Finding the Activation Energy Between Hydrochloric Acid an Sodium Thiosulfate

Topics: Temperature, Observational error, Measurement Pages: 10 (2375 words) Published: May 22, 2013
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

Method

1. Label two boiling tubes A and B. Mark a dark spot on the side of a 400cm3 beaker, then ½ 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. 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

Reason for the uncertainty on the initial time: as it was very hard to determine the exact time of the end point I feel that the time difference between each time I determined the end point of the reaction would have been plus or minus two seconds.

Raw Data

|Temperature ±2 (°C) |Trials |Time ± 1 (seconds) |Average time (seconds) |Uncertainty on Average time | | | | | |(seconds) | |4 |1 |271 |273.6666666=274 |±4 | | |2 |276 |...