The first diagram shows that only a certain amount of particles have enough energy to react, but when the catalyst is added, the new route allows more particles to react because of the fact that the activation energy is lower. This therefore shows that if there is no catalyst present in the reaction (zero concentration of yeast (control of my experiment)) the rate of oxygen produced will be lower than if there was a catalyst present.
In my investigation I will be changing the concentration of yeast (Catalase) in the reaction and my hypothesis states clearly that as concentration of yeast increases so must the rate of oxygen produced. I am now going to try and prove this again by Collision Theory, but this time not of catalysts but of concentration.
In Collision Theory it is well known that if you increase the concentration of particles in a certain area, the number of collision will increase. This can be shown by two very simple diagrams:
These two diagrams show that if the concentration is increased the number of collisions will increase, in the low concentration diagram there are two collisions and in the high concentration diagram there are four collisions. Since there are more collisions there will probably be more successful reaction and therefore a higher rate of reaction which will mean a faster rate of oxygen produced. Another way of putting this would be to say that if the concentration of the enzyme is increased there is more Catalase and so more active sites that can catalyse the substrate at one time, which means that the reaction rate would be higher leading to a faster rate of oxygen being produced.
Also I expect to get a straight line graph of rate of oxygen produced against concentration of the enzyme because the higher the concentration of the enzyme, the higher the amount of active sites which can catalyse the substrate molecules to produce oxygen. However this would only happen if there was always excess substrate because if all the substrate was catalysed then the rate of oxygen produced would level off.
Method
In this experiment I will measure the volume of oxygen produced in cm3, within a certain time, when the Catalase breaks down the H2O2 by using a gas syringe. The gas syringe will be able to measure the volume of gas produced because the equipment will be set up so it is air tight when the reaction start (set up as above).
I did some preliminary work on this experiment in order to find the best concentrations and best time range I could use for my own reactions. All the H2O2 that I used and I will use was and will be 100% solution 20 volume. First I tried to find the right concentration by just seeing how much gas the reaction was giving off in 10 seconds. I used for these test 5 cm3 of H2O2 with 10 cm3 of yeast, 10 cm3 of H2O2 with 10 cm3 of yeast and 5 cm3 of H2O2 with 5 cm3 of yeast. I found from these tests that the reaction with 10 cm3 of yeast was too vigorous because the froth it made from the gas rising in the liquid rose all the way into the gas syringe, and so therefore the best concentration was 5 cm3 of H2O2 with 5 cm3 of yeast. After having done this I began to find the best possible time range. I did this by reacting 5 cm3 H2O2 with 5 cm3 of Catalase and measuring the volume of gas given off at different times. I found that 15 seconds after the start of the reaction was a good time range because it gave the reaction plenty of time and the volume of gas given off was about 75 cm3 which means that for my own reactions if I have anything that gives off more gas within this time will not go off the scale of the gas syringe.
In the reactions there will permanently be 5 cm3 of H2O2 and then I will use 5 cm3 (100% yeast), 4 cm3 (80% yeast), 3 cm3 (60% yeast), 2 cm3 (40% yeast) and 1 cm3 (20% yeast) of the yeast solution, but since I have to have a total volume of 10 cm3 I will then add the corresponding amount of water to the total volume in order to get 10 cm3. (For example 5 cm3 of H2O2 + 3 cm3 of Yeast + 2 cm3 of water = 10 cm3 total volume). Also, as a control I will have 0 cm3 (0% yeast) of yeast with 5 cm3 of water and 5 cm3 of H2O2. This control experiment will show me that pure water will not react or catalyse the hydrogen peroxide. I will also being doing 3 repeats of each reaction in order to get a set of more reliable and accurate results because they results will vary. The reactions will be as follows:
1) 5 cm3 of H2O2 and 5 cm3 of yeast (100% of yeast)
2) 5 cm3 of H2O2, 4 cm3 of yeast and 1 cm3 of water (80% of yeast)
3) 5 cm3 of H2O2, 3 cm3 of yeast and 2 cm3 of water (60% of yeast)
4) 5 cm3 of H2O2, 2 cm3 of yeast and 3 cm3 of water (40% of yeast)
5) 5 cm3 of H2O2, 1 cm3 of yeast and 4 cm3 of water (20% of yeast)
6) 5 cm3 of H2O2 and 5 cm3 of water (0% of yeast) (Control of experiment)
This is how the experiment will be carried out after setting up the apparatus:
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Use syringe to add y cm3 of yeast into the conical flask.
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Use a second syringe to add the x cm3 of water depending on how much yeast is going to be used. (Look at the list above) Make sure volume of gas in syringe is still 0 cm3.
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Use a third syringe to put 5 cm3 of H2O2 into the conical flask and use the other hand to start the timer on the stop watch at the same time as the H2O2 is added. (Leave syringe in the tube so it stays air tight).
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Note the volume of gas given off in cm3 when the timer reaches 15 seconds.
- Rinse conical flask and then repeat with 80%, 60%, 40%, 20%, 0% of yeast (use the list of reactions above for volumes)
- Repeat each reaction 3 times
Safety
Always wear lab coats, glasses and gloves because H2O2 is a substance that very corrosive and can cause lots of damage to the skin and it can also blind if there is contact with the eyes. If there are any spillages onto the skin or it seems you have a burning sensation on your skin, rinse immediately and inform teacher. If there is any contact between eyes and H2O2 inform the teacher quickly and rinse until the teacher attends to you.
Variables and Fair Test
In this experiment I will have to keep certain variable the same in order to get accurate and reliable results. In total there are 7 variables in this experiment and they are the following:
- pH
- Temperature
- Enzymes Concentration
- Substrate Concentration
- Total volume
- Time
- Oxygen produced
Out of all of these variables there is one “Dependent Variable”, this is oxygen produced variable because this is what I am measuring by altering one of the other variables. Apart from the dependent variable all of the rest are independent variables and so therefore any one of these could alter the dependent variable giving me unreliable results. Therefore I will now explain what these variables can do and what to do with them in order to keep them constant.
The variable pH is important because if the pH of the solution in which the enzyme is present is too low or too high the enzyme will not function properly because the pH will damage the shape of the enzyme and the shape of the active site making it so that the substrate molecule will not be able to fit into the active site, thus not allowing the enzyme to break it down. This will mean that the rate of oxygen produced will slow down. In my experiment I will keep the pH about 7 because this is Catalase’s optimum pH and this maintaining of pH is not difficult because all the substance that I am going to use will be neutral pH. This damaging of the enzyme is called “Denaturing” and is permanent.. Enzymes as I said earlier were proteins and this is why they can be denatured very easily. Temperature is important because it is able to increase the rate of oxygen produced and also decrease the rate of oxygen produced. I know it can increase the reaction rate because if there is a higher temperature the particles would have more kinetic energy and therefore there would be a faster rate of collisions and a higher % of successful collisions because since the particles would have more energy it would be easier to reach the activation energy and so more reactions (this would be the opposite if it was a lower temperature). The rate of oxygen produced could be decreased by the fact that proteins can easily be damaged by heat so if there is a high temperature the enzyme can be denatured and if there is a low temperature they can become inactive. My experiment will be done at room temperature.
Enzyme concentration is not a problem because this is the variable which I am changing to find out the effects it has on the dependent variable. Substrate concentration can alter my results in two ways. One is that it will make my results quicker than they should be, if I put a higher volume in some and lower in others because from Collision Theory I have learnt that if there is a higher concentration of a substance in a solution it will have more collision which means more reactions and therefore this will make my results quicker than if I used a lower concentration of it. I can solve this problem by using the same volume of H2O2 in all my reactions. The second way that substrate concentration can affect my results is that it can level off the rate of oxygen made, if all the substrate is catalysed within my time range. I can solve this problem by simply using excess substrate but at the same time I will still have to keep the volume the same because of the problem I discussed earlier. Total volume must also be kept the same because otherwise the experiment would be unfair due to collision theory. This is because if the numbers of particles are changed, the number of collisions would be changed and therefore this means the number of reactions would change leading to a different amount of gas produced in my experiment. I can solve this by adding water when I use a smaller amount of enzyme.
Time is another one of the variable that must be kept the same because if not the reaction would have more or less time to react and so producing different volumes of gas. I can solve this by always letting the reaction go on for the same amount of time; this will be 15 seconds in my experiments.
