• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12

The Decomposition of Hydrogen Peroxide.

Extracts from this document...


The Decomposition of Hydrogen Peroxide Aim - To investigate how changing the temperature of the conditions that the enzyme catalase works in affects the decomposition of hydrogen peroxide. Hydrogen Peroxide is a toxic waste product produced from chemical reactions that take place in living cells. As an unstable molecule, it decomposes in light and so has to be stored in brown bottles. It can, however, stay in a stable state for a number of years. Therefore, in nature, it is necessary that the decomposition of hydrogen peroxide to water and oxygen is sped up in order to prevent it from intoxicating cells. The enzyme catalase, which is readily found in potatoes, liver, celery and yeast, catalyses the hydrogen peroxide and greatly increasing the rate of decomposition. It is also possible to speed up the decomposition in the lab using manganse oxide as the catalyst. Enzymes are tiny protein molecules that are made in cells that work as biological catalysts. Catalysts are substances that speed up reactions, but remain unaffected by the reaction, therefoe they can be used again and again. A catalyst works by lowering the activation energy required for a reaction - the amount of energy needed to break the bonds in the reactents. Less energy is required to break the bonds and so the reaction takes less time. Enzymes are specific and each enzyme catalyses a certain reaction - this is because each type of enzyme is a different shape. Hydrogen Peroxide is the substrate (the substance an enzyme helps to react) in this reaction. Enzymes exactley fit the shape of the substrate and the part of the enzyme that the substrate molecule fits into is known as the active site. This is known as the 'lock and key model' The above diagram shows a catabolic reaction because large molecules are being split into smaller ones. The decomposition of hydrogen peroxide is a catabolic reaction. ...read more.


* 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. * 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. * 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. * 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). * 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. * 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. ...read more.


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). ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Life Processes & Cells section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Life Processes & Cells essays

  1. Marked by a teacher

    The aim of this investigation is to find out what effect pH has on ...

    4 star(s)

    But let's look at why the pH changes the structure of the enzyme. The reason is to do with ionic concentrations. Enzymes are very complex macromolecular entities and so can only function within very specific pH range. pH of the media is responsible for the charged state of the charge-bearing

  2. Marked by a teacher

    Investigating the breakdown of hydrogen peroxide by the enzyme catalyse in potatoes.

    4 star(s)

    This diagram shows how the equipment was assembled together to obtain the results. Range of Independent Variables After settling on our independent variable, we chose the ranges which would entitle us to have the most suitable results for the experiment.

  1. To investigate the effect of enzyme concentration on the activity of catalase in potato ...

    Putting the potato pieces under distilled water in a petri dish prevents the potato from being contaminated or dehydrated. Handling potato pieces with a pair of tweezers to prevent contamination. The conical flask must be clean after each trial so to prevent any unwanted contamination.

  2. Factors affecting the activity of potato catalase on hydrogen peroxide.

    0.0242 40% R=0.0294 60% R= 0.0202 60% R=0.0371 80% R= 0.0276 80% R=0.0449 100% R= 0.0280 100% R=0.0460 R= rate of reaction. From the graphs I made I could see a general trend. As time passes the level of oxygen given out increases.

  1. Inhibition of the decomposition of hydrogen peroxide by the catalase enzyme using copper sulphate.

    Temperature can alter the active site at high temperatures as it changes the shape of the enzyme and destroys the active site. This means the substrate can no longer be catalysed and therefore the rate of reaction decreases. Up until the temperature denatures the enzyme, every 10�c increase in temperature doubles the rate of reaction.

  2. My hypothesis is that the higher the concentration of hydrogen peroxide the more catalase ...

    Surface area Increasing surface area would increase the rate of reaction because there are more particles exposed to the other reactant so there is a greater chance of a collision. This will not affect my experiment as enzymes only have one site to where the substrate can react.

  1. What influence does pH have on the enzyme Catalase?

    Therefore it becomes neutral and cannot form an ionic bond with the substrate and the enzyme cannot properly work. 3 This is shown in the two following diagrams: Therefore, there are fewer reactions between the enzyme and substrate and thus the rate of reaction decreases.

  2. An investigation to determine how the rate of an enzyme-catalysed reaction is affected by ...

    Hypothesis As the surface area of the potato chips increases, more active sites of catalase will be exposed to the substrate, hydrogen peroxide. Catalase-hydrogen peroxide complexes will form at a faster rate and hence the product (water and oxygen) will be produced at a faster rate also.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work