The Iodide-Persulphate Reaction: the Effect of Concentration on Reaction Rate
790 Words4 Pages
This experiment will measure the rate of oxidation of iodide ions by persulphate ions to derive the rate law for the reaction. Starch will be added to the reaction to facilitate the measure of time during the reaction. The reactant solutions will contain (NH4)2SO4 and KI, represented as:
(NH4)2S2O8 + 2KI -> I2 + (NH4)2SO4 + K2SO4
This can be simplified to:
S2O82- + 2I- -> I2 + 2SO42-
These equations can only be carried out and be visible after the iodine has completely reacted with thiosulphate added – two moles of thiosulphate for every mole of iodine. Once all the thiosulphate has been used up in the reaction, the colour will start to appear.…show more content… This shows that the exponent m for [S2O8-2] equals 1, as [I-] is held constant through these values.
Graph of –log∆t versus log[I-] as calculated in trials 1,4,5
The slope of this graph, calculated using the average rise / run is also 1. As [S2O8-2] is held constant during these trials, the exponent n for [I-] equals one. The k constant can now be calculated.
The average rate constant for runs 1-5 =
(0.0104+0.00532+0.00641+0.00368+0.00268)/5 = 0.0570 Lmol-1sec-1
Compare k – values.
The k value in run 2 is nearly double that of run 6. This is directly related to the ∆t found for each, with ∆t2 being half that of ∆t6. This is a result of the H2O being present in run 6 while (NH4)2SO4 being present in run 2.
The k value in run 7 is 75% of that in run 5. This will be directly related to the addition of KNO3 in run 5 and H2O in run 7. These will make different ionic concentrations, resulting in the change in rate constant
A difference in temperature will change the reaction rate of the chemical reaction