Determining activation energy (Ea) of a reaction potassium between peroxodisulphate and iodine

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Ho Ka Wing (9) Group: 3 Date: 09-11-09

Determining activation energy (Ea) of a reaction potassium between peroxodisulphate and iodine

Objective

To determine the activation energy for the reduction of peroxodisulphate(VI)ions. S2O82- (aq) by iodide ions I- , using ‘clock reaction’.

Theory

Activation Energy (Ea)-The minimum amount of energy required to start a reaction.

A certain amount of activation must be supplied to initiating a reaction (even if the reaction is exothermic) because this energy must be absorbed to weaken or even break the bond holding the atoms in the reactant particles before the bond formation. The activation energy is used to overcome the energy barrier in the reaction. If the reactants do not gain energy that is greater or equal to the activation energy, the reaction won’t occur.

The equation for the reduction of peroxodisulphate(VI)ions by iodide ions : S2O82- (aq)+2I-(aq) -> 2SO42-(aq) +I2(aq)

The equation of the reaction between thiosulphate ions and iodine formed in the above equation: 2S2O32-(aq) +I2 (aq) ->S4O62-(aq) +2I-(aq)

Controlled Variables:

1. Concentration of potassium peroxodisulphate(VI)solution, potassium iodide solution,sodium thiosulphate solution and starch solution

2. Volume of potassium peroxodisulphate(VI)solution, potassium iodide solution, sodium thiosulphate solution and starch solution

Dependent variable:

- Time taken for the appearance of the blue-black color of the iodine-starch complex

Independent variables:

-temperature of the mixture

Small amount Na2S2O3of is used to restrict the main reaction (the first reaction) to its early stage when the concentration time curve is almost linear. The Na2S2O3 (aq) acts as a gatekeeper.

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If Na2S2O3 was absent. After mixing the solution, the reaction will take place immediately. The iodine produced will immediate react with starch to form blue complex .The time taken for the appearance of blue colour of each 5 temp reading will be would be too quick and immeasurable

The reaction between peroxodisulphate (VI) ions and iodide ions in solution forms sulphate (VI) ions and iodine. The equation for this reaction is as follows:

S2O82- (aq) + 2I- (aq) → 2SO42- (aq) + 2I2 (aq)

The iodine formed immediately reacts with sodium thiosulphate.

2S2O32-(aq)+I2(aq)->S4O62-(aq)+2I-(aq)

When all the sodium thiosulphate solution is used, free ...

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