Determining the Activation Energy of a Reaction

If these formulas are followed and the appropriate time and in turn rate of reaction are used, the values of k and ln k can be found. These values are presented in Table 3. For the experiments that were taken with 2 trials, an average time was used. The time was inversed in order to graph ln k = -(Ea/R)(1/T) + ln A

Average time sample calculation:  (676.4 s +710.9 s) / 2 = 693.65 s

Sample calculation of k (at 2.5 °C)

k = Initial Rate of Reaction

       [KMnO4][H2C2O4]

k = 9.50977129 x 10-6 mol dm-3 s-1

      (0.02 mol dm-3)(0.5 mol dm-3)

k = 9.50977129 x 10-4 mol-1 dm-3 s-1

Table 2 – Determining Average time and Initial Rate of Reaction

Table 3 – Determining k, ln k, and 1/T

Therefore, the points ln k vs 1/T were graphed. They were graphed using Microsoft Excel. A line of best fit was included in order to find the slope ( - EA /R)

Line of best fit equation:

y = -7387.4x + 19.922

∴   ln k = -7387.4 (1/T) + 19.922

To find EA:

-7387.4 = -EA / R

 

EA = -7387.4 K-1 mol-1  x 8.134 J K-1mol-1

EA = 60089.1116 J

EA =  60 kJ

Conclusion

Since the literature value for the activation energy of KMnO4 could not be found, the values of other student’s calculations which followed the same procedure were obtained. In order to conclude if the EA obtained was an acceptable value, the standard deviation was found.

Activation Energies of other students (in kJ)

1. 60
2. 52
3. 63
4. 58
5. 56
6. 48
7. 57
8. 50
9. 42
10. 53

These activation energies were put into a Texas Intruments (TI) 83-Plus, into a List. The Statistics for the list were then found. The following screenshots are the results.

Inserting all values (10) into a list.                Finding Statistics on list (L1)

Statistics                                        Statistics cont.

Since Q1 and Q3 have been found, any outliers may be found. All values are in kJ

Interquartile Range = Q3 – Q1

                        = 58  – 50

                        = 8

Therefore, outliers would be found below:

Q1 – (1.5 x IQR) = Outliers

50 – (1.5 x 8) =38

Since minX = 42 , there are no outliers below Q1

Outliers found above:

Q3 + (1.5 x IQR) =Outliers

58 + (1.5 x 8) = 70

Since maxX = 63 , there are no outliers above Q3

The standard deviation of the sample, δx = 5.9  and the mean = 53.9

To find out if 60 kJ (the value obtained) is included in standard deviation,

53.9 + 5.9 = 59.8 kJ

             = 60 kJ
Therefore it is part of the standard deviation and a suitable activation energy,

Evaluation

There are a few limitations in the experiment. For example, it was very difficult keeping the temperature of the reaction constant in water baths of 2.5, and 12.5. The rate at which the reaction took place at these temperatures was very low, and therefore it was difficult to keep the temperature constant because the temperature would rise once the ice in the water baths would melt. The errors caused by this could have been reduced slightly by constantly replacing the ice and trying to keep the temperature constant.

Since the reaction at 2.5 degrees Celsius was only calculated once, there could have been discrepancies with the time that it reacted. In order to reduce the error, the reaction should have been repeated 3-4 times to get an average time. The other reactions at different temperatures could also have been repeated 3-4 times to get a better average time.