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The effect of changing temperature on the rate of decomposition of hydrogen peroxide
with a catalyst.
Exploration
Reason for interest
In my lessons on chemical kinetics, I learned about the importance of catalysts in many chemical industrial processes,
as they lead to faster and more energy-efficient reactions without being used up. This means that the use of catalysts
underpins the success of major world economies and may be key to the future of achieving more sustainable energy
and food supplies. This sparked my curiosity, so I wanted to investigate the potential of catalysts, especially as
catalysts are likely to be crucial to facing challenges such as decreasing non-renewable resources and increasing
global consumption. Thus, my investigation will explore how temperature affects a catalysed reaction, and this will
allow me to determine the effect of the catalyst, as I can work out the activation energy of the catalysed reaction and
compare this with the literature value of the uncatalysed reaction.
Research question
How does changing the temperature affect the rate of decomposition of hydrogen peroxide with the catalyst
manganese (IV) oxide in the reaction 2H2O2 (aq) → 2H2O (l) + O2 (g), as shown by the time taken to produce 50 cm3
of oxygen gas? And hence, what is the activation energy?
Background information
Theory
According to collision theory, particles must collide with each other with enough energy and at the right orientation to
react1. They must collide with energy equal to or greater than the activation energy, which is the minimum amount of
energy colliding particles must have before they can react. This is required to overcome internuclear and electron
repulsions so that bonds can be broken, and atoms can be rearranged.
The mathematical equation – Arrhenius’ equation: k = Ae-Ea/RT – shows how the rate constant increases exponentially
with temperature, since the value of - Ea/RT becomes less negative as temperature increases, causing the value of the
rate constant, k, to increase2. Thus, this means the rate of reaction increases exponentially with temperature, as the rate
constant is directly proportional to the rate of reaction through the rate equation: rate = k[A]m[B]n.
One reason for this relationship is that when the temperature increases, particles will have more kinetic energy, thus
move faster, since the average kinetic energy is proportional to the absolute temperature (in Kelvin)3. This means the
frequency of collisions will increase, causing a small increase in the rate of reaction. The main reason for the effect of
increasing temperature on rate can be explained using the Maxwell-
Boltzmann distribution in Fig 1, which depicts the distribution of
kinetic energy among particles in a sample of gas4. Only particles to the
right of the line (indicating activation energy) will have sufficient
energy to react successfully. As temperature increases, the distribution
curve flattens out, so there is a higher proportion of particles that have
kinetic energy equal to or greater than the activation energy. This means
there is a greater chance that a collision will result in a reaction, thus the
frequency of successful collisions increases, and the rate of reaction
increases5. Fig 1: Maxwell-Boltzmann distribution
at different temperatures6
As hydrogen peroxide is a relatively stable substance, its rate of decomposition is very slow at moderate temperatures.
Thus, often a catalyst is added to increase the rate of decomposition. A catalyst is a substance that increases the rate of
reaction by providing an alternative reaction pathway which has a lower activation energy, without being chemically
1
J. Clark, ‘6.1.7: The Collision Theory’, Chemistry LibreTexts, 2013
<https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/
Kinetics/06%3A_Modeling_Reaction_Kinetics/6.01%3A_Collision_Theory/6.1.7%3A_The_Collision_Theory> [accessed 6 April 2023].
2
V. H. Carvalho-Silva, N. D. Coutinho and V. Aquilanti, ‘Temperature Dependence of Rate Processes beyond Arrhenius and Eyring: Activation and Transitivity’,
Frontiers in Chemistry, 7 (2019) < https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6548831/> [accessed 6 April 2023].
3
J. Clark, ‘The Effect of Temperature on Rates of Reaction’, Chemguide, 2018 <https://www.chemguide.co.uk/physical/basicrates/temperature.html> [accessed 8
April 2023].
4
A. Maley, ‘3.1.2: Maxwell-Boltzmann Distributions’, Chemistry LibreTexts, 2013
<https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/
Kinetics/03%3A_Rate_Laws/3.01%3A_Gas_Phase_Kinetics/3.1.02%3A_Maxwell-Boltzmann_Distributions> [accessed 8 April 2023].
5
Clark, ‘The Effect of Temperature’, 2018, [accessed 13 April 2023]
6
ChemNinja, ‘6.2 Collision Theory’, ChemNinja, 2020 <https://ibchemninja.weebly.com/62-collision-theory.html> [accessed 18 April 2023].
, Personal identifier code:
changed or used up in the reaction7. The Maxwell-Boltzmann graph can be used to demonstrate this, since decreasing
the activation energy means the line showing activation energy shifts more to the left, so a higher proportion of
particles have kinetic energy equal to or greater than the activation energy and so the frequency of successful
collisions increases. In this experiment, I will use manganese (IV) oxide, which is a heterogeneous catalyst since it is
in a different phase from the reactant. The manganese (IV) oxide will allow the reaction to take place on its surface
and will weaken the chemical bond within hydrogen peroxide, so the bond breaks more easily. Thus, the hydrogen
peroxide decomposes at a lower activation energy. I decided to use powdered manganese (IV) oxide as it has a large
surface area for contact between the catalyst and hydrogen peroxide particles, thus increases the rate of reaction
significantly.
Deciding the method
I had considered using an organic catalyst instead of manganese (IV) oxide (an inorganic catalyst), such as the enzyme
catalase. However, after researching, I realised that this would limit the temperature I could conduct this experiment
at, since the organic catalyst would denature at higher temperatures and so limit the range of temperature I could
investigate8. Thus, I decided using manganese (IV) oxide was more appropriate because it allows me to investigate the
rate of reaction at higher temperatures without my results being affected by denaturation.
In the preliminary experiments, I tried to conduct the entire experiment in the water bath in order to maintain constant
temperature, however I found that this was impractical as there were issues with spilling and limitations with setting
up my apparatus, as the water baths were shared between many classmates. Therefore, I had to adjust my method so
that I would heat up the H2O2 solution in the water bath and then transfer it to my workstation quickly. While this
would result in heat loss to the surroundings, and so the temperature would not remain constant throughout the
experiment, I decided this effect could be minimised by measuring the initial and final temperature, then working out
the mean temperature at which the reaction occurred.
During my preliminary experiment, I made up my concentration of hydrogen peroxide solution by adding 5 cm3 of 1.7
mol dm-3 of hydrogen peroxide (ie. 20 volume of H2O2) with 25 cm3 of distilled water.
5
• Concentration of H2O2 = 30 × 1.7 = 0.28 mol dm-3
However, I found that the reaction occurred too quickly when I added 0.50g of MnO2 to the aforementioned H2O2
solution. I was not able to accurately measure the time taken for 50 cm3 of O2 gas to be produced, since the reaction
finished within 10 seconds at around 30°C. Therefore, for my actual experiment, I decided to use 0.10g of MnO2 and
to dilute the H2O2 solution further by adding 5 cm3 of 1.7 mol dm-3 of H2O2 with 45 cm3 of distilled water. These
changes allowed for the reaction to occur more slowly, so I could measure the time taken more accurately.
5
• Concentration of H2O2 = 50 × 1.7 = 0.17 mol dm-3
• Volume of oxygen produced from 0.17 mol dm-3 of hydrogen peroxide
50
o No. mol of hydrogen peroxide = 1000 × 0.17 = 8.5 × 10-3 mol
8.5×10−3
o No. mol of oxygen = 2 = 4.25 × 10-3 mol
o Volume of oxygen produced at room temperature = (4.25 × 10-3) x 24 = 0.102 dm3 = 102 cm3
• In my preliminary experiment, I measured the volume of O2 produced every 10 seconds at room temperature
(20°C) and found that about 90 cm3 of the gas produced was collected. Then I plotted a graph of volume of O2
gas produced against time. I found that there was a linear relationship from 0 cm3 to 65 cm3 of O2 produced,
thus, measuring time taken for 50 cm3 of O2 gas to evolve is an appropriate measure which can be used to
calculate the initial rate of reaction. This will allow me to plot the graph to find the activation energy of
hydrogen peroxide.
Designing the experiment
Independent variable
The independent variable is the temperature. The rate of reaction will be measured at around the following
temperatures: 20°C, 30°C, 40°C, 50°C, 60°C, 70°C.
I decided not to investigate temperatures higher than 70°C because the reaction is already very fast at this temperature
with the mass of MnO2 I am adding, so a higher temperature would make it very difficult to measure the time taken
accurately.
7
‘Kinetics - the Maxwell– Boltzmann Distribution and Catalysts (A-Level Chemistry)’, Study Mind <https://studymind.co.uk/notes/the-maxwell-boltzmann-
distribution-and-catalysts/> [accessed 14 April 2023].
8
‘Difference between Enzymes and Inorganic Catalysts’, Major Differences <https://www.majordifferences.com/2013/09/difference-between-enzymes-and.html>
[accessed 23 April 2023]
