Thiosulfate clock reaction: judging first-order behaviour
A student investigates the rate of reaction between sodium thiosulfate and dilute hydrochloric acid by timing how long a black cross takes to disappear. Each run uses 50.0 cm3 total volume made by mixing thiosulfate solution with water and then 5.0 cm3 acid. The thiosulfate stock is 0.100 mol dm−3. Results: run A uses 10.0 cm3 thiosulfate plus 35.0 cm3 water, time 38 s; B uses 20.0 plus 25.0, time 19 s; C uses 30.0 plus 15.0, time 13 s; D uses 40.0 plus 5.0, time 10 s; E uses 50.0 plus 0.0, time 8 s. One time is suspect because the cross was faint. The teacher asks whether the rate is proportional to thiosulfate concentration and whether the experiment supports a first-order rate law. The student also notices the first run became cloudy slowly at the edges, making the endpoint subjective.
1. Calculate the thiosulfate concentration in each reaction mixture and the relative rate 1/t. Decide whether the suspect run should be discarded.
Read the worked answer
The total volume is 50.0 cm3 = 0.0500 dm3, so [S2O3^2−] in the mixture is (0.100 × V/1000)/0.0500 = 0.00200 × V mol dm−3 when V is in cm3.
| Run | V thiosulfate/cm3 | [S2O3^2−]/mol dm−3 | t/s | (1/t)/s−1 |
|---|---|---|---|---|
| A | 10.0 | 0.0200 | 38 | 0.0263 |
| B | 20.0 | 0.0400 | 19 | 0.0526 |
| C | 30.0 | 0.0600 | 13 | 0.0769 |
| D | 40.0 | 0.0800 | 10 | 0.100 |
| E | 50.0 | 0.100 | 8 | 0.125 |
- B compared with A: concentration doubles and 1/t doubles exactly.
- C compared with A: concentration is 3.00 times larger; 1/t is 0.0769/0.0263 = 2.92 times larger.
- D compared with A: concentration is 4.00 times larger; 1/t is 3.80 times larger.
- E compared with A: concentration is 5.00 times larger; 1/t is 4.75 times larger. The suspect point is most likely A, because the endpoint was subjective and slow. However, A lies on the trend rather than far from it, so it should not be discarded without evidence. The correct judgement is to repeat A using the same endpoint criterion; if the repeat agrees, keep it. E is slightly slower than predicted from exact proportionality, but the deviation is small enough to be timing or endpoint uncertainty.
2. Explain whether the experiment supports a first-order rate law, and why the total volume and the acid volume must be kept constant.
Read the worked answer
For a first-order reaction, rate = k[S2O3^2−]. If the time t to reach the same visible sulfur endpoint is inversely proportional to the initial rate, then 1/t should be proportional to [S2O3^2−]. The data show 1/t rising roughly in proportion to concentration: doubling concentration halves the time, and the higher runs are within about 5% of the same proportionality. This supports, but does not prove, first-order behaviour with respect to thiosulfate.
Keeping the total volume constant at 50.0 cm3 is essential because it fixes the concentration of every species for a given volume of stock. If the total volume changed, the dilution would change both the thiosulfate concentration and the acid concentration, so the rate comparison would no longer isolate the effect of thiosulfate. Keeping the acid volume constant at 5.0 cm3 also fixes [H+], which affects the rate. Temperature should also be controlled because it changes k. The endpoint is subjective, so using the same cross, same observer and same lighting is important; a light sensor or colorimeter would make the endpoint more objective.