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Chemistry-Sodium Thiosulphate

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Introduction

GCSE Chemistry Coursework How does the concentration of sodium thiosulphate affect the rate of reaction of with hydrochloric acid? In this coursework I aim to find out how the concentration of sodium thiosulphate affects the rate of reaction of with hydrochloric acid. I aim to see if higher or lower concentrations make the rate of reaction faster or slower. The aim of the experiment is to test different concentrations to see which works best. The aim is to find the best concentration by seeing which concentration makes the experiment work faster. This is when the reaction between sodium thiosulphate and hydrochloric acid is at its fastest. I will find out the rate of reaction, which is done by diving 1 by the average time taken. I will test this by using 30ml of sodium thiosulphate and 0ml of water, and gradually getting more dilute. I will go on to use 25ml of sodium thiosulphate and 5ml of water, keeping the solution constant at 30ml. I will always use 5ml of hydrochloric acid, as this is an independent variable that I am not testing the concentration of. I predict that when the concentration of the sodium thiosulphate is higher, the rate of reaction will be quickest. This means that both graphs in my analysis will have positive correlation (my time taken and rate of reaction graphs). Positive correlation is when the plotting of the graphs goes down in time as the concentration increases and/or vice versa. I think this will happen because of what is stated in the collision theory. ...read more.

Middle

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/dm�. 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. ...read more.

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. Chris Collins ...read more.

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