Investigating the kinetics of reaction between sodium thiosulphate and dilute hydrochloric acid.

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Chemistry Coursework

Investigating the kinetics of reaction between sodium thiosulphate and dilute hydrochloric acid.

Introduction:

The reaction to be studied is as follows:

Na2S2O3   +   2HCl                       2NaCl   +   H2O   +   SO2   +   S

There are several possible ways of finding out the reaction rate, as there are four by-products formed in the above reaction.

The fact that sodium chloride is in solution accounts for why it is not suitable to be collected. Water is hard to measure, as there are too many moles of it in comparison to the salt, which is in aqueous solution. Sulphur dioxide gas would be a suitable factor for measuring reaction rate, and it would not have to separated from any other gases, as no other gases are given off. However, it will react and dissolve into the water, and it is also fairly toxic, making it a safety hazard. This leaves us with sulphur, which is left as a solid.

Sulphur is the most suitable substance to measure reaction rate, as when the two colourless, aqueous solutions are mixed together, a cloudy precipitate is formed. This is the sulphur, which is insoluble in water. Therefore, an experiment can be designed to measure the amount of time for the solution to completely precipitate.

There are four factors, which affect reaction rates: surface area, use of a catalyst, temperature and concentration.

It is impossible to vary surface area, as both reactants are liquids, and in order for this to work, they would both have to be solids. Use of catalysts is a hit-and-miss variable, as it involves putting in random chemicals and seeing if they react. Temperature and concentration are both suitable for variation, but I have decided to choose varying the concentration, as varying temperature is often too quick, and

can therefore be inaccurate.  


Knowledge:

Na2S2O3   +   2HCl                       2NaCl   +   H2O   +   SO2   +   S

  • As we are using Na2S2O3 as a variable, we need to ensure that the HCl is constantly in excess. Therefore, I need to use the following calculation to find out how much HCl I should use:
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Moles = concentration x volume

        =        0.23             x          0.05

        =                0.0115

Volume = (0.0115 x 2)         = 11.5 cm3                

                           2                

                        Amount of precipitate formed

  • Reaction Rate =                        time

                              Extent of reaction

                 =                        time

  • When Na2S2O3 reacts with HCl, a precipitate forms. The time taken for the precipitate to form is an indicator for the reaction rate.

  • In order for a reaction to occur, reactant particles must collide with a certain minimum ...

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