Rates of reaction - kinetics (A2)
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I'm doing the iodine clock for my individual investigation and I'm really confused. If I have used 25 cm3 potassium iodide (1 M) in a reaction mixture with a total volume of 50 cm3 (including sodium thiosulphate, starch and potassium peroxodisulphate) is the concentration of iodide ions in the mixture 1 M or is it 0.5 M (50/25 = 2, 1/2 = 0.5)? Also, how do you work out the concentration of peroxodisulphate ions in each reaction mixture when varying the concentration of potassium peroxodisulphate (0.04M)? I need to find the concentrations of the ions in each mixture to be able to plot a conc against log(rate) graph. For this type of graph is it ok to use 1/time for the rate? And when finding the gradient, can you just count the graph squares to find change in y over change in x or do you have to use the units used for your axis. Sorry there are so many questions!
Igloo says
The simple answer to your question is that the concentration of your iodide ions in the mixture is one-half of what it was, i.e. 0.5 M
The calculation which you carried out in your question was perfectly sound, but I expect that you’re still puzzled as to how you carry out the calculation generally. I’ll try to help.
You say that your solution of potassium peroxodisulphate has a concentration of 0.04 M.
Let us assume that you measure out 4.0 cm3 of this solution, but that by the time you have added all the other ingredients the total volume is 50 cm3. Divide this total volume by the volume of potassium peroxodisulphate solution used (50/4.0). This gives a value of 12.5. So the solution must have been diluted 12.5 times in the mixture. Since its original concentration was 0.04 M, the new concentration in the mixture must be 0.04/12.5 = 0.0032 M
Hopefully you will now be able to calculate all the other concentrations.
When calculating rate in this context you are quite correct to use (1/time) and to use these values in your plot.
As for determining the gradient, a very common mistake is to do just what you’ve mentioned, i.e. count vertical squares, count horizontal squares, and divide the first value by the second. But this is totally incorrect! By all means count the squares, but, using the scale on the relevant axis, work out the quantity represented by this particular number of squares before dividing the “y quantity” by the “x quantity”.
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updated: 18 December 2006
