1.reaction principle ：H 2O2 + 2 I − + 2 H + = I 2 + 2 H 2O
Rate = k[ I − ]2 [ H 2O2 ]
PV = nRT
2.1 prepare a KI solution with a concentration that is about 0.3M.
Use an analytical balance to accurately weigh 24.9g of KI. Use a volumetric flask to dilute to
2.2 prepare a H2O2 solution with a concentration that is about 0.1M.
Dissolve about 6 ml of 30% H2O2 in 500ml distilled water.
2.2 prepare a 500ml pH buffer solution with 0.1mM of acetic acid.
3.1 At room temperature, a certain volume of H2O2 solution was put into a beaker, stirred under a
3.2 A certain volume of KI solution was added to the beaker, the adjust the pH.
3.3 Measure the change in pressure per unit time and calculate the amount of I2 in unit time.
The effect of the amount of H2O2 at room temperature
The effect of the amount of KI at room temperature
The effect of the pH at room temperature
The effect of the amount of the temperature
Measure the pressure calculate the produce rate of I2 .
Research Question: What is the activation energy (kJmol-1) of the decomposition of hydrogen
peroxide (H2O2) to oxygen (O2) and water (H2O) by catalase (0.1%), by measuring the time
taken for 10cm3 of oxygen gas to be evolved (s) at different temperatures (K)?
When we studied catalysts as part of chemical kinetics, I was fascinated by how enzymes function as
biological catalysts and I was drawn into the roles enzymes play in biological systems. I found that a
particular enzyme, catalase, which is found in animals, catalyses the decomposition of hydrogen peroxide
(H! O! ) in the blood. H! O! is secreted by white blood cells as a defence mechanism against external
pathogens. Hence, in order to reduce the exposure of body (somatic) cells to the toxic hydrogen
peroxide, catalase decomposes H! O! . (GMO Compass, 2010).
2H2O2 (aq) → 2H2O (l) + O2 (g) (in the presence of catalase)
This sparked my curiosity about this particular reaction. I then decided to delve further into reactions
involving catalase and found that catalase is also used to preserve egg products by producing oxygen gas
when catalysing the decomposition of hydrogen peroxide (GMO Compass, 2010). This oxygen is utilised
by glucose oxidase in the egg to catalyse the acidification of glucose to gluconic acid, reacting with all the
available glucose in the process. Glucose, in egg products, leads to browning because of its reactions with
amino acids present in the albumen of the eggs (Tucker, 1995). Given the importance of the
decomposition of hydrogen peroxide by catalase, I questioned the value of the activation energy of the
catalysed reaction (E! ). This led me to my research question; What is the activation energy (kJmol-1) of
the decomposition of hydrogen peroxide (H2O2) to oxygen (O2) and water (H2O) by catalase (0.1%), by
measuring the time taken for 10cm3 of oxygen gas to be evolved (s) at different temperatures (K)?
2.1: Reaction under study
2H2O2 (aq) → 2H2O (l) + O2 (g) (in the presence of catalase) at 298.0K, 300.5K, 303.0K, 305.5K and 308K.
2.2: Background Information
Previous research has shown that the rate expression for decomposition of H! O! in the presence of
catalase is 𝑟𝑎𝑡𝑒 = 𝑘 H! O! catalase (Tao, 2009), where k is the rate constant. The rate refers to the
rate of reaction, which is defined in this investigation as the change in the concentration of H! O! per
The E! of a reaction is the minimum amount of energy with which reactant molecules need to collide
successfully, forming the transition state in the process. It is axiomatic that the E! of a reaction would be
significantly reduced if a catalyst was present, because a catalyst, such as the aptly named catalase,
provides an alternative reaction pathway by bringing reactant molecules closer together. Through a
series of stochastic collisions, H! O! molecules move into the active site of catalase molecules. Therefore,
H! O! molecules are brought closer together by catalase. In the process, it provides an alternative
reaction pathway with a lower E! . Hence catalase, acts as a biological catalyst, reducing E! , as illustrated
on the Maxwell-Boltzmann Distribution and the enthalpy change diagram on the next page.
Figure 1: A Maxwell-Boltzmann distribution displaying how there are an increased number of particles with energies
more than or equal to the EA of the catalysed reaction (Gems, 2011)
Figure 2: An enthalpy change diagram illustrating the reduction in activation energy for an exothermic reaction
when a catalyst is used
The E! of the decomposition was found through a clock reaction. A stopwatch was started when the
reaction began and was stopped when 10cm3 of oxygen gas was evolved. The number of moles of oxygen
evolved was ascertained through the use of the ideal gas law, which is 𝑃𝑉 = 𝑛𝑅𝑇, where P is pressure in
Pascal, V is the volume of oxygen evolved in m3, T is the temperature of the surrounding air in Kelvin (K),
𝑛 is the number of moles of oxygen evolved and R is the gas constant (8.3145JK-1mol-1) (John, 2013).
Using this equation, the number of moles of oxygen evolved at each temperature was found. With this in
mind, the number of moles of H2O2 consumed was determined, using the molar ratio between O2 and
H2O2, which is 1:2, as seen from the equation: 2H2O2 (aq) → 2H2O (l) + O2 (g). The number of moles of
H2O2 consumed was subsequently divided by the time taken to evolve 10cm3 of oxygen gas as well as the
volume of the solution in dm3 to produce a value for the rate of reaction in moldm-3s-1. Using the
equation 𝑅 = 𝑘 H! O! catalase , the rate of reaction was divided by the concentrations of H! O! and
catalase to produce a value for the rate constant (k) in dm3mol-1s-1.
To find the activation energy, the Arrhenius equation, which is shown below, was used, where k is the
rate constant of the reaction, A is the frequency of successful collisions, R is the gas constant (8.3145JK-1
mol-1) and T is temperature in Kelvin. Please note that ln 𝑘 is simply the natural logarithm of k (log ! 𝑘).
ln 𝑘 = ln 𝐴 −
This equation was plotted using the k values found at each temperature, with ln 𝑘 on the y-axis and on
the 𝑥-axis. The graph obtained was similar to that shown below.
Figure 3: A graph displaying the shape of an Arrhenius graph (𝑙𝑛 𝑘 = 𝑙𝑛 𝐴 −
Using Microsoft Excel, the line of regression for this graph was sketched and its equation was found, to
provide a value for the gradient of the line, which is equal to −
, the coefficient of in the Arrhenius
equation. The gradient was then multiplied by – 𝑅, such that the product of this multiplication was equal
to E! .
Independent Variable: Temperature (K). This is because, use of the Arrhenius equation, requires values
of k at different temperatures, in order to plot a graph of ln 𝑘 against , whose gradient is used to find
the value of EA. Therefore, the independent variable that was chosen was relevant to the investigation,
whose purpose is to find the EA of the catalysed decomposition of hydrogen peroxide. The temperature
values that were tested were 298.0K, 300.5K, 303.0K, 305.5K and 308.0K. The use of 5 different
temperature values increased the reliability of the results, because it increased the number of data
points on the graph, allowing for a more accurate representation of the linear relationship between and
ln 𝑘. The values chosen also do not exceed 313K, because previous studies have shown that catalase
starts to denature (undergo an irreversible conformation change) at temperatures exceeding 313K (410C)
(Abuchowski, 1977). The conformation change results in the deterioration in the shape of the active site,
such that fewer H! O! molecules can “lock” into it. Therefore, if the experiment were conducted at
temperatures in excess of 313K, the investigation would yield inaccurate values for k, because the
concentration of reacting catalase would be lower than the value used in data processing.
Dependent Variable: The time taken (s) for 10cm3 of oxygen gas to be evolved. This was selected as the
dependent variable, since it allows for the quantification of the rate of reaction (moldm-3s-1). By using the
equation 𝑅 = 𝑘 H! O! catalase , we can find the value of the rate constant at each temperature, by
dividing the rate of reaction found by the product of the concentrations of hydrogen peroxide and
catalase. k is essential for use in the Arrhenius equation, where ln 𝑘 is used to find EA, hence the
dependent variable chosen is fully relevant to the investigation. It is important to note that the units for
the rate were chosen to be moldm-3s-1 because the units for the rate constant for a second order rate
expression (this is the order of the reaction under study) are dm3mol-1s-1. Therefore, to ensure the rate
constant found is found in terms of dm3mol-1s-1, the units of the rate of reaction must be moldm-3s-1.
1. pH: The pH was kept constant at pH 7 using a sodium hydroxide buffer solution. This was to
ensure that the catalase in each experiment was operating at its optimum pH (Su), allowing for
an accurate basis for comparison in data processing. A buffer is a solution that resists changes in
pH when small amounts of acid or base are added, therefore, it allowed the pH of the mixture to
remain constant, with neglible changes. This also ensured that the pH was not a factor that
affected the differences in the rate constant at different temperatures.
2. Concentrations of reactants: The concentration of H! O! and catalase in the experiment to
determine the activation energy of the catalysed decomposition were kept at 0.01moldm-3 and
0.1% respectively to ensure that the changes in the rate of reaction when different temperatures
were compared were only caused by temperature, and not concentration, which is another
factor that affects the rate of reaction. To do so, samples of 1.5moldm-3 H! O! were diluted to
reduce their concentration to 0.01moldm-3.
3. Volume of reactants: The volume of H! O! , the pH 7 buffer and catalase were also kept constant
at 10cm3, 5cm3 and 5cm3 respectively. These quantities were measured and added using a
graduated pipette. A low volume and concentration of catalase was chosen because each
catalase molecule can react with approximately 4 ⋅ 10! molecules of H! O! (RSC, 2007).
Therefore, a low volume and low concentration of catalase was chosen, so the progress of the
reaction would be easily observable.
4. Pressure: Data collected by Singapore’s National Environmental Agency (NEA) has shown that
pressure in Singapore, both indoors and outdoors undergoes small fluctuations around 101kPa
(NEA, 2015). Therefore, pressure can be considered a controlled variable, since previous statistics
and research have shown that pressure in Singapore stays relatively constant at 101kPa (NEA,
2015). This would have affected the calculations for the number of moles of oxygen gas evolved,
because the value of P in the ideal gas equation would fluctuate.
Memmert Water Bath (±0.1K)
25 cm3 pipette (±0.06cm3)
250cm3 volumetric flask
50cm3 glass gas syringe (±0.1cm3)
1 Test Tube
20.5cm rubber tubing
6.67cm3 of 1.5 moldm-3 H! O!
125cm3 of 0.1% catalase solution
393.3cm3 of distilled water
4.2: Photograph of set-up
A photograph taken by myself using an iPhone 6, on 26/04/2016, that displays the experiment in progress in the
Memmert water bath, with the mixture of catalase (0.1%) and H2O2 (0.01moldm ) in the test tube, connected to a
glass gas syringe held by a retort stand
4.3: Experimental Procedure
1. Prepare a standard solution of H! O! with a concentration of 0.01 moldm-3 by diluting a
1.5moldm-3 sample of H! O! in a volumetric flash. Immediately, seal the volumetric flask to
reduce the risk of H! O! decomposing immediately. The concentration of H! O! was kept constant
at 0.01 moldm-3 because the decomposition of hydrogen peroxide with catalase present is a very
fast reaction, hence a low concentration of H2O2 was chosen to allow for the progress of the
reaction to be more easily observed, therefore reducing systematic error.
2. Set the Memmert water bath to 298.0K (±0.1K).
3. Use a graduated pipette (±0.06cm3) to measure exactly 5cm3 of the catalase solution and place it
into a test tube. For this and all remaining measurements with the pipette, read the pipette at
the meniscus to ensure that the volumes of solutions added to the test tube are accurate.
4. Use a graduated pipette (±0.06cm3) to transfer 5cm3 of the pH 7 buffer to the test tube
containing the catalase solution.
5. Place the test tube holding the catalase solution into the water bath for exactly 10 minutes, with
the lid closed, to allow the temperature of the test tube and its contents to equalise.
6. Connect a 20.5 cm rubber tube to a 50cm3 glass gas syringe (±0.1cm3).
7. Use a graduated pipette (±0.06cm3) to transfer 10cm3 of the prepared H! O! solution into a test
tube and allow the tip of the pipette to touch the surface of the solution to allow for cohesion
between any H! O! that remains in the pipette and the solution, to ensure that exactly 10cm3 of
H! O! is added to the solution.
8. Immediately, cover the test tube with the rubber bung connected to the 25-cm3 gas syringe and
start a digital stopwatch (±0.01s).
9. Record the time taken (s) for 10cm3 of oxygen gas to be evolved using the stopwatch.
10. Repeat steps 4-9 for a total of 4 additional times to reduce the impact of random error on the
results and allow for the collection of sufficient data.
11. Repeat steps 3-10 at the following temperatures; 300.5K, 303.0K, 305.5K and 308.0K (±0.1K).
4.4: Risk Assessment
Safety Considerations: H2O2 (aq) is a powerful bleaching agent and “causes skin
irritation…discolouration, swelling and the formation of papules and vesicles (blisters).” (Fisher Scientific,
2000) Therefore, to ensure a high level of safety during the experiment, latex gloves and goggles were
worn throughout the duration of the investigation.
Ethical Considerations: There were no ethical considerations to be taken into account.
Environmental Considerations: There were no environmental considerations to be taken into account.
5: Raw Data
Table 1: A raw data table showing the time taken by each replicate to produce 10cm3 of oxygen gas (s)
at each temperature (K) for the catalysed decomposition of hydrogen peroxide (0.01moldm-3)
The cancelled values (indicated by a line through the value) were excluded from further calculations as
they are anomalous points, as substantiated by the fact that the standard deviation (s) and variance (s2)
of the sets of data they belong to decrease significantly following their removal.
5.1: Qualitative Observations
1. As temperature increased, the vigour of the effervescence observed in the test tube visibly
2. The colour of the solution remained constant; a very light green colour.
3. The gas syringe indicated 5cm3 of oxygen gas within 20 seconds, whereas more than 20 seconds
was required to produce the remaining 5cm3.
6: Processed Data
The average time taken to evolve 10cm3 of oxygen gas was found by the following formula
time taken to evolve 10cm! of oxygen gas for each replicate
number of replicates
Example calculation displaying how the average time taken to evolve 10cm3 of oxygen gas was calculated
52.32 + 50.82 + 51.68 + 52.73 + 50.85
To calculate the number of moles of oxygen evolved, the ideal gas law equation (𝑃𝑉 = 𝑛𝑅𝑇) was used.
Example Calculation displaying how the number of moles of oxygen evolved was calculated for 298K
𝑃𝑉 = 𝑛𝑅𝑇 = 101,000
= 𝑛 8.3145 298
= 4.08 ⋅ 10!! moles
8.31 ⋅ 298
The remaining values of n were found in a similar manner.
The number of moles of H2O2 consumed was found by multiplying the number of moles of oxygen
evolved by 2, because according to the equation for the reaction, the molar ratio of O2 to H2O2 is 1:2.
Example Calculation displaying how the number of moles of H2O2 consumed was calculated for 298K
2 4.08 ⋅ 10!! = 8.16 ⋅ 10!! moles
The rate of reaction was then calculated by dividing the number of moles of H2O2 consumed by the
volume of the solution (0.02dm3), and subsequently by the average time taken for 10cm3 of oxygen gas
to be evolved.
Example Calculation displaying how the rate of reaction was calculated for 298K
8.16 ⋅ 10!!
= 7.89 ⋅ 10!! moldm!! s !!
(0.02 ⋅ 51.68)
Table 2: A processed data table showing the average time taken to evolve 10cm3 of oxygen gas (s)
(±0.01s), number of moles of oxygen evolved (mol), the number of moles of H2O2 consumed (mol) and
the rate of reaction (moldm-3s-1) for each temperature (K) (±0.1K)
Rate of reaction
taken to evolve
moles of H2O2
oxygen gas (s)
7.89 ⋅ 10!!
8.08 ⋅ 10!!
8.16 ⋅ 10!!
8.41 ⋅ 10!!
8.81 ⋅ 10!!
Please note that full values were used in calculations, but the displayed values are shown in a manner
that is consistent with the uncertainty of the apparatus used.
Temperature !! values were established by calculating the reciprocal of each temperature value that
Example calculation displaying how 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 !! was calculated for 298K
Temperature !! =
= 3.36 ⋅ 10!! K !!
k (rate constant) values were found using the rate equation for the decomposition of hydrogen peroxide
( 𝑅 = 𝑘 H! O! catalase ). The rate of reaction at each temperature was found divided by the
concentration of H! O! and subsequently by the concentration of catalase (3.03 ⋅ 10!! moldm-3). The
concentration of H! O! was assumed to remain constant at 0.01moldm-3, due to the small number of
moles of H! O! consumed in the clock reactions (please see table 2).
The units for the concentration of catalase were converted to moldm-3 from % to generate the rate
constant. In the calculation of this concentration, a critical assumption was made; that 100g of water
(H2O) has a volume of 0.1dm3. Hence, the concentration of catalase (percentage by mass) was divided by
the molecular mass of catalase (33,000) (RSC).
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑎𝑠𝑒
÷ 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑎𝑠𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑎𝑠𝑒
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑎𝑠𝑒
𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑎𝑠𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
The concentration of catalase remains unchanged, because as a catalyst, it is simultaneously regenerated
as it is being used to provide an alternative reaction pathway. This is substantiated by my observation
that the colour of the solution (a light green, because of the catalase) remained constant throughout the
Example calculation displaying how the concentration of catalase (moldm-3) was found
Assuming we have a sample weighing 100g
÷ 33000𝑔𝑚𝑜𝑙 !! =
÷ 33000𝑔mol!! = 3.03 ⋅ 10!! 𝑚𝑜𝑙𝑑𝑚 !!
k at each temperature was then found by dividing the rate of reaction (R) at each temperature by the
product of the concentrations of H! O! and catalase.
Example calculation displaying how k at 298K was found
H! O! catalase
7.89 ⋅ 10!!
= 2603.96dm! mol!! s !!
(0.01)(3.03 ⋅ 10!! )
ln 𝑘 was calculated by taking the natural logarithm of the calculated k values.
Example calculation displaying how 𝑙𝑛 𝑘 at 298K was found
ln 𝑘 = log ! 2603.96 = 7.86
Table 3: A processed data table displaying the rate constants of the reaction (moldm-3s-1) at different
temperature (K) (±0.1K) and 𝐭𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞 !𝟏 (𝐊 !𝟏 )
(10 K )
(moldm s )
Table 4: A processed data table displaying how ln k varies with 𝐭𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞 !𝟏 (𝐊 !𝟏 )
(10 K )
An Arrhenius graph was then plotted, with Temperature on the x-axis and ln 𝑘 on the y-axis.
Graph 1: An Arrhenius graph displaying how 𝐥𝐧 𝒌 varies with 𝐓𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞 !𝟏
(10-3 K-1) with the equation of the line of regression and its R2 value indicated
y = -0.9746x + 11.126
R² = 0.88267
10^(-3) * 1/Temperature (1/K)
The gradient of the line is given by the coefficient of x on the line of regression, which is
-0.9746. The gradient for an Arrhenius graph, which is the type of graph shown above, is equal to −
Hence, the gradient was multiplied by – 𝑅 to provide a value for E! in Jmol .
E! = −0.9746 ⋅ −8.3145 = 8.10Jmol!!
The point (3.25, 7.98) does not follow the line of regression as closely as the remaining points, hence it
was discarded as an anomalous data point, and E! was recalculated using the graph below.
Graph 2: An Arrhenius graph displaying how 𝐥𝐧 𝒌 varies with 𝐓𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞 !𝟏
(10 K ) with the equation of the line of regression as well as its R2 value indicated and the anomalous
y = -0.6667x + 10.1
R² = 0.99999
10^(-3) * 1/Temperature (1/K)
E! = −0.6667 ⋅ −8.3145 = 5.54Jmol!! = 0.00554kJmol!!
The systematic error of the experiment is another factor that must be taken into account, hence it was
calculated in the section below.
7: Calculation of Random Error
The average percentage uncertainty of the time taken for 10cm3 of oxygen gas to be evolved across all
replicates was found by finding the percentage uncertainty of each measurement for the time taken for
10cm3 of oxygen gas to be evolved for each replicate and subsequently dividing this value by 5 (as there
were 5 replicates for each temperature).
Example calculation displaying how the average percentage uncertainty of the time taken for 10cm 3 of
oxygen gas to be evolved across all replicates at 298K was calculated
⋅ 100% + ⋯ +
The percentage uncertainty of the volume of oxygen evolved, the total volume of solution added to the
test tube for each replicate and the temperature the water bath was set to for each replicate were found
by the following formula;
!”#$%&'(“&) !” !””!#!$%&
!”#$%&”‘ !”#$% !”#$% !!! !””!#!$%&
Example calculation displaying how the average percentage uncertainty of the volume of oxygen evolved
across all replicates was calculated
⋅ 100% = 1%
The total uncertainty for each temperature was ascertained by performing the sum of the average
percentage uncertainty of the time taken for 10cm3 of oxygen gas to be evolved across all replicates, the
percentage uncertainty of the volume of oxygen evolved, the total volume of solution added to the test
tube for each replicate and the temperature the water bath was set to for each replicate.
Example calculation displaying how the total uncertainty at 298K was calculated
0.1940% + 1.0000% + 0.3000% + 0.0336% = 1.5276%
Next, the random error of the investigation was calculated, by adding the total uncertainties at each
temperature and dividing the sum by 5.
Example calculation displaying how the random error of the investigation was calculated
1.5276% + 1.5333% + 1.5371% + 1.5441% + 1.5568%
The uncertainty of the E! value calculated was found by multiplying the random error of the experiment
(%) by the E! value calculated in the manner shown below.
1.5398% ⋅ 0.00554Jmol!! = ±0.0000853kJmol-1
Table 5: A table showing the error from each apparatus and the total random error for each replicate
at each temperature
Percentage of the total
of the time
error of the
bath was set
to for each
The E! of the catalysed decomposition of H! O! (0.01moldm-3) in the presence of catalase (0.1%) was
successfully found in the course of this investigation to be 0.00554kJmol-1 ±0.0000853kJmol-1. This value
is in general agreement with previous research conducted on the reaction under study. A literature value
(0.00658kJmol-1) (Su) was used in order to calculate the experiment’s total error.
𝑙𝑖𝑡𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑣𝑎𝑙𝑢𝑒 − 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒
0.00658 − 0.00554
⋅ 100% =
⋅ 100% = 19.9%
The systematic error can now be found by subtracting the random error of the experiment from the total
error of the experiment.
Total error =
𝑆𝑦𝑠𝑡𝑒𝑚𝑎𝑡𝑖𝑐 𝑒𝑟𝑟𝑜𝑟 = 𝑇𝑜𝑡𝑎𝑙 𝑒𝑟𝑟𝑜𝑟 − 𝑟𝑎𝑛𝑑𝑜𝑚 𝑒𝑟𝑟𝑜𝑟 = 19.9% − 1.5398% = 18.3602%
Despite the relatively high systematic error, I have high confidence in my results due to their precision (as
can be seen by the low standard deviation and variance amongst replicates) as well as the low total error
of the experiment. The experiment is also is in agreement with the current scientific consensus, such as
the increase in the rate constant as temperature increases, which is substantiated by my observation that
at higher temperatures, the effervescence observed in the reaction was more vigorous. This indicates an
increase in the rate of reaction as temperature increased, which stemmed from an increase in the rate
constant. My observation that the rate of reaction started to decline after 5cm3 of oxygen gas was
produced is supported by the work of P. George, who found that the rate of decomposition of H! O!
slowly declined over the course of the reaction, but only marginally (George, 1947). Despite the age of
this study, it is reliable because of the stature of the author, a Professor at the University of Cambridge.
Because of his position, he had access to extremely precise apparatus and conducted multiple repeats;
hence, the conclusions he drew were of low uncertainty and are therefore reliable.
The experiment had low random error (1.5398%) due to low uncertainty of the apparatus used,
increasing the certainty of the conclusion drawn in the section above. In addition, the low standard
deviation (s) and variance (s2) in the times taken for 10cm3 of oxygen to be evolved at each temperature
across all replicates were very low, after anomalous points had been removed. This delineates the fact
that my results are extremely precise. Furthermore, the high R2 value indicated in graph 1 (0.99999)
demonstrates the accuracy of the processed data because this R2 value indicates a strong correlation
between Temperature !! and ln 𝑘, which is the ideal description of an Arrhenius graph. My processed
data is thus consistent with established scientific theories. The use of a water bath was also a strength of
the experiment because it allowed for the uniform distribution of thermal energy in the solution. Hence
temperature, as an independent variable, was effectively controlled.
However, the experiment had a number of weaknesses.
The data used to find the E! is limited because of the exclusion of the value of the rate constant at
298.0K, decreasing the number of data points on the Arrhenius graph (Graph 2). This had the effect of
increasing the potential impact of random error on the investigation, as substantiated by the relatively
high systematic error of 18.3602%. Therefore, the investigation is limited because of the use of only 4
data points for the Arrhenius graph, decreasing the certainty of the conclusion drawn. This can be
rectified by repeating the experiment at 298.0K and 295.5K in order to increase the number of data
points on the graph, which would decrease the impact of random error on the results.
New solutions of H! O! were only made once every day, hence increasing the possibility that, before
being transferred to the test tube, a small amount of the H! O! had possibly decomposed. This possibly
reduced the concentration of H! O! , a change that was not accounted for in the calculations, hence
resulting in the values of the rate constants calculated not being accurate representations of the true
rate constants. This could have potentially affected the accuracy of the value of E! that was calculated.
This can be rectified by preparing standard solutions of H! O! just before the addition of H! O! to the test
tube, allowing for more accurate rate constant values to be calculated. A more accurate value for E!
could then be calculated.
Futhermore, the temperature used in the calculation of the number of moles of oxygen evolved may not
have been a representation of the oxygen’s true temperature, since its temperature was assumed to be
the same as that of the water bath following equalisation. Therefore, it is possible that value for the
number of moles of oxygen used in the calculation of ln k was unreliable, impeding the reliability of the
E! value calculated in this experiment. This can be rectified by inserting a thermometer into the gas jar
following the collection of the 10cm3 of oxygen, such that the actual temperature of the oxygen
produced can be measured.
The small range of the independent variable was also a weakness, because it limits the accuracy of the
gradient value calculated by Microsoft Excel, as a result of fewer coordinates on the graph. This weakness
possibly had an effect on the final E! value, as the gradient calculated may not have been a
representation of the true gradient, consequently affecting the final E! value calculated. To reduce the
impact of this limitation, the experiment could have been repeated at 5 additional temperatures, all
lower than 298.0K and none higher than 308.0K, as catalase would denature at temperatures higher than
Furthermore, it is possible that the rate calculated does not reflect the initial rate, because the first 5cm3
of gas was evolved in less time than the remaining 5cm3 in all the experiments conducted, indicating that
the rate calculated was the average rate, hence the concentration values employed in the rate equation
to calculate the different values of k were likely not reflections of the true values. However, if the
stopwatch were stopped at 5cm3, such that the rate calculated would be the initial rate, the random
error of the investigation would increase, as the percentage uncertainty of the volume of gas measured
increases from 1% to 2%, thereby increasing the random error of the investigation. Considering this
possible increase in uncertainty, it is pellucid that the calculation of the average rate was accurate, as it
significantly reduced random error relative to if the investigation calculated the initial rate of reaction.
In addition, the buffer solution used could have potentially affected the accuracy of the final E! value
produced, because it contained sodium hydroxide, whose dissacoiated sodium ions could have
potentially caused in a conformation change in the catalase, due to its positive charge, hence the rate at
which the H! O! decomposed was possibly reduced. Conversely, research conducted by Eyster found that
the presence of sodium ions had a neglible effect on the rate at which the catalysed decomposition of
H! O! occurs in the presence of catalase (Eyster, 1953), hence this limitation had a minor effect on the
A possible extension to this investigation would be to deduce the difference in the activation energy of
the catalysed reactions, in the presence of different catalysts, such as transition metal ions and iodide
ions, to find which catalyst can reduce the activation energy of the reaction to the greatest extent. This
would uncover the catalyst would best suited in the preservation of egg products. Another investigation
could also be carried out to assess if other chemical reactions can produce more oxygen per unit time,
relative to the catalysed decomposition of H! O! . This knowledge will be helpful in maximising the
efficiency of the preservation of egg products.
8.5: Limitations of the scope of the investigation
However, the investigation is limited because it does not calculate the E! of the uncatalysed
decomposition of H! O! . This limits the extent to which the investigation examines the magnitude of the
difference between the activation energy of the uncatalysed and catalysed decomposition of H! O! . This
limitation can be rectified by extending the investigation to conduct the same experiment in the absence
of catalase, to find the E! of the uncatalysed decomposition of H! O! .
1. Tucker, G.A. and Woods, L.F.J., 1995. Enzymes in food processing. Springer Science & Business
Media, Retrieved from
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2. Tao, Z., Raffel, R.A., Souid, A.K. and Goodisman, J., 2009. Kinetic studies on enzyme-catalyzed
reactions: oxidation of glucose, decomposition of hydrogen peroxide and their
combination. Biophysical journal, 96(7), pp.2977-2988.
3. Avogadro Chemistry., Material Safety Data Sheet – Hydrogen Peroxide. ACC#11189 Sections 1-3.
Retrieved from http://avogadro.chem.iastate.edu/MSDS/H2O2_30pct.htm, date of access
4. GMO Compass., 2010. Catalase. GMO. Retrieved from http://www.gmocompass.org/eng/database/enzymes/89.catalase.html, date of access 5/11/15, 3:23pm
5. Abuchowski, A., McCoy, J.R., Palczuk, N.C., van Es, T. and Davis, F.F., 1977. Effect of covalent
attachment of polyethylene glycol on immunogenicity and circulating life of bovine liver
catalase. Journal of Biological Chemistry,252(11), pp.3582-3586.
6. Eyster, H., 1953. Effects of certain inorganic ions on corn leaf catalase.The Ohio Journal of
Science, 53(2), pp.102-104.
7. Su., Catalase Kinetics. MIT
8. Gems., 2011. 6.25-6.27. Gems Chemistry Blog . Retrieved from
http://gemschemistry12.blogspot.sg/2011/05/625-627.html, date of access 27/4/2016, 12:27pm
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11. George, P., 1947. Reaction between catalase and hydrogen peroxide. Nature, 160, pp.41-43.
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