81. 18| | 2| 0. 00002| 64. 90| 69. 69| 3| 0. 00004| 68. 15| | 4| 0. 00006| 67. 12| | 5| 0. 00008| 66. 56| | 6| 0. 00012| 62. 86| | 7| 0. 00016| 59. 66| | 8| 0. 00020| 57. 59| | From these results I can determine that between 0. 00020 mols per dm3 and 0. 00100 mols per dm3 is where the optimum catalyst concentration lies.

The 2 results for the first and second preliminary tests shows that there is no further effect when the concentration is raised from 0. 00100 mols per dm3 to 0. 01000 mols per dm3. I needed to repeat mixture 2 because the result didn’t show the trend of the other results so I repeated this run to remove the outlier and replace it with a more accurate result. Results seconds:| 1/result=rate| 44. 98| 0. 02223| 45. 71| 0. 02188| 63. 82| 0. 01567| 81. 18| 0. 01232| 64. 90| 0. 015408| 68. 15| 0. 01467| 67. 12| 0. 01489| 66. 56| 0. 015024| 62. 86| 0. 01591| 59. 66| 0. 01676| 57. 59| 0. 01736| Order:

I will determine the order of this reactant Iron (III) sulphate (the catalyst to this reaction) with respect to the reaction by using the data that I have collected and converting it to its reciprocal number and plotting this onto a graph, from this graph I will be able to see what order this reaction is because if the line of best fit is curved the reaction is second order, but if the line of best fit is parallel to an axis then the order will be zero order, if the line of best fit is straight but showing a trend with respect to reaction time and concentration (or temperature) the reaction is first order, finally the reaction may go from first order to zero order, this will occur at the point when the concentration of the variable makes no further difference to the reaction rate/time. In respect to this reaction this catalyst Iron (III) sulphate displays an order of second; I have determined this by plotting the reaction rate against the concentration this gives a curve so this is 1st order. Outliers:

With my set of results I found that I only produced one outlier this was the run with a 0. 00001 mol dm-3, even though I may have produced an outlier I believe this is down to a mixture of things, mainly this could have been caused by equipment set up and solution mixing where I could have added slightly less catalyst than I was supposed to this is down to the very minute quantities being used despite this the outlier is on the correct side of the line of the results as it took longer than the other results. On my graph of reaction rate against concentration I found an outlier but I believe this was caused by insufficient addition of the catalyst. Graph here. Graph here. Potassium iodide:

These are the results are for the concentration change variable of potassium iodide being altered between the concentrations of 1. 000 mols per dm-3 down to a concentration of 0. 125 mols per dm-3. Mixture:| Concentration of potassium iodide mol dm3:| Results: (seconds)| 1| 1. 000| 22. 37| 2| 0. 875| 31. 56| 3| 0. 750| 40. 66| 4| 0. 675| 52. 84| 5| 0. 500| 69. 66| 6| 0. 375| 97. 08| 7| 0. 250| 245. 09| 8| 0. 125| 570. 92| Result:| 1/Result = rate| 22. 37| 0. 044703| 31. 56| 0. 03169| 40. 66| 0. 02459| 52. 84| 0. 01893| 69. 66| 0. 01436| 97. 08| 0. 01030078| 245. 09| 0. 004080134| 570. 92| 0. 001752| From this second table I will determine the order by producing a graph of reaction rate against concentration. Order:

I will determine the order of this reactant potassium iodide with respect to the reaction by using the data that I have collected and plotting this onto a graph, from this graph I will be able to see what order this reaction is because if the line of best fit is curved the reaction is second order, but if the line of best fit is parallel to an axis then the order will be zero order, if the line of best fit is straight but showing a trend with respect to reaction time and concentration (or temperature) the reaction is first order, finally the reaction may go from first order to zero order, this will occur at the point when the concentration of the variable makes no further difference to the reaction rate/time. In respect to this reaction this reactant potassium iodide displays an order of first order, I can tell this because the results produce a straight line of best fit on my graph representing rate (1/seconds) against concentration. This result isn’t surprising as this reactant of potassium iodide is integral to the reaction and the reaction rate is dependent on this reactant. Graph here. Graph here Potassium peroxodisulphate:

These are the results are for the concentration change variable of potassium peroxodisulphate being altered between the concentrations of 0. 040 mols per dm-3 down to a concentration of 0. 008 mols per dm-3. Mixture:| Concentration of potassium peroxodisulphatemol dm3:| Results: (seconds)| 1| o. o40| 28. 84| 2| 0. 036| 31. 68| 3| 0. 032| 34. 66| 4| 0. 028| 32. 25| 5| 0. 024| 33. 45| 6| 0. 020| 52. 99| 7| 0. 012| 66. 87| 8| 0. 008| 131. 91| Result:| 1/result= rate:| 28. 84| 0. 03467| 31. 68| 0. 03156| 34. 66| 0. 02885| 32. 25| 0. 0310078| 33. 45| 0. 02989| 52. 99| 0. 01887| 66. 87| 0. 01495| 131. 91| 0. 0075809| I will use the table above to plot a graph to determine the reaction order; the graph will plot concentration and rate against each other. Order:

I will determine the order of this reactant potassium peroxodisulphate with respect to the reaction by using the data that I have collected and plotting this onto a graph, from this graph I will be able to see what order this reaction is because if the line of best fit is curved the reaction is second order, but if the line of best fit is parallel to an axis then the order will be zero order, if the line of best fit is straight but showing a trend with respect to reaction time and concentration (or temperature) the reaction is first order, finally the reaction may go from first order to zero order, this will occur at the point when the concentration of the variable makes no further difference to the reaction rate/time. The order of this reactant potassium peroxodisulphate is first order with respect to the reaction. I expected this because this reactant is also an integral part of the reaction but in the second step rather than potassium iodide in the first step Outliers:

I think the occurrence of these 3 outliers was caused by doing these runs on different days and this could have been caused by on these separate days the room could have been warmer increasing reaction rate and also the chemicals used could have been replaced by slightly different concentrations when the solutions where re-made up. Graph here. Graph here Temperature change: These results show the variation of reaction rate and how it is dependant and variable due to temperature, I will do reaction runs at temperatures of 10 degrees Celsius up to 60 degrees Celsius (reactants starting temperatures). Mixture:| Variable change in temperature. (degrees Celsius)| Results: (seconds)| 1| 10| 92. 06| 2| 15| 59. 37| 3| 20| 40. 68| 4| 25| 26. 44| 5| 30| 18. 57| 6| 40| 14. 03| 7| 50| 7. 98| 8| 60| 5. 92|

Graph here. Arrhenius equation: I will use the arrhenius equation to calculate the activation enthalpy of my reaction between potassium iodide and potassium peroxodisulphate, I will do this by using the results I collected for my temperature change variable. This is the logarithmic of the arrhenius equation below: (6) K: is the rate constant Ea is the activation enthalpy R is the gas constant 8. 31j?? mol?? T is temperature in Kelvin Step 1: find the rate constant To find the rate constant I must convert the results of time to the reciprocal value (1/time in seconds). Time taken in seconds:| 1/time in seconds (to 4 significant figures):| 92. 6| 0. 010862| 59. 7| 0. 01675| 40. 68| 0. 02458| 26. 44| 0. 03782| 18. 57| 0. 05385| 14. 03| 0. 07128| 7. 98| 0. 1253| 5. 92| 0. 1689| These numbers are each now ready to be made into the InK part of the logarithm this is done by typing in each number and pressing the In button on the scientific calculator. This step is done below. 1/time in seconds (to 4 significant figures):| Ink (found using k (the other column) and pressing the In button on a scientific calculator)(4 significant figures). | 0. 010862| -4. 522| 0. 01675| -4. 0893| 0. 02458| -3. 7058| 0. 03782| -3. 274| 0. 05385| -2. 922| 0. 07128| -2. 641| 0. 1253| -2. 07704| 0. 1689| -1. 778|

The next step is to also convert the temperatures in to Kelvin and then in to the reciprocal of the Kelvin. Temperature degrees Celsius:| Temperature in Kelvin:| 1/temperature Kelvin(to 4 significant figures):| 10| 283| 0. 003533| 15| 288| 0. 003472| 20| 293| 0. 003413| 25| 298| 0. 003356| 30| 303| 0. 00330033| 40| 313| 0. 003195| 50| 323| 0. 0030959| 60| 333| 0. 003003003003| I will then plot a graph of 1/temperature Kelvin against Ink with Ink on the y-axis and 1/teperature Kelvin on the x-axis. From this I need to measure the gradient of the line of best fit that I draw the gradient will be negative. Turn over to see the graph. -2. 0- -4. 00. 003063-0. 003480 =-4796. 163 Graph here.

This figure is then multiplied by the gas constant 8. 31j?? mol?? and then further multiplied by 1000 to get the answer in KJ mol-1 -4796. 163? -8. 31=39856. 11453 39856. 11453? 1000=39. 856 Therefore the activation energy for this reaction is 39. 856 KJmol-1 Percentage error: Percentage error is calculated by the error of the equipment divided by the quantity of chemical used and multiplied by 100. Percentage error is an important factor to consider when testing and evaluating the reliability and accuracy of the results, higher precision equipment causes a lower the percentage error higher reliability and accuracy of the experiment as a whole.

Percentage error for making solutions: I used 3 decimal place scales, to weigh out my raw chemicals, I then used a 100cm3 beaker with 75cm3 of solution appr0x and then transferred the solution to a 100cm3 volumetric flask and filled it up to the 100cm3 line. What percentage error I will need to measure for each different solution I made: Scales: 0. 005/quantity in grams of measured quantity? 100=….. 100cm3 beaker: 0. 1/volume of solution? 100=…. 100cm3 volumetric flask: 0. 05/volume of solution (100cm3)? 100=…. Total percentage error= addition of all percentage errors. Percentage error for making iron (III) sulphate solution: 0. 005/0. 001? 100=5% error 0. 1/75? 00=0. 13% error 0. 05/100? 100=0. 05% error Total percentage error=5. 18% Percentage error for making potassium iodide solution: 0. 005/16. 6? 100=0. 03012 %error 0. 1/75? 100=0. 13% error 0. 05/100? 100=0. 05% error Total percentage error=0. 21012% Percentage error for making potassium peroxodisulphate solution: 0. 005/1. 0812? 100=0. 4624% error 0. 1/75? 100=0. 13% error 0. 05/100? 100=0. 05% error Total percentage error= 0. 6424 Percentage error for making sodium thiosulphate solution: 0. 005/0. 248? 100=2. 0161% error 0. 1/75? 100=0. 13% error 0. 05/100? 100=0. 05 Total percentage error=2. 196% Percentage error for making starch solution:

The percentage error of this isn’t relevant for this because the starch solution is of a un measured concentration and doesn’t make a difference to the reaction and just needs to be present. Total percentage error=0% Conclusion: My investigation aim was to perform numerous kinetics reactions varying 4 variables, these being the use of an Iron (III) sulphate catalyst concentration of potassium iodide, changing concentration of potassium peroxide and varying the temperature. I was able to perform these experiments and I obtained results from them. From my experiment where I used an Iron (III) sulphate catalyst I saw a change in reaction rate, as the catalysts quantity was increased the reaction rate increased, this only happened to an extent because there was no further need for more catalyst after 0. 00020 mol dm-3 .

From my results I produced a graph and this showed that the order of the reaction with respect to Iron(III) sulphate was first to zero order as the catalyst stops being needed as all the reactants are on an active site so more active sites are unnecessary. My results from my temperature variation reaction support the collision theory and also back up as evidence that the Maxwell Boltzmann distribution curve is correct. I found that with the temperature variable in thisexperiment produced smooth continuous curve on the graph which represented the variable to be second order. I also used the results that I obtained from this experiment to discover the reaction enthalpy, I found that it was 39. 856 KJmol-1.

From my experiment runs where I changed the concentration of potassium iodide I found that the rate of reaction increased, making the reaction first order. When the potassium iodide was at a 0. 125 molar solution and took 570 seconds but with a 1 molar solution the reaction only took 22. 37 seconds. This is a huge decrease shows that this is a great variable to alter to increase the reaction rate. When I changed the concentration of potassium peroxodisulphate I found that the reaction was first order and as the concentration increases the reaction time decreases. When I used a 0. 0400 mol dm-3 solution the reaction took 27. 87 seconds and it took 131. 31 seconds for 0. 008 mol dm-3 solution this shows an increase but it is not a very strong increase as it was in the case of potassium iodide.

I also used a catalyst, this was Iron (III) sulphate and this is a known catalyst of this reaction, it is a catalyst because it changes oxidation state under the right conditions. In my set of runs with this catalyst I found that even with an incredibly small quantity of Iron (III) sulphate solution in the mixture that the reaction took place almost twice as fast as before, but the catalyst is limited and stops increasing reaction rate after a certain point as no further active sites on the catalyst are needed because all of the reactants are being consumed. From my results I was able to find that the concentration of the catalyst affected the reaction rate to an extent the limit of the concentration making a difference was 0. 0033 mol per dm3. knowing this means that the optimum concentration of catalyst to use would be 0. 00033 without using excess catalyst. Because of the differences in rate change from these different variables it makes it easy to make a decision about which variables are best to change. I believe that the use of the catalyst Iron (III) sulphate in this reaction to increase the rate is a good choice as it works at room temperature and lowers the reaction enthalpy so no energy needs to be given to the reaction this means in industry the fuel costs are kept to a minimum for this same reason I wouldn’t use a higher temperature because of the fuel costs.

But I would use a higher concentration of potassium iodide because it reduced the reaction time by a minute and this would lead to a higher yield of the products being produced. I wouldn’t use a higher concentration of potassium peroxodisulphate because it doesn’t make a big enough difference to reaction rate. If I was a company I would use a high concentration of potassium iodide and a catalyst because the catalyst is recyclable and can be reused so essentially will save the company money on energy and will not need to be renewed much. Evaluation: I believe that I carried out my experiment well and produced reliable and accurate data, but I know that there is a degree of error in my results this is due to numerous factors.

These limitations include glassware, I used b- class glassware and this is quite accurate but it would have been even more accurate if I could have used a-class glassware but fortunately because of my reaction not needing to be incredibly precise the glassware won’t matter too much, I would also of liked to use a proper water bath that is calibrated and maintains a consistent temperature but I wasn’t able to use 1 unfortunately, but my temperature results turned out well and followed a clear pattern that was expected. I also didn’t have control of the room temperature which could have produced a slight fluctuation in the results of some runs that I did on different days. Also when I made up my solutions the weighing boat may have still kept some of the chemicals making the weighing slightly inaccurate. My solutions in my 100cm3 volumetric flask may have contained less than 100 cm3 because the solution may have been warmer than room temperature. I have also been inhibited by myself and human error, this has probably caused the biggest errors in my results, I have been careful through all steps of production of chemicals and the reaction it’s self but

I could still have made the solutions inaccurately because of the parallax effect of not being perfectly level this could cause a higher or low concentration to be achieved than wanted. I could have less accurate results as the reactants may have decomposed between reaction sessions because of the time elapsed they also could have been contaminated, I know this happened once as my starch solution had turned a clear blue so I believe this was caused not by potassium iodide but by potassium peroxodisulphate as I noticed this happen when I miss mixed a solution. From my investigation I gained sets of data that have allowed me to measure rate and activation enthalpy of my reaction. My data has also enabled me to compare and contrast the effects of the variables.

The biggest cause of inaccuracies in my results was my fault caused my human error from the point of making up the solutions to the point of stopping the stop watch at the first sign of a blue/black complex forming but these inaccuracies are nearly impossible to compare unless the results are measured by a machine which I obviously didn’t have access to. Unfortunately I would also have liked to vary the temperature of my reactants with the use of my catalyst as well to determine the activation enthalpy of my catalyst. I would also have liked to use more catalysts, one of which would have been Iron (II) sulphate because it does the same steps as the Iron (III) catalyst that I used but in the opposite order and I would have been interested in if this minor change would have made a difference to reaction rate I would assume it could but as it is overall doing the same just in a different order that it would take the same amount of time.

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