Hydrogen Peroxide Reaction Lab Report

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.

Peroxidase is an enzyme found in potatoes that catalyzes the breakdown of hydrogen peroxide, H2O2, into O2 gas and water. We examined the different pH environments that can affect the enzyme activity during the breakdown of H2O2. In order to do this, we added different levels of pH, low, medium, and high, into different test tubes with the enzyme and H2O2, and we then inverted the tube. The amount of O2 gas produced was then measured and recorded. The result was that the higher pH produced more gas, followed by medium pH, then low pH. The enzymes were more active in the pH of about 10. It increased

8.When the reaction is completed, record the volume of gas in the graduated cylinder. Record observations about which reactant was the limiting reactant.

=0.353g O_2 1.002×〖10〗^(-2) mol O_2×(32g O_2)/(1 mol O_2 ) =0.321g O_2 percent yield=(0.321g O_2)/(0.353g O_2 ) ×100%=90.9% Post-lab questions What is the purpose of waiting for some bubbles to escape from the gas collection tube before collecting the oxygen

The purpose of this experiment was to simply measure oxygen production rates released from decomposed hydrogen peroxide under different conditions (concentration of enzymes, temperature, and PH level).

This experiment will measure the rate of oxidation of iodide ions by persulphate ions to derive the rate law for the reaction. Starch will be added to the reaction to facilitate the measure of time during the reaction. The reactant solutions will contain (NH4)2SO4 and KI, represented as:

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.

In this above reaction, oxygen is released and is used for other cellular purposes, but when it occurs in a test tube, similar to this experiment, the oxygen gas bubbles producing a layer of foam on the surface of the peroxide. The amount of foam and the speed it is produced are forms of measuring the catalase activity. In the next experiments, one would determine the degree of catalase action by calculating the thickness of the foam layer. It is hypothesized that when reacting with: potato, apple, steak, or liver, the plants and animal tissues will react differently.

The purpose of this experiment is to see how the enzyme peroxidase performs under different conditions. An enzyme is a protein molecule that is a biological catalyst. A catalyst is a substance that speeds up a chemical reaction, while also lowering the activation energy of a reaction. The activation energy of a reaction is the initial amount of energy that is necessary to bring reactants together with the proper amount of energy and placement so that products can be formed. Enzymes have a unique 3-D shape, which enables it to stabilize a temporary association between substrates (the reactant molecules that binds to the active site of an enzyme and undergoes chemical modification).

The purpose of this lab was to explore how certain quantities of hydrogen peroxide would affect the amount of oxygen gas produced in the following lab. The hypothesis of the lab in question was that, if the amount of Hydrogen peroxide added to the catalase (liver) solution increased, then the amount of oxygen gas produced would be much greater. This is because the greater amount of hydrogen peroxide in a catalase solution will result in there being more solution to deform into oxygen. Evidently after this experiment the hypothesis stated has been proven correct, as the quantity of hydrogen peroxide increased the amount of oxygen gas produced also increased. This can proven as at 10 ml there was 45.2cm^3 and at

2) Calculate the a mean rate constant using orders of reactions and the rate equation allowing for the overall order or reaction to be found.

Testing the Effects of Temperature on the Decomposition of Hydrogen Peroxide with the Enzyme Catalase

The chemical hydrogen peroxide(H₂O₂) is broken down by the enzyme catalase. Hydrogen peroxide is a byproduct formed in cellular reactions that, if not broken down, could inflict severe damage to the cell. Catalase is an enzyme that breaks down hydrogen peroxide in to water and oxygen. How efficient and strong the enzymes reaction to break down H₂O₂ determines largely on temperature and pH level. An enzyme only functions within a set pH and temperature range. Beyond that it becomes denatured, rendering it useless. The purpose of this lab is to determine at which temperature and pH level the enzyme catalase reacts best. Catalase in chicken and beef livers will be used to do the lab because enzymes still function after death as long as they are kept refrigerated at a low temperature.

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.

After gathering all the data, we solved for the rate of diffusion of each using the formula.