Chemical Kinetics

Determination of Rate of Chemical Reaction for an Iodine Clock Reaction

Datoya Brown CHEM 160: General Chemistry II

26 July 2012

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Abstract

The factors that affect the rate of a chemical reaction are important to understand due to the importance of many such reactions to our health, well-being and comfort. It would be advantageous to slow down some of these reactions such as food spoilage and rust formations, while in the cases of reactions such as the Tums-stomach acid reaction and the conversion of organic matter to fossil fuels, it would be beneficial to them speed up. The purpose of this lab is to measure the effect of concentration upon the rate of a reaction of peroxydisulfate ion with iodide ion: ; to determine the order of the reaction with respect to the reactant concentrations; and to obtain the rate law for the chemical reaction:

Introduction

The full purpose of this experiment is to deal with the laws of chemical kinetics, and by doing the experiment, compare the experimental results with the theories and see if they were consequently followed. From the kinetics studies, it is obvious that the rate of a reaction increases as the temperature of the reaction increases and as the concentration of the reactants increases. Also, the catalyst increases the rate of the reaction and decreases the activation energy. So, this experiment is divided into three sections and the dependence of the reaction rate from different factors is observed in step by step fashion. The idea of the first part of this experiment is to find the reaction orders with respect to each reactant and the rate constant, k. The method of initial rates to calculate the order with respect to each reagent will be used. With different times, the concentrations will vary in each trial. In this experiment, the following type of reaction is considered:The peroxodisulfate ion, S2O82-, oxidizes iodide ion reasonably slowly at room temperature in accord with the equation: (Equation 1)

The rate of reaction can be expressed simply in terms of the decrease of any one of the reactants with respect to time or the increase of one of the products with respect to time and is written as follows:

Rate = - Δ [S2O82-]/Δt = - [I-]/2Δt = Δ/ [I2] Δt = [SO4

2-]/2Δt (Equation 2)

The first equality states that the reaction rate is equal to the decrease in peroxodisulfate ion concentration for a given time interval. Equation 1 indicates that two iodide ions are used for every peroxodisulfate ion that reacts, and the second equation involving the iodide ion also shows this fact.

S2O82−+2 I−→ I 2+2SO4

2−

Δ [ S2O82− ]

Δt=k [ S2O8

2− ]x[ I− ] y

S2O82−+2 I−→ I 2+2SO4

2−

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One of the main purposes of a kinetics experiment is to find the rate law for the reaction, the expression that relates rate to the concentration(s) of the reactant(s). In the above reaction the rate depends both on the concentration of I- and S2O8

2- ions and takes the general form: Rate = k [I-] m

[S2O82-] n (Equation 3), where k is the rate constant and m and n are the

reaction orders.

In this experiment you will study the kinetics of the reaction of iodide ion with peroxodisulfate ion and determine the rate law. The method for measuring the rate of the reaction involves what is frequently called a “clock” reaction. This is a second reaction (or set of rapid consecutive reactions) that gives a signal, such as a color change, when a particular quantity of one of the reactants has been consumed. Our “clock” will be based on the facts that thiosulfate ion, S2O3

2-, reacts rapidly and quantitatively with iodine, I2: I2 + 2S2O3

2- → 2I- + S4O62- (Equation 4)

and an intense blue-black coloration is formed almost instantly when exceedingly small traces of I2 are in the presence of starch and I- (hence, an indicator that detects iodine). Thus, the reaction of peroxodisulfate ion with iodide ion is started in the presence of a known trace amount of thiosulfate ion and the starch indicator. As the reaction progresses, the iodine, as it is formed, immediately reacts with the thiosulfate and the reaction mixture remains colorless. When the trace quantity of thiosulfate is used up, iodine produced by the oxidation of iodide ion now remains in solution and causes the starch indicator to turn blue-black in color. Therefore, at time t, the time after mixing when the solution turns blue-black, for every mole of thiosulfate originally present, 1/2 mole of iodine has been produced and used up, and 1/2 mole of peroxodisulfate also has been used up. At this point, the change in concentration of S2O8

2- must be equal to half the original S2O3

2- concentration, and thus, where a rate has been expressed in terms of -Δ[S2O8

2-]/Δ t, or Rate = - Δ[S2O82-]/Δt = 1[S2O3

2-]orig/2Δt (Equation 5)

The calculation of the rate constant ( k) and the orders of the reaction (m and n) with respect to I- and S2O8

2- follow from the dependence of the reaction rate on reactant concentrations. (Compare equations 3 and 5.)

The factors that affect rates of reaction are as follows:o Nature of Reactants -The reactants must come into direct contact with

each other and must collide with sufficient energy to result in a reaction

o Concentration - Changes the number of particles per unit volumeo Temperature - Changes the kinetic energy of the particles (i.e.

velocity) which changes the number of collisions in a period of timeo Catalyst - Lowers the activation energy of the reaction while

remaining chemically unchanged

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oThe Order of Reaction is as follows:

The rate of the reaction may be expressed by:

o [A] and [B] are the molar concentrations of A and Bo x and y are the powers which the concentration must be raised to

describe the rateo k is the specific rate constant for a reaction

Independent of concentration and depends only on temperature Once the rate is known the value of k can be calculated

o Suppose we found that x = 2 and y = 1 for the reactiono The rate law would then be:

If [B] is doubled (keeping [A] the same)the rate will doubleo If [A] is doubled (keeping [B] the same)the rate will quadrupleo The powers in the rate law are the order of the reactiono The reaction above would be:

Second order in A First order in B The overall order is the sum of the exponents or third-order

overall

MaterialsApparatusBurets (2) 25-mL pipet1-mL pipet 50-mL pipetStopwatch test tubes (8)Pipet bulb 250-mL Erlenmeyer flasks (4)Buret clamp 100-mL beakers (4)Ring stand

Chemicals0.2 M KI 0.4 M Na2S2O3 (freshly prepared)1% starch solution, boiled 0.1 M solution of Na2H2EDTA0.2 M (NH4)2S2O8 0.2 M KNO3

(freshly prepared)

ProceduresPart A: Preliminary Experiments

1. Dilute 5-mL of 0.2 M KI solution with 10-mL of distilled water, add 3 drops of starch solution, mix thoroughly, and then add 5 mL of 0.2 M (NH4)2S2O8. Mix thoroughly, wait, and observe color change.

A+B→C+D

rate=k [A ]x [B ] y

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2. Repeat step one. When the solution changes color, add 4 drops of 0.4 M Na2S2O3, mix and note the effect on the color.

Part B: Kinetics Experiment Solution preparation. Prepare the four reaction solutions assigned to:

Solution 1 Solution 2 25-mL KI 25-mL KI 1-mL starch solution 1-mL starch solution 1-mL Na2S2O3 1-mL Na2S2O3 48-mL KNO3 23-mL KNO3 1-drop EDTA 1 drop EDTA

Total volume = 75.0 mL Total volume = 50.0 mL

Solution 3 Solution 4 50-mL KI 12.5-mL KI 1-mL starch solution 1-mL starch solution 1-mL Na2S2O3 1-mL Na2S2O3 23-mL KNO3 35.5-mL KNO3 1 drop EDTA 1 drop EDTATotal volume = 75.0 mL Total volume = 50.0 mL

Rate measurements.

Prepare solution 1 in a 250-mL flask. Transfer 25-mL of (NH4)2S2O8 solution into the flask. Swirl the solution strongly. Be ready to begin timing the reaction when the solutions are mixed. The reaction starts the moment the solutions are mixed. Note the time you begin mixing to the nearest second. The instant the blue-black color appears, note the time and immediately add 1-mL aliquot of Na2S2O3 (aq), which should be measured in 1-mL increments in the test tubes, to the solution; the color will disappear. Repeat this process for a total of 7 trials. Solutions 2, 3, and 4 should be treated in the same manner, using the following volumes of (NH4)2S2O8:

Solution mL (NH4)2S2O8 2 50.0 mL 3 25.0 mL 4 50.0 mL

Results

A. Preliminary Experiments1. What are the colors of the following ions? K+ colorless ; I- colorless

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2. The color of the starch · I2 complex is blue-black

B. Kinetics ExperimentSolution 1. Initial [S2O8

2-] = 0.05 M; initial [I-] = 0.05 M

Aliquot no.

Tim (s) between

appearances of color

Cumulative Time (s)

Total moles of S2O8

2-

consumed

1 188 188 2.0 x 10-4

2 190 378 4.0 x 10-4

3 200 578 6.0 x 10-4

4 215 793 8.0 x 10-4

5 235 1028 10.0 x 10-4

6 255 1283 12.0 x 10-4

7 274 1557 14.0 x 10-4

Solution 2. Initial [S2O82-] = 0.10 M; initial [I-] = 0.05 M

Aliquot no.

Tim (s) between

appearances of color

Cumulative Time (s)

Total moles of S2O8

2-

consumed

1 91 91 2.0 x 10-4

2 95 186 4.0 x 10-4

3 97 283 6.0 x 10-4

4 99 382 8.0 x 10-4

5 100 482 10.0 x 10-4

6 100 582 12.0 x 10-4

7 102 684 14.0 x 10-4

Solution 3. Initial [S2O82-] = 0.05 M; initial [I-] = 0.10 M

Aliquot no.

Tim (s) between

appearances of color

Cumulative Time (s)

Total moles of S2O8

2-

consumed

1 122 122 2.0 x 10-4

2 126 248 4.0 x 10-4

3 136 384 6.0 x 10-4

4 143 527 8.0 x 10-4

5 153 680 10.0 x 10-4

6 160 840 12.0 x 10-4

7 170 1010 14.0 x 10-4

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Solution 4. Initial [S2O82-] = 0.10 M; initial [I-] = 0.025 M

Aliquot no.

Tim (s) between

appearances of color

Cumulative Time (s)

Total moles of S2O8

2-

consumed

1 192 192 2.0 x 10-4

2 198 390 4.0 x 10-4

3 203 593 6.0 x 10-4

4 209 802 8.0 x 10-4

5 209 1011 10.0 x 10-4

6 218 1229 12.0 x 10-4

7 221 1450 14.0 x 10-4

Calculations

1. Rate of reaction, Δ[[S2O82-]/Δt, as calculated from graphs (that is, from

slopes of lines): Solution 1 9.2 x 10 -6 mol/L-s Solution 2 2.0 x 10 -5 mol/L-sSolution 3 1.5 x 10 -5 mol/L-s Solution 4 1.1 x 10 -5 mol/L-s

2. What effect does doubling the concentration of I- have on the rate of this reaction?From one and three, rate increases by factor of 1.6 (should double)

3. What effect does changing the [S2O82-] have on the reaction?

From one and two, rate increases by a factor of 2; doubling concentration doubles rate.

4. Write the law for this reaction that is consistent with your data.Rate law: rate = k[I-][S2O8

2-]5. From your knowledge of x and y in the equation (as well as the rate in

a given experiment from your graph), calculate k from your data. Rate = k[S2O8

2-]x[I-]y

Solution (1) k = 9.2x 10−6mol /L−s¿¿

= 3.7 x 10-3 L/mol-s

Solution (2) k = 2 x10−5

(0.1 )(0.05) = 4.0 x 10-3 L/mol-s

Solution (3) k = 1.5 x10−5

(0.05 )(0.10) = 3.0 x 10-3 L/mol-s

Solution (4) k = 1.0 x10−5

(0.10 )(0.025) = 4.0 x 10-3 L/mol-s

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188 378 578 793 1028 1283 14000.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

Solution 1Initial [S2O82-] = 0.05 M; initial [I-] = 0.05 M

Time (s)

Mol

es S

2O82

- x 1

04

Slope = (8.0−2.0 )mol x10−4

(880−230 ) s = 9.2 x 10-7 mol/s

Rate = 9.2 x 10-6 mol/L-s

Solution (1) k = 9.2x 10−6mol /L−s¿¿

= 3.7 x 10-3 L/mol-s

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91 186 283 382 482 582 6840.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

Solution 2Initial [S2O82-] = 0.10 M; initial [I-] = 0.05 M

Time (s)

Mol

es S

2O82

- x 1

04

Slope = 0.020 x 10-4 mol/sRate = 0.20 x 10-4 mol/L-s

Solution (2) k = 2 x10−5

(0.1 )(0.05) = 4.0 x 10-3 L/mol-s

122 248 384 527 680 840 10100.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

Solution 3Initial [S2O82-] = 0.05 M; initial [I-] = 0.10 M

Time (s)

Mol

es S

2O82

- x 1

04

Slope = 0.015 x 10-4 mol/s

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Rate = 0.15 x 10-4 mol/L-s

Solution (3) k = 1.5 x10−5

(0.05 )(0.10) = 3.0 x 10-3 L/mol-s

192 390 593 802 1011 1229 14500.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

Solution 4Initial [S2O82-] = 0.10 M; initial [I-] = 0.025 M

Time (s)

Mol

es S

2O82

- x 1

04

Slope = (14.0– 2.0 )mol x10−4

(1420−220 ) s = 1.0 x 10-6 mol/s

Rate = 1.0 x 10-5 mol/L-s

Solution (4) k = 1.0 x10−5

(0.10 )(0.025) = 4.0 x 10-3 L/mol-s

Discussion

According to the data the purpose was achieved. In the study of the effect of concentration on the rate of a reaction, it was found that when increasing the concentration of a reactant, the rate also increases. This was seen in the data. This result is expected when taking the collision theory into consideration. According to the collision theory, in order for reactants to form products there must be successful collisions between reactant molecules. Successful collisions are those that have enough energy to break and reform bonds and have the correct collision orientation or geometry. The correct orientation is necessary to form the activation complex, a short-

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lived transition between reactants and products. When there are a greater number of reactant molecules available to react, there is a greater chance of a successful collision due to probability arguments. The more KIO3 that was present per unit volume, the greater opportunity there was for a successful collision to happen. There were more successful collisions as evidenced by more product being formed in a shorter period of time (color change faster).

There were many possible sources of errors in the lab procedure. Reaction times when attempting to record the time it took for color to change are extremely variable. Since the procedure included the use of data from other lab groups, this would have a large affect on the results. Since the time it takes for the color change to first appear is used to calculate the rate values, this error will have a direct impact on the accuracy of the rate values.

It is possible that the solutions that were provided were not exactly the same molarity. If the same solution was not used each time, the reaction times may be faster or slower than expected. Concentration will affect the reaction rate calculated. Also, the solutions may not be well mixed thereby changing their effective concentrations. Again, since the concentration values were used in order to calculate rate, this error will affect the results substantially. In the calculation of rate, the given molarity is used in the calculation. If the actual concentration was lower than the given one, the reaction time would be slower than expected. Conversely, if the actual concentration was higher than the one given, the reaction time would be larger than it should be.

Conclusion

According to the data, the purpose was reached. The rate of a reaction is seen to increase when concentration is increased considering experimental error. On can notice high concentration of certain reactants, high temperatures, and the presence of catalyst simply speeds up a reaction. The goal of the experiment was achieved through various trials of reactions. In the end the experiment could have been complete with more accuracy but do to haphazard human error in measuring for concentration purposes it was slightly off from other group that conducted the experiment as well.

Acknowledgements

Thanks to my professor/lab instructor, Dr. Shubo Han, and also lab partners.

References

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Nelson, John H and Kenneth C Kemp. Laboratory Experiments to Accompany Chemistry The Central Science 11E. Upper Saddle River: Pearson Education, Inc., 2009