Investigation- To investigate the factors involved in the following reaction:Na2S2O3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + SO2(aq) + S(s)

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Caroline Earp

Investigation- To investigate the factors involved in the following reaction:

Na2S2O3(aq) + 2HCl(aq) --> 2NaCl(aq) + H2O(l) + SO2(aq) + S(s)

PLANNING STAGE

Variables:

The following are variables which will affect my investigation.

Temperature, Pressure, Concentration of Na2S2O3, Concentration of HCl, Stirring.

The key variables which I'm going to investigate are Concentration of Na2S2O3 and Concentration of HCl.

What I'm going to find out:

I'm going to find out what effect varying the above stated variables has on the rate at which the reaction between Na2S2O3 and HCl takes place.

What I expect to happen:

I predict that when I double the concentration of the Na2S2O3, the rate of reaction will double. The relationship will be directly proportional and the graph that I will get will be a straight line thought the origin. The origin will be a point on the graph since if there is no Na2S2O3 then there will be no reaction. The relationship will only be directly proportional up to a certain point, when the rate will level off and increasing the concentration further will have no added effect.

I predict that when I double the concentration of the HCl, the rate of reaction will double. Again the relationship will be directly proportional, and the graph a straight line through the origin. If there was no HCl then there would be no reaction and so the origin is a point on the graph. Again the graph will level off when a further increase in the concentration has no more effect.

Why I think this will happen:

For the reaction to take place the thiosulphate ion must be broken down. The ion will be broken down when it collides with another molecule successfully i.e.: with enough energy to overcome the energy barrier or the activation energy. Once a successful collision has taken place the reaction can happen.

When I double the concentration of X (either the Na2S2O3 or the HCl, which ever is being varied.)then I am doubling the number of X molecules in a given volume. If you keep the number of Y molecules (either the Na2S2O3 or the HCl depending on what X is) constant then what was originally a ratio of X:Y now becomes 2X:Y. Therefore the chance of the Y molecules colliding with the X molecules has doubled. I will assume that the number of successful collisions doubles proportionally. This will mean that the rate will double. Since the movement of the molecules is random and all the collisions happen by chance, then you can't say the rate will double exactly. Therefore the points will lie almost on a straight line through the origin (not taking into account experimental error). When you reach a certain point the number of X molecules will be in such an excess that doubling the number further will have such a small effect on the probability of Y molecules colliding with X molecules. This will be shown by a line on the graph levelling off.

Relative effects:

I predict that both variables will double the rate of the reaction and so neither will have more effect than the other.
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How I will make my investigation a fair test:

To make my investigation a fair test I must only change one variable at a time and keep everything else constant. The following will be kept constant in all my experiments.

Temperature: The whole investigation will be done at lab temperature. I will have to look over the fact that the temperature will be different on different days, although the effect on my results should only be very small.

Pressure: The investigation will be done at normal air pressure

Amount Stirred: I will stir ...

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