14 Clay Kaleia - Grade: B PDF

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CHE 113 Experiment 14: Rate law Constant...

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Kaleia Clay Nathaniel George Alexis Chandler CHE 113-011 05 October 2020

Determining the Rate Law and Rate Constant for the Decomposition of Hydrogen Peroxide

Introduction In a chemical reaction, bonds are broken and regenerated to make other molecules. All molecules have kinetic energy. The amount of kinetic energy must be equal or greater than the activation energy, Ea, to break the bonds of the two molecules that collided in the chemical reaction. Greater kinetic energy means that the molecule moves at a faster rate, and a slower molecule will have less kinetic energy. If the total amount of energy released from the molecules that struck each other is small, they will simply bounce off each other and the bonds will not break (French, et al.). In this experiment, hydrogen peroxide was decomposed and catalyzed for 4 different trials using various concentrations of hydrogen peroxide (H2O2), potassium iodide (KI) and water (H2O), with the purpose of calculating the rate law and rate constant of H2O2. The hypothesis is that with the addition of a catalyst, and an increase in concentration and temperature, the chemical reaction will occur at a faster rate due to the molecules moving faster. The reaction rate relies on certain circumstances: “the nature of the reaction, the concentration of the reactants, the temperature of the reaction and the presence of a catalyst” (French, et al.). An increased temperature causes the molecules to separate so that they have more room to collide

and move freely, and an increase in concentration means that there are more molecules in the solution, so there will be more molecular collisions. A catalyst will always increase the speed of a reaction because it decreases the activation energy, When the activation energy is lowered, the reactants can more easily form the products by overcoming the energy barrier. In order to determine the reaction rate, the rate law equation must be used:

rate=

Δ [ C] Δt

If the equation “aA + bB  cC” is given, the difference in concentration of

the product as a function in time will give the rate of the reaction.

The rate law= k[A]x[B]y where the superscripts come from the order of the reaction according to A and B.

The decomposition of hydrogen peroxide comes from the reaction: 2H2O2(aq)  2H2O(l) + O2(g) Potassium Iodide (KI) was used as the catalyst for this reaction. When KI is combined with hydrogen peroxide (H2O2), the peroxide breaks down into water and oxygen very quickly. The speed of the reaction depends on the concentration of hydrogen peroxide used.

The ideal gas law, (P)(V)=(n)(R)(T), was used in this experiment to change the beginning rate

from

kPa s

equation to:

to

mol . To find the rate of formation of O2 in l∗s

mol , we must rearrange the l∗s

P s M = s RT

, so the equation will look like

torr s mol = l∗s L∗torr ∗K K∗mol

The Arrhenius equation, k=Ae-Ea/RT, was used to calculate the activation energy required to overcome the energy barrier. Methods Materials: 1. Gas pressure probe kit 2. Pipets 3. Thermometer 4. MeasureNet 5. 3% Hydrogen Peroxide, H2O2 solution 6. 0.5 M Potassium Iodide, KI, solution 7. Two 6in. test tubes 8. Pipet bulb 9. 400 mL beaker 10. Other equipment as needed Procedure: 1. Obtain and put on protective gear, which includes, but is not limited to lab scrubs, goggles and gloves. 2. Put 4 mL of 3% hydrogen peroxide, H2O2, and put the test tube in the water bath.

3. Obtain a pipette and draw 1 mL of the KI solution into the pipette and transfer the KI to the H2O2 solution quickly. The reaction will begin once the KI solution is in the peroxide solution. 4. Hold the stopper firmly inside the test tube and begin the reaction. Press “START” on the MeasureNet workstation and leave it running for at least 120 seconds. 5. Once the reaction has taken place, repeat the above steps for the second and third trials with the appropriate concentrations of the solutions and save the data. 6. For the fourth trial, obtain a warm water bath in a 400 mL beaker. 7. Create the hydrogen peroxide solution with 4 mL H2O2 and 1 mL KI solution in a test 2tube. 8. Place the test tube in the warm water bath and put the stopper in the test tube. 9. Press “START” on the MeasureNet workstation and leave it running for at least 120 seconds. 10. Be sure to save the data.

Discussion The purpose of this lab was to generate the decomposition of hydrogen peroxide in order to find the rate constant and rate law of a reaction. The hypothesis is that the addition of a catalyst and increase in temperature and concentration of solutions will all increase the rate of the reaction. The molar concentration (M) of 3.0% H2O2 was 0.00235 M. The rate of decomposition of H2O2 was as follows for each trial: Trail 1= 4.9E-4 mol/L*s; Trial 2= 1.633E-4 mol/L*s; Trial 3= 1.76E-4 mol/L*s; Trial 4= 1.69E-3. The rate of the reaction was rate=k[H2O2] [KI]2 because for trials 1 and 4, the H2O2 the concentration of the reaction doubles, which means

it is a first order reaction with respect to the reactant (French, et al.). For the KI solution the concentration of the quadruples, which means it is a second order reaction with respect to the reactant. The activation energy was calculated at 113.29 kJ/mol. My results support my original hypothesis because in the fourth trial, the concentration of H2O2 and KI was much higher than the rest of the trials and the test tube was placed in a warm water bath, which increased the temperature. This means that the rate of the reaction should have been much higher than the previous trials. My results support this because when comparing the first and fourth trials, Trial 1’s rate of decomposition of H2O2 was 4.9E-4 and Trial 4’s rate of decomposition of H2O2 was 1.69E-3. The rate of reaction for trial four was much greater than the previous trials due to the increase in temperature and concentrations. Three problems that could have occurred in this lab were that the concentrations could have been wrong. If the procedure calls for 4 mL of solution, but 6 mL was put in, this will throw off the data by speeding up the reaction because of the increase in concentration. The second error is that the stopper may not have been held firmly enough on top of the test tube. This will throw off the data because the pressure increases as the stopper is pushed more firmly, so if it is not held firmly enough the data may not have collected the correct measurement. The third source of error is that the fourth trial was not placed in a warm water bath. This experiment was performed to see if the temperature will increase the rate of the reaction, so if the temperature was not increased, the data will not change.

Conclusions I learned that the rate of a reaction is dependent upon the temperature and concentrations of a given solution. This scientific concept occurs in everyday life because when baking, the temperature of the oven determines the speed the food will cook. The lower the temperature, the

slower the food will bake. The higher the temperature, the faster the food will cook. This is very important to understand in day-to-day lives.

Work Cited French, April N., Allison Soult, Stephen Testa, Pauline Stratman, Merval Savas, Francois Botha, Charles Griffith, Carolyn Brock, Donald Sands, Darla Hood, Robert Kiser, Diane Vance, Penny O’Conner, William Plucknett, and William Wagner. “Experiment 14: The Decomposition of Hydrogen Peroxide.” General Chemistry Laboratory Manual. Plymouth HaydenMcNeil Publishing, 2020; p. 81-85. Web. 05 October 2020. https://www3.chem21labs.com/labfiles/36194_45_Exp%2014_FrenchA%2021871%20W20..pdf?rf=8236...