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Purpose - The purpose of this lab is to use a chemical titration to measure and them calculate the rate of conversion of hydrogen peroxide (H 2 O 2) to water and oxygen gas using the enzyme catalase. Hypothesis - The rate of enzymatically-catalyzed reactions can be determined by taking time reactions at different time intervals. When the rates are plotted on a graph, the rate of the reaction is the slope of the linear portion of the curve. To determine a rate, any two points on the straight-line portion of the curve is picked out. Divided the difference in the amount of product formed between these two points by the difference in time between them. The result will be the rate of the reaction.
I believe that the reactions rate will be that highest during the first interval (0 - 10 seconds) because there are larger amounts of substrate molecules than enzyme molecules, but as times increase the lower the reaction rate because there will be more product molecules as result of the chemical reactions between the enzyme and the substrate. Materials: ring stand double clamp 2 burettes beakers test tube 1. 5 % hydrogen peroxide (H 2 O 2) distilled water liver potassium permanganate (K MnO 4) graduate cylinders 1. 0 % sulfuric acid (H 2 SO 4) Procedure: See handout Observations: See handout Analysis - My hypothesis proved to be correct. The reaction rate was the highest during the intervals of (0 - 10 seconds) and also at the beginning of 10 - 30 seconds because there is a large amount of substrate molecules in comparison to the number of enzyme molecules and there will be a maximal number of collisions between the enzymes and the substrate. As the number in seconds increased the reaction rate decreased.
The lowest reaction rate was during the last interval (120 - 180 seconds). As the number of substrate molecules decreased and the number of product molecules (O 2) increases the number of collisions between the substrate decreases. Eventually, the enzyme will break all of the substrate down, since 02 escapes from the system. Conclusion - In this lab catalase is the enzyme. It will react with hydrogen peroxide (H 2 O 2), the substrate. The products formed should be water and oxygen.
H 2 O 2 -- catalase -- > H 2 O + O 2 In Exercise 2 A the bubbles coming form the reaction mixture are O 2, which results from the breakdown of H 2 O 2. By capturing the evolved gas in a test tube and inserting a glowing splint one could show that the gas evolved is in fact 02. The split will burst into flames in presence of oxygen. When the catalase solution was added to the H 2 O 2 the reaction is not as active and apparent as the unboiled catalase. Very little oxygen is generated because boiling has denatured the enzymes. The enzymes in the boiled catalase have been changed by the rise of temperature and can't let the substrate in to form a reaction.
Large amounts of oxygen gas are released rapidly from the piece of liver. Macerating the liver will increase the rate of the reaction. Blood from the liver will produce an even more explosive rate of oxygen gas production. However, if the liver were boiled before being added to the H 2 O 2 there would be no reaction. The heat denatures the catalase in both the plant and animal tissues. Add 10 ml of H 2 O 2 to 7 different beakers, Label the 7 beakers, 0 sec (control), 10 sec, 30 sec, 60 sec, 120 sec, 180 sec, and 360 sec.
Add 1 ml of catalase to each of the beakers. Allow the reaction to occur for the time labeled on each of the beakers. After the time has elapsed add 10 ml of H 2 SO 4 to stop the reaction. Potassium permanganate was then titrated into each beaker in order to determine the presence of peroxide left in the solution.
The longer the reaction was left to sit; the last K MnO 4 was used in titration. In Exercise 2 C, when peroxide was left out to sit uncovered for twenty-four hours, it decomposed 98. 1 %. The baseline we established was 8. 7 ml K MnO 4. The first and second intervals proved to be the highest when I tested two points on the graph within that time period.
I divided the difference in the amount of product formed between these two points by the difference in time between them. In addition, the graph shows that in reaction to the longer time given for it to sit, there was more peroxide-solution within it. Two important influences are pH and temperature. When an enzyme's conformation is significantly altered because of pH or temperature variation, the enzyme may no longer catalyze reactions.
An enzyme is denatured when it loses its functional shape.
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Research essay sample on H 2 O 2 O 2
