Fair test:
To make my experiment a success it must be fair and accurate. I firstly had to measure as accurately as I possibly could. The whole test was based on the sodium thiosulphate, hydrochloric acid and water, which were all measured. The measuring had to be correct and accurate for the reaction to happen successfully. If the measuring were a few millilitres off, it would have changed the results a lot, making the test inaccurate. This had to be done correctly, and I feel I did this quite well. Also to keep the test fair I must try and use the same standard for the ‘X’ not being visible. I could not decide on different visibilities, as it would be unfair on different concentrations. I tried my best to use the same standard of looking at the visibility, and I feel I did this well also. I must certainly not mix up the measuring cylinders, as there would be reactions with substances before they enter the beaker. If this happened the results again would be inaccurate. I am certain that the measuring cylinders did not get mixed up. I used distilled water, not just tap water. This is so the water did not react with any other substances. This use was very accurate and made the test very fair and reliable. All of these precautions will make my final results more reliable and keep anomalies at a minimum so making the whole experiment more successful.
Results:
Experiment:
I firstly set up the equipment; I put the beaker over the card and fetched the measuring cylinders. I measured out 30ml of sodium thiosulphate and poured it into the beaker. I then used the pipet to accurately measure 5ml of hydrochloric acid. I put this in the beaker and started the stopwatch, with a slight delay. I had to use my eyes to decide when the ‘X’ was not visible and to stop the stopwatch. In the reaction sulphur is made. It is this that causes the ‘X’ to not be seen. It collects and gathers to create the cloudy affect and as the reaction happens more, and more of the sulphur is gathered the cloudier the solution becomes. I did this three times and worked out my average time, which can be seen above. I got an average of 36.68 seconds for 30ml of sodium thiosulphate and no water. My next step was to dilute the sodium thiosulphate to see if there were any time changes. I measured the 25ml of sodium thiosulphate and poured into beaker. But this time I added the distilled water. I added 5ml of this to dilute the sodium thiosulphate. I then added the hydrochloric acid and started the stopwatch. Looking at my results you can see that the time taken was longer, it had gone up by roughly 6 seconds on average. The average for this was 43.02 seconds. I repeated this adding less sodium thiosulphate, and more distilled water. The averages got higher and higher, increasing by 14 seconds, then by 26 seconds and finally by 81 seconds. I gathered the results in the table above and rounded them to two significant figures after the decimal point.
Dilution:
I can work out the dilution of sodium thiosulphate by working out the molarity of sodium thiosulphate as I took less sodium thiosulphate away from the reaction every time. Firstly I must work out 0.2 moles of sodium thiosulphate. The formulae for sodium thiosulphate is Na2S2O3.The relative mass of Na is 22.99 grams, of S is 32.06 and O is 15.9999. I rounded these to 23 grams, 32 grams and 16 grams. As there are more than 1 atom I must calculate the total number of relative mass in the substance. Na has 2 atoms, so that equals 46 grams, S has 2 atoms so that equals 64 grams and O has 3 atoms so that has 48 grams. These all added together will give me the relative mass of 1 mole of sodium thiosulphate. The total mass is 158 grams. But sodium thiosulphate has 0.2 moles so I must divide this total by 5 to work out 0.2 moles, which is 31.6 mol/dm³. I can now work out the dilution of sodium thiosulphate. The equation to work out how many moles are in a solution is; Moles equal mass over RFM. The RFM is 156 grams as I worked out earlier. 0.2 moles are diluted in 1000ml of solution, but I am only using 30ml of it. So to work this out I will divide it by 1000 and times it by the volume. This will mean it will be 0.0312 times the volume of sodium thiosulphate. When I have worked out the mass of sodium thiosulphate in 30ml, I must times that to work out the mass in 1000ml. The mass is then used in the equation, to work out the moles. I repeated the same process by just changing the amounts of sodium thiosulphate from 25 to 20 to 15 and finally to 10. These are what resultsI found:
Analysis:
Looking at my first graph there is positive correlation proving my prediction correct. The graph slopes downwards in time as the solution goes up, which shows the reaction is happening much quicker. It starts at 168 seconds at concentration of 10ml of sodium thiosulphate, and as the concentration goes up the time decreases. By the time the concentration is at 30ml, the temperature has hugely decreased to less than 40 seconds. My prediction says that the more concentrated, the less time to react. So my prediction was correct. Looking at my rate of reaction graph I can see positive correlation again. This graph shows a great accuracy in my results, as the line of best fit is almost straight. My rate of reaction is measured in seconds times 10 . At 10 ml the rate of reaction is really slow at 0.6 seconds. The line increases as the concentration increases, like my time graph. By the time it reaches 30ml the rate of reaction has increased to 2.78 seconds. This shows that the rate of reaction is increasing greatly and again proving my prediction correct. The fact that the line is almost straight shows accuracy, because it shows there is a constant ratio between the different concentrations and making it a reliable test and trustworthy.
Conclusion:
Looking at my results and my graphs I can positively say my prediction was correct. It proved that when the concentration of the sodium thiosulphate is higher, the rate of reaction would be quickest. This meant that both graphs in my analysis would have positive correlation, my time taken and rate of reaction graphs. This happened because of what is stated in the collision theory. For a reaction to happen particles must collide with each other. The particles must overcome a energy barrier and only particles with enough energy will overcome this barrier. The minimum amount of energy that a particle must have to overcome the energy barrier is called the activation energy. If the frequencies of collisions are increased the rate of reaction will increase. The frequency of collisions can be increased in a number of ways, but I looked at the concentration. When the concentration of a solution is increased there are more reactant particles per unit volume, hence the number of collisions per unit time in that volume. This increased the number of reactant particles colliding with each other. The more particles in the reaction, the more successful collisions meaning increased rate of reaction. The variable is continuous and independent. I tested this variable. So therefore I predicted that when I lower the concentration, there are less particles to react and the rate of reaction will decrease as the time taken increases. This was correct, as at 10ml of sodium thiosulphate the reaction time was longest, proving less reacting particles. It took a lot longer than the more concentrated solution.
Evaluation:
The test was not completely accurate. To improve the accuracy there are many things I could do to in future experiments, if I was to repeat it. I could firstly use more concentration tests. I only used 5 concentration amounts and got a graph which showed that the higher the concentration, the faster reaction time. If I used 10 concentration amounts I could tell if there would be too high a concentration or if there was a difference in my results.
It was also inaccurate when looking at the ‘X’, as it was my own eyes testing when the ‘X’ was not visible. This could have become unstable. This was one of the biggest inaccuracies as I felt the vision of the ‘X’ was varied at different parts of the experiment. In future experiments I would have to improve this for my experiment to be a success.
When doing my experiment we were short on time. We did not wash them out before each testing. This meant there was already solution in it, which may have caused a difference in the results. I also did not wash the beaker, only emptied it. This meant there was dilute solution at the bottom. This may have caused anomalies. To improve my test, with more time, I should wash the beaker after every experiment and wash the measuring cylinders.
Different things except the concentration affect the rate of reaction. The surface are change sin heterogeneous reactions which changes the number of collisions between the fluid phase and solid surface. Once again, if the number of collisions changes then the rate will also change. Although this could not be used in our experiment, as ours was homogeneous, it could be part of further studies.
In a future test I could change the temperature of the reaction-taking place. I could have a Bunsen burner under the beaker, which would heat up the solution and reaction when it entered. The temperature, as shown in the collision theory, increases the rate of reaction to a certain extent. If I did this I would get a faster reaction time. Unfortunately if I did that experiment, in my evaluation I would have more inaccuracies of the Bunsen burner having to burn at the same temperature for each experiment. So this option would have both advantages and disadvantages.
Another possibility is to do the test more times than 3. If I did the experiment more times I would get a better and more accurate average. If I did this results and graphs would be more accurate making the test more successful. With better results I could be certain to make a clear and proven conclusion.
I could also add a catalyst. A catalyst is a substance which speeds up the rate of reaction but is chemically unchanged at the end of it. Because the rate of reaction is based on successful collisions, the particles must have more energy than the activation energy, to overcome the barrier. A catalyst added to a reaction has an affect on the activation energy. A catalyst provides an alternate route for the reaction , and this alternate route has a lower activation energy. A catalyst also provides a surface on which the reaction can take place, this increases the number of collisions between the particles of the reacting substance.
The reaction is Na2S2O3 +2HCl ----> S + SO2 + H2O + 2NaCl. The reaction makes H2O. This is water, a liquid. So because it is a liquid it dilutes the solid, so it will take longer for the reaction to happen and the solution to become cloudy. If I could possibly get the liquid to be reduced or even removed, the reaction will happen quicker, improving the accuracy.
The sodium thiosulphate would have become saturated. Saturation in chemistry is when the solution of a substance can dissolve no more of that substance. The saturation point literally means the maximum concentration point. The molecule size of the solute will determine how much of it can be dissolved before the substance becomes saturated. If the molecules are large then less can be dissolved whilst small molecules can be dissolved more. I could not test this, as my concentrations were not high enough. To make my test more accurate I could have found an optimum temperature, before it was saturated and at its quickest point.
