Thiosulphate Clock Reaction: An Anomalous Run
A student investigates how the concentration of sodium thiosulphate affects the rate of its reaction with dilute hydrochloric acid. Each trial uses 5.0 cm³ of 1.0 mol dm⁻³ HCl and enough distilled water to make a total volume of 50.0 cm³, with sodium thiosulphate solution of concentration 1.0 mol dm⁻³. The flask is placed over a printed cross, and the time for the cross to disappear is recorded. Results: A: 20.0 cm³ thiosulphate, 25.0 cm³ water, 40 s. B: 15.0 cm³, 30.0 cm³, 54 s. C: 10.0 cm³, 35.0 cm³, 80 s. D: 5.0 cm³, 40.0 cm³, 160 s. E: 7.5 cm³, 32.5 cm³, 130 s. The student claims that rate is directly proportional to thiosulphate concentration. A classmate notes that the acid used in trial E came from a bottle that had been left open in a warm room for a week.
1. Use the results to test the student's claim that rate is directly proportional to thiosulphate concentration. Show the processed data.
Read the worked answer
Because the total volume is 50.0 cm³ in every trial, the thiosulphate concentration is: [S₂O₃²⁻] = 1.0 × volume of thiosulphate / 50.0.
For a fixed endpoint, 1/t is a comparative measure of rate. The processed values are:
| Trial | V(thiosulphate)/cm³ | [S₂O₃²⁻]/mol dm⁻³ | t/s | 1/t /s⁻¹ | (1/t)/[S₂O₃²⁻] |
|---|---|---|---|---|---|
| A | 20.0 | 0.40 | 40 | 0.0250 | 0.0625 |
| B | 15.0 | 0.30 | 54 | 0.0185 | 0.0617 |
| C | 10.0 | 0.20 | 80 | 0.0125 | 0.0625 |
| D | 5.0 | 0.10 | 160 | 0.00625 | 0.0625 |
| E | 7.5 | 0.15 | 130 | 0.00769 | 0.0513 |
For A–D, (1/t)/[S₂O₃²⁻] is constant at about 0.062 s⁻¹ dm³ mol⁻¹ (within about 1%). That is the signature of a directly proportional relationship: rate = k[S₂O₃²⁻]. The claim is supported by A–D.
Trial E does not fit. Its value, 0.0513, is about 18% below the constant, so E is an outlier and should not be used to support or reject the proportionality until it is repeated.
2. Explain whether the old acid bottle can account for trial E, and state the effect on the student's conclusion.
Read the worked answer
Trial E is slower than expected, not faster. The expected time at 0.15 mol dm⁻³ is about 1/(0.0625 × 0.15) ≈ 107 s, but 130 s was observed.
Hydrochloric acid is volatile; a bottle left open in a warm room can lose HCl gas, lowering the acid concentration. The rate depends on both thiosulphate and acid concentrations (rate = k[S₂O₃²⁻][H⁺] under these conditions). If [H⁺] fell in trial E, the rate would fall and the time would rise, which is the direction observed. So the old acid is a plausible explanation, but it is not proved by this experiment alone; the acid should be renewed and trial E repeated.
The student's conclusion should be based on A–D and should exclude E as invalid. If E were included, it would falsely suggest that the relationship bends downwards at low concentration. The correct conclusion is that rate is proportional to thiosulphate concentration when the acid concentration is held constant.
3. Suggest one improvement that would make the comparison fairer and one reason the disappearing-cross method may be unreliable for very fast reactions.
Read the worked answer
Improvement: use a light sensor/colorimeter to measure the absorbance of the sulphur suspension at a fixed wavelength, or use a datalogger, so that the endpoint is judged objectively. Also thermostat all solutions at the same temperature (e.g., 25 °C) and use freshly prepared acid; temperature and acid concentration are the main variables that must be controlled.
For fast reactions (times of only a few seconds), mixing, starting the clock and judging the cross by eye introduce large percentage errors. The reaction may also not be at a uniform temperature immediately after mixing. A fast run can be slowed by using more dilute solutions or a lower temperature, or measured with an instrument with a faster response. Comparing 1/t is only valid if the endpoint corresponds to the same amount of sulphur each time; an objective sensor helps ensure that.