In this experiment I will investigate the effect of using a catalyst (enzyme), called catalase found in yeast, on the decomposition of H2O2. To measure the rate of this reaction I will collect the O2 given off when the yeast and H2O2 react. I will then see how temperature effects the reaction. Safety 1. Goggles will be worn at all times to prevent any splashes from irritating the eye. 2. Any spills will be cleared up straight away. 3. H2O2 is flammable so the substance will be kept away from naked flames. 4. The gas syringe will be placed in a clamp to prevent it being knocked over. 5.
The gas syringe has a string attaching the two pieces together so that if the rear part of the gas syringe gets pushed out it will not fall on the floor and smash. Pre-tests We set up an experiment as shown in the diagram below. The diagram also shows the equipment we used. Pre-test Results First we tried 5cmi?? of H2O2 at 40i?? C for 3 minutes. We collected a result after every 30 seconds. The results were as follows: Time (sec) O2 Produced (cmi?? ) This amount of Hydrogen Peroxide didn’t give very good results so we re-did the experiment with 10cmi??of H2O2 instead.
These are the results we got: Time (sec) O2 Produced (cmi?? ) 30 17 60 32 90 44 120 52 150 57 180 59 These results are much better but the space between each result will not be easy to plot on a graph and will not give a very smooth curve, which is what I am expecting. So I will take the results every 20 seconds instead of 30 seconds in the actual investigation. Method The method will be: 1. Set up the equipment as shown on the first page but use 10cmi?? of H2O2 instead of 5cmi?? 2. Make sure the gas syringe is at 0 3. Pour the H2O2 into the conical flask.
4. Add the yeast 5. Put the bung in quickly to stop the initial burst of oxygen from escaping 6. At each temperature the amount of oxygen given off will be measured at 20-second intervals for 3 minutes. This will be read off the gas syringe 7. I will do tests between 20i?? C and 70i?? C at 10i?? C intervals 8. The temperature will be altered by varying the temperature of the water bath, which surrounds the conical flask with the solution in it. 9. At each temperature I will repeat the experiment 2 times and then take the mean of these 2 results. How the Experiment is “fair”.
Table of variables Factor Why it affects the reaction Diagram (if applicable) How it is controlled Amount of Yeast The more catalyst (yeast) there is the quicker the hydrogen peroxide decomposes because there is a higher concentration and there will be more collisions. Each time I will only use0. 1g of yeast. Volume of H2O2 If there is more H2O2 then the reaction will be able to go on longer because as enzymes do not get used up. Also the collisions would happen more often because there is a higher concentration of the substrate. Each time I will only use 10cmi?? of H2O2. Stirring.
If you stir the solution the reaction would happen more quickly due to the catalase colliding with the substrate more often. I will not stir the solution. Time measure The longer you time the more oxygen is going to be given off until all the H2O2 has decomposed. I will time for 3 minutes and take a result after every 20 seconds. Temperature of the water bath The hotter the water bath is the quicker the reaction will be because the substrate and enzymes will be moving more quickly and thus will collide more often. This will only happen up to a certain point (until the enzyme becomes denatured).
This is the variable so I will control this as said in the method. Prediction I predict that as the temperature rises so will the rate of the decomposition of the H2O2. I think this will happen because in the yeast is an enzyme called catalase. This is an enzyme, which is a biological catalyst. All reactions need activation energy (the energy required to break the bonds of the substance in the reaction) to take place. A catalyst lowers the activation energy therefore making the reaction happen quicker because the bonds break more easily.
As you increase the heat, the particles vibrate more vigorously which makes the bonds even easier to break. The way the catalyst speeds up the break down of the H2O2 is that it collides with the hydrogen peroxide molecules. This is stated in the collision theory “Chemical reactions occur when particles of the reactants collide. They must collide with a certain minimum energy, called the activation energy. ” If you look at catalase as an enzyme rather than a chemical catalyst then the temperature will still increase the rate of the reaction.
This is because enzymes are a Biological catalyst. An enzyme is not a surface for the reaction to take place on as with a chemical catalyst but it breaks down the substrate with a lock and key mechanism. This theory says that the enzyme acts as the lock and every time the substrate fits into the lock the substrate is broken down i. e. In Mackean’s it states “A 10i?? C rise will double the rate of reaction”. So if we get 10cmi?? of oxygen produced after 20 seconds at 20i?? C then we should get 20cmi?? of oxygen produced after 20 seconds at 30i?? C.