The purpose of this part of the experiment is to find out the course of an enzymatic reaction in a reaction. In order to do this, a measurement of the amount of substrate disappearing over time increments of 10, 30, 60, 90, 120, 180, and 360 seconds had taken. In this experiment we see that as time progressive the enzymatic rate lowered. The highest rate is the first in the first time interval (0-10) . It was the highest because iit had the highest catalysis amount and the most amount of H2O2 to decompose. The lowest rate was the last time interval (180 -360) there are 2 possible reason why it is the lowest. One is the H+ content makes the solution more basic. This moves the solution away from its optimal ph, thus causing the enzyme to denature. Denaturing is when the enzyme becomes biologically inactive because the proteins begins to unfold. Another reason could be is that all the catalysis amount is at the lowest because all the enzymes are already being used. This causes an inhibiting effect on reaction. Inhibiting is when the reaction is stopped or slowed down. If we were to lower the temperature it would still cause the enzyme to denature. Like ph, enzymes also have an optimal temperature if the temperature gets too low or too high it will denature. Testing the concept of the amount of substrate decomposed in these time amounts helped to determine the enzyme reaction. The dependent variable was the amount of hydrogen peroxide, sulphuric acid and water that were combined in the beaker while the independent variable was the amount of time the reaction was allowed to take place before the KMnO4 was added to stop the reaction. This experiment measures the disappearance of substrate which is Hydrogen Peroxide H2O2. In this experiment the level of KMnO4 the burette dropped from 5 ml to 0 ml. This means that the level of KMnO4 dropped 5.0 ml. These results are important because the baseline is used in every experiment. The baseline is used to test the amount of H2O2 in a 1.5 solutions. At low concentrations, the graph will show an increasing rate of reaction as concentration increases, levelling off at higher concentrations. The shape is explained by the concentration of substrate directly affecting the rate of reaction until another limiting factor becomes more important. Reaction that were involved in this experiment was :
5 H2O2 + 2 KmnO4 +3 H2SO4 → K2SO4 + 2 MnSO4 + 8 H2O + 5 O2
2 H2O2 → 2 H2O + O2
Without catalase this reaction occurs spontaneously but very slowly. Catalase speeds up the reaction notably. This reaction was studied by measuring the disappearance of the substrate. For example, hydrogen peroxide (H2O2) is converted to water (H2O) and oxygen (O2) gas. Catalysis speeds up the reaction and sulfuric acid (H2SO4) stops the reaction by lowering the pH and denaturalizing the enzyme. The rate is the highest in the first ten seconds because the rate decreases as the concentration of the catalase decreases over time.The rate is lowest during the last time period of 360 seconds because the most time has passed. The catalase concentration has been reduced and the product amount has increased, blocking the enzymes from reacting with the hydrogen peroxide. Miscalculations involving numbers and amounts of solutions would have a severe effect upon the results. Mathematical errors may also have of occurred. When the catalase arrived, it had melted. Because it is to remain on ice at all times, this may have caused errors. The age of the hydrogen peroxide effected results. For example, when calculating the percent of hydrogen peroxide spontaneously decomposed after 24 hours, new hydrogen peroxide yielded a much higher percentage than the aged hydrogen peroxide.
Conclusion

By performing this experiment ,the quantity of a substance in different solutions is determined through the titration method. The rate of how quickly the catalsye enzyme was able to convert Hydrogen Peroxide to water and oxygen gas also recorded, which helped to expand our knowledge of the importance enzymes and how they function. From the graph , catalase, or enzymes, drastically increases the rate of hydrogen peroxide decomposition but the rate of reaction is decreasing by time. This is due to the