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Experiment 8: Kinetics of the Reaction Between Iodide Ion and Peroxodisulfate Ion(The Iodine Clock Reaction)

For each procedure, the instructor will assign students to perform each experiment (A1-C3) enoughtimes to generate a complete and good data set.

I. Introduction

The objective of this lab is to determine the rate law and activation energy for the redox reaction betweenthe iodide ion (I–) and the peroxodisulfate (S2O8

2–) ion in water, for which the overall reaction is shownbelow (1).

Overall Reaction:3 I–(aq) + S2O82–(aq) → I3

–(aq) + 2 SO42–(aq) (1)

The general expression for the rate law, given this overall reaction, is:

rate of reaction = k [I–]m [S2O82–]n

If reaction (1) is an elementary reaction, then the rate law is:

rate = k[I–]3 [S2O82–]

Because this is a fourth order rate law (tetramolecular?) and would involve four molecules colliding, it isunlikely that reaction (1) is an elementary reaction. One of the objectives of this experiment is, therefore, todetermine the rate law for reaction (1).

The Method of Initial Rates

The way in which the rate law will be determined is using the “Method of Initial Rates”. In this method, aseries of experiments is carried out in which the initial concentration of one of the reactants (let’s say I–) isvaried systematically while the concentration of the other reactant (S2O8

2–) is left constant. This allows oneto determine one of the coefficients (“m”). Then the process is repeated in a second series of experimentsby systematically varying the other reactant’s concentration.

In the method of initial rates, it is necessary to determine the initial rate of reaction for each experiment.The rate of reaction that is being measured is an average rate of reaction, and it’s a little hard to understandhow it works. Let’s have a go:

The triiodide ion (I3–) is identified by an intensely colored starch-triiodide complex. Because we add a few

drops of starch to the reaction solution, if there is also any triiodide ion in solution, the solution will be verydark blue. Triiodide is rapidly reduced by thiosulfate ion (reaction 2, below) and the reaction of triiodideand thiosulfate will be used to monitor the overall reaction (reaction 1).

Monitoring Reaction: I3–(aq) + 2 S2O3

2–(aq) → 3 I–(aq) + S4O62–(aq) (2)

Because reaction (2) is much faster than reaction (1) but depends upon the product of reaction (1) as areactant of reaction (2), reaction (2) can only go as fast as reaction (1) can go. Therefore, the rate ofreaction (2) is the same as the rate of reaction (1). (Read that a few times and see if it makes sense).

Also because reaction (2) is much faster than reaction (1), there is an insignificant concentration of triiodideion (and insignificant starch-triiodide complex) in solution until the thiosulfate is completely consumed.When the thiosulfate is completely consumed, the triiodide reacts with the starch to turn the solution verydark blue.

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To summarize: The monitoring reaction uses thiosulfate to react away triiodide so that there is no coloruntil there is no thiosulfate left in solution. Once the concentration of thiosulfate is zero, the solution turnsvery dark blue.

We know the initial concentration of thiosulfate, [S2O32–]i, based on the concentration put into the reaction

solution. We also know that the final concentration of thiosulfate, [S2O32–]i, is zero. The initial time is zero

and the final time is just how long since the reaction started. We can then calculate the initial rate ofreaction (see Sample Calculation A).

Once “m” and “n” have been determined, the rate constant can be determined by plugging the rate andinitial concentrations back into the rate law expression and solving for k.

It is important to note that the concentration of thiosulfate is much less than the concentration of eitheriodide or peroxodisulfate in the reaction mixture. Therefore, very little of reaction (1) has been completedwhen the change in color occurs. For this reason, we assume that the rate of reaction derived from thisexperiment is an initial rate of reaction.

Determining the Activation Energy

Another goal of this lab is to determine the activation energy for reaction (1). The activation energy is theenergetic barrier that has to be overcome between the reactants and the products. At the top of thisenergetic barrier, there is a transition state complex that forms instantaneously before the products areformed.

The only way that we know of to determine the activation energy is to prepare an Arrhenius plot. AnArrhenius plot has an x-axis of 1/T, where T is temperature in Kelvin, and a y-axis of ln k, where k is therate constant. Reaction rates will be determined in three different temperature ranges: room temperature,above room temperature and below room temperature.

Adding a Catalyst

A third experimental goal of this lab is to see the effect that a catalyst can have on the rate of reaction forreaction (1). With the catalyst, the reaction can be written:

Cu2+

Overall Reaction:3 I–(aq) + S2O82–(aq) → I3

–(aq) + 2 SO42–(aq) (1)

and the general form of the rate law becomes:

rate of reaction = k [I–]m [S2O82–]n [Cu2+]p

Though it is not necessarily true, we will assume that the exponents “m” and “n” don’t change with theaddition of the catalyst.

A catalyst is a substance that increases the rate of reaction (both in the forward and reverse directions)without being consumed in the reaction. Catalysts often allow the reactants to react using a reactionpathway (or mechanism) that is lower in activation energy than without the presence of the catalyst. Oftentimes the catalyst plays a role in making the transition state complex lower in energy. The catalyst beingused in this experiment is Cu2+ ion in copper sulfate.

Three different experiments will be carried out with the same initial concentrations of reactants as one ofthe experiments that was done without a catalyst (A6). As I’ve often exclaimed after seeing how muchfaster the experiment goes with the catalyst: “Behold! The power of the catalyst!”

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Copper ion is not even that good of a catalyst. It lowers the activation energy, but not nearly as much asmost biological enzymes lower the activation energy. For comparison purposes, we’ll also performexperiments to determine the activation energy with the catalyst.

Diluting with Ions

In experiment A1 of this lab, 10.00 mL of KI are placed into “Beaker 1”, but in experiment A2 of this lab,5.00 mL of KI and 5.00 mL of KCl are placed into “Beaker 1”. Experiment A2 uses only half the volume(and is therefore only half as concentrated) of reactant KI that experiment A1 uses. The KCl is used as adiluent, instead of water, to keep the number of ions in solution the same even though the KCl does notreact or do anything else except add ions to solution.

When the number of ions is kept constant, we say that the solutions “ionic strength” is constant. Whileionic strength is a fascinating concept, we will not be discussing ionic strength as part of this class.However, it is taught in Chemistry 410: Quantitative Chemistry here at SCC.

Recording Reaction Times and Accuracy

Most students will record the reaction times using stop watches that report time to the nearest onehundredth of a second (0.01 sec). Considering that there is at least half a second when the chemicals arebeing poured and mixing and that there is at least half a second error in the determination of when thesolution turns blue, we will report our times to the nearest second. In general, it’s a safe practice to alsoreport the averages and standard deviations to the nearest second too.

A Friendly Reminder about Dilution

Because all of the experiments in today’s lab involve the mixing of three different beakers of solutions,each chemical will be diluted. It is important to consider this fact in your calculations.

Writing Reaction Mechanisms

The last question for this lab requires that you write a reaction mechanism. A reaction mechanism is thesequence of molecular scale events (or elementary steps) that occurs in going from reactants to products. Agood reaction mechanism follows two simple rules: (i) the elementary steps add up to the overall reaction.In this case, our overall reaction is reaction (1) and (ii) the rate law for the reaction mechanism is the sameas the experimentally-determined rate law. In this experiment, we will be determining the experimental ratelaw and we already know the overall reaction.

Hint: The simplest way to write a reaction mechanism is to use the experimentally-determined rate law toconstruct the first, rate determining step in a two step reaction mechanism. Second hint: the reactionmechanism can include species that we’ve never seen before (dare we say that we’ve made-up?).

II. Sample Calculations

A. Determination of the Reaction Rate

Assume that at the start of the reaction, the concentration of S2O32– is 1.00 × 10–3 M (not the

correct concentration as your calculations will show). As reaction (1) proceeds, the I3– is

immediately consumed in reaction (2) until there is no S2O32– remaining. At this point, the reaction

mixture becomes blue and the time of reaction is noted. If the time is noted to be 220 seconds,then:

reaction rate = − 1

2

∆[S2O32−]

∆t= − 1

2

[S2O32− ]f −[S2O3

2 −]i

tf − ti= −1

2

(−1.00 ×10−3 )220 sec

= 2.27×10–6

Msec

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It is very important to point out that it is the concentration of S2O32– (that’s a THREE) that is used

for this calculation and NOT the concentration of S2O82–.

B. Determination of the Rate Law

Experiment [I–]0,M

[S2O82–]0

M[CuSO4] 0

MInitial Rate of reaction,

M/sA1 8.00 × 10–2 8.00 × 10–2 0 2.28 × 10–6

A2 8.00 × 10–2 4.00 × 10–2 0 1.38 × 10–7

Table 1: Some sample (incorrect) data for experiments A1 and A2.

The general form of the rate law for reaction (1) is

rate = k[I− ]m [S2O82− ]n

where k is the rate constant and “m” and “n” are the exponents associated with the reactantconcentrations. Assuming that temperature is constant (and therefore k is constant), by holding [I–]constant, the effect of [S2O8

2–] on reaction rate can be determined from two experiments withdifferent values of [S2O8

2–]. In the following expression, everything cancels out except thedifferent rates and the different [S2O8

2–].

rateexpt1 = k[I− ]m [S2O82− ]expt1

n

rateexpt2 = k[I− ]m [S2O82− ]expt2

n → cancel out k[I–]m →

rateexpt1 = [S2O82− ]expt1

n

rateexpt2 = [S2O82− ]expt2

n

2.28× 10−6

1.38× 10−7=

(8.00×10−2 )n

(4.00 × 10−2)n

16.5 =2( )n

1( )n→ log(16.5) = n ⋅ log

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→ n =4.05

For each determination of an exponent (m, n, or p for the catalyst), there are three possiblecombinations of experiments that can be used to estimate the exponent. Do the full blowncalculation for all of them as shown above, average the three values of the exponent andround the average to the nearest whole number.

C. Determination of the rate constant

Once the rate law is determined, any experiment can be used to determine the rate constant k bysimply plugging in the appropriate values for an experiment. Since there are A1–A6 experiments,do at least three determinations to find the average k value. Also report the standard deviation of k.

D. Determination of the activation energy

From the k values at each of three different temperatures, a plot of ln k versus 1/T should lead to alinear plot, with the slope being –Ea/R. Determine by using Excel the slope of the best fit line tothis plot. Do not use the two-point slope formula!

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III. Experimental Procedure

A. Equipment needed

Set-up in lab: color-coded burets, beakers and funnels with each of the five main chemicalsneeded to perform this labChemicals in lab: approximately 0.20 M KI, 0.20 M KCl, 0.2 M (NH4)2S2O8, 0.200 M(NH4)2SO4 and 0.00250 M Na2S2O3, 1% starch solution (exact concentrations need to be noted inyour notebook)From stock room: stopwatches, bucket of ice, crystallizing dishes

B. Disposal

All solutions must be disposed of in the proper waste container.

C. Experimental considerations:

1. The volumes of KI, (NH4)2S2O8 and Na2S2O3 are critical for this lab. It is these that will be usedto determine the reaction order (specifically the last). These should be measured carefully with aburette, pipette or an micropipette. The volumes of KCl and (NH4)2SO4 are not as critical, butshould be measured using a graduated cylinder.

2. Many of the chemicals in this lab have similar chemical formulae. Please be careful to obtainthe correct chemical!

3. For each procedure, the instructor will assign students to perform each experiment (A1-C3)enough times to generate a complete and good data set.

D. Procedure 1. Determination of the Reaction Rate Law.

1. Note the exact concentrations of each of the chemicals used in class in your notebook.

2. Fill burettes with KI, KCl, (NH4)2S2O8 and (NH4)2SO4 if it has not been done already.

3. The instructor will assign each pair of students an experiment to perform.

4. For your assigned experiment A1-A6 (see below)), prepare the solutions in 50 mL beakerslabeled “Beaker 1”, “Beaker 2” and “Reaction Beaker”. For each experiment, the Reaction Beakerwill contain 10.00 mL of 0.00250 M Na2S2O3, three drops of 1% starch solution. At t=0, mix allthree solutions together in the Reaction Beaker. Record the time (to the nearest second) at whichthe solution turns blue. Repeat the first experiment twice to ensure accurate results. Repeat furtherexperiments as needed to ensure quality results. Record the data in Table 4 (Cut out Table 4 andput it in your lab notebook.).

5. Monitor and record the temperature of the reaction solution during each experiment.

Experiment Beaker 1 Beaker 2A1 10.0 mL KI 10.0 mL (NH4)2S2O8

A2 5.0 mL KI5.0 mL KCl

10.0 mL (NH4)2S2O8

A3 2.5 mL KI7.5 mL KCl

10.0 mL (NH4)2S2O8

A4 10.0 mL KI 5.0 mL (NH4)2S2O8

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5.0 mL (NH4)2SO4

A5 10.0 mL KI 2.5 mL (NH4)2S2O8

7.5 mL (NH4)2SO4

A6 5.0 mL KI5.0 mL KCl

5.0 mL (NH4)2S2O8

5.0 mL (NH4)2SO4Table 2: Experimental parameters for determining the reaction rate law.

B. Procedure 2. The Effect of Temperature on the Reaction Rate.

1. Repeat experiment A6 at a lower temperature, between 0–4°C. Prepare an ice water bath in acrystallizing dish and prepare all three necessary solutions. Cool all three solutions in the ice waterbath until all three are at 0–4°C. At t = 0, quickly take the solutions out of the ice bath, pour thesolutions together and place the Reaction Beaker back in the ice bath. This is Expt. B1.

2. Repeat experiment A6 at a higher temperature, above 40°C, using a hot water bath. This will beExpt. B2. Record the exact temperature at which you perform the experiment.

C. Procedure 3. The Effect of Copper Ion as a Catalyst.

Repeat experiment A6 except add the following amounts of copper sulfate solution using amicropipette.

Experiment Beaker 1 Beaker 2C1 5.0 mL KI

5.0 mL KCl5.0 mL (NH4)2S2O8

4.5 mL (NH4)2SO4

0.50 mL CuSO4

C2 5.0 mL KI5.0 mL KCl

5.0 mL (NH4)2S2O8

4.75 mL (NH4)2SO4

0.250 mL CuSO4

C3 5.0 mL KI5.0 mL KCl

5.0 mL (NH4)2S2O8

4.875 mL (NH4)2SO4

0.125 mL CuSO4Table 3: Experimental parameters for determining the effect of copper ion as a catalyst on the reaction rate law.

D. Extra Credit Procedure 4. Determine the Activation Energy with Copper as a Catalyst.

We’ll discuss how to do this in class. Do all necessary experiments and write down all data forthese experiments before you leave class.

IV. Formatting Your Lab Notebook

A. Experiment title, date, names of partners. Page numbers for each page.

B. Introduction: written in paragraph form1. Write out the overall reaction and all of the different kinetic values that will be determined inthis lab (for example, “m”, “n”, etc …).2. Describe in your own words how the reaction rate is determined using the monitoring reaction.

C. Procedure1. Handwritten procedure with spaces left open for each measurement and data table. Data tablesmay be cut out and taped into the lab notebook.2. For each experiment (A1-C3), average and standard deviation of the time of reaction. Roundeach average and standard deviation to the nearest second.

D. Experimental Summary: in notebook before you leave lab

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1. Apply the Q-test to any possible “outliers”. Were there any that could be discarded?2. How many times faster was the reaction with the catalyst? To show this take the ratio of thetime of reaction for experiment A6 divided by the time of reaction for experiment C1.

E. Results/Calculations: at home1. Calculations for “m”, “n”: Calculations to determine rate law without the catalyst. Report therate constant with its standard deviation.

For the exponent of I–, use experiments A1/A2, A1/A3 and A4/A6 to estimate theexponent three times and average the values you find. Then, choose the closest wholenumber. Also, do three estimates of the exponent of S2O8

2–.2. Calculations for “p” with the catalyst. Calculations to determine rate law with the catalyst.Report the rate constant with its standard deviation.3. Plot and calculations to determine activation energy without catalyst. You must prepare yourplot in such a way (using Excel) that the computer fits a best-fit line through the data and reportsthe slope for you to determine Ea.3. Plot and calculations to determine activation energy with catalyst.

F. Experimental Summary: after all calculations are complete 1. By how many kJ/mol does the activation energy change when the catalyst is added?2. Write a good reaction mechanism for reaction (1) without the catalyst. It will have two steps

V. Data Table for Notebook

Expt. [I–]0

M[S2O8

2–]0

M[CuSO4] 0

MTime of

Reaction, secInitial Rate ofreaction, M/s

A1 0

A2 0

A3 0

A4 0

A5 0

A6 0

B1Low T:

0

B2High T:

0

C1

C2

C3

Table 4: Data table for determining the rate law and activation energy. Cut out this table and put it in your lab notebook or recreate itin your notebook.