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Effect of Concentration on the Rate of Reaction Paper

Words: 655, Paragraphs: 16, Pages: 3

Paper type: Essay, Subject: Chemistry

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 A 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:

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S2O82- + 2I- A 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 A 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 A 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
appear will allow us to calculate the rate of the reaction. In this
experiment, 5 mL was also added in order to provide a more accurate
measure of the time at which the colour first appears. Starch is
helpful because it forms a blue complex with free iodine. Once we have
the time elapsed for each run, we can calculate the rate of the
reaction by applying the equation:

Rate = -a??? S2O8-2 / a??? t

The change in S2O8-2 is simply half the concentration of S2O3-2
because in reaction (2), the consumption of iodine and persulphate has
a 1:2 ratio. Thus, the consumption of iodine can be seen as half the
consumption of persulphate (S2O3-2). After calculating the rate of the
reaction, the rate constant can be found by using the equation:

Rate = k [ S2O82-]m[I-] n

By comparing 2 sets of data at a time from 2 different runs, the order
exponents m and n can be calculated, and thus, we can write the
rate law for the iodide-persulphate reaction.

We should also expect that the expected relationships between the
concentration of S2O8-2 and the rate of reaction and rate constant
might not always be extremely accurate in this experiment. When
dealing with ions, we must always consider the ionic strength of the
ions involved. The rate of the reaction will increase as the ionic
strength gets stronger, thus, we will not always see a perfect linear
relationship between [S2O8-2] and the rate of the reaction and the
rate constant.

The purpose of this experiment was to perform many trials of the same
experiment, varying only the concentrations of certain ionic compounds
in order to determine the affect of concentration on the rate of a
reaction. By varying the concentration of different compounds for each
run while keeping other factors constant, we were able to obtain
experimental data that would give us a relationship between
concentration and reaction rate.

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Effect of Concentration on the Rate of Reaction. (2016, Nov 07). Retrieved from https://paperap.com/paper-on-40632/

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