Decomposition of Hydrogen Peroxide.pdf

Please see the attached file 

 

Note from professor 

(heres something from the prof.

it might help

as promised, i searched for and found the document i wrote to help you make excel plots for your Ea determination in the Kinetics lab…it’s posted to Canvas and attached here.***reminder….when you are determining orders and then using them to calculate the rate constants, don’t round the order values when you use them to do the rate constant calculations (use them to 2 or 3 decimal places from the solution of the equations that look like rate x / rate y = 2^a or 2^b). your rate constants will come out much better this way.in order to find units for k you will have to round to the nearest integer and this is ok.)

Experiment 2 Decomposition

of Hydrogen Peroxide
The decomposition of hydrogen peroxide in aqueous solution proceeds very slowly. A bottle of
3% hydrogen peroxide sitting on a grocery store shelf is stable for a long period of time. The
decomposition takes place according to the reaction below.

2 H2O2(aq) → 2 H2O + O2(g)

A number of catalysts can be used to speed up this reaction, including potassium iodide,
manganese (IV) oxide, and the enzyme catalase. If you conduct the catalyzed decomposition of
hydrogen peroxide in a closed vessel, you will be able to determine the reaction rate as a function
of the pressure increase in the vessel that is caused by the production of oxygen gas. If you vary
the initial molar concentration of the H2O2 solution, the rate law for the reaction can also be
determined. Finally, by conducting the reaction at different temperatures, the activation energy,
Ea, can be calculated.

OBJECTIVES
In this experiment, you will

· Conduct the catalyzed decomposition of hydrogen peroxide under various conditions.
· Calculate the average rate constant for the reaction at room temperature.
· Determine the rate law expression for the reaction.
· Calculate the activation energy for the reaction.

The rate law for this reaction can be determined using the observed rates of reactions from a
series of different experiments. The concentration of one reactant is held constant between two
different experiments, acting as the control, while the concentration of the second reactant is
different between the two experiments. The rate of reaction is measured in each experiment so
the impact of changing the concentration of the second reactant can be determined. The order of
the reaction with respect to each reactant is determined in this fashion and once the order of each
reactant is know the rate law can then be written.

(A) Sample Exercise for Determining Reaction Order

Consider the following reaction: (CH3)3CBr(aq) + OH-(aq) → (CH3)3COH(aq) + Br-(aq)
A series of experiments is carried out with the following results:

Exp 1 Exp 2 Exp 3 Exp 4 Exp 5
[(CH3)3CBr] 0.50 1.0 1.5 1.0 1.0
[OH-] 0.050 0.050 0.050 0.10 0.20
Rate (M/s) 0.0050 0.010 0.015 0.010 0.040

Find the order of the reaction with respect to both (CH3)3CBr and OH-.

Advanced Chemistry with Vernier 1

2 Advanced Chemistry with Vernier

To find the order of the reaction with respect to (CH3)3CBr, choose two experiments, 1 and 3 for
example, where [OH-] is constant. A similar approach can be used to find the order of the
reaction with respect to OH-, comparing experiments 2 and 5, where [(CH3)3CBr]

(1) Order with Respect to (CH3)3CBr:

Rate exp 3 = k([(CH3)3CBr]exp3)m ([OH-]exp3)n
Rate exp 1 k([(CH3)3CBr]exp1)m ([OH-]exp1)n

0.015 M/s = k([1.5 M])m (0.050 M)n simplifies to: 3.0 = (3.0)m
0.0050 M/s k([0.50 M])m (0.050 M)n

Using the natural log applied to both sides and solving for “m”

Ln 3.0 = (Ln 3.0) (m) (Ln 3.0) / (Ln 3.0) = 1.10 / 1.10 = 1 = m

Since “m” = 1 the reaction is first order with respect to (CH3)3CBr.

(2) Order with Respect to [OH-]:

Rate exp 5 = k([(CH3)3CBr]exp5)m ([OH-]exp5)n
Rate exp 4 k([(CH3)3CBr]exp2)m ([OH-]exp2)n

0.040 M/s = k([1.0 M])m (0.20 M)n simplifies to: 4.0 = (2.0)n
0.010 M/s k([1.0 M])m (0.10 M)n

Using the natural log applied to both sides and solving for “m”

Ln 4.0 = (Ln 2.0) (m) (Ln 4.0) / (Ln 2.0) = 1.39 / 0.69 = 2.01 = n

Since “n” = 2 the reaction is second order with respect to (CH3)3CBr.

Now that the order of the reaction for each reactant has been determined the rate law can be
written for this equation:

Rate = k [(CH3)3CBr] [OH-]2

(B) Sample Exercise for Determining Molarity of a Diluted Solution

When two solutions are mixed in an experiment the total volume of the solutions is increased and
the concentration of each solution is diluted. Recall:

Moles solute before dilution = moles solute after dilution

so; (Molarity)conc x (Volume) conc = (Molarity)dil x (Volume)dil

The Decomposition of Hydrogen Peroxide

Advanced Chemistry with Vernier 3

What is the concentration of each compound when 6.0 mL of a 0.60 M solution of H2O2 is mixed
with 2.0 mL of a 0.25 M potassium iodide solution?

(1) First consider the total volume of the solution after mixing:

6.0 mL H2O2 + 2.0 mL KI = 8.0 mL total volume of diluted solution

(2) Determine the concentration of H2O2 after mixing:

[0.75 M]conc x (6.0 mL) conc = [H2O2]dil x (8.0 mL)dil

(0.60 M)(6.0 mL) / (8.0 mL) = 0.45 M = [H2O2]dil

(3) Determine the concentration of KI after mixing:

[0.25 M]conc x (2.0 mL) conc = [H2O2]dil x (8.0 mL)dil

(0.40 M)(2.0 mL) / (8.0 mL) = 0.10 M = [KI]dil

The concentration of the H2O2 decreased from 0.60 M to 0.45 M and the concentration of KI
decreased from 0.25 M to 0.10 M after the two solutions are mixed. The solutions will be
diluted as the reactions begin so be sure to use the diluted concentrations of each reactant when
determining the reaction orders or rate constants.

(C) Sample Exercise for Converting Pressure Rate Data to Molarity Units

The data rate recorded in this experiment is in kilopascals per second (kPa/s). Since pressure is a
unit of concentration in the gas phase it can be used to determine rate when the change of
pressure is measured over the change in time. However, in this experiment the reactants are in
the aqueous phase and the concentration unit for the solutions is in molarity (M). kPa/s can be
converted to M/s by using a derivation from the ideal gas law:

Molarity = P / RT

In which P = kPa/s, R = 8.314 L * kPa / K*mol (one of ideal gas constants), and T = temperature
on Kelvin.

In one experiment the rate of O2 production was 0.22 kPa/s when the temperature of the water
bath was 19.0 oC. What is this rate in M/s?

Molarity = (0.22 kPa/s) / (8.314 L*kPa/K*mol)(292.0 K) = (0.22 kPa / 2.43 x 103 L*kPa / mol)

Molarity = 9.05 x 10-5 M/s

The units kPa and Kelvin cancel leaving mol/L*s or M/s.

(D) Sample Exercise for Converting Percent Solutions (mass/volume) to Molarity Units

Assumed 100 mL of total volume of solution in a percent (mass/volume) solution and the percent
value will represent the grams of solute dissolved in water to make a total volume of 100 mL.
For example a 10% NaCl solution will have 10 g of NaCl dissolved in 100 mL of solution.

What is the molarity of this 10% NaCl solution?

4 Advanced Chemistry with Vernier

Molarity = moles of solute / liters of solution

If you assume 10 g NaCl and 100 mL of solution and the molar mass of NaCl is 58.5 g / mol
then:

(1) Moles NaCl = (10 g NaCl)(1 mole NaCl / 58.5 g NaCl) = 0.17 moles NaCl

(2) Liters solution = (100 mL solution)(1 L solution / 100 mL solution) = 0.100 L solution

(3) Molarity = (0.17 moles NaCl / 0.100 L solution) = 1.7 M

Diagram of the equipment set for this experiment:

Figure 1

MATERIALS
Vernier computer interface 3% hydrogen peroxide, H2O2, solution
distilled water 0.5 M potassium iodide, KI, solution
Vernier Gas Pressure Sensor 18 × 150 mm test tube
thermometer or Temperature Probe two 10 mL graduated cylinders
one-hole rubber stopper with stem graduated plastic Beral pipet
tubing with two Luer-lock connectors 1 liter beaker
solid rubber stopper (#1) ~800 mL room temperature water

Procedure
Part I Decompose 3% H2O2 solution with 0.5 M KI solution at ~20°C.

The Decomposition of Hydrogen Peroxide

Advanced Chemistry with Vernier 5

1. Obtain and wear goggles.

2. Prepare the reagents for temperature equilibration.
a. Obtain room temperature water to set up a water bath to completely immerse the test tube.

Use a thermometer or a Temperature Probe to measure the temperature of the bath. Record
this temperature in your data table for Parts I-III; presume that the water bath temperature
remains constant throughout.

b. Measure out 4 mL of 3% H2O2 solution into the test tube. Seal the test tube with the solid
rubber stopper and place the test tube in the water bath.

c. Measure out 2 mL of 0.5 M KI solution in a graduated cylinder. Draw about 1 mL of the
KI solution into a graduated Beral pipet. Invert the pipet and immerse the reservoir end of
the pipet in the water bath.

3. Connect a Gas Pressure Sensor to Channel 1 of the Vernier computer interface. Connect the
interface to the computer using the proper cable.

4. Use the plastic tubing to connect the one-hole rubber stopper to the Gas Pressure Sensor, as
shown in Figure 1. About one-half turn of the fittings will secure the tubing tightly.

5. Start the Logger Pro program on your computer. Open the file “12 Peroxide” from the
Advanced Chemistry with Vernier folder.

6. Prepare to run the reaction and collect pressure data.
d. Remove the test tube from the water bath and remove the solid stopper.
e. Remove the plastic Beral pipet from the water bath and quickly transfer 1 mL of KI

solution into the test tube. Tap or lightly shake the test tube to mix the reagents.
f. Seal the test tube with the one-hole stopper connected to the Gas Pressure Sensor.
g. Place the test tube back in the water bath.

7. Click to begin data collection. Data will be gathered for three minutes. If necessary,
gently hold the test tube so that it stays completely immersed in the water bath.

8. When the data collection is complete, carefully remove the stopper from the test tube to
relieve the pressure. Dispose of the contents of the test tube as directed.

9. Examine the graph of Part I. Select a linear region just beyond the initial flat portion of the
graph that covers one minute of the reaction. Click the Linear Regression button, , to
calculate the best-fit line equation. Record the slope as the initial rate of the reaction in your
data table. Store the results from the first trial by choosing Store Latest Run from the
Experiment menu.

10. Rinse and clean the test tube for the second trial.

Part II Decompose 3% H2O2 solution with 0.25 M KI solution at ~20°C

11. Measure out 4 mL of 3% H2O2 solution into the test tube. Seal the test tube with the solid
rubber stopper and place the test tube in the water bath.

12. Add 1 mL of distilled water to the remaining 1 mL of KI solution in the graduated cylinder.
Swirl the mixture gently to mix the solution.

13. Draw 1 mL of the KI solution into a plastic Beral pipet. Invert the pipet and immerse the
reservoir end of the pipet in the water bath. Allow both the test tube and the Beral pipet to
remain in the water bath for at least two minutes before proceeding.

14. Repeat Steps 6–10 to complete Part II. Remember to store the data.

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Part III Decompose 1.5% H2O2 solution with 0.5 M KI solution at ~20°C

15. Prepare a 1.5% H2O2 solution by mixing 2 mL of distilled water with 2 mL of 3% H2O2
solution. Transfer the resulting 4 mL of the 1.5% H2O2 solution to the test tube, seal the test
tube with the solid stopper, and place the test tube in the water bath.

16. Rinse and clean the graduated cylinder that you have used for the KI solution. Add a fresh
2 mL of 0.5 M KI solution to the graduated cylinder.

17. Draw 1 mL of the KI solution into a plastic Beral pipet. Invert the pipet and immerse the
reservoir end of the pipet in the water bath. Allow both the test tube and the Beral pipet to
remain in the water bath for at least two minutes before proceeding.

18. Repeat Steps 6–10 to complete Part III, and store the data.

Part IV Decompose 3% H2O2 solution with 0.5 M KI solution at ~30°C

19. Conduct Part IV identically to the procedure in Part I, with one exception: set the water bath
at 30°C.

Part V Decompose 3% H2O2 solution with 0.5 M KI solution at 5-10°C

20. Conduct Part V identically to the procedure in Part I, with one exception: use ice to cool the
water bath to the 5-10 °C range – record the temperature of the bath using a thermometer.

PRE-LAB EXERCISE – to be completed before coming to class
The hydrogen peroxide solution that you are using in this experiment is labeled as a 3% solution,
mass/volume. However, in order to complete the calculations, the concentration must be in
molarity. Calculate the molarity of a 3% mass/volume H2O2 solution (Part I, II, and IV) and a
1.5% mass/volume H2O2 solution (Part III) and record these values in the table below. This table
should be completed in your lab notebook as part of the pre-lab assignment for this experiment.

Part Volume H2O2
(mL)

[H2O2]
before mixing

Volume K

I

(mL)

[KI]
before mixing

I 4 1 0.50 M
II 4 1 0.25 M
III 4 1 0.50 M
IV 4 1 0.50 M
V 4 1 0.50 M

Show the calculations used in determining the [H2O2] in the carbonless notebook as part of your
pre-lab assignment.

The Decomposition of Hydrogen Peroxide

Advanced Chemistry with Vernier 7

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.88M

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.88M

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.44M

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.88M

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.88M

DATA TABLE
Type up the following two tables and include them in your formal lab report.

Part Reactants Temperature
(°C)

Initial rate (kPa/s)

I 4 mL 3.0% H2O2 + 1 mL 0.5 M KI

II 4 mL 3.0% H2O2 + 1 mL 0.25 M KI

III 4 mL 1.5% H2O2 + 1 mL 0.5 M KI

I

V 4 mL 3.0% H2O2 + 1 mL 0.5 M KI

V 4 mL 3.0% H2O2 + 1 mL 0.5 M KI

DATA ANALYSIS

Part
Initial rate
(mol/L-s)

[H2O2]
after mixing

[I–]
after mixing

Rate constant
k

I

II

III

I

V

V

Show the calculations used to answer questions 1 – 4 below after the data section of your formal
lab report and explain the chemical principles behind each calculation in your discussion section.
You should also explain the chemistry behind your answer to question 5 and define the two
different roles described in 5(a) and 5(b).

1. What is the reaction order with respect to hydrogen peroxide?

2. What is the reaction order with respect to potassium iodide?

8 Advanced Chemistry with Vernier

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21.9

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21.9

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21.9

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30.4

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10.0

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.20775

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.09819

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.1007

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.4012

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.08099

3. Write the rate law expression for the catalyzed decomposition of hydrogen peroxide and
explain how you determined the order of the reaction in H2O2 and KI.

2. Calculate the rate constant, k, for each part of this experiment. Calculate an average rate
constant at ~room temperature. Do any of the parts of the experiment have rate constants
that are equal or almost equal?

3. Compare the rate of the reaction in part 1 at room temperature with the rate of the reaction in
part 4 at approximately 10oC above room temperature. Approximately how much does rate
increase with a 10oC increase in temperature? Use your values to explain your answer to this
question.

4. Use the Arrhenius equation to determine the activation energy, Ea, for this reaction.
Temperature needs to be expressed in Kelvin when using the Arrhenius equation. This
determination requires analysis of an Arrhenius plot (ln k vs 1/T) as discussed in class and
detailed in your textbook. Use a graphing program to complete the graph and have the program
determine the best straight-line fit to the data (it is not acceptable to do this graph by hand).
From the equation of the line, you can calculate Ea. Show all calculations. Print your graph and
attach it to your lab report when you turn it in…no lab is complete without this plot.

5. The following mechanism has been proposed for this reaction:

H2O2(aq) + I–(aq)→ IO–(aq) + H2O(aq)

H2O2(aq) + IO–(aq) → I–(aq) + H2O(l) + O2(g)

If this mechanism is correct, which step must be the rate-determining step? Please explain your
choice.

(a) What is IO-(aq) called in this reaction? (What type of role the ion is playing in the reaction not
its actual ionic name, hypoiodite.)

(b) What is I-(aq) called in this reaction? (What type of role the ion is playing in the reaction not
its actual ionic name, iodide.)

The Decomposition of Hydrogen Peroxide

Advanced Chemistry with Vernier 9

Sora
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Note : Ea should be in J/mol or KJ/mol
convert ‘C to ‘Kelvin

Discussion Section
The discussion section should relate the results of this experiment to the impacts of concentration
and temperature to the rate of a chemical reaction. For example your discussion should include
an explanation of why or how changing the concentration of a reactant impacts the rate of a
reaction. You should also include why or how changing the temperature of the solutions impacts
the rate of the chemical reactions observed. Cite your data/results as evidence when you discuss
the impact of concentration (and temperature) changes on the rate of the reaction.

Since any experiment done in lab involves procedures and measurements the discussion section
should also include analysis of errors associated with these. You should include a paragraph or
two describing the errors you may have made in this experiment (any procedural errors) and the
errors inherent to any measurements (measurement errors) made in the experiment. Explain how
each type of error may have influenced the result of the experiment. Since graphical decisions
were crucial in this experiment (and can also contribute significantly to the error) be sure to
explain how you determined the rate of each reaction (why you chose a certain region of the
graph), why it is important to be consistent in this process, and the role this decision plays in the
confidence you have in your results. You are welcome to include a small drawing to aid in your
discussion if it helps to make an idea more clear.

Conclusion
Write a paragraph with a brief conclusion summarizing the results of your experiment and
describing what you learned in the process of doing this experiment. Be sure to include the
results for the quantities you determined (average rate constant at room temperature, the rate law,
and the activation energy) and indicate the largest source of error in your experiment.

10 Advanced Chemistry with Vernier

Sheet1

k

						this data is from problem 14.95 in the 11th edition textboo	k
T(deg)	T(K)	1/T (K^-1)	ln k
300	573.15	0.001744744	-24.1652852131	3.20E-11
320	593.15	0.0016859142	-20.7232658369	1.00E-09
340	613.15	0.0016309223	-17.3220684553	3.00E-08
355	628.15	0.0015919764	-15.2426269136	2.40E-07
how to plot in Excel to get the info needed to calculate the activation energy
1. make the table above with your data – you will have to input your rate constants as the k values and your T values in degrees C
– write the formulas as shown in the cells – if you click on the box, the formula will display
– names for cells are ‘column row’ so this cell you are now reading would be alled A12
2. select cells that are light blue and click on the chart button (excel wants data in two columns, left column is x and right column is y)
3. select XY scatter, click next twice to get the the options box
4. fill in chart options with title inluding your name, x axis name as 1/T (K^-1), y axix name as ln(k)
5. click next until this box closes and you see your plot like the one above (exept with your data)
6. left click on a data point to select the data series, then right click on the data point to get the menu
7. choose ‘add trendline’ (type is linear, under options click to add the trendline and R squared correlation to the chart)
8. now use your trendline to calculate the activation energy from the slope of the line

Sheet1

1/T (K^-1)
ln (k)
carney’s graph

Sheet2

Sheet3