Decomposition of Hydrogen Peroxide, Heat Produced in Chemical Reactions-Calorimetry | CHEM 111, Lab Reports of Chemistry

Download Decomposition of Hydrogen Peroxide, Heat Produced in Chemical Reactions-Calorimetry | CHEM 111 and more Lab Reports Chemistry in PDF only on Docsity! CHEM 111 LABORATORY EXPERIMENT DATE: NAME PART I: QUANTITIES IN CHEMICAL REACTIONS: DECOMPOSITION OF HYDROGEN PEROXIDE (A modification of experiment written by Joe Eridon and published in Central New Mexico Community College (CNM) Laboratory Manual for Introduction to Chemistry Lab (CHEM 1492) PART II: HEAT PRODUCED IN CHEMICAL REACTIONS - CALORIMETRY OBJECTIVES PART I. 1. To practice calculations related to quantities in chemical reactions with a gas product. 2. To measure amount of gas based on its volume. 3. To determine percent concentration of hydrogen peroxide in a commercial solution. PART II. 3. To practice calculations related to heat produced in chemical reactions. INTRODUCTION PART I. DECOMPOSITION OF HYDROGEN PEROXIDE A household hydrogen peroxide is a solution of H2O2. On contact with a tissue it readily decomposes according to the following equation producing oxygen, O2, gas. 2 H2O2(aq) –> 2 H2O(l) + O2(g) A rapid decomposition of hydrogen peroxide is caused by enzyme catalase present in tissue. The released oxygen has disinfecting properties and, therefore, a hydrogen peroxide solution is used as an disinfectant. We will look at a quantitative aspect of this reaction. The amount of produced oxygen gas is proportional to the amount of hydrogen peroxide in the aqueous solution. Therefore, if we determine the oxygen amount, we can calculate the percent by mass concentration of hydrogen peroxide in the solution. We will add a homogenized animal liver to a measured amount of hydrogen peroxide solution to initiate the decomposition. The produced oxygen will be collected in a graduated cylinder so the volume of the gas can be measured. The graduated cylinder will be filled with water, turned upside down, and immersed in water. Oxygen will replace water in the cylinder. A formula called “ideal gas law” will be used to calculate the mole amount of oxygen based on its volume. This, in turn will enable us to calculate the percent by mass concentration of hydrogen peroxide in the solution. Below is the explanation how to use the ideal gas law formula: The ideal gas law: PV = nRT P = atmospheric pressure measured in atmospheres (atm). We will use barometer to obtain this value. V = volume of the gas (as measured in a graduated cylinder) in liters (L). n = number of moles of gas (oxygen). This is the value we need to calculate. R = gas constant: 0.0821 atm L/(mol K) Page 1 of 8 T = temperature expressed in kelvins (K). K = degree Celsius + 273.15 We have to re-arrange the equation to isolate n: n = PV/(RT) The pathway for the calculations is: mL O2 –> L O2 –> mol O2 (using n = PV/(RT) –> –> mol H2O2 (using conversion factor from the equation) –> g H2O2 –> % H2O2 in a sample The following is the example of calculations that will be used in today’s lab. PROBLEM 1: A 5.20 g sample of hydrogen peroxide solution was decomposed with catalase. 69.0 mL of oxygen, O2, was collected. The atmospheric pressure = 0.910 atm and temperature = 23.5 degree Celsius. Calculate the percent by mass of hydrogen peroxide, H2O2, in the sample. SOLUTION: Step 1, unit conversion. 69.0 mL = 0.0690 L K = 23.5 degree C + 273.15 = 296.7 K Step 2, calculate moles of oxygen mol O2 = PV/(RT) = 0.910 atm x 0.0690 L/(0.0821 L atm/(mol K) x 296.7 K) = 0.00258 mol O2 Step 3, calculate grams of H 2O2 0.00258 mol O2 2 mol H2O2 34.0 g H2O2 0.175 g H2O2 1 mol O2 1 mol H2O2 Step 4, calculate the percent by mass of H 2O2 in the sample 0.175 g H2O2/5.20 g sample x 100 % = 3.37 % H2O2 PART II. CALORIMETRY Many chemical reactions produce or absorb heat. The reactions which absorb heat are called endothermic, and the reactions which produce heat are called exothermic. In the following experiment, we will measure heat produced in acid-base neutralization reaction. The produced heat is “trapped” in the solution in a coffee-cup calorimeter. By measuring the temperature change of the solution we can calculate the amount of heat. EXAMPLE: 40.0 mL of 1.00 M NaOH is neutralized with 20.0 mL of 1.00 M H2SO4 in a coffee-cup calorimeter Page 2 of 8 e. How many grams of NaCl is produced in the reaction? Assume that the calorimeter loses only a negligible quantity of heat, density of the solutions is 1.00 g/mL, and the specific heat of the solutions is 4.184 J/g degree C. EXPERIMENT 2. In this experiment, NaOH will be neutralized with H3PO4. Pour 60.0 mL of 2.00 M NaOH solution in a coffee-cup calorimeter. Measure and record the temperature of the solution (since both NaOH and H3PO4 solution are in the same room, we assume that both have the same temperature, which is the initial temperature). Add 20.0 mL of 2.00 M H3PO4 to the coffee cup calorimeter and gently mix the solutions with thermometer. Wait until the temperature stabilizes and record the final temperature. a. FIRST, write the balanced equation for the reaction. b. How much heat is produced in the reaction (in the coffee-cup calorimeter)? c. How much heat would be produced if we would use 1.00 mol of NaOH? d. How much heat would be produced if we would use 1.00 mol of H3PO4? e. How many grams of Na3PO4 is produced in the reaction? Assume that the calorimeter loses only a negligible quantity of heat, density of the solutions is 1.00 g/mL, and the specific heat of the solutions is 4.184 J/g degree C. CHEM 111 LABORATORY REPORT DATE: NAME PART I. DECOMPOSITION OF HYDROGEN PEROXIDE DATA: Weight of large test tube with holding beaker (weight of container):___________________________ Weight of large test tube with holding beaker + H2O2 solution:_______________________ Net weight of H2O2 solution:____________________________ Volume of the collected O2:_________________________ Temperature:_____________________ Atmospheric pressure:____________________ CALCULATIONS: Calculate the % by mass of H2O2 in the hydrogen peroxide solution. Show all calculations. Follow the example. Page 5 of 8 PART II. CALORIMETRY EXPERIMENT 1. NaOH + HCl. Initial temperature (accuracy to 0.1 degree C)___________ Final temperature (accuracy to 0.1 degree C)_______________ a. Balanced equation: b. Calculations of heat produced in the reaction (100.0 g of the solution was used): q = m x C x delta T c. Calculations of heat produced per 1.00 mol NaOH: d. Calculations of heat produced per 1.00 mol HCl: e. Calculations of the amount of NaCl produced in the experiment. Page 6 of 8 EXPERIMENT 2. NaOH + H3PO4. Initial temperature (accuracy to 0.1 degree C)___________ Final temperature (accuracy to 0.1 degree C)_______________ a. Balanced equation: b. Calculations of heat produced in the reaction (80.0 g of the solution was used): q = m x C x delta T c. Calculations of heat produced per 1.00 mol NaOH: d. Calculations of heat produced per 1.00 mol H3PO4: e. Calculations of the amount of Na3PO4 produced in the experiment. Page 7 of 8