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

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)

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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.

To practice calculations related to quantities in chemical reactions with a gas product.
To measure amount of gas based on its volume.
To determine percent concentration of hydrogen peroxide in a commercial solution. PART II.
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 H 2 O 2. On contact with a tissue it readily decomposes according to the following equation producing oxygen, O 2 , gas.

2 H 2 O 2 (aq) –> 2 H 2 O(l) + O 2 (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)

T = temperature expressed in kelvins (K). K = degree Celsius + 273. We have to re-arrange the equation to isolate n: n = PV/(RT) The pathway for the calculations is: mL O 2 –> L O 2 –> mol O 2 (using n = PV/(RT) –> –> mol H 2 O 2 (using conversion factor from the equation) –> g H 2 O 2 –> % H 2 O 2 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, O 2 , was collected. The atmospheric pressure = 0.910 atm and temperature = 23.5 degree Celsius. Calculate the percent by mass of hydrogen peroxide, H 2 O 2 , 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 O 2 = PV/(RT) = 0.910 atm x 0.0690 L/( 0.0821 L atm/(mol K) x 296.7 K) = 0. mol O 2 Step 3, calculate grams of H 2 O 2 0.00258 mol O 2 2 mol H 2 O 2 34.0 g H 2 O 2 0.175 g H 2 O 2 1 mol O 2 1 mol H 2 O 2 Step 4, calculate the percent by mass of H 2 O 2 in the sample 0.175 g H 2 O 2 /5.20 g sample x 100 % = 3.37 % H 2 O 2 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 H 2 SO 4 in a coffee-cup calorimeter

0.0200 mol H 2 SO 4 1 mol Na 2 SO 4 142.05 g Na 2 SO 4 2.84 g Na 2 SO 4 1 mol H 2 SO 4 1 mol Na 2 SO 4 2.84 g Na 2 SO 4 is a theoretical yield obtained from H 2 SO 4 Both yields are identical. There is not a limiting reactant. Both reactants are in the ideal stoichiometric ratio. PROCEDURE. PART I. DECOMPOSITION OF HYDROGEN PEROXIDE Place a large test tube in a small beaker (the beaker serves as a holder to keep the test tube upright) and weigh both - this is a weight of container. Add to the test tube about 6 mL of hydrogen peroxide solution and weigh the beaker + test tube with the solution again. Calculate the net weight of the hydrogen peroxide solution. Put into the large test tube a small test tube half-filled with the liver homogenizate. Make sure that the small test tube is not contaminated with the homogenizate on the outside. Close the large test tube with a specially prepared rubber stopper. The stopper has a hole and attached tubing to it. Make sure the stopper fits tight. Fill a tub with water. Submerge a 100 mL graduated cylinder in the water to fill it. You should have no air bubbles in the cylinder. Invert the cylinder upside down while the lower part of it is in the water. Under the water in the tub, place the end of the tubing (coming out from the large test tube) inside the graduated cylinder and hold it there. Now gently tilt the large test tube in such a way that the content (liver homogenizate) of the small test tube inside mixes with the solution of hydrogen peroxide. However, make sure that the mixture doesn’t flow into the tubing coming out of the stopper. The solution should produce gas bubbles. Let the gas flow into the inverted cylinder until the bubbling completely stops. Read the volume of oxygen collected in the inverted graduated cylinder. Measure the temperature of the water in the tub. The instructor will measure the atmospheric pressure. Comments: two students need to participate in the procedure. One person will hold the large test tube and tilt it when ready. Second person will hold the graduated cylinder and the tubing inside it, under the water in the tub. The instructor will demonstrate the procedure before the students’ work. PART II. CALORIMETRY EXPERIMENT 1. In this experiment, NaOH will be neutralized with HCl. Pour 50.0 mL of 1.50 M NaOH solution in a coffee-cup calorimeter. Measure and record the temperature of the solution (since both NaOH and HCl solution are in the same room, we assume that both have the same temperature, which is the initial temperature). Add 50.0 mL of 1.50 M HCl 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 HCl?

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 H 3 PO 4. 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 H 3 PO 4 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 H 3 PO 4 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 H 3 PO 4? e. How many grams of Na 3 PO 4 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 + H 2 O 2 solution:_______________________ Net weight of H 2 O 2 solution:____________________________ Volume of the collected O 2 :_________________________ Temperature:_____________________ Atmospheric pressure:____________________ CALCULATIONS: Calculate the % by mass of H 2 O 2 in the hydrogen peroxide solution. Show all calculations. Follow the example.

EXPERIMENT 2. NaOH + H 3 PO 4. 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 H 3 PO 4 : e. Calculations of the amount of Na 3 PO 4 produced in the experiment.

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

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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

PART II : HEAT PRODUCED IN CHEMICAL REACTIONS -

CALORIMETRY

OBJECTIVES

PART I.

2 H 2 O 2 (aq) –> 2 H 2 O(l) + O 2 (g)

The ideal gas law: PV = nRT

R = gas constant: 0.0821 atm L/(mol K)