Experiment 24 PDF

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Natalie Garcia February 11, 2019 Lab Partners: Aneesh Paulson, Vernette Spence

Experiment 24: Molar Mass of a Solid

The purpose of this experiment was to determine the rate law for a chemical reaction. As well as, determine the order of each reactant in the reaction and the activation energy for the reaction utilizing a graphical analysis of experimental data.

Abstract: The purpose of this experiment was to determine the rate law for a chemical reaction. Additionally, graphical analysis was utilized to determine the order for each reactant in the reaction and determine the activation energy of the reactants. From basic principles of rate law and activation energy, it can be hypothesized that as the concentration and temperature for the reactants increase, the reaction rate will increase as well. This was tested by alternating the concentration levels of seven different solutions and timing the reactions to see how long it took for them to occur. Based upon the collected data from the experiment, it can be concluded that Introduction: Reaction rates measure the change in the concentration of a reactant per unit of time. Many times, a change in a physical property such as color (absorbance), temperature, and pH can also be used to monitor the change in concentration of a species within an experiment. The relationship between rate and concentration can be determined experimentally. The expression for this relationship between these two factors is called the rate law. The basic rate law is expressed as: rate=k[A]p [B]q. The value of k, the reaction rate constant, generally varies with temperature and is independent upon the concentration of the reactants. The subscripts p and q indicate the order of the reactants and are always determined experimentally. In this experiment, the rate law of hydrogen peroxide and potassium iodide is determined. A quantitative statement is made from how the changes in reactant concentrations affect the reaction rate at room temperature. The rate of the reaction, governed by the molar concentrations of I– , H2O2 , and H3O + , based upon the chemical equation of the reaction, is expressed by the rate law: rate = k [I– ] p [H2O2 ] q [H3O + ] r. A buffer is a solution that resists changes in pH when an acid or other chemical is added to it. It is typically a weak acid and a common salt. A buffer made from acetic acid-sodium acetate solution is added to maintain the [H3O + ] pH of .5 throughout the experiment. Through several kinetic trials with iodide and hydrogen peroxide, the reaction rate is observed. Because reaction rates are temperature dependent, the higher the temperature, the greater the kinetic energy between molecules. This increase of kinetic energy between molecules means that they collide harder and quicker with one another. This causes bonds to break, atoms to rearrange, and new bonds to form faster. The level of energy required for all of aforementioned actions to occur is the activation energy of the reaction. This also results in a relationship with the reaction rate and the activation energy through the equation: k  = Ae–Ea /RT. Materials and Methods:

In Part A of the experiment, the reaction times were determined for the various solutions. 7 different solutions were prepared with boiled, deionized water in 20-mL beakers. The volumes of KI and Na2S2O3 solutions were measured with clean pipets. Next, Solution A was created. Solution was made by pipetting 3.0 mL of 0.1 M H2O2 into a clean, 10-mL beaker. To begin the reaction, H2O2 (solution B) is added to solution A. The reaction was timed (in seconds) until the beaker changed into the deep-blue color. This procedure was repeated for the seven remaining trials. In part E, the activation energy was determined for the different solutions. Two additional solution A and solution B were prepared. Once combination of solution A/solution B was set in an ice bath. The other set was placed in a warm water (-35C) bath. Let them sit for 5 minutes. Once thermal equilibrium was established, solution B was quickly poured into solution A and the timer was started. Once the deep blue color appeared, the timer was stopped. The time lapse and the temperature of the bath was recorded. Results: Table 1: Determination of Reaction Times Molar concentration of Na2S2O3

0.02M

Volume of Na2S2O3 (L)

0.001 L

Ambient temperature (°C)

25

Molar concentration of KI

0.3M

Molar concentration of H2O2

0.1M

Total volume of kinetic trials (mL)

10 mL

Table 2: Calculations for Determining the Rate of Law Kinetic Trial

1

2

3

4

5

6

7

8

Time for color change, ∆t

220

172

94

36

56

62

216

39

Moles of S2O32consumed (mol)

0.00002

0.00002

0.00002

0.00002

0.00002

0.00002

0.00002

0.00002

∆(mol I3 –) produced

0.00001

0.00001

0.00001

0.00001

0.00001

0.00001

0.00001

0.00001

∆(mol I3 – )/∆t (mol/s)

4.55E-8

5.81E-8

1.06E-7

2.78E-7

1.79E-7

1.61E-7

4.63E-8

2.56E-7

log ∆(mol I3 – )/∆t

-7.3

-7.2

-7.0

-6.6

-6.7

-6.8

-7.3

-6.6

Volume KI (mL)

1

2

3

4

1

1

1

3

[I– ]0 (mol/L)**

.03

.06

.09

.012

.03

.03

.03

.09

log [I– ]0

-1.5

-1.2

-1.0

-0.9

-1.5

-1.5

-1.5

-1.0

Volume H2O2 (mL)

3

3

3

3

5

7

2

4

[H2O2]0 (mol/L)**

.03

.03

.03

.03

.05

.07

.02

.04

log [H2O2]0

-1.52

-1.52

-1.52

-1.52

-1.30

-1.15

-1.70

-1.40

Tables 3: Determination of the Reaction Order, p and q, for Each Reactant Value of p from graph

.8867

Value of q from graph

.2508

Table 4: Determination of k´, the Specific Rate Constant for the Reaction Kinetic trial

1

2

3

4

5

6

7

8

Value of k´

1.96E6

1.98E-6

2.33E-5

1.97E5

3.18E-5

4.47E-5

6.13E6

1.16E-5

Average value of k´

1.09x10-6

Standard deviation of k´

1.14x10-6

Relative standard deviation of k´ (%RSD)

105%

Table 5: Determination of Activation Energy Time for color change

Reaction rate

Calc. k´

ln k´

Temperatur e

1/T(K)

Trial 4

36

3.57E-6

1.89E-6

-13.1

25

1/298K

Cold

300

1.33E-7

7.06E-8

-16.5

3

1/276K

Warm

9

1.11E-5

5.89E-6

-12.04

35

1/315K

Graphs Graph 1: log (∆mol I3 – /∆t) versus log [I– ]0.

Graph 2: log (∆mol I3 – /∆t) versus log [H2O2]0.

Graph 3: Activation Energy, Ea

Calculations: Moles of S2O320.02M = x/0.001L = 0.00002mol  ∆(mol I3 –) produced Moles of S2O32-/2 = 0.00002/2 = 0.00001mol  ∆(mol I3 – )/∆t (mol/s) 0.00001mol/220=4.55E10-8  log ∆(mol I3 – )/∆t log[4.55E10-8]=-7.3  [I–]0 (mol/L)** (0.3M*Volume KI)/10=0.3*1/10=.03 log [I– ]0 log[.03]=-1.5  [H2O2]0 (mol/L)** (0.1M*Volume H2O2) /10=0.1*3/10=.03  log [H2O2]0 log[.03]=-1.52 Value of k´ ∆(mol I3 – )/∆t =k´ [I–].8667 [H2O2].2508 k´=(7.3x10-9)(1/0.3^1.12)(1/0.3^8.36)  k´= 1.96x10-6 Average value of k´ (1.9628x10-6+...+1.1606x10-6)/8=1.09x10-6  Standard deviation of k´ sqrt(1.9628x10-6+...+1.1606x10-7)/8=1.14x10-6  Relative standard deviation of k´ (%RSD)

stdev/avg*100=1.14x10-6/1.09x10-6*100=105% 

Discussion By using eight trials of solutions in which one reactant molar concentration was held constant for four trials while the other was changed. By graphing the rate against the concentration of the reactant whose concentration had changed over four trials, the values of p and q were determined and therefore the rate law was also determined as rate = k[I-] 0.7656 [H2O2] 0.9482 . While the rate law was successfully found, the significance of the equation was compromised by the average value of k over the eight trials, which was 1.0930E-5 M -1 s -1. After performing three trials with constant molar concentrations of both hydrogen peroxide and potassium iodide, with the independent variable being temperature, using room temperature, warm water, and cold water as the experimental groups. The trial with warm water had the fastest reaction rate because as temperature increases, the rate of the reaction also increases. After calculating the k values of each trial, a graph was made comparing the inverse of temperature to the natural log of each k value, with the slope of the linear relationship used to calculate the total energy of activation of the reaction, which was found to be 19.242 kJ/mol. While this figure could be accurate, the figure seems unusually high given that the significance of the value is that about nineteen kilojoules are required for each mole of reactant to reach the energy threshold to go to completion and form the products. A possible source of error could be not immediately detecting the color change of the solution due to the fact that it occured very quickly. Conclusion The purpose of this experiment was to determine the rate law for a chemical reaction and to utilize a graph analysis of experimental data to determine the order of each reaction and the activation energy for the reactant. The hypothesis that an increase in temperature and concentration will result in an increased reaction time is accepted. This is due to the increased level of kinetic energy the molecules in the solutions possess. Due to the increase in kinetic energy, the molecules collide at a faster rate, causing a faster reaction time.

As a result, it can be concluded that the overall order of the reaction is 2. The activation energy of the reaction was determined to be 19.242 kJ/mol. This means that the minimum amount of energy that reactant particles must possess in order to react successfully is experimentally determined to be 19.242 kJ/mol. The rate constant k of the chemical reaction was found to be 1.0930E-5 M -1 s -1. Post Lab Questions 1. Part A.4. Describe the chemistry that was occurring in the experiment between the time when solutions A and B were mixed and STOP TIME. a. Solution A and solution B were reacting with one another in order to create a compound with both starch and iodide. Once the solution had formed, a deep blue color was shown. 2. Part A.4. For kinetic trial 2, Alicia was distracted when the color change occurred but decided to record the time lapse read from her watch. Will this distraction cause an increase or decrease in the slope of the log (rate) versus log [I– ]o? Explain. a. If you have a reaction involving A, with in order of n With respect to A, the rite equation says: Rate = k [A]n. If you take the log of each side of the equation, you get: Log(rate) = log k + n log [A]. This second equation would plot as a straight line with slope n. If you measure the slope of this line, you get the order of the reaction. Its rate is independent of initial concentration, so the slope will remain constant. 3. Part A, Table 24.1. a. When doing the kinetic trials, Susan forgot to include the deionized water. Will this omission hasten or delay the formation of the blue color in the trials (exclusive of Trial 6)? Explain. i. It will hasten the formation of the blue color because omitting the water, which is used for dilution, will cause the solution to become more concentrated, causing the rate of the reaction to increase. b. When doing the kinetic trials, Oscar mistakenly omitted the sodium thiosulfate solution. How will this omission change the appearance of the resultant solution (from the mixing solutions A and B) from that of a correctly completed experiment? Explain your reasoning. i. There will be no color change because the sodium thiosulfate plays an important role in causing the chemical change when it mixed with the hydrogen peroxide. c. When doing the kinetic trials, Peyton mistakenly omitted the starch solution from the kinetic trials. How will this omission change the appearance of the resultant solution (from the mixing solutions A and B) from that of a correctly completed experiment? Explain your reasoning.

i.

Without the starch solution there would be no color change in the reaction, which would completely change the reaction d. Of the three chemists above, which chemist will have the most accurate results? Explain. i. Susan 4. Part C.2. Review the plotted data. a. What is the numerical value of the y-intercept? b. What is the kinetic interpretation of the value for the y-intercept? i. It is the point in which the color change begins to start, where there no other added solution. c. What does its value equal in equation 24.8? i. C  equals log k  + log [I– ] p 5. State the effect that each of the following changes has on the reaction rate in this experiment—increase, decrease, or no effect. (Assume no volume change for any of the concentration changes.) a. An increase in the H2O2 concentration. Explain. i. An increase in the concentration of H2O2 would increase the rate of the reaction because it is a strong reactant and more molecules in the solution means that there is more of a chance that they will react. b. An increase in the volume of water in solution A. Explain. i. An increase in the volume of water in solution will cause a decrease in the rate of reaction because the solution is being more diluted. c. An increase in the Na2S2O3 concentration. Explain. i. An increase in the Na2S2O3 concentration will have no effect on the reaction rate because in reaction rate=k[I-][H2O2], S2O3 is not in either concentrations. d. The substitution of a 0.5% starch solution for one at 0.2%. Explain. i. A substitution of 0.5% starch solution for one of 0.2% will not change the rate of reaction because it is not directly proportional to the rate reaction law 6. If 0.2 M KI replaced the 0.3 M KI in this experiment, how would this affect the following—increase, decrease, or no effect? a. The rate of the reaction. Explain. i. Will decrease because it depends on the amount of substance b. The slopes of the graphs used to determine p and q. Explain. i. The slopes would not change because it represents the rate contestant, therefore, it is independent of the concentration of reactants. c. The value of the reaction rate constant. Explain

i.

The value of the reaction rate would not change, for it is represented in the slope, therefore, rate constant does not change with decreasing concentration. 7. Part E.2. The temperature of the warm water bath is recorded too high. How will this technique error affect the reported activation energy for the reaction—too high or too low? Explain. a. The activation would be too high, because the energy needed to make the reaction happen would be less. 8. Part E.4. Arnie’s data plot has a greater negative slope than Bill’s. Which student will record the higher activation energy for the reaction? Describe your reasoning. a. Arnie’s data plot will have the higher activation energy because negative slope will have greater the activation energy slope as opposed to a positive slope.

References: Bern’s, J.A., Laboratory Manual for Principles of General Chemistry, Texas A&M University, 10 the edition: 2014, pgs 281-292...