• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Determining the Activation Energy of a Reaction

Extracts from this document...

Introduction

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.

Middle

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.

Conclusion

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.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Lab - Measuring an Activation Energy

    and we clearly know this is not the case. From the graph above, visually we can see that if we were to draw tangents at each of the red points, each one would have a different value, as it should. Graph 2: Temperature (K) vs. Rate (Ksec-1) Again from the graph, we can see that the temperature (K)

  2. Lab 1 - Determining Hydrate Formulas

    Mass given by weighing machine in grams (g) � 0.005g *Percent of Uncertainty = (absolute uncertainty) � actual weight) � 100 {0.005g (absolute uncertainty) � 17.92g (actual weight)} � 100 = 0.028% Percent of Uncertainty = 17.92g � 0.028% No# 4: Mass of crucible, lid, and Zinc(II)

  1. Determining the activation energy of a reaction, By using the experimental data and the ...

    The burette used for measuring the solution's quantity also had an associated uncertainty, which was ml�0.05. Lastly, uncertainties could be found in data collection. This investigation required the experimenter to determine the time of colour change and to press the stopwatch.

  2. Aim: To calculate the activation energy (EA) for the reaction between Br- and BrO3- ...

    the rate of the reaction decreases. Procedure: This experiment was done in pairs. First 10 cm3 of 0.01 mol dm-3 of a bromide/bromated (V) solution was measured with a measuring cylinder and then poured into a boiling tube. Then 4 drops of methyl red were added to the boiling tube with the bromide/bromated (V)

  1. Surface area vs Rate of Reaction

    and 8 ml of 2M HCl over 3 trials and the mean, in intervals of every 10 seconds. The uncertainty is 0.05 ml as the readings are analog and the smallest unit is 0.1 ml Time/s Volume of CO2 produced /ml (�0.05) Trial 1 /ml (�0.05) Trial 2 /ml (�0.05)

  2. Group 4

    the impact is distributed over a much larger area of the body, resulting in less severe injuries. The area that hits the airbag is shown in orange. Conclusion:The law of inertia, is demonstrated in a car collision and it is Newton's first law which states that: objects moving at a

  1. Reaction Rate

    RESULTS Results Table: Time Taken To Produce 20mL of Hydrogen Gas (sec) From a Reaction Between Magnesium Metal (Mg) and Different Concentrations (M) of Sulfuric Acid (H2SO4) Concentration Of H2SO4 (M) Time Taken to Produce 20mL of Hydrogen Gas (sec)

  2. Measuring the fatty acid percentage of the reused sunflower oil after numerous times of ...

    Many cell types can use either glucose or fatty acids for this purpose. In particular, heart and skeletal muscle prefer fatty acids. Although this information can lead to a thought that an increase in fatty acids may be beneficial for human body, during the process of lipolysis which is the

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work