Investigate the effect of the concentration of the enzyme catalase on the decomposition of hydrogen peroxide.

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Helen Bewick 10C

Investigate the effect of the concentration of the enzyme catalase on the decomposition of hydrogen peroxide.

        During my investigation I hope to find out if the concentration of the enzyme catalase will have an effect on the decomposition of hydrogen peroxide. I will do this by changing the surface area of a potato cube and reacting it with hydrogen peroxide.

What are enzymes?

Enzymes are proteins that are involved in every chemical reaction in living cells. Enzymes are biological catalysts that speed up reactions but do not get used up in the reaction. An enzyme has an active site that is specific to its substrate. The substrate fits into the active site where it is broken down. The lock and key diagram shown below illustrates this.

   

     Substrate fits into enzyme’s specific active site.             Enzyme breaks down substrate.    

In my investigation I will be using the enzyme catalase in a potato cube. Hydrogen peroxide is decomposed by the enzyme catalse into oxygen and water. The equation for the reaction is:

   Catalyses

Hydrogen peroxide                         water + oxygen

2H2O2 (aq)                 2H2O (I) + O2

(Equation source - Chemistry for You written by Lawrie Ryan)

 

Prediction

I predict that as the surface area increases, thereby increasing the concentration of catalase, oxygen will be produced more quickly as the rate of reaction will increase. As the concentration of catalase is increased the rate of reaction will increase because the collision theory states that in a higher concentration there is a larger number of particles in the same volume. For the particles to react with each other they need to collide. So in a higher concentration there will be more collisions, therefore there will be more successful collisions, which increases the rate of reaction. This means that in a higher concentration the number of enzymes will increase thereby increasing the number of active sites causing the hydrogen peroxide to be decomposed more quickly and produce oxygen more quickly.

I also predict that as the concentration doubles the rate of reaction will also double. This is because there will be double the amount of hydrogen peroxide particles in a given volume so there will be double the amount of successful collisions and therefore the rate of reaction should double.

When I reach a certain concentration the maximum amount of oxygen will have been produced and after that concentration the same amount of oxygen will keep being produced each time. This is because the enzyme catalase decomposes the hydrogen peroxide into water and oxygen and the enzyme is then re used. Once all the hydrogen peroxide has been decomposed by the catalase and the maximum amount of oxygen has been produced the reaction will stop. This is shown on my predicted graph, the amount of oxygen stays the same after the experiment reaches a certain high concentration.

Predicted graphs

Preliminary plan

        The aim of my preliminary experiment is to test my method. I want to find out what volume of hydrogen peroxide I will need to use. It will have to cover the entire surface of the potato. I will also need to find out what the best way of cutting the potato is and how long I will need to time the experiment for to gain a wide range of results because I do not know how fast the reaction is.

Join now!

In this investigation I could use two variables to see if the concentration of the enzyme catalase will have an effect on the decomposition of hydrogen peroxide:

  • I could measure the amount of oxygen released after a set amount of time.
  • I could measure the amount of time taken for a set amount of oxygen to be released.

I will find the amount of oxygen produced after 3 minutes by measuring the volume of gas produced using downward displacement of water.

Method

  1. I will set up my apparatus as shown below.
  2. I ...

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