Catalytic Decomposition Of Hydrogen Peroxide

This process is called decomposition as two hydrogen peroxide molecules are broken down into oxygen and two water molecules.

The topic of this lab is on biochemistry.This experiment was conducted to show how cells prevent the build of hydrogen peroxide in tissues. My group consisted of Lekha, Ruth, and Jason. There were used two different concentrations of hydrogen peroxide through this experiment , 1.5% and 3%. By testing two different types it is easier to understand how the H2O2 and catalase react with one another. To do this both the yeast, which was our catalase, and H2O2 were mixed together in a beaker. Each concentration was tested out twice for more accurate results . 1.5% concentrated H2O2 had an average reaction rate of 10.5 seconds while 3% concentrated H2O2 had an average reaction rate of 7.5 seconds. From this experiment we learned that by increasing the concentration of H2O2 and chemically combining it with a catalase it will speed up the reaction. Enzymes speed up chemical reactions . The independent variable in this experiment was the concentration of the H2O2. Some key vocabulary words are Catalase, enzyme, hydrogen peroxide ( H2O2), and concentration.

This would be a suitable independent variable as it would be easy to change ourselves. If the concentration of the substrate increases the faster the rate of reaction, as more hydrogen peroxide molecules can collide with catalase molecules, so more reactions will take place. As you increase the concentration of substrate this will continue to occur up to a point where all the active sites on the enzyme are being used.

If another enzyme like lactase is used, no reaction would take place because the substrate, hydrogen peroxide, wouldn’t fit into the active site.

How does changing the substrate concentration affect the rate of a catalase reaction in an enzyme? Hydrogen peroxide was used as the substrate and the rate was measured by oxygen production.

If different amounts of enzyme solution are added to the hydrogen peroxide, then the highest amount of enzymes will have the greatest reaction rate because enzymes catalyze reactions, meaning more oxygen will be produced quicker.

In this week’s lab experiment, the rate of decomposition of hydrogen peroxide forming oxygen gas will be observed and studied. Since the rate of a chemical reaction is dependent on two things; the concentrations of the reactants and the temperature at which the process is performed, the rate can be measured at which a reactant disappears or at which a product appears. When measuring the rate, the rate law will be applied. The objective of this lab is to demonstrate how the rate changes with varying initial concentrations of hydrogen peroxide by measuring the rate at which oxygen is evolved.

According to the results of the lab, it appears that hydrogen peroxide can be broken down by catalysts other than enzymes; such as magnesium dioxide. In the lab, hydrogen peroxide barely reacted with the sand (thus indicating its lack or absence of catalase). However, when the hydrogen peroxide was paired with magnesium dioxide, a significantly greater reaction occurred. Therefore indicating that hydrogen peroxide can in fact be broken down by (inorganic and organic) catalysts other than enzymes.

Lab Report Lesson 1 – Plan Your Experiment Aim: How does changing the mass of liver placed inside hydrogen peroxide affect the amount of oxygen formed? Identify factors, which affect: ______the amount of oxygen formed_______. Hypothesis: If the mass of liver increases when put in to the same amount of Hydrogen Peroxide, then more oxygen will be produced in the same amount of time.

What does the changes in the amount of substrate on an enzyme’s reaction effect on?

The very word, photo “synthesis” gives away what kind of reaction it is…a synthesis reaction. In this case, you have two products ( ) in the presence of light energy

If the temperature is too hot or too cold, then the reactivity and reaction rate of which the enzyme catalase breaks down hydrogen peroxide will decrease.

BACKGROUND: Catalase (the enzyme) is found in yeast, it breaks down hydrogen peroxide (the substrate) into water and oxygen according to this equation. 2H2O2(aq) -------------------> 2H2O(l) + O2(g) + catalase(aq) One molecule of catalase can break 40 million molecules of hydrogen peroxide each second. Factors that affect the rate of reaction § Increasing the temperature increases the kinetic energy at which the enzyme and substrate collide.

The presence of catalyst will cause hydrogen peroxide to break down into water and oxygen. The increase in the pH of the substrate will cause the increase in the rate of reaction. It will optimize at pH 7 and will diminish/moderate after pH 9.

Hydrogen peroxide is a toxic byproduct of cellular functions. To maintain hydrogen peroxide levels the catalase enzyme deconstructs hydrogen peroxide and reconstructs the reactants into oxygen gas and water. The catalase enzyme is found inside cells of most plants and animals. Regulating the levels of hydrogen peroxide is crucial in homeostasis and analyzing it’s optimal conditions for performance is just as important. To understand the optimal environment for this enzyme, they are put into different environments based off protein activity (enzymes are proteins). Catalase samples will be put into different hydrogen peroxide environments based off pH and temperature. The more active the enzyme, the more oxygen and water it will produce. Enzyme activity can be seen through the release of oxygen in the hydrogen peroxide. Since oxygen cannot be accurately measured, the data will consist of the longevity of the reaction in different environments. If the pH is higher than 7, then the reaction rate will increase due to the ample amount of hydrogen ions in the hydrogen peroxide. However the pH level cannot be higher than 10 or else there will be too many hydrogen atoms in the peroxide for the enzyme to be able to deconstruct them. If the temperature is increased, then the reaction rate will increase due to the ample amount of energy and movement in the hydrogen peroxide and enzyme.