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The Effect of Concentration on Reaction Rate

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The Effect of Concentration on Reaction Rate


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 à 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- à 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 à 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 à 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 = -∆S2O8-2 / ∆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.

Experimental Procedure:

Refer to Experiment 3, pages 3-1 to 3-4 within First Year Chemistry:
Chem 123L Laboratory Manual

Experimental Observations:

Table 1 - Data Sheet

Add to Erlenmeyer

Add to Beaker

Run #

0.1 M (NH4)S2O8 (mL)

0.1 M (NH4)2SO4 (mL)

0.2 M

KI

(mL)

0.2 M KNO3 (mL)

0.01M Na2S2O3 (mL)

H2O

(mL)

0.2% Starch (mL)

Elapsed Time (s)

1

20

0

20

0

10

0

5

113.5

2

10

10

20

0

10

0

5

218.2

3

5

15

20

0

10

0

5

476.3

4

20

0

10

10

10

0

5

228.2

5

20

0

5

15

10

0

5

431.9

6

10

0

20

0

10

10

5

183.4

7

20

0

5

0

10

15

5

574.6

Results and Calculations:

1. Sample Calculation: [S2O8-2], [I-], [S2O3-2], -∆S2O8-2 =? mol L-1

Run #1:

[S2O8-2] = 20 mL (NH4)S2O8 x 0.1 mol (NH4)S2O8 = 3.64 x 10-2 mol L-1

55 mL total 1 L (NH4)S2O8

[I-] = 20 mL KI x 0.2 mol KI = 7.27 x 10-2 mol L-1

55 mL total 1 L KI

[S2O3-2] = 10 mL Na2S2O3 x 0.01 mol Na2S2O3 = 1.82 x 10-3 mol L-1

55 mL total 1 L Na2S2O3

-∆S2O8-2 = 1/2 [S2O3-2] = -9.09 x 10-4 mol L-1

2.

[IMAGE]

[IMAGE]

From plots 1 &2:

Since; -∆S2O8-2 = k [S2O8-2]m[I-] n

∆t

and since -∆S2O8-2, k and [I-]n are constant in runs 1, 2 and 3
(referring to plot 1), we can rewrite the equation to yield:

1 = B [S2O8-2]m where B = k [I-]n

∆t ∆S2O8-2

Thus we can find the exponent the order "m" with respect to S2O8-2 by
writing the equation of the line in the form of y = mx + b where m=
slope of plot 1= exponent

i.e. - log ∆t = log B + m log [S2O8-2]

Similarly, we can find the order "n" with respect to I- by using the
equation

1 = A [I-]n where A = k [S2O8-2]m

∆t ∆S2O8-2

and by rewriting the equation of the line in the form of y = nx + b
where n= slope of plot 2 = exponent

i.e. - log ∆t = log A + n log [I-]

slope 1: -log ∆t vs. log [S2O8-2]

m= rise = -2.678 - (-2.055) = -0.623 = 1

run -2.041 - (-1.439) -0.602

slope 2: -log ∆t vs. log [I-]

n= rise = -2.635 - (-2.055) = -0.58 = 1

run -1.74 - (-1.138) -0.602

3. Sample Calculations:

Rate of Reaction for run #1:

Rate = -∆S2O8-2 / ∆t = -9.09 x 10-4 M / 113.5 s = 8.01 x 10-6 M s-1

Rate constant (k) for run #1:

k = Rate / ( [S2O8-2]m [I-]n )

k = 8.01 x 10-6 M s-1 / {(3.64 x 10-2 M)1(7.27 x 10-2 M)1}

k = 3.03 x 10-3 s-1

Ionic Strength (µ) for run #1:

µ = 0.5 Σ CiZi2

µ = 0.5{([NH4]x(+1)2) + ([S2O8]x(-2)2) + ([K]x(+1)2) + ([I]x(-1)2) +
([Na]x(+1)2) + ([S2O3]x(-2)2)}

µ = 0.187 mol L-1

Table 2 - Calculations Summary Table

Run #

[S2O8-2] (M)

[I-]

(M)

[S2O3-2] (M)

-∆S2O8-2 (M)

∆t

(s)

Rate

(M s-1)

Rate Constant, k (s-1)

Ionic Strength (M)

1

3.64 x 10-2

7.27 x 10-2

1.82 x 10-3

-9.09 x 10-4

114

-8.01 x 10-6

3.03 x 10-3

0.187

2

1.82 x 10-2

7.27 x 10-2

1.82 x 10-3

-9.09 x 10-4

218

-4.16 x 10-6

3.15 x 10-3

0.187

3

9.09 x 10-3

7.27 x 10-2

1.82 x 10-3

-9.09 x 10-4

476

-1.91 x 10-6

2.89 x 10-3

0.187

4

3.64 x 10-2

3.64 x 10-2

1.82 x 10-3

-9.09 x 10-4

228

-3.98 x 10-6

3.01 x 10-3

0.187

5

3.64 x 10-2

1.82 x 10-2

1.82 x 10-3

-9.09 x 10-4

432

-2.10 x 10-6

3.18 x 10-3

0.187

6

1.82 x 10-2

7.27 x 10-2

1.82 x 10-3

-9.09 x 10-4

183

-4.96 x 10-6

3.75 x 10-3

0.133

7

3.64 x 10-2

1.82 x 10-2

1.82 x 10-3

-9.09 x 10-4

575

-1.58 x 10-6

2.39 x 10-3

0.133

Questions:

1. The average rate constant for this reaction for runs 1 to 5 can be
calculated using the equation:

Avg. rate constant = (k1 + k2 + k3 + k4 + k5)/ 5 = 3.05 x 10-3 s-1

Therefore, the average rate constant is approximately 3.05 x 10-3 s-1.

2. The k value for run #2 is close to the k value of run #6 - it has a
deviation of approximately 16%. However, the k value for run #5 is
still similar to the k value for run #7, but it has a higher deviation
than the first comparison - it has a deviation of approximately 25%.
We can say that all the k values for all 7 runs are similar to one
another since the difference between each trial is not much. However,
we can account for run #5 and run #7 having a larger deviation between
their k values because it has a larger difference in volume. Runs #2
and #6 only differ with 10 mL of 0.1 M (NH4)2SO4. An extra addition of
10 mL of (NH4)2SO4 to the reaction will cause a slight decrease in the
rate of the entire reaction as we can see in Table 1. Run #2 has a
faster rate of reaction than run #6 because when we added more (NH4)2SO4,
it hindered the product of more iodine, which will in turn take longer
to react completely with the sodium thiosulphate, our iodine
indicator. Since we calculated our rate of reaction based on the
disappearance of the sodium thiosulphate colour, if there is hindering
of the production of iodine, it will result in a slower rate of
reaction. Runs #5 and 7 have k values that have a larger deviation
because unlike runs #2 and 6, they differ by 15 mL of 0.2 M KNO3. Run
#5 had the extra 15 mL of KNO3 and resulted in a faster reaction rate
because it pushes the reaction in the direction that produces more
iodine, and this in turn will result in a faster rate of reaction for
the same reasons above.

3. It is very important when reporting a rate constant for a reaction
that the ionic strength and temperature be stated. Ionic strength and
temperature of the solution and/or surroundings can have immense
effects on the rate of the reaction. This can affect the rate constant
heavily because these are two factors that affect the rate of
reaction, which is linearly proportional to the rate constant. For
example, if we were to raise the temperature of the solution, the rate
of the reaction would increase. Ionic strength can also have a great
influence on the rate constant and the rate of reaction; at low
concentrations of reactants, the majority of effects on the rate of
reaction can be from ionic strength, disregarding the actual chemical
identity of the ion.

4. If the S2O8-2 concentration were doubled, the rate of reaction and
rate constant would both be affected. The rate constant depends on the
concentration of S2O8-2; they have a linear relationship. That means
that if the concentration of S2O8-2 is doubled, then - assuming
everything else remains constant - the rate constant will double as
well. It is clear that if the rate constant changes than the rate of
reaction should change as well. If every other component is constant
and the only variation occurs with the concentration of S2O8-2
concentration, the rate of reaction is linearly proportional to the S2O8-2
concentration. That means that if the concentration doubles than the
rate of the reaction should double, more or less. This can be quite
clear when comparing runs #1 and 2.

Discussion:

By measuring the amount of iodine consumed after a certain amount of
time elapsed, we were able to calculate the rate of the overall
reaction. Knowing that the concentration of S2O8-2 was linearly
proportional to the rate constant, we can see in the results that the
relationship is very close to being linear, but there is some
variation. This can be explained by the ionic strength factor. Since
the reaction involves 2 negatively charged ions, the rate is greatly
dependant on ionic strength; the greater the ionic strength, the
faster the rate of reaction. In order to optimize the conditions for
this experiment, there must be an addition of an electrolyte such as
(NH4)2SO4 or KNO3 when the reactant ions are being lowered to keep the
ionic strength constant. For example in runs #1 and 6, the
concentration of the persulphate ion has decreased by a half in run #6
and the iodide ion is constant; this should give the rate constant in
run #6 a decrease by a half. However we can see that this is not the
case. Since the ionic strength has also decreased, it has some effect
on the resulting rate constant and therefore skews the results a bit.
The rest of the results seem to agree with the logical way the
experiment should have occurred. For example, the runs with the longer
elapsed times had the slower reaction rates and vice versa with the
runs with the shorter elapsed times. This makes sense due to the
linear relationship between reaction rate and time.

Some sources of error in this experiment may have been a mistake in
mixing certain reactants, or inaccuracy with measuring volume of the
solutions. It was more likely that there was inaccurate measuring of
the solutions because it was quite difficult to always use the Mohr
and transfer pipettes precisely.

Conclusions:

The purpose of this experiment was to determine how concentration of a
certain reactant in a reaction can affect the rate of the entire
reaction. The experiment was overall a success because we could see
that when we varied the concentrations of certain compounds, the
reaction rate was affected accordingly. Overall we know that the rate
of the reaction is linearly proportional to the concentration of your
reactant. However, if your reaction can exist in equilibrium and you
increase the concentration of a product, the reaction will favour in
the left direction, and if you are measuring rate of product
formation, this will result in a decrease in reaction rate.

Reference(s):

Chemistry Department, First Year Chemistry: Chem 123L Laboratory
Manual. University of Waterloo: 2004

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