Iodine Clock Reaction Introduction

In this experiment you will determine the Rate Law for the following oxidation-reduction reaction:

2 H+ (aq) + 2 I

(aq) + H2O2(aq) I 2 (aq) + 2 H2O (l) (1)

The rate or speed of the reaction is dependent on the concentrations of iodide ion (I-) and hydrogen peroxide,

H2O2. (The spectator ions are left off the reaction.) Therefore, we can write the Rate Law (concentration

dependence) for the reaction as:

Rate = k [I]x [H2O2]

y (2)

Where: x is the order of the reaction in I-, y is the order of the reaction in H2O2, and k is the rate constant.

The temperature dependence of the rate is seen in k – that is, there is a separate value of k for each temperature

at which the reaction takes place. The temperature must therefore be held constant to accurately calculate x, y and k.

Since the Rate Law is empirical, we have to go to the lab to make measurements that will enable these values to be

calculated. The rate will be measured for the reaction near time = 0, so that few products been formed and there will

be no reverse reaction. The concentrations of iodide and hydrogen peroxide will be varied and the rates compared to

find each order (i.e., the exponents x and y). This is the Method of Initial Rates and it will be used to find x, y and k.

As with a lot of kinetics, the concentration of reactants or products at any instant is difficult to measure

directly, so in this lab the rate will be determined indirectly. We have a very handy test for the presence of one of the

products, iodine (I2), namely starch. Starch reacts with iodine to form a blue/black colored complex. Unfortunately as

soon as any iodine is produced it will react to make the complex and the solution will turn blue/black instantaneously.

Thus, using starch as an indicator by itself would not be of much help. It confirms that some amount of I2 is being

formed, but it tells us nothing about what we are trying to measure - the rate (how long it takes to produce a given

quantity of I2.)

To get around this problem we will introduce a side reaction that will remove the initial I2 that is produced by

our main reaction. This will prevent the solution from turning black for long enough so that we can make some good

time measurements. We will use the following side reaction:

I2 (aq) + 2 S2O3 2—

(aq) 2 I (aq) + S4O6 2—

(aq) (3)

S2O3 2—

, thiosulfate ion , reacts with I2 which prevents the solution from turning blue/black. How will this help?

Since we have carefully measured the amount of the thiosulfate (a small amount that will run out fairly quickly), we

know exactly how much iodine it will take to react with this thiosulfate. As soon as all of the thiosulfate has reacted,

iodine will then be left over and then the reaction will change color. By putting in this "time delay", we can now

calculate the rate at which I2 is being formed. The length of time required for the reaction to change color will be the

time that you will measure, and the amount of I2 produced in that time is the amount of I2 that is needed to react with

the thiosulfate. So now we have the rate of formation of I2 and depletion of thiosulfate, and can calculate the final

rate value we are looking for in this lab, which is the moles of hydrogen peroxide reacted per second.

Rate = moles H2O2 reacted/time in seconds (4)

1 mole H2O2 reacts to produce 1 mole I2 which reacts with 2 moles S2O3 2-

. (See reactions 1 and 3)

So 1 mole H2O2 = 2 mole S2O3 2-

. Since the moles of S2O3 2-

in solution is known (we can calculate it), we know that

the number of moles of H2O2 that react will be ½ of the moles of S2O3 2-

in the solution. The rate is then this number

of moles of H2O2 divided by the time. Calculate the moles of S2O3 2-

from the volume and molarity of Na2S2O3

(2.5 mL of 0.010 M). Don’t forget that molarity is moles/Liter. Use this to calcuate the moles of H2O2 reacted.

Include this number in the final data sheet for Parts I and II

Equipment

Three 125 or 250 mL Erlenmeyer flasks Three 100 or 150 mL beakers

10 ml and 5 ml Pipettes Thermometer

Stop watch (cell phone ?) Hot water bath (for Part II) Ice bath (for Part II)

One bin of chemicals per group that will contain:

0.050 M KI 0.050 M KCl 0.010 M Na2S2O3 0.050 M H2O2

1.0 M H2SO4 (Spill : B1) 1% starch solution

Disposal: All mixtures Spill/Disposal B1 (down the sink)

Procedure:

Part I

1. Clean and mostly dry three Erlenmeyer flasks.

2. Label them 1, 2 and 3.

3. Use fresh pipettes for each solution. Rinse pipette twice with the solution that you will be measuring and keep

this prepared pipette with the corresponding solution.

Add the following amounts of the solutions below to prepare each flask. The chemicals must be added in the

order listed (Left to Right)

Flask # 0.050 M KI 1% Starch 0.010 M Na2S2O3 1 M H2SO4 0.050 M KCl

1 15.0 mL 5.0 mL 2.5 mL 5.0 mL 0

2 15.0 mL 5.0 mL 2.5 mL 5.0 mL 0

3 7.5 mL 5.0 mL 2.5 mL 5.0 mL 7.5 ml

4. Rinse and mostly dry 3 beakers and label as 1, 2 and 3.

5. Prepare the following solutions in clean beakers:

Beaker # 0.050 M H2O2 Deionized H2O

1 15.0 mL 0

2 7.5 mL 7.5 mL

3 15.0 mL 0 6. Add the contents of beaker 1 to flask 1.

7. Start the stopwatch as soon as you mix the solutions. Swirl the flask to mix and note the time it takes for the color

to change. This is Run 1. Record the temperature.

8. Add the contents of beaker 2 to flask 2. Repeat procedure in step 7. This is Run 2.

9. Add the contents of beaker 3 to flask 3. Repeat procedure in step 7. This is Run 3.

Make sure that the shade of blue/black you observe is the same in each of the three runs.

10. The KCl solution and deionized water are added so that the ionic strength and final volume of the various

solutions (42.5 mL) is kept constant in each run.

11. Create a table so that the method of initial rates can be used. The initial concentrations of iodide and hydrogen

peroxide are not 0.050 M. We have to remember that when the solutions were mixed, they were diluted

Thus; (Volume 1)(Molarity 1) = (Volume 2 )(Molarity 2)

Run [ I—

] (initial) [H2O2] (initial)

1 (Flask 1 + beaker 1)

(15.0 mL)(0.050M) = (42.5 ml)(M2)

Solving for M2, we get [ I—

] (initial)= 0.018 M

2 (Flask 2 + beaker 2)

3 (Flask 3 + beaker 3)

12. Fill in the rest of the table as well as the table on the Pre-lab Part I, the Preliminary Data Sheet and the

final data sheet for Part I

Part II (Temperature Dependence)

1. Clean and mostly dry two 250 mL Erlenmeyer flasks. Prepare the solutions according to the following table

(Label the flasks as 1-2)

Flask # 0.050 M KI 1% Starch 0.010 M Na2S2O3 1 M H2SO4 0.050 M KCl

1 15.0 mL 5.0 mL 2.5 mL 5.0 mL 0

2 15.0 mL 5.0 mL 2.5 mL 5.0 mL 0

2. Clean and mostly dry two more 250 mL Erlenmeyer flasks. In each flask, take 15 ml of the 0.05 M H2O2. (Label

the flasks as 3 – 4.)

Flask # 0.050 M H2O2

3 15.0 mL

4 15.0 mL

3. Place Flasks 1 and 3 in an ice water bath. Let both of them cool until the temperature of both solutions is < 5°C.

(The final temperature should be about 10°C lower than your room temperature from Part I.) Record the exact

temperature, mix the contents of the two flasks and note the time it takes until the color changes.

4. Place Flasks 2 and 4 in a hot water bath. Let both flasks warm up until the temperature of the two solutions is

around 40°C. (The temperature should be about 10°C higher than your room temperature from Part I.) Record the

exact temperature, mix the contents and note the time it takes until the color changes.

5. For the room temperature run, use your data from trial 1 in part I of the experiment.

Disposal: All contents of the reaction flasks may be disposed of into the sink.

Calculations:

Part I

Review the method of initial rates in your text. You will need the values for the initial concentrations of each reactant

as well as the rate.

Hypothetical data for reaction: A + 2B → C + 2D Run [A]0 [B]0 Initial [C]/t (Molarity/sec)

1 0.150 M 0.150 M 8.00

2 0.150 M 0.300 M 16.0

3 0.300 M 0.150 M 32.0

Rate = k[A]x [B]

y. We need to calculate x, y and k

To get x, we have to hold [B] constant and just see how [A] affects the rate.

Take the ratio of Run 3/Run 1

Run 3 Initial [C]/ t = k[A]x [B]

y 32.0 M/sec = k(0.300M)

x (0.150M)

y

____ ___________ ________ __________ ___________________

Run 1 Initial [C]/ t = k[A]x [B]

y 8.0 M/sec= k(0.150M)

x (0.150M)

y

Simplifying: 4.0 = (0.300M) x / (0.150M)

x or 4.0 = (0.300/0.150)

x or 4.0 = 2

x Or x = 2

Do the same for B (holding [A] constant) and discover that y = 1.

You can calculate k as well. Using run 1:

8.00 M/sec = k (0.300M)1 (0.150M)

1 and on solving for k, we get k = 178 M

-1 s

-1 .

For your data:

Find x and y as explained above. Calculate k for each run and average the three values at the end. If your values for

x and y do not turn out to be whole numbers, round to the nearest whole number. Write the Rate Law using your

values for x, y and k:

Rate = k [I]x [H2O2]

y

Part II:

The Arrhenius Equation will be used to determine the Temperature dependence. Using x and y from Part I and your

rates from Part II, calculate k for each temperature in Part II. Using Excel, plot ln k versus 1/T . If the reaction

exhibits Arrhenius behavior, a straight line (y = mx + b) will be obtained where:

y = m x + b

ln k (Ea/R)(1/T) ln A

where Ea is the activation energy of the reaction, T is the temperature in Kelvin, k is the rate constant at this

Temperature, and R is the ideal gas constant (8.314J/mol K). Use Excel to add a best fit line and show the equation.

The slope (m) will equal -Ea/R. The activation energy can therefore be calculated from the slope for the best fit line

Please attach the graph to your lab report.

Iodine Clock reaction - Preliminary Data Sheet Name ______________________

Part I

Run Initial [I- ](M) Initial [H2O2](M) Reaction Time (secs) Temperature (

C)

1

2

3

Part II

Run Temperature (C) Temperature (K) Reaction Time (secs)

1 (cold)

2 (room temp)

3 (hot)

Partners Name _____________________________________________________

Iodine Clock reaction - Final Data Sheet for Part I

Name:_______________________________________________________

Moles H2O2 reacted: ______________________

Molarity of H2O2 reacted = moles of H2O2 reacted / 0.0425 L = ______ M

(Remember that the total volume of each solution was 42.5 ml.)

Temperature: Run 1 ______ Run 2 ________ Run 3 ________ Avg. T ________

Run Initial [ I- ] M Initial [H2O2] M Reaction

Time (in s)

Reaction Rate (in Ms-1

) = Molarity of H2O2 reacted ÷ reaction time in s

1

2

3

Show calculations for x and y and k (Include appropriate units)

x = ______ y = _______ Rate Law : __________________________

k (run 1) = __________ k (run 2) = _____________ k (run 3) = _________

Average k = __________

Partners Name _____________________________________________________

Iodine Clock reaction - Final Data Sheet for Part II

Name:________________________________________________________

Moles H2O2 reacted: ______________________

Molarity of H2O2 reacted = moles of H2O2 reacted / 0.0425 L = ______ M

Run Initial

[I- ] M

Initial

[H2O2] M

Reaction

Time (secs)

Reaction Rate (Ms-1

)=

(Molarity of H2O2 reacted

÷ reaction time in sec)

Rate constant ‘k’

(include proper units)

Cold

Room temp

Hot

Using Excel, Plot a graph of ln k vs 1/T. See if the reaction exhibits Arrhenius Behavior and if the equation:

ln k ( Ea/R) (1/T) ln A

can be used to fit the data. (y = m x + b)

Run Temperature

(in ˚C)

Temperature

( in Kelvins) Rate constant ‘k’ (include

proper units)

1/T in Kelvin ln (k)

Cold

Room temp

Hot

Plot ln (k) vs 1/T, insert a trendline, display the equation on the chart and use the equation for the line to

calculate the activation energy for the reaction. Attach your graph with the lab report.

Show your calculations for the activation energy below:

Ea = __________________________

Partner’s names: __________________________________

Iodine Clock reaction : Post Lab questions Name: _________________________________

1. Use your Rate Law from Part I to calculate what the rate would be if [I-] = 0.0125 M and [H2O2] = 0.010 M.

2. Using your Arrhenius Plot and the best fit equation, calculate the value for your rate constant at 50.0 °C.

3. If the value for the activation energy should have been 53 kJ/mole, using your calculated value for Ea, calculate

the percent error.

4. Suppose that while mixing the solutions, you accidentally spilled some. Would this have affected your

experiment? How?