Effect of Concentration on the Rate of Reaction

Introduction:

In this experiment, we utilized the ability for the iodide ion to
become oxidized by the persulphate ion. Our general reaction can be
described as:

(NH4)2S2O8 + 2KI Ã   I2 + (NH4)2SO4 + K2SO4 (1a)

However, we know that in an aqueous solution, all of these compounds
except iodine will dissociate into their ionic components. Thus we can
rewrite the equation in a more convenient manner:

S2O82- + 2I- Ã   I2 + 2SO42- (1b)

It is important however to note that the NH4 and K ions are still in
the solution, they are just unreactive. In order to measure the rate
of the reaction, the conventional method would be to measure the
species in question at certain times. However, this would be
inconvenient, especially for a three hour laboratory period. Since the
iodide ion can be oxidized by the persulphate ion, we can use sodium
thiosulphate to be an indicator of the presence of iodine in the
solution. For this experiment, we can simply calculate the rate of the
reaction by timing the amount of iodine being produced in several
runs. The reaction between iodine and sodium persulphate can be
depicted as:

I2 + 2Na2S2O3 Ã   2NaI + Na2S4O6 (2a)

Similarly, this reaction above can also be simplified due to
dissociation of all the ions except for iodine and persulphate.

I2 + 2S2O3 Ã   2I- + S4O62- (2b)

An interesting property of reaction (1) is that it produces a
brilliant violet colour. However, this violet colour only results in
the presence of iodine, or in other words, when iodine is being
produced in the reaction. If sodium thiosulphate is added to reaction
(1), than as long as there are two moles of thiosulphate for every
mole of iodine, the solution will be colourless because the iodine is
being used up in reaction (2). However, as time passes, the
thiosulphate must run out at some point, and when it does, the violet
colour will appear. Timing how long it takes for the violet colour to...