The Study of Chemical Kinetics: Decomposition of Hydrogen Peroxide
Meena Bansal
Ethan Wolf
Maia Popova
Yasmin Jessa
CHM 145 Section A
March 3, 2015
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Abstract
The objectives of this experiment was to determine the kinetic order of a reaction
based of the rate on reactant concentrations, to understand the importance of the rate
limiting step of the reaction and understand that it was determined by the kinetic order of
the reaction, and to determine the energy of activation (Ea), the minimum energy required
to cause a chemical reaction. Solutions of H2O2 and KI were prepared. Four different trials
were needed to determine the kinetic order for each reactant by changing one condition at
a time. Three additional trials were conducted at different temperatures to determine Ea.
The results indicated that H2O2 had a kinetic order of 1 and that KI had a kinetic order of 1,
and the energy of activation amounted to -58422J or -58.422kJ. KI was determined to be a
catalyst that sped up the rate of the reaction. Reaction mechanism, a sequence of single
steps that are possible in a reaction were looked at for the decomposition of H2O2 and the
chemical components in the rate-limiting step of a reaction were studied.
Introduction
Chemistry investigates properties of substances and the ways in which they interact,
combine and change to form products. Rate is referred to how quickly a chemical reaction
occurs. Some reactions can take just a few seconds to complete, while some can take up to
thousands to millions of years. Chemical kinetics studies the speeds, or rates, of reactions.
Reaction rate is expressed as the change in the concentration of a substance divided by the
time interval per unit of time. Hydrogen peroxide, H2O2 is a stable chemical compound, yet
does decompose slowly over time. Hydrogen peroxide is an important oxidizing agent in
manufacturing bleach, textile, leather and even used in sewage treatment facilities in order
to destroy bacteria.
2H2O2(aq)  2H2O(l) + O2(g)
Catalysts are substances that increase the speed of a chemical reaction without itself
undergoing a net change. The decomposition of H2O2 produces O2. Measurement of the
change in pressure allows the kinetics of the reaction to be monitored. A plot of pressure
vs. time will be made to measure the initial rates of reactions. The slope of the plot will
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measure the rate of the reaction. The rate law is the core of a kinetic study, which shows
how rate depends on concentration. In this experiment, the following rate law was given:
Rate = k[H2O2]a[KI]b
The rate constant is known as k, which determines how temperature affects the rate of the
reaction. Exponents “a” and “b” are derived experimentally and usually have a value of 0, 1,
or 2.
During this experiment the reaction order will be determined with respect to H2O2
and KI in a series of experiments and the exponents “a” and “b” will be determined
graphically by observing the plot of pressure vs. time. The initial rate measurements will
have units of pressure/time, (kPa/s) and will be converted to concentration/time (M/s)
with the use of the ideal gas law.
PV = nRT and M = n/V
M = P/(RT)
As temperature increases, the rate of most chemical reactions increases. The Arrhenius
equation gives the dependence of the rate constant on the absolute temperature.
k = Ae-Ea/RT
Activation energy (Ea), is the minimum quantity of energy needed to undergo a reaction. A
(frequency factor) is a constant term that includes factors relating to the frequency of
collisions.
The following equation is found when the natural logarithm (ln) of both sides is taken:
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ln k = -(Ea/RT) + lnA
This equation takes the form of the equation for a straight line (y=mx+b). A plot will be
made with lnk on the y-axis versus 1/T on the x-axis. The plot will consist of a linear line
with a slope equal to –Ea/R and a y-intercept equal to ln A. With four trials conducted with
differing temperatures measuring the rate constant, the activation energy (Ea) can be
found.
The following two-point form is used when the rate constants are known at two different
temperatures:
ln(
𝑘2
𝐸𝑎 1
1
) = −( )( − )
𝑘1
𝑅 𝑇2 𝑇1
Methods
Preparation of H2O2 and KI
Approximately 10 mL of 3% of 0.88 M of H2O2 were placed into a clean, labeled and
dry test tube and placed in a large beaker with room temperature water of about 293K. A
thermometer was inserted to keep track of the temperature of the water bath. 50 mL of
0.50M aqueous solution of KI was prepared and transferred into a clean, labeled and dry
Erlenmeyer flask. Approximately 10 mL of the KI solution was transferred in a large test
tube and placed in the water bath along with the H2O2 solution. (Jessa, 2015).
Determination of the Rate Law Procedure
A temperature probe replaced the thermometer and a 25 mL Erlenmeyer flask was
clamped in the room temperature water bath. 2 mL of the incubated H2O2 solution was
transferred into the 25 mL Erlenmeyer flask with a disposable pipet. 1 mL of the incubated
0.5 M KI solution was added to the same Erlenmeyer flask using a disposable pipet. A
vernier LabQuest was set up. A table was made and drawn in lab notebooks to keep track of
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summary of the results for all trials. The equation below was used in order to the find the
final concentration after mixing of H2O2 and KI. (Jessa, 2015).
M1V2 = M1V2
Trial
Number
Vol. H202
(mL)
Original
Conc.
H202 (M)
Vol. KI
(mL)
Initial
Conc. KI
(M)
Temp
(K)
Final
Conc. KI
(M) after
mixing
293
Final
Conc.
H202 (M)
after
mixing
0.59
1
2.0
0.88
1.0
0.50
2
1.5
0.66
1.0
0.50
293
0.44
.17
3
1.0
0.44
1.0
0.50
293
0.29
.17
4
0.5
0.22
1.0
0.50
293
0.15
.17
5
2.0
0.88
0.5
0.25
293
0.59
.083
6
2.0
0.88
0.8
0.40
293
0.59
.32
7
2.0
0.88
0.6
0.30
293
0.59
.18
8
2.0
0.88
1.0
0.50
280
0.59
.17
9
2.0
0.88
1.0
0.50
273
0.59
.17
10
2.0
0.88
1.0
0.50
303
0.59
.17
.17
Calculations of the Rate Constant, k for trials 1-10
The rate constant k, was calculated for each trial from the observed rate (M/s).
𝑘=
rate (𝑀/𝑠)
[H2 O2 ]𝑎 {KI]𝑏
Results
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Figure 1. A plot of the pressure (kPa) of the reactant H2O2 as a function of Time (s).
Figure 1 depicts the pressure (kPa) versus time(s) graph for the reactant H2O2 with an
original concentration of 0.88M. The linear fit was taken from data attained in the time
range of 181 s to 500 s. The slope was determined by linear least-squares analysis and
represents the rate of the reaction.
Trial #
1
2
3
4
5
6
7
8
9
10
Table 1. Data Analysis for Reagents H2O2 and KI
Original
Initial Conc. Average
Reaction
Conc. H2O2
KI (M)
Temperature Rate (kPa/s)
(M)
(K)
0.88
0.50
296
.01263
0.66
0.50
295
.08269
0.44
0.50
295
.07760
0.22
0.50
296
.01263
0.88
0.25
295
.002873
0.88
0.40
295
.08270
0.88
0.30
295
.07750
0.88
0.50
280
.02050
0.88
0.50
273
.01012
0.88
0.50
303
.1146
Reaction
Rate (M/s)
5.13E-6
3.37E-5
3.16E-5
5.13E-6
1.17E-6
3.37E-5
3.16E-5
8.81E-6
4.46E-6
4.55E-5
Table 1 shows the original concentrations (M) of H2O2 and KI, the average temperature, and
the slope in both kPa/s and M/s for all 10 trials.
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Figure 2. A plot of the rate (M/s) of the reagent H2O2 as a function of Concentration (M).
Figure 2 depicts the rate (M/s) versus concentration (M) of the reagent H2O2 .with a kinetic
order of 1. The rate law for the reagent H2O2, can be denoted as rate = k[H2O2]. A plot of a
reagent with a kinetic order of 1 would be linear. As observed in this plot, the line is
relatively linear, making the reagent H2O2 have a kinetic order of 1.
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Figure 3. A plot of the rate (M/s) of the reagent KI as a function of Concentration (M).
Figure 3 depicts the rate (M/s) versus concentration (M) of the reagent KI with a kinetic
order of 1. The rate law for the reagent KI can be denoted as rate = k[KI]. A plot of a reagent
with a kinetic order of 1 would be linear. As observed in this plot, the line is relatively
linear, making the reagent KI have a kinetic order of 1.
The rate law that applies to the data presented is: Rate = k[H2O2][KI]
Trial #
1 (H2O2)
2 (H2O2)
3 (H2O2)
4 (H2O2)
5 (KI)
6 (KI)
7 (KI)
Average
Table 2. Values of Rate Constant (k) for Trials 1-7
Rate Constant
5.1E-4
4.5E-4
6.4E-4
2.0E-6
2.4E-5
6.4E-6
3.0E-4
2.8E-4 ±2.7E-4
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Table 2 shows rate constant values for the 7 trials that were conducted at room
temperature and the average of the rate constants. The standard deviation is reported
along with the average to show variation among the trials.
Trial #
Table 3. Values of the Rate Constant (k) for Trials 8-10
Rate Constant
8 (280K)
8.8E-5
9 (273K)
4.5E-5
10 (303K)
4.5E-4
Table 3 shows the rate constant values for the 3 temperature trials. In these trials, the
temperatures were varied of ~10 K.
Figure 4. A plot of the inverse of the temperature (1/T) vs. the natural log of the rate
constant (ln k).
Figure 4 represents a plot of the inverse of temperature (1/T) vs. the natural log (ln k) of
the rate constant from the four trials that were performed under similar conditions at
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different temperatures. A linear fit was taken of the data. The slope of this line is equal to –
Ea/R.
Discussion
Hydrogen Peroxide (H2O2) in aqueous solution decomposes very slowly under
ordinary conditions, and occurs by a one-step collision between reactants A and B.
A+BC
Therefore, an uncatalyzed reaction has a large activation energy value and a relatively low
rate. The equation for the decomposition of H2O2 is the following:
2H2O2 (aq)  2H2O(l) + O2(g)
Potassium iodide (KI) is a catalyst that lowers the activation energy, which increases the
rate constant, thus increasing the rate of the reaction. In the catalyzed reaction, reactant A
would collide with the catalyst to form the intermediate C in step one of the reaction.
A + catalyst  C
Then, the intermediate C would react with the reactant B in the second step of the reaction
to form a product and to regenerate the catalyst.
C +B  product + catalyst
Conducting a catalyzed decomposition of H2O2 enables the determination of the reaction
rate based on the pressure increasing from the production of O2.
There are two elementary steps in the decomposition of H2O2:
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H2O2 (aq) + I-(aq)  IO-(aq) + H2O-(l)
H2O2 (aq) + IO- (aq)  I-(aq) + H2O-(l) + O2 (g)
Solid KI is dissolved in an aqueous solution of H2O2. O2 gas is produced and I- catalyzes the
decomposition of H2O2. I- is the catalyst and IO- is the intermediate. The first reaction
determines the rate of the reaction, which shows the decomposition of H2O2 slower than
the second reaction. I- is reactant and a product; therefore IO- is never consumed and or
interferes with the reaction. IO- acts a catalyst, speeding up the reaction. A catalyst such as
KI has the ability to significantly speed up the rate of the reaction without interacting with
the reaction itself.
A balanced equation does not tell us how a reaction is preceded. A reaction
mechanism is a series of equations that sums up how the reaction actually undergoes. Each
equation or reaction in a reaction mechanism has different rates. However, there is one
reaction in the series of reactions that is slower than the other reactions. This is the ratelimiting step. This is the one-step that limits the speed of the overall reaction.
The first step is known as the slow step, which is the rate determining equation. The rate
law for the slow reaction is- Rate1 = k1[H2O2][I-].
The second step is known as the fast step. The rate law for the fast reaction is- Rate2 =
k2[H2O2][IO-].
The reaction rate is consistent with my results because the reaction rate was determined
experimentally to have an order of 1 for both H2O2 and KI. Because the coefficients of both
reagents in step one are 1, the order of the proposed rate law of this mechanism matches
the order of the experimentally determined rate law.
Rate = k[H2O2][KI]
Conclusion
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As a result of performing this experiment, I learned a couple of things. I learned
how to apply the rate mechanics in order to formulate the rate law. I had used rate laws in
high school, but I did not understand the chemistry behind it. However, during the course
of this experiment I learned that this is the study of chemical kinetics, and that a reaction
rate is a change in the concentration of reactants or products per unit of time. Furthermore,
I knew what a catalyst did but I never fully put terms such as catalysts, activation energy
and rate constant together to get the full understanding of what actually happens when a
catalyst is added to the reagents of the reaction. Important experimental results I found
were kinetic orders of the two reagents: KI and H2O2, the rate constants, and the activation
energy. These ultimately lead to the finding of the rate law. A skill I learned during the
course of the two weeks of this experiment was the manipulation of variables. I worked on
using the Pv = nRT equation to switch the units from kPa/s to M/s. It was challenging at
first, but I was able to switch the units in the end by working with the units. Finally, I
learned the importance of reading and following the procedure to ensure the best results.
References
Jessa, Y.; Novak, M. General Chemistry Laboratory Manual; Hayden-McNeil Publishing:
Plymouth, MI, 2015; p. 38-47.
Silberberg, Martin S., and Patricia Amateis. "16.7 Catalysis: Speeding up a Reaction."
Chemistry: The Molecular Nature of Matter and Change. 7th ed. New York: McGrawHill Education, 2015. 712-13. Print
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