# Finding the Activation Energy Between Hydrochloric Acid an Sodium Thiosulfate

9 September 2016

Finding the Activation Energy of the reaction between Hydrochloric Acid and Sodium Thiosulfate The equation for the reaction is: S2O32- (aq) + 2H+(aq) ? SO2 (g) + S(s) + H2O (l) Equipment – 2 boiling tubes – 400 cm3 beakers – Marker pen – Stand and clamp – Timer – Bunsen burner, tripod and gauze – 0 – 100 oC thermometer – 2 x 10 cm3 measuring cylinders – Access to a fume cupboard. Method 1. Label two boiling tubes A and B. Mark a dark spot on the side of a 400cm3 beaker, then ? fill it with water.

Clamp tube A and immerse in the water bath as shown in the diagram above. 2. Using a measuring cylinder, transfer 10cm3 of sodium thiosulfate solution to tube A. 3. Using a clean measuring cylinder, transfer 10cm3 of hydrochloric acid to tube B and place the tube in the beaker of water. 4. Allow both solutions to reach thermal equilibrium with the water in the beaker for a few minutes. 5. Add the solution from tube B to that in tube A, starting a timer as you do so. Mix the solution in A by gently stirring using the thermometer.

## Finding the Activation Energy Between Hydrochloric Acid an Sodium Thiosulfate Essay Example

Read and record the temperature. 6. Observe the spot on the side of the beaker by looking at it through the solution in A. Record the time at which the spot can no longer be seen due to it becoming obscured by the sulphur precipitate formed in A. 7. Dispose of the mixture in tube A as directed. Rinse out tube and wash and dry the thermometer. 8. Using a Bunsen burner to gently heat the water bath, and/or use ice to cool the water bath repeat steps 2 – 7 until you have 5 sets of results at five different temperatures.

The first will be at room temperature and the other four evenly spaced between zero degrees celcius and about 50oC (try not to exceed this temperature). 9. Record your results in a suitable manner. Processing Data To work out Average Time: Add all trail results for one specific temperature together and divide by the number of trials done for this drop height (in this case 3 for every temperature) eg. the average Time height for the temperature of 44°C is (29+29+30)? 3 = 29. 3333333

To work out Uncertainty on the Time: Find the highest and lowest values of the time taken for each temperature. Add the uncertainty on the time taken to the highest value and minus the uncertainty on the time taken from the lowest value. Minus the new smallest value from the highest value to find the range, divide this range by two to find the uncertainty eg. The highest and lowest time values taken for the 52°C temperature was 22 and 20 seconds. This then becomes 23cm and 19cm when you add and minus the uncertainty on the time taken from the maximum and minimum values respectively.

Then largest value minus smallest value: 23-19= 4, diving this value by two= 4? 2= 2 which is the uncertainty on the averaged time taken for the 52°C temperature. To round the average time taken: The averaged time taken must be rounded to the same number of decimal places as the uncertainty eg. The time taken for the temperature of 52°C is 20. 33333333seconds and the uncertainty is 2seconds, the averaged time taken is therefore rounded to 20. The complete value is now 20±2 seconds

Another random error could have been the approximate nature of the rate as any small changes in the concentrations of either reactant would have changed the results of the experiment as I assumed that the concentrations remained constant, I did not take this error into account, though it would be a minor random uncertainty. The percentage difference between what I got as the activation energy and the activation energy (47kj) is equal to (47-41)/47×100 which equals 12. 7659…. % which equals 13%. As this percentage is bigger than the random uncertainty percentage on the activation energy I calculated. This indicates that there is significant systematic errors in the experiments that I ran. Conclusion and Evaluation Errors One of the main errors would have been deciding consistently between experiments when the cross is obscured. This has a strong impact on the time taken for the experiment.

As the result I got was less than the expected value it indicates that not only did I not identify the same point when the cross was obscured for each experiment, but I also in order from the set of experiments at the hottest temperature to the set of experiments at the coldest temperature I first observed the obscuring of the cross as much later then the actual obscuring of the cross. From there with each set of experiments I conducted I observed the obscuring of the cross a little less later than I did in the previous set of experiments. If I hadn’t done this than the gradient of my plotted line would have been much greater which would have increased the activation energy I calculated for this equation.

Another major systematic error in this experiment would have been my inability to keep the temperature of the experiments constant with each set of experiments. This could also be included as a random error as the temperature of the experiment at the time of the reaction does not stay at the same temperature as the water bath. Yet another systematic error would have been the uncertainty on the exact time the reaction started and the starting of the stop watch. This would impact on the time taken of the reaction, as the activation energy I calculated was less than the book value this indicates that for each set of experiments I altered the difference between the starting of the reaction and the time I started the stop watch. Chemistry behind Reaction and Experiments

At higher temperatures the experiment speeds up and reaches an endpoint quicker because an increase of temperature means an increase in average kinetic energy of the particles involved in the reaction. This in turns increases the chance of particles colliding with enough speed and force to set off the reaction and more collisions means a quicker reaction. Improvements on the Experiment One major issue in this experiment is how to determine when the cross is obscured. To combat this in the next time doing this experiment I would suggest the use of a calorimeter, to calculate the exact moment the cross is obscured. To improve the temperature issue I would measure the temperature of the reactants of the experiment rather than the temperature of the water bath around it, this would make the temperature of the experiment more accurate.

To help with the random uncertainty of the accuracy of the equipment, I would use a burette to measure the volumes of the reactants as a burette has less of an uncertainty on it than a measuring cylinder. To increase the over all accuracy of the Arrhenius plot of the reaction I would repeat the experiment for a greater range of temperatures-though as this would lead to safety issues I would use a fume cupboard for any experiments above a temperature of 55 degrees Celsius. I would also rather than doing only five trials at each temperature I would increase this to ten to lower the uncertainty on the experiment and gain a more accurate average time taken for each trial at each temperature.

A limited
time offer!
Save Time On Research and Writing. Hire a Professional to Get Your 100% Plagiarism Free Paper