Some students investigated the rate of reaction between sodium thiosulfate solution and dilute hydrochloric acid. The chemical equation for this reaction is:
Na2S2O3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+S(s)+SO2(g) \text{Na}_2\text{S}_2\text{O}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{S}(\text{s}) + \text{SO}_2(\text{g}) Na2S2O3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+S(s)+SO2(g)The precipitate of sulfur causes the solution to become opaque over time.
To measure this, the students used the following method:
The students attempted to keep the total volume of the sodium thiosulfate solution and water constant at 50 cm350\text{ cm}^350 cm3 across all runs. They performed all trials at a constant temperature of 22∘C22^\circ\text{C}22∘C.
Their results are presented in the table below:
| Run | Volume of sodium thiosulfate solution in cm3\text{cm}^3cm3 | Volume of water in cm3\text{cm}^3cm3 | Time taken in s\text{s}s |
|---|---|---|---|
| 1 | 50 | 0 | 14.2 |
| 2 | 40 | 10 | 18.5 |
| 3 | 30 | 20 | 24.8 |
| 4 | 25 | 25 | 30.1 |
| 5 | 20 | 30 | 38.2 |
| 6 | 10 | 40 | 78.4 |
| 7 | 5 | 55 | 155.0 |
Explain why the result of Run 7 should not be included in the final analysis of their data.
The rate of reaction for each run can be modeled using the formula:
rate of reaction=500time taken \text{rate of reaction} = \frac{500}{\text{time taken}} rate of reaction=time taken500Calculate the rate of reaction for Run 3's experiment. Give your answer to one decimal place.
In a follow-up analysis, a graph of the rate of reaction against the concentration of sodium thiosulfate was plotted. The resulting graph shows a straight line passing through the origin (0,0)(0,0)(0,0).
Describe the relationship between the concentration of sodium thiosulfate solution and the rate of reaction shown by this graph.
Use collision theory to explain why increasing the concentration of sodium thiosulfate has this effect on the rate of reaction.
