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Investigate the effect of surface area on the rate of reaction of the enzyme catalase.

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Introduction

Plan Catalase is an enzyme commonly found in living organism. It speeds up the breakdown of hydrogen peroxide into water and oxygen. The chemical equation for this process is: 2 H2O2 � 2 H2O + O2 + Energy I am going to find out the effect of surface area on the activity (rate of reaction) of the enzyme catalase. I will do this by cutting potato, which contains catalase, into smaller pieces. I will then drop the small pieces of potato into hydrogen peroxide solution and observe how the rate of reaction changes as I am varying the size of the potato. I predict the rate of reaction will be quicker when the potato is in smaller pieces because they give a larger surface area, thus it will increase the collision between the catalase and the hydrogen peroxide. As a result, the rate of reaction will be increased as the small pieces of potato are giving greater surface areas. I know the pH of hydrogen peroxide is 7, which means neutral. In order to collect fair results, I must make sure that no other substances are mixed into the hydrogen peroxide I will be using for my experiments. The amount and the concentration of hydrogen peroxide must also be kept the same for both tests. It is because if they have different amounts or concentrations of hydrogen peroxide, one test might react faster or slower than the predicted result and therefore the experiment will not be accurate and reliable. Enzymes can only work within a certain range of temperature and they work best at an optimum temperature of 37�C, which is close to the body temperature. ...read more.

Middle

3 16 Surface Area 5 (SA 5) 4 10 I then repeated the experiment for 2 more times so I can get more results for analysis. These 2 repeated tests would be done in exactly the same way as the first one. 2nd attempt The number of time(s) has the potato rod been cut Time taken for the liquid to move up 10cm. (Seconds) Surface Area 1 (SA 1) 0 41 Surface Area 2 (SA 2) 1 35 Surface Area 3 (SA 3) 2 22 Surface Area 4 (SA 4) 3 17 Surface Area 5 (SA 5) 4 13 3rd attempt The number of time(s) has the potato rod been cut Time taken for the liquid to move up 10cm. (Seconds) Surface Area 1 (SA 1) 0 38 Surface Area 2 (SA 2) 1 32 Surface Area 3 (SA 3) 2 28 Surface Area 4 (SA 4) 3 27 Surface Area 5 (SA 5) 4 20 In all three of the tables I have recorded the time taken for the liquid to move up 10cm from each test (to 2 significant figures) Analysis In order to find what is the total surface area of my potato rods. I have worked out a formula that can calculate the total surface area of them: (Radius of a potato rod = 0.35cm) (Length of the potato rod = 3.6cm) (2 + 2n) x (? x 0.35 2) + (? x 0.7 x 3.6) * n = Number of time(s) the potato rod has been cut In order to convert the time taken for a reaction into the rate of reaction, I must first calculate the mean time by adding up the total time taken from all 3 attempts and then divide it by 3. ...read more.

Conclusion

So I will be using a finer manometer next time if I am going to repeat this investigation. Potato: The main reason of why I have got some anomalous results was because I changed my potato at later part of my investigation. So next time if I am going to do the same investigation again, I will prepare a larger potato for it and so it will give me more of the same potato for testing. The main reason of why the later part of my investigation (see the results which are highlighted in red in table 2) didn't go as I had predicted was because I had used some new potato, which had a cooler temperature. Since temperature is an important factor that can vary the rate of reaction. I could extend the investigation by testing two potatoes that are in different temperature. This extended investigation can hopefully help me to understand more about the importance of temperature control and I can also research more on the knowledge about the rate of reaction, so next time I can plan my experiment more carefully. For the experiment, I would set up 2 experiments in exactly the same way. One would be using a potato that is in room temperature (approx. 24 degrees Celsius) and the other one would be taken out from the fridge thus it would be cooler. The experiments would be carried out just like I did before and the results will be measured with stopwatch. The manometer used this time will be finer than my previous investigation so it can improve my accuracy in obtaining my results. I predict the rate of reaction for the warmer potato would be faster because there would be more kinetic energy for the Catalase to collide with the hydrogen peroxide. ...read more.

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