Activation Energy - Lab Report PDF

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A Spectr Spectrophotometri ophotometri ophotometricc Deter Determination mination of the Act Activation ivation Energ Energyy for a Complex Complexation ation React Reaction: ion: The Kinetics of Formation of a C Cr(III)/EDTA r(III)/EDTA Complex

Name: Shaili Batsri Date: 10/05/2021 Lab Partners: Chris Zalayes

Introduction: For first order reactions, the rate depend only on one reactant to the first power. The rate law becomes to be: Rate = k [A], and the integrated rate expression is: ln([A] / [A]o) = − kt. In this experiment we will use the first order rate equation to determine the rate constant and the activation energy. Activation energy is the minimum amount of energy that must be provided for compounds to result in a chemical reaction. The formula for activation energy is ln k = −Ea /RT + ln A. By graphing this equation, the activation energy can be determined by calculating the slope of the linear function of ln k vs. 1/T. In this lab, we are going to determine the activation energy for the formation of Cr(III)/EDTA complex using the first order and the activation energy equations.

Procedure: Part I 1. Set up a spectrophotometer at 545 nm wavelength. 2. Set up three 13 by 100 mm test tubes and add 2.0 mL EDTA and 1.0 mL Cr(III). Mix well. 3. Heat the tubes using boiling water for 10 minutes, then cool the tubes and measure the absorbance using the spectrophotometer. 4. Calculate a mean value for Af.

Part II 1. Obtain a water bath in a Styrofoam container at the desire temperature using cold water and hot water from part I. *Repeat step 1 for each temperature. 2. Label three new 13 by 100 mm test tubes with the numbers 1, 2, and 3. 3. Obtain an additional test tube with approximately 4 to 5 mL of Cr(III) chloride. 4. Insert the 4 test tubes in the water bath for 2 to 3 minutes and record the exact temperature in one test tube. 5. When the solution is at the equilibrium temperature, add 1.0 mL of Cr(III) chloride to tube #1 and mix well. 6. Record At and time when color is achieved. 7. Record tube #2 after 30 more seconds. 8. Record tube #3 after 30 more seconds. *Obtain at least three replicate values for At at a given temperature. 9. Calculate the mean rate constant and its standard deviation for each temperature.

Date Sheet: Part I – Calculated by another group: Mean A f 0.765

Part II – o Temperature Temperature: 35oC , 308oK o Reaction time: overall reaction time was 32 minutes / 1920 seconds. o Data table:

Comments

First check Second check for better results (the calculations are based on this data) o

Test Tube

Absorbance (At)

Time (minutes)

Time (seconds)

fc

K (s-1)

1 2 3 1 2 3

0.0975 0.048 0.182 0.298 0.336 0.343

4 8 12 24 28 32

240 480 720 1440 1680 1920

0.390 0.439 0.448

3.433x10-4 3.441x10-4 3.095x10-4

Show your calculations for determining fc and the rate constant K: Test Tube

fc

1 2 3

o Mean K K: 3.323x10-4

o Average K of the 35oC group group:: 2.8615x10-4

K (s-1)

Class Data: Temperature (oC)

Temperature (oK)

Rate Constant, K

65 55 45 35

338 328 318 308

7.00x10-3 3.47x10-3 4.80x10-4 2.86x10-4

o Plot:

o Slope: -12026 o Correlation Coefficient: - 0.9711079 o Ea in KJ/mol: 99.984

o Frequency factor, A: A = e 30.669 = 2.086x1013

ln K 0.0029586 0.0030488 0.0031447 0.0032468

-4.9618451 -5.6636007 -7.6417245 -8.1595187

Post-Lab Questions:

1. Do not need to answer this question.

2. Define activation energy. What is the physical basis for the activation energy in this experiment? Activation energy is the minimum amount of energy that must be provided for compounds to result in a chemical reaction. The physical basis of activation energy is the energy needed to break specific bonds so the reaction will be able to occur.

3. Which of the six steps of mechanism given determines th the e act activation ivation energy for this reaction? The rate-determining step is the slowest reaction in the mechanism. Therefore, step A determines the activation energy for this reaction. [Cr(H2O)6]3+ → [Cr(H2O)5]3+ + H2O.

4. What happens to the rate constant as the temperature iincreases? ncreases? With increase in temperature, the rate of the reaction and the rate constant increases. We can also see that in our experiment results. According to the activation energy formula,

If the temperature is increased then T2 > T1 and the left side of the equation is negative. So, k2 > k1. Therefore, increasing the temperature increase the rate constant.

5. The rate determining step of the mechanism involv involves es breaking a Cr – O bond. What would happen to the rate of the reaction if the Cr – O bond in this step were stronger? How would this affect the magnitude of Ea? Since the physical basis of activation energy is the energy needed to break specific bonds so the reaction will be able to occur if the Cr – O bond will be stronger the activation energy will be higher. We will need to invest more energy in order to start the reaction. As result, the rate of the reaction will be slower because there are fewer molecules that have enough energy to react and turn into a product.

Conclusion: In this lab, we examined the activation energy of the Cr(III)/EDTA complex. The objectives of the experiment were to determine the activation energy for the formation of Cr(III)/EDTA complex using the first order and the activation energy equations. The results showed that the activation energy for this complex is 99.984 KJ/mol. In addition, the results also showed that with increasing temperature, the rate constant, K, is also increasing. In this lab I learned how to find the value of activation energy, Ea, of a reaction using the slope of the linear graph ln K vs. 1/T. In addition, I also learned how to calculate the rate constant, K, using absorbance values measured by the spectrophotometer. In this lab I also learned hoe to calculate the frequency factor A, and how to calculate the correlation coefficient....