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# Bubble troubel - The problem posed is how to speed up the reaction of hydrogen peroxide to water and oxygen using the enzyme catalase.

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Introduction

Bubble Trouble - GCSE Coursework Raj Patel Introduction The problem posed is how to speed up the reaction of hydrogen peroxide to water and oxygen using the enzyme catalase. The catalase will be the independent variable in the experiment, and the amount of oxygen produced is the dependent variable. Using 5 different concentrations of the enzyme for the experiment we aim to measure progress by how big the foam column is to see how much oxygen is produced. I am going to vary the catalase and in order for a fair test I will keep the others constant. Hydrogen Peroxide ---...Catalase...---> Water and Oxygen Apparatus Test tube Water Bath Timer Planning the investigation I will vary the catalase in this experiment. This will be my independent variable. The dependant variables (e.g. height of foam) cannot be changed to keep this a fair test. I predict that the more catalase added to the hydrogen peroxide, the more oxygen will be produced (for the more catalase you add the more there will be to react with the hydrogen peroxide). The graph shows that the amount of catalase that was added, the more oxygen was given off. This leads me to think that the 'quantitative prediction', is true, so the amount of catalase added and the amount of oxygen given off is directly proportional. So, the more catalase molecules there are, the more chance of a reaction with the hydrogen peroxide, so therefore it is directly proportional. This is the collision theory. So, when I double the amount of catalase added, I will double the amount of oxygen produced. Inevitably, the graph will come to an end for all the oxygen molecules have been reacted with so there will be nothing else left to react which brings the end of the graph to a halt. This is due to the lock and key theory, when all the hydrogen peroxide has been reacted with, for the enzymes act as a lock and the hydrogen peroxide as a key. ...read more.

Middle

Once all of the hydrogen peroxide is broken down, there will be nothing for the catalase to react with, so this is why I predict the glass to level out. <--Lock & key--> Method Having cut 5 lots of potatoes from the same source (5 because we will have 5 different concentrations of catalase). The more potato discs the more catalase present. We're going to use 5, 10, 15, 20, 25 discs for the experiment. We added each group of discs to a group of five test tubes. Each test tube contained ph9 buffer solution. Then with another 5 test tubes containing 5ml of 20 vol hydrogen peroxide solution, we put both groups of 5 test tubes (all together 10 test tubes: 5 containing 5ml of ph9 buffer solution and different amounts of catalase, the other containing 5 ml of 20 vol hydrogen peroxide solution), and place them each in a water bath for them to equilibrate. As soon as they are all the same temperature, we spill the contents of one group of test tubes into another, and wait. After 10 minutes we measure the foam to see and differentiate between how much catalase is used and how much oxygen is produced. This is the same method that was used in our preliminary work. We kept other variables constant by the use of water baths (for the temperature to equilibrate), Results These are the results I got from the experiment. Time taken for successive 2ml of oxygen to be generated in seconds. The more slices of catalase the higher the concentration of catalase Slices 1st 2ml 2nd 2ml 3rd 2ml Average 6 79 78 80 79 8 63 65 64 64 10 49 50 49 49 12 40 39 43 41 14 25 23 26 25 Group results Slices 1st 2ml 2nd 2ml 3rd 2ml Average 6 93 77 79 83 8 63 67 70 66 10 49 56 48 51 12 42 40 45 42 14 25 27 22 25 Slices 1st ...read more.

Conclusion

1st 2ml 2nd 2ml 3rd 2ml Average 6 60 62 58 60 8 48 49 46 48 10 40 38 42 40 12 33 28 30 30 14 25 22 23 23 Conclusions From the tables and graphs I conclude that the higher the concentration of catalase which is added, the faster the hydrogen peroxide is broken down into water and oxygen. There is an obvious connection between the time taken and the concentration of catalase added shown in the graphs. The pink line (which is the averages of the time taken), clearly shows negative correlation when the other line shows positive correlation, so therefore it shows that the more catalase that was used the faster the hydrogen peroxide was broken down. If there had been more readings taken on the graph, eventually it would all have levelled out due to all the catalase molecules having broken down all the hydrogen peroxide molecules. This is related to collision theory. When the potato discs and the hydrogen peroxide reacts, this is where the collision theory is put into place. So with the help of the theory, I state that the more catalase added was responsible for the more water and oxygen produced, since the more catalase molecules meant that there was more potential and probability for a reaction to take place. With the lock and key theory, the more catalase there is added the more chance of one of the enzymes to lock onto a hydrogen peroxide molecule and break it down, and this is why the graph would probably level out at the end for all the catalase will break down all the hydrogen peroxide molecules and so there will be no increase in the speed of the reaction. Overall my results were pretty good, and even showed some quantative qualities, for when you double the amount of potato discs you halved the time it took for water and oxygen to be produced. The repetitions of my results give a suitable degree of accuracy. ...read more.

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