Investigating the rate of reaction between sodium thiosulphate and hydrochloric acid

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Investigating the rate of reaction between sodium thiosulphate and hydrochloric acid

Background information

In this reaction two aqueous substances, one is sodium thiosulphate and the other is hydrochloric acid are reacted together. to form sulphur, sulphur dioxide, salt and water.

Sodium thiosulphate + Hydrochloric acid = Sulphur + sulphur dioxide

+ Sodium chloride + water

Na2 S2 O3 (aq) + 2HCl (aq) = S (S) + SO2 (G) + 2NaCl (aq) + H2O (L)

This reaction is no different to any other reaction in that it has several variables in which the reaction can be controlled. In this experiment I will control the rate of the reaction. In order to control the rate of reaction I must only change one variable and keep the others at a constant. This is so I can take fair and accurate measurements.

There are several factors which affect how fast a reaction goes. The first is pressure. The more the pressure, the faster the reaction will go, this is because the atoms are pushed closer together so they hit each other more often. This would obviously be too hard to try and control without expensive equipment; also this variable only applies to gases. The next is surface area. The larger the surface area, the faster the reaction. This is because the particles in the solution have more area to work on so there are more reactions resulting in a faster reaction. This would be an unwise variable to control because the parietals are already in solution, which provides the most surface area, and this would be very difficult to alter into powders or solid lumps Also this only applies to solids so again I cannot use this variable. The next is temperature. You could alter temperature and this could be a possibility for a variable because it is relatively easy to obtain and you can get fairly accurate results from it. The third is a catalyst. This speeds up a reaction but is not used up and reacted with. If you added a catalyst to this reaction you could only get one set of results so there is not much point in having one; Also the reaction is fairly fast already and does not need to be speeded up anymore. The fourth factor is concentration. Altering the concentration either speeds up or slows down the reaction depending on the concentration. If the solution is more dilute there are less molecules of the chemical in a set area and therefore less to collide with each other. If the solution is more concentrated then there are more molecules in a set area so there are more to react and collide with each other. This is the variable I will be using because it is one of the easiest to control along with concentration and I have happened to choose this variable over temperature.
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Prediction

I predict that the higher the molarity of the solution, the faster the reaction will be because there are more molecules to collide with each other. The lower the molarity, the slower the reaction will be because there are fewer molecules to collide. Using the collision theory I can explain why the rate will be increased also. A certain amount of activation energy is required to start a reaction and when a molecule collides with another they will only react if there is enough activation energy. If there are more molecules colliding then there is a ...

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