Experiment Two

Experiment Two - Experiment Two The Iodide-Persulphate...

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Unformatted text preview: Experiment Two The Iodide-Persulphate Reaction: The Effect of Concentration on Reaction Rate Stephanie Bruno 20313719 Partner: Shireen Misri T.A: Jong Leu Lab Section: 006 Date Due: February 24, 2009 Introduction: The prediction of the relationship between the rate of reaction and the concentration of reactants is called the rate law for a reaction. (Petrucci, 2007) The overall rate law question is: rate of reaction = k [A] m [B] n The exponent’s m and n are the orders of reaction of A and B respectively. (Petrucci, 2007) In this experiment the rate law can be determined by measuring the rate of oxidation of iodide ion by persulphate ion. (Stathopulos, 2008) There are a few equations that are needed to determine this rate law for this particular reaction. For the first reaction, which is the reactant solution, contains (NH 4 ) 2 S 2 O 8 and KI and is shown as: (NH 4 ) 2 S 2 O 8 + 2KI I 2 + (NH 4 ) 2 SO 4 + K 2 SO 4 (1a) For this above reaction, all of the compounds, excluding I 2 are dissociated into ionic species, in aqueous solution. (Stathopulos, 2008) In the equation below the ammonium and potassium ions are not shown because they are not reacting in the solution. (Stathopulos, 2008) This equation is shown as: S 2 O 8 2- + 2I- I 2 + 2SO 4 2- (1b) There are two ways to measure the rate of this reaction, but for this experiment only one way will be used because the second way is difficult and time-consuming. Iodine is being produced in both equation 1a and 1b, so the rate at a specific time can be determined by measuring the amount of iodine produced in a specific time. (Stathopulos, 2008) The time it would take to produce a given amount of iodine is the same thing as finding the amount of iodine produced at a given time to determine the rate at that time. (Stathopulos, 2008) If this measurement of the time it would take to produce a given amount of iodine happens the reaction that will occur is: I 2 + 2Na 2 S 2 O 3 2NaI + Na s S 4 O 6 (2a) Again, in the aqueous solution all of the compounds, excluding I 2 , are completely dissociated into ionic species, therefore the equation is shown as: I 2 + 2S 2 O 3 2- 2I- + S 4 O 6 2- (2b) The colour of the different reactions helps to determine when the reaction is complete. In reaction 1 the production of iodine produces a violet colour. (Stathopulos, 2008) Reaction 2 is colourless because for every one mole of iodine there are two moles of thiosulphate, which takes away the violet colour. (Stathopulos, 2008) When the thiosulphate is used up the reaction turns back to the violet colour. Since the initial amount of thiosulphate, the amount of iodine that would react with the amount of thiosulphate is also known....
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Experiment Two - Experiment Two The Iodide-Persulphate...

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