Minzak 1
Ac#va#on Energy Lab of the Reac#on Between Potassium Permanganate
and Oxalic Acid
Anna Minzak
Mr. Centritto
SCH3UC-02
06/10/2023
Minzak 2
Qualita#ve Data
Table 1: qualitative observation of the reaction at 0ºC
Before Reaction
During Reaction
After Reaction
à The solution of
KMnO by itself is a
deep purple
à The solution was dark purple
when the potassium permanganate
and oxalic acid were first
combined.
à At 12 minutes, the solution
became maroon.
à At 13 minutes, the solution
turned an orangish colour.
àAt 14 minutes the solution
turned a yellowish colour.
à at 18 minutes the solution
turned a yellowish-brown colour.
à After the reaction was
complete and the solution
turned yellow/brown, the
solution remained
yellow/brown.
4
à The solution of C H O
is colourless and
transparent.
2
2
4
Table 2: Raw Data with Uncertainties
Concentrations (M)
Volume with uncertainties
(mL)
H2C2O4
0.02
20 ± 0.05
KMnO4(aq)
0.5
10 ±
Table 3: Raw Data
Temperature Trial #1
Reaction Time
(± 𝟎. 𝟐)ºC
(min)
4
22.34
10
9.40
20
4.55
30
2.02
Trial #2
Reaction Time
(min)
22.38
10.40
5.33
2.05
Trial #3
Reaction Time
(min)
18.53
5.32
2.19
Trial #4
Reaction Time
(min)
17.16
5.38
2.17
Calcula#ons:
Overview:
This section displays the calculation and techniques used to calculate the activation energy (Ea)
for the reaction between potassium permanganate (KMnO4) and oxalic acid (H2C2O4). The
observed reaction is given by the rate equation: k[KMnO4] [ H2C2O4.] = rate. To calculate the
activation energy, the average reaction time between the four trials at each temperature was first
calculated. The average reaction time for each temperature was then used to compute the
matching k value. The natural logarithm of each estimated quantity was then plotted against the
Minzak 3
temperature reciprocal (1/T) in kelvin. Finally, because its linear function corresponds to the
Arrhenius equation, this graph was used to calculate the Ea of the reaction.
Uncertainty Conversion:
percentage uncertainty =
𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦
× 100%
𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦
Uncertainty temperature for 4ºC
0.2º𝐶
× 100%
4
= ±0.05%
=
See Appendix A for uncertainty calculations for 10ºC, 20ºC, and 30ºC
Uncertainty for volume (ml) of KMnO4
=
0.05 𝑚𝑙
× 100%
10 𝑚𝑙
= ±0.05%
Uncertainty for volume (ml) of H2C2O4
0.05 𝑚𝑙
× 100%
20 𝑚𝑙
= ±0.25 %
Calculating average reaction time for each temperature
=
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒
(𝑡𝑟𝑖𝑎𝑙 1 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 ) + (𝑡. 2 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒) + (𝑡. 3 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒) + (𝑡. 4 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒)
=
4
Calculating average reaction time of trials for 4ºC
=
(22.57 ± 0.05%) + (23.63 ± 0.05%) + (18.88 ± 0.05) + (17.27 ± 0.05%)
4
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 = 20.59 min ±0.5%
See Appendix B for calculations of average reaction time of 10ºC, 20ºC, and 30ºC
Determining the k value for each temperature
Rate = k[KMnO4] [H2C2O4]
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Concentration: C1V1 = C2V2
Initial concentration of KMnO4
𝐶! =
=
𝐶" 𝑉"
𝑉!
(0.02)(0.01 ± 0.5%)
(0.01 ± 0.5% + 0.02 ± 0.5%)
𝐶! = 0.333
𝑚𝑜𝑙
± 0.5%
𝐿
Initial concentration of H2C2O4
=
(0.5)(0.02 ± 0.5%)
(0.01 ± 0.5% + 0.02 ± 0.5%)
𝐶! = 0.333
Determining the k value for 4ºC
𝑘=
𝑚𝑜𝑙
± 0.5%
𝐿
𝑟𝑎𝑡𝑒
[KMnO4][H2C2O4]
1
20.59 ± 0.5%
=
[0.00667 ± 0.5%][0.333 ± 0.5%]
𝑘 = 21.87 𝑚𝑜𝑙 #" 𝑠 #"
See Appendix C for k value of 10ºC, 20ºC, and 30ºC.
Determining ln k value for each temperature
Determining ln k value for 4ºC
ln 𝑘
= 𝑙𝑛 21.87 ± 1%
= 3.09
Determining ln k value for 10ºC
"
𝑟𝑎𝑡𝑒 = $%&'
Minzak 5
= 𝑙𝑛 44.27 ± 1%
= 3.49
Determining ln k value for 20ºC
= 𝑙𝑛 82.01 ± 1%
= 4.41
Determining ln k value for 30ºC
= ln 207.5 ± 1%
= 5.34
Figure 1: ln k vs. 1/Temperature graph
ln k vs 1/Temperature
6
5
y = -6980.3x + 28.339
ln k
4
3
2
1
0
0.00325
0.0033
0.00335
0.0034
0.00345
0.0035
0.00355
0.0036
0.00365
1/Temperature (K)
Calculating Activation Energy (Ea) for Reaction:
ln 𝑘 = `−
𝐸𝑎 1
d ` d + ln 𝐴
𝑅 𝑇
Where
𝑘 = 𝑅𝑎𝑡𝑒 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐴 = 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝐹𝑎𝑐𝑡𝑜𝑟 (𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝑠 𝑎𝑛𝑑 𝑜𝑟𝑖𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛)
𝐸𝑎 = 𝐴𝑐𝑡𝑖𝑣𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦
𝑅 = 𝐼𝑑𝑒𝑎𝑙 𝐺𝑎𝑠 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝐽/𝑚𝑜𝑙/𝐾)
𝑇 = 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
Minzak 6
`−
𝐸𝑎
d = −7242.8
𝑅
𝐸𝑎 = −7242.8(−𝑅)
𝐸𝑎 = −7242.8(−8.31)
𝐸𝑎 = 60 187
𝐸𝑎 = 60.2
𝐽
𝑚𝑜𝑙
𝑘𝐽
𝑚𝑜𝑙
Figure 2: Literature Value vs. Experimental Value
Experimental Value vs.Literature Value
80
Activation Energy (KJ/mol)
70
60
50
40
30
20
10
0
Experimental Value
Literature Value
Table 4: All Processed Data Within the Experiment
Value
Initial concentration of KMnO4
0.00667 ± 0.5%
Initial concentration of H2C2O4
0.333 ± 0.5%
Average reaction time for 4ºC
20.59 𝑚𝑖𝑛
Average reaction time for 10ºC
10.17 𝑚𝑖𝑛
Minzak 7
Average reaction time for 20ºC
5.47 𝑚𝑖𝑛
Average reaction time for 30ºC
2.18 𝑚𝑖𝑛
k value for 4ºC
21.87 𝐿 𝑚𝑜𝑙 #" 𝑚𝑖𝑛 #" ± 1.0%
k value for 10ºC
44.27 𝐿 𝑚𝑜𝑙 #" 𝑚𝑖𝑛 #" ± 1.0%
k value for 20ºC
82.01 𝐿 𝑚𝑜𝑙 #" 𝑚𝑖𝑛 #" ± 1.0%
k value for 30ºC
207.5 𝐿 𝑚𝑜𝑙 #" 𝑚𝑖𝑛 #" ± 1.0%
ln k value for 4ºC
3.09
ln k value for 10ºC
3.79
ln k value for 20ºC
4.41
ln k value for 30ºC
5.34
Activation energy of reaction
60.2
𝑘𝐽
𝑚𝑜𝑙
Conclusion
The purpose of this experiment was to determine the activation energy of the reaction of KMnO4
and H2C2O4. The reaction was observed at various temperatures, and as the temperature
increased the reaction rate dropped, since increasing the temperature leads molecules to collide
more frequently and with more force, reducing the time for the reaction to occur. The activation
energy is the amount of energy that two colliding molecules must possess before a reaction may
occur. In this experiment, it was measured by determining the average reaction time for each
temperature variation, calculating the k value, which is the rate constant, and graphing it against
the inverse of temperature on a graph. This results in an activation energy of 59.31± 3.61
kJ/mol, which is lower than the literature value of 76.56 kJ/mol. This provides a systematic error
of 22.53%, indicating that there were faults in the technique other than the random error that
caused this value to be big, which were procedural problems inside the experiment.
Table 5: Procedural errors within experiment
Procedural Errors
Improvements
Minzak 8
This approach used four trials for each temperature
fluctuation, but only two for the 10°C variation. As a
result, the lack of trials may have resulted in inaccurate
values. If this were done again, other reaction time
findings could have been produced, which could have
been significantly different from the initial study. The
absence of trials leads to inaccuracy in temperature
changes and, as a result, imprecision in the value of the
change in enthalpy value.
If more time had been provided to
confirm the accuracy of the data
obtained, more trials might have
been conducted as an improvement.
Stirring was not part of the procedure. We were not
instructed when or how many times to stir during the
reaction, so everyone did it differently, and the kinetic
energy for each trial varied. When you stir the solution,
you give it energy, which raises the overall temperature
and causes it to react faster, which could explain the
shorter timeframes.
Adding stirring into the procedure, in
respect to time. For example, stating
that stirring every 5 minutes for one
minute; would improve the lab as
every trial would be obtain the same
amount of energy from stirring.
Procedural errors related to temperature control can
significantly affect the accuracy of the results. The
reactants were chilled individually to 4ºC with an ice
bath for the first trial, but they were removed after the
solutions were mixed. As the atmosphere was warmer
than the initially cooled reactants, they would have
warmed up as the reaction progressed, increasing the
rate of the reaction. The same occurs for the 30ºC trial,
which would have cooled at room temperature, thereby
increasing reaction time. If the reaction time is wrong
because of this, the activation energy may be less exact
and have a higher difference between the literature
value.
Ensure that we maintain a consistent
and precise temperature throughout
the experiment by conducting by
continuing the experiment in the ice
bath or hot plate.
We were not given a specific shade or colour shift that
was intended to occur, indicating that the reaction was
complete. As a result, some persons may have used the
period when the solution turned a pale yellow, while
others may have used the time when the solution
became darker. As a result, because the colour is so
broad, people may have seen it differently, and it is
difficult to establish when the solution is exactly
yellow/brown, which varies reaction time greatly.
There would be less procedural error
if there was a photo of the colour the
solution turned when the reaction
was completed, or if there was a
more detailed explanation of what
hue the solution should have become
after the reaction.
Strengths
This experiment was carried out meticulously to obtain an accurate result. This is why numerous
factors were controlled: the volume of H2C2O4 and KMnO4 solution, measuring instruments, and
Minzak 9
the identical thermometers used throughout. Before beginning the experiment, all the equipment
was thoroughly cleaned and washed to eliminate any components that could interfere with the
experiment, such as chemically stained equipment that could produce undesirable chemical
reactions. Finally, one of the experiment's strengths was that several images of the experiment
were taken, and notes were recorded while the procedure was being completed, in order to recall
components that may have been lost when completing the written portion of the lab, such as the
qualitative observations.
Appendix A
Uncertainty temperature for 10ºC
=
0.2º𝐶
× 100%
10
= ±0.02%
=
0.2º𝐶
× 100%
20
= ±0.01%
Uncertainty temperature for 20ºC
Appendix B
Calculating average reaction time of trials for 10ºC
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
(𝑡𝑟𝑖𝑎𝑙 1 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 ) + (𝑡𝑟𝑖𝑎𝑙 2 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒)
2
= (9.67 + 10.67)/2
= 10.17 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
* for this reactions time (10ºC) there were only 2 trials done
Calculating average reaction time of trials at 20ºC
=
(4.92) + (5.55) + (5.53) + (5.97)
4
= 5.49 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
Calculating average reaction time of trials for 30ºC
Minzak 10
=
(2.03) + (2.08) + (2.32) + (2.28)
4
= 2.18 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
Appendix C
Determining the k value for 10ºC
1
10.17 ± 0.5%
𝑘=
[0.00667 ± 0.5%][0.333 ± 0.5%]
𝑘 = 44.27 𝑚𝑜𝑙 #" 𝑠 #" ± 1.0%
Determining the k value for 20ºC
1
5.49 ± 0.5%
𝑘=
[0.00667 ± 0.5%][0.333 ± 0.5%]
𝑘 = 82.01 𝑚𝑜𝑙 #" 𝑠 #" ± 1.0%
Determining the k value for 30ºC
1
2.17 ± 0.5%
𝑘=
[0.00667 ± 0.5%][0.333 ± 0.5%]
𝑘 = 207.5 𝑚𝑜𝑙 #" 𝑠 #" ± 1.0%