The effect of Concentration on Rates of Reaction

The following equation shows the reaction between Sodium Thiosulphate and Hydrochloric Acid:

Na2S2O3(aq) + 2HCL(aq) 2NaCl(aq)+H2O(l)+SO(g)+S(s)

Sodium + Hydrochloric Sodium + Water + Sulphur + Sulphur

Thiosulphate Acid Chloride dioxide

In order to find how the concentration affects the rate of reaction, I will be varying the concentration of Sodium Thiosulphate and I will be keeping constant the amount of Hydrochloric Acid.

The equipment that I will use for the experiment will be:

Conical Flask

50ml of Sodium Thiosulphate

25ml of Hydrochloric Acid

Stopwatch

Sheet of Paper with cross

Water

Measuring Cylinder (100ml) for measuring the Sodium Thiosulphate

Measuring Cylinder (100ml) for measuring the water

Measuring Cylinder (10ml) for measuring the Hydrochloric Acid

Beaker

This is how I will carry out my experiment:

* To begin with I will measure out the appropriate quantities of Sodium Thiosulphate and water to make the required concentration - (as shown in the table below) and pour into a conical flask.

* I will draw a cross on a piece of paper and place the conical flask over the paper.

* I will measure out 5ml of Hydrochloric Acid and pour into the measuring cylinder.

* Now I will pour 5ml of Hydrochloric Acid into the conical flask and immediately, whilst looking over the flask I will start the stopwatch.

* When the cross disappears I will stop the stopwatch and record the results in the table below.

(The diagrams were taken from: http://www.leekhigh.staffs.sch.uk/science/chemistry/chim/RRfolder/RR2.htm) and then edited.

I predict that the more the concentration of the Sodium Thiosulphate the faster the reaction will take place. This is because the precipitate of Sulphur will form faster and the cross will disappear in a faster time. I predict that this will happen due to the fact that more Sodium Thiosulphate particles will be present in the same amount of space and the Hydrochloric Acid particles will have more particles to collide with. This will mean the Activation Energy level will be reached faster and the reaction will take place quicker.

Therefore I would expect my results to illustrate a graph similar to the one below:

(The graph was taken from: www.gcsechemistry.com)

The following table shows my results:

Sodium Thiosulphate (ml)

Water (cm3)

Hydrochloric Acid (cm3)

Concentration (mol/dm3)

% Of Sodium Thiosulphate

Time taken for cross to disappear

0

40

5

0.03

20

205

20

30

5

0.06

40

05

30

20

5

0.09

60

68

40

0

5

0.12

80

52

50

0

5

0.15

00

43

In order to try and compare my results with a "perfect" set of results I will have to draw a graph and observe to see if any anomalous results have been produced. This would be very likely and to illustrate this I have drawn a graph:

The graph that I drew according to my results is very similar to the graph that I included in my prediction. Therefore I can say that my prediction was correct and that an increase of concentration in the same space results in a faster rate of reaction.

The table of results also represents this statement as it shows that the least concentrated solution of the dilute Sodium Thiosulphate (0.03 mol/dm3) resulted in the slowest time of 205 seconds. The highest concentrated solution of the Sodium Thiosulphate (0.15mol/dm3) resulted in the fastest time of 43 seconds. The other results also followed this simple pattern and as stated I can confirm that as concentration was increased, the time taken for the cross to disappear decreased.

Now that I have discovered how the concentration of Sodium Thiosulphate affected the rate of reaction between the Sodium Thiosulphate and the Hydrochloric Acid I will try to scientifically explain why changing the concentration affects the behaviour of the time taken for the cross to disappear.

For the reaction to take place there is a fixed amount of energy known as the Activation Energy level. This is the minimum amount of energy required to create the bonds between the Hydrochloric Acid and the Sodium Thiosulphate particles. In order to create these bonds this minimum amount of energy or a higher amount of energy must be applied. To do this there were 5 factors, which could have been tested. I chose the concentration factor and tested this as the variable. The higher the concentration the more the energy being applied, therefore as the concentration was being increased the energy being applied was becoming greater. This meant that the Activation energy level was reached faster and therefore the reaction happened quicker.

In terms of the collision theory, when the Concentration of Sodium Thiosulphate is increased in the same amount of space there is more Sodium Thiosulphate particles to react with the Hydrochloric Acid particles in the same space. This means that as there are now more particles in the same space, so the particles have less space between them. This results in more collisions, and finally more successful collisions. My results support the collision theory as they prove that the rate of reaction depends on how the particles must collide with each other in order to react.

) 2)

) Here there are few Sodium Thiosulphate particles. This means that there is a lot of space in between them and the Hydrochloric Acid particles. The particles will only occasionally successfully collide.

2) Here there are more Sodium Thiosulphate particles. This means that there is only little space between these particles and those of the Hydrochloric Acid. Now the particles will Successfully collide more often.

The graph produced finally gave a smooth curve to illustrate the statement "The higher the concentration of the Sodium Thiosulphate the quicker the cross will disappear."

However at first one anomalous result was achieved. This was for the concentration of 0.12 mol/dm3. The first time given for this concentration level was 53 seconds. This result did not fit the smooth curve and the graph suggested that the result should have been closer to 49 seconds. The second time round I obtained a result of 50 seconds, which fitted the curve and completed the pattern.

There could be many reasons for why I gained an anomalous result and the main reason, which I think affected the results, was that it is virtually impossible to stop the stopwatch at the same opaqueness all the time. Therefore repeating some results could improve the accuracy.

Also the amount of light in the room is also a major factor. For example, doing one of the results at night and the others in daylight would mean that the experiment is not fair and the cross will disappear quicker in the dark. Therefore using a light meter could be useful, as it would measure the amount of light in the room.

OMER ANJUM

0LJ