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Investigating the Breakdown of Hydrogen Peroxide To Oxygen and Water By Catalase Enzyme.

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Investigating the breakdown of hydrogen peroxide to oxygen and water by catalase enzyme. AIM To investigate the how increasing the mass of potato [and therefore the amount of catalase enzyme] effects the volume of oxygen produced. There are a few variables: * The concentration of hydrogen peroxide. * The volume hydrogen peroxide. * The amount of catalase which is in the cubes of potato. * The surface area of the potato. * The temperature at which the reaction takes place. These are variables because: * If you increase the concentration of hydrogen peroxide, there are more H2O2 molecules per cm3. Therefore there is a greater oxygen content, so more oxygen will be produced. * If you increase the volume of hydrogen peroxide, then more oxygen will be produced by complete decomposition because there are more O2 molecules in the solution. * Increasing the amount of catalase means that the reaction will go faster. The yield of oxygen will not change [because the number of O2 molecules is not changed], but more oxygen will be produced in a give time because the rate of decomposition is greater. * If you increase the surface area of the potato, then it will have the effect of increasing the rate of reaction as a larger amount of catalase is presented to the H2O2 molecules in solution. ...read more.


PREDICTION As the number of potato cubes increases, so does the mass of potato and therefore amount of catalase. Catalase is an enzyme [a biological catalyst], which speeds up the decomposition of hydrogen peroxide to water and oxygen. As the rate of reaction is speeded up, more oxygen will be produced in a given time [because the rate is increasing]. Therefore, I predict that as the number of potato cubes increases, the amount of time it takes for the test tube of water to be displaced will decrease, as more oxygen is produced in a given time. I predict that my graph of results will look like this: I also predict that as number of potato cubes increases, so will the rate of reaction, because more catalase molecules will be presented to the same number of hydrogen peroxide molecules (as both concentration and volume of hydrogen peroxide are constant). Table of results Number of cubes Result No.1 Time taken to collect 60cm3 of oxygen ResultNo.2 Result No.3 Rate of reaction no.1 Rate of reaction no. 2 Rate of reaction no. 3 1 36 40 35 1.7 1.5 1.7 2 30 30 33 2 2 1.3 3 19 20 19 3.2 3 3.2 4 8 10 10 7.3 6 6 5 5 7 5 12 8.6 12 (Each cube 1cm3) ...read more.


The only difficulty I found with this method was that it was impossible to both insert the bung and start the timer in the same instant. Therefore the time figures would be very slightly inaccurate in each case because I started the timer shortly after having inserted the bung. However, I do not think my results and conclusions are less valid because of this. I do not have any anomalous results, I can tell this by my graph. This reflects well on the way I conducted my experiment. If I were to continue the investigation I would vary the volume or concentration of the hydrogen peroxide and measure the total volume of oxygen produced. This would enable me to see how the number of oxygen molecules changes with either volume or concentration, as opposed to the how rate effects the amount of oxygen produced in a given time. I would be investigating how much oxygen is produced as opposed to the time it takes to be produced. I think that as volume/concentration increase, so would the total volume of oxygen produced. I could also vary the temperature, the PH, the volume of substrate etc. These would all have an effect on the rate of reaction. ...read more.

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