Hydrogen Peroxide molecules are broken down to form water and oxygen molecules. The chemical and word equaitons for this reaction are shown below.
Hydrogen peroxide → Water + Oxygen
2H2O2 → H20 + O2
Measuring the rate at which oxygen gas is produced once the hydrogen peroxide has come into contact with the catalysing enzyme catalase will be the simplest way of measuring the rate of the reaction. There are a number of possible methods that I could use to measure the rate at which gas is produced, such as counting the bubbles given off from a delivery tube submerged in water, collecting the gas in a syringe or a measuring cylinder. However, all of these methods have flaws as they are likely to be very inaccurate.
I have chosen to collect the gas by means of an inverted burette as shown above as this is the most accurate equipment available to me. It will allow me to record the exact volume of oxygen produced by the reaction at given intervals to the nearest cm3. However, from previous experience I am aware that bubbles of gas with large volumes often move up the burette. This means that there may be some slight anomalies in my results depending on which side of the time limit the bubble of gas reaches the top of the burette but I will take these into account.
I am using a syringe to insert the hydrogen peroxide into the test tube as this is the best way of mixing the substrate (hydrogen peroxide) and the catalyst (the enzyme catalase in the potato) without loosing any of the oxygen produced to the surrounding atmosphere (e.g. if I started with the hydrogen peroxide in the tube and put the potato in quickly then inserted the bung I wouldn’t necessarily be able to insert the bung before some oxygen had been produced and lost).
In this experiment the independent variable will be the temperature of the enzyme catalase in the potato and the dependant variable is the rate that oxygen is produced by the decomposition of the hydrogen peroxide.
The factors or Variables that are likely to have an affect on my experiment are the mass of the potato used, the surface area of the potato, the amount of catalase in the potato that is used (thus which potato and the type of potato may be variables), the pH of the hydrogen peroxide, the volume of hydrogen peroxide that is to be decomposed, the amount of light the experiment is exposed to, and the concentration of the hydrogen peroxide.
The mass of the potato will alter how much of the enzyme catalase is available to catalyse the decomposition and the surface area of the potato will affect how much of this enzyme is immediately exposed to the hydrogen peroxide. The more of the enzyme there is in contact with the hydrogen peroxide the faster the rate of the reaction will be. For this reason I will always use a 1.5 cm piece of potato that has been cored using the same corer as this should keep the mass and the surface area of the potato exactly the same in each test.
Different potatoes may have varying amounts of the enzyme that will catalyse this reaction in them so all of my tests will be done using pieces of the same potato as hopefully the spread of the enzyme through this one potato will be fairly even.
Hopefully the mixture of the potato and the hydrogen peroxide will always have a similar pH because as discussed earlier, the pH of the environment has an affect on the rate at which the enzyme works.
The volume and the concentration of the hydrogen peroxide will be kept the same in each test because altering either would change the number of hydrogen peroxide particles available to be catalyzed and the distance between each particle. A more concentrated solution of hydrogen peroxide would increase the chance of an H2O2 particle connecting with and being catalysed by an enzyme thus increasing the rate of reaction. I will always use 5cm3 of 10VOL hydrogen peroxide.
I will not change the amount of light that the experiment is exposed to as I know that hydrogen peroxide decomposes in light and so varying light amounts would affect its rate of decomposition.
I have planned to keep all variables other than the independent variable - the temperature – that I am investigating, the same as this is the only way of making my experiment fair. If I did not do this then my results would not be accurate or reliable enough to support a conclusion.
Equipment List: Test tube with a two tube bung
Large beaker (for water bath)
10VOL hydrogen peroxide
Basin of water
5 cm3 syringe
Delivery tube
Thermometer
Clamp stand
Stopwatch
Burette
Potato
Borer
Kettle
Ruler
Knife
Ice
Method
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Fill a 50cm3 burette with water and invert it in a basin of water with your thumb over the end so as not to spill any of the water.
- Clamp the burette vertically using a clamp stand leaving enough space at the bottom in the water basin to insert a delivery tube.
- Core a large potato using the borer and cut a 1.5cm piece with the knife using a ruler.
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Half fill a beaker with cold water put in the thermometer, then boil the kettle and add enough boiling water so that the temperature in the beaker reaches 50oC.
- Put the piece of potato in the test tube and put the test tube into the water bath then leave it in there for two minutes so that the potato reaches a similar temperature as the water.
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Measure out 5cm3 of hydrogen peroxide in the syringe and insert the syringe into one side of the two tube bung.
- Connect the delivery tube to the other side of the two tube bung and insert the bung into the test tube then insert the other end of the delivery tube into the inverted burette.
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Run the water in the inverted burette down so that the bottom of the meniscus is level with 50cm3 mark (or level with another mark that you have recorded).
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Check that the temperature in the water bath is still 50cm3 and if it had dropped at slightly more boiling water so that it reaches the correct temperature.
- Start the stopwatch at the same time as pressing down the syringe so as to mix the hydrogen peroxide and its catalyst.
- Record the level of the water in the burette after every 30 seconds for 5 minutes in an appropriate table, always reading from the bottom of the meniscus.
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Follow this method exactly for 0oC, 10oC, 20oC, 30oC and 40oC except for adding ice instead of boiling water to those temperatures that are lower than the temperature of tap water, and adding less boiling water than for 50oC to reach those temperatures above the temperature of tap water. I am using this range of temperatures as hopefully the results will give a good representation of the prediction I have made.
- I will repeat this whole experiment at least twice so that I get three sets of results for each temperature that I can use to check the reliability of my results and repeat any results that appear anomalous. I will be careful to keep all variables the same as my prior experiments.
Safety
Hydrogen peroxide is toxic. It will cause irritation if it comes into contact with skin and must not be swallowed. To reduce the risk of irritation, I will clear up all spills immediately. I will also wear goggles to protect my eyes from the hydrogen peroxide and follow all lab rules such as tying back hair. I will also take particular care when handling boiling water.
Preliminary experiment
I followed the above method to carry out a trial test to see if all the measurements that I have planned to use are appropriate. Very little happened using a cored piece of potato and I have concluded that this is because the surface area of the potato (the amount of catalase) exposed to the hydrogen peroxide was minimal. As a result I am going to make one alteration to my method. Instead of using a piece of potato 1.5cm in length that has been cut with a borer, I will grate one whole potato using a grater and weigh out exactly 5 grams of the grated potato for each test using measuring scales accurate to 0.01 grams. This is a much more accurate way of ensuring the same mass of potato is used in every experiment and also allows a much larger amount of the enzyme catalase to mix with the hydrogen peroxide.
Analysis
The results produced from this experiment show that as the temperature of the conditions that the enzyme catalase and hydrogen peroxide are in increases, so does the rate at which the hydrogen peroxide decomposes.
The evidence for this conclusion is shown in the graph below.
The graph clearly shows that as the temperature of the conditions increases, the rate at which oxygen gas is produced by the decomposition increases. The gradient of each line is representative of the rate at which gas is produced. The higher the gradient (steeper line) the more gas is produced per second. The increase in the rate at which the Oxygen gas was produced shows that the rate of the reaction increases as the temperature increases given that the rate reaction is equal to the volume of gas produced per second. Oxygen gas is produced as a result of the hydrogen peroxide decomposing to form oxygen and water. The bonds holding the hydrogen and oxygen molecules are broken with the assistance of the catalyst enzyme catalase, and new bonds are formed to produce the oxygen and water molecules. The oxygen atoms escape the solution in the form of O2 gas.
As suggested in my prediction, if the temperature of the conditions is increased, the enzymes in the potato are working in favourable conditions and so work faster, increasing the rate of the reaction as more of the hydrogen peroxide bonds are broken as a result of being held to the enzymes. However, contrary to my prediction, I did not see a decrease in the rate of reaction once the conditions reached a temperature of 50oC as a result of the enzymes being denatured and I believe this was because of a fault in my method. It is likely that the water surrounding the potato was at a temperature of 50oC but that the potato itself was not given time to reach this temperature and so the enzymes were still working in their optimum temperature of just over 40oC.
The gradient of the lines on this graph represent the rate at which the gas is given off. As I predicted, the curvature of these lines shows clearly that the rate the gas is produced at decreases as the length of time that the reaction has been taking place for increases. This is because the concentration of hydrogen peroxide particles in the solution decreases as the reaction continues because many of the hydrogen peroxide particles have decomposed to water and oxygen leaving fewer particles in the solution per cm3. Thus there are fewer hydrogen peroxide particles to decompose and so the rate of reaction decreases.
I know that the
Rate of reaction = amount of products made
Time taken
I have produced a table of the rate of the reaction at different temperatures for the first 30 seconds after the enzyme and hydrogen peroxide have been mixed as this is the point when the rate of the reaction is fastest.
I have plotted a graph of the rate of the reactions (vol. of gas/time) which is shown below.
This graph clearly shows that as the temperature increases, the rate at which hydrogen peroxide decomposes also increases. This provides enough evidence to support my conclusion that as the temperature of the conditions that the enzyme catalase works in increase up to around 45oC, the rate at which it catalyses the decomposition of hydrogen peroxide also increases. As the temperature increases so does the amount of energy on average that each particle has. The particles therfore move around more and the probability of there being collisions between the enzyme and the reactent particles is greatly increased. Thus, a larger number of bonds are broken and the rate of the reaction is faster. This very clearly supports the prediction that I made in my plan.
However, contrary to my prediction, I did not see a reduction in the rate of the reaction once the temperature exceeded 45oC, which I would have expected due to the denaturing of the enzymes at too high a temperature. This is probably due to a flaw in my method and it is likely that the decrease in the rate of decomposition would have become apparent had I carried out the experiment at temperatures above 50oC.
Evaluation
My experiment went well and I did not experience any significant problems with my method. My results showed a clear pattern and the closeness of each test result to its repeat test results suggests that these results were accurate and are reliable enough to provide evidence for my conclusion. My results for the average volume of oxygen gas produced after every thirty seconds and the average rate of reaction both produced good graphs with no particular anomalous results.
There were no obvious anomalies in my results but it is likely that not all of the results are entirely accurate because of the fact that the bubbles of oxygen gas produced by the decomposition of hydrogen peroxide travelled into the burette in volumes of up to 4cm3 at a time. This meant that if a gas bubble with a volume of 4cm3 reached the top of the water in the burette at 29 seconds in the first test, but 3 seconds in the second, the results produced would be extremely different as the volumes would be recorded on either side of the 30second mark. This is the reason why, for example, the third test of 20 oC appears to produce oxygen so much slower than the other two tests for that temperature, the bubbles in this experiment were just a bit slower than those in the other two experiments.
The one significant fault in my experiment was the fact that I did not plan to test the activity of the enzyme catalase in conditions over 50oC. My method did not allow for the fact that however long the potato was left in the water bath, it was not necessarily going to reach the same temperature as the water itself. It is the likely that the catalase enzyme in the 50oC water bath only reached temperatures of around 45oC and so was working at its optimum rate as opposed at a temperature where a large number of the enzymes had denatured. If the actual temperature of the enzymes in the potato had reached 50oC then my predicted decrease in the rate of the decomposition of hydrogen peroxide would have been seen. If I had the chance to carry out my experiment again, I would continue testing the activity of catalase to at least 60oC to ensure that my prediction was in fact correct. I have also learnt since carrying out my experiment, that using a buffer in the mixture of potato and hydrogen peroxide would have helped to keep the pH of the conditions the same in every experiment. The rest of my planned procedure was suitable for the investigation.
If I had time to extend my investigation further, I would like to investigate the optimum working temperatures of other enzymes, not just catalase. I would also be interested to see what affect changing the concentration of the hydrogen peroxide would have on the rate at which it decomposes with an equal mass and surface area of potato (exposed catalase).
