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Determining the Activation Energy of a Reaction

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Determining the Activation Energy of a Reaction Submitted by: First Name, Las Name Submitted to: Teacher Name Date Submitted: Date Course: IB HL Chemistry Data Collection and Processing Temperature (�C � 0.2�C) Reaction Time - 2 trials (s � 0.2 s) 2.5 2103.1 - 12.5 676.4 710.9 20.5 362.6 405.4 33.5 131.3 138.6 To find the Activation Energy, the value for "k" needs to be found. Since: Initial Rate of Reaction = k[KMnO4][H2C2O4] And: Initial Rate of Reaction = ?[KMnO4] ?T Sample Calculation Reaction at 2.5 �C � 0.2�C Initial Rate of Reaction = ?[KMnO4] ?T = (0 mol dm-3 - 0.02 mol dm-3) / 2103.1 s = 9.50977129 x 10-6 mol dm-3 s-1 Note: Significant figures will be taken in final equation (finding activation energy) 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. ...read more.


k (mol-1 dm-3 s-1) ln (k) 1/T (K-1) 275.5 9.50977129 x 10-4 -6.958020545 0.0036297641 285.5 0.0028832985 -5.84882033 0.003502627 293.5 0.0052083333 -5.257495372 0.003407155 306.5 0.0148203038 -4.211757159 0.0032626427 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. ...read more.


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. ?? ?? ?? ?? ...read more.

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