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Introduction Sulphate ions and iodine can be formed from reactions between peroxodisulphate ions and iodine ions. S2O82-aq + 2I-aq -----------> 2S042-aq + I2aq The reactants and the sulphate ions are colourless so the achievements of the experiment must be determined by the iodine produced. If starch is added to the reaction iodine is more clearly visible as a strong complex of blue-black colour is formed. For this experiment I will be measuring the time taken for the iodine to appear in the solution. To make this more clear and accurate thiosulphate ions are added to the mixture which will emphasize the changes. This chemical also returns iodine to iodine ions. 2S2O32-aq + I2aq -----------> S4062-aq + 2I-aq As the reactants of this reactions are colourless it is difficult to tell when the colour change will happen, and when it does it is quite sudden. This happens because the thiosulphate is being used up in the reactions, and when it is all used up the starch-iodine complex is formed resulting in the colour change. Equipment and apparatus needed: "¢ Test tubes "¢ Boiling tubes "¢ Thermometer 0-110°C "¢ Stopwatch "¢ Burettes "¢ 15cm3 Potassium iodide solution,KI, 1.00 mol dm-3 "¢ 10cm3 Potassium peroxodisulphate solution,K2S2O8, 0.0400 mol dm-3 "¢ 10cm3 Sodium thiosulphate solution, Na2S2O3, 0.0100 mol dm-3 "¢ 5cm3 fresh Starch solution "¢ Test tube rack Method "¢ Read and understand carefully the table below illustrating the different mixtures of concentrations for individual experiments. mixture Volume of KLaq/ cm3 Volume of water / cm3 Volume of Na2S2O3aq / cm3 Volume of starch solution/cm3 Volume of K2S2O8aq / cm3 1 5 0 2 1 2 2 4 1 2 1 2 3 3 2 2 1 2 4 2 3 2 1 2 5 1 4 2 1 2 "¢ When making the mixtures, make sure that all chemicals are added to a boiling tube, except the potassium peroxodisulphate which must be added separately as this is what will trigger the reaction. "¢ Make up mixture 1 and place a thermometer in the boiling tube. Add the potassium peroxodisulphate at begin the timing with the clock "¢ Stop timing when the solution turns blue-black "¢ Constantly stir the solution. "¢ Take a temperature reading "¢ Repeat procedure for each mixture of different concentrations. Results A Table mixture Concentration of I-aq / mol dm-3 Clock time / seconds Rate / mol dm-3 s-1 Temperature / °C 1 0.5 28.8 3.45 X10-7 23 2 0.4 40.9 2.44 X10-7 23 3 0.3 58.0 1.72 X10-7 26 4 0.2 136.7 7.32 X10-8 26 5 0.1 455.5 2.20 X10-8 24 Questions B Rate = concentration of I2 Time = [ I2 ] = mol dm-3 s-1 S [ I2 ] = 1.0 X10-5 Mixture: 1. = 1.0 X10-5 = 3.45 X10-7 28.8 2. = 1.0 X10-5 = 2.44 X10-7 40.9 3. = 1.0 X10-5 = 1.72 X10-7 58.0 4. = 1.0 X10-5 = 7.32 X10-8 136.7 5. = 1.0 X10-5 = 2.20 X10-8 455.5 C The Iodine ions are in excess in the reaction mixtures. The reactant that is not in excess in the reaction will be used up and it is this which will determine the total amount of iodine produced. D Moles = mol dm-3 x dm3 = 0.04 x 2/1000 = 8x10-5 mols of S2O8 2- E i 2cm3 of Na2S2O3 0.0100 mol dm-3 Moles of Na2S2O3 = 0.01 x 2/1000 = 2x10-5 moles of S2O3 ii Moles of = 1.0 X10-5 I2 iii Percentage % of reaction = 1.0 X10-5 x100 8.0 X10-5 = 12.5 %
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Introduction Sulphate ions and iodine can be formed from reactions between peroxodisulphate ions and iodine ions. S2O82-aq + 2I-aq -----------> 2S042-aq + I2aq The reactants and the sulphate ions are colourless so the achievements of the experiment must be determined by the iodine produced. If starch is added to the reaction iodine is more clearly visible as a strong complex of blue-black colour is formed. For this experiment I will be measuring the time taken for the iodine to appear in the solution. To make this more clear and accurate thiosulphate ions are added...
mixtures. The reactant that is not in excess in the reaction will be used up and it is this which will determine the total amount of iodine produced.

D Moles = mol dm-3 x dm3

= 0.04 x 2/1000

= 8x10-5 mols of S2O8 2-

E i 2cm3 of Na2S2O3 0.0100 mol dm-3

Moles of Na2S2O3 = 0.01 x 2/1000

= 2x10-5 moles of S2O3

ii Moles of = 1.0 X10-5

I2

iii Percentage % of reaction = 1.0 X10-5 x100

8.0 X10-5

= 12.5 %

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