Morrow 1

Justin Morrow

Mrs. Keep

AP Chemistry

21 December 2012

Iodine Clock Reaction Lab

Purpose

The purpose of this experiment is to determine the rate law of the iodine clock reaction by using the method of initial rates.

Theory

Sodium meta-bisulfite (Na2S2O5) undergoes the following transformation when dissolved in water where it dissolves and reacts with the water to form bisulfite (HSO3

-):

Na2S2O5 + H2O 2HSO3- + 2Na+

When mixed with potassium iodate (KIO3) the following reaction mechanism is proposed leading to the observed color change:

IO3- + 3HSO3

- I- + 3H+ + 3SO42- Slow

I- + IO3- I2 + O3

2- Fast

I2 + HSO3- + H2O 2I- + SO4

2- + 3H+ Fast

If we assume the reactions following the first are much faster in comparison, the rate of the first reaction can be determined by varying the initial amounts of reactions and measuring the rate.

Assuming all of the sodium meta-bisulfate is consumed by the time the color change takes place,

Rate = -Δ[HSO3] / Δt

This rate can be equated to:

Rate = k [ IO3- ]x [ HSO3

- ]y

Mathematically, the variables can be determined with sufficient data.

Procedure

Morrow 2

1. Prepare a series of 4 solutions called solution B by mixing differing amounts of 0.10 M sodium meta-bisulfite, starch solution, and distilled water.

2. Prepare a second series of 4 solutions (solution A) by diluting 0.20 M potassium iodate solution with distilled water.

3. Pour solution B into solution A quickly while stirring and time the reaction using the second hand of a watch (or a digital timer). Note the initial concentrations of potassium iodate and bisulfate ion, and time for color change.

Results

Calculation Example 1 – Initial Concentration of sodium meta-bisulfite (Na2S2O5) and potassium iodate (KIO3)

Run Volume 0.1M Na2S2O5

Total Volume

1 10.mL 280.mL2 10.mL 280.mL3 10.mL 280.mL4 5.mL 280.mL

MdVd=McVc

Md(280mL)=(.10M)(10.mL)Md=.0036M Na2S2O5

Run Volume 0.20M KIO3

Total Volume

1 100.mL 280.mL2 50.mL 280.mL3 25.mL 280.mL4 100.mL 280.mL

MdVd=McVc

Md(280.mL)=(.20M)(100.mL)Md=0.071M KIO3

Run Concentration Na2S2O5

Concentration KIO3

Morrow 3

1 0.0036M 0.071M2 0.0036M 0.036M3 0.0036M 0.018M4 0.0018M 0.071M

Calculation Example 2 – Initial [IO3-] and [HSO3

-]

[Na2S2O5] * 2 = [HSO3-] (as taken from the balanced reaction equation)

(0.0036M Na2S2O5) * 2 = 0.0072M HSO3-

[KIO3] * 1 = [IO3-] (as taken from the balanced reaction equation)

(0.071M KIO3) * 1 = 0.071M IO3-

Calculation Example 3 – “Initial” (Average) Rate

Rate = -Δ[HSO3] / Δt

0.071M / 4.2s = 0.017 mol L-1 s-1

Run Time Initial [IO3-] Initial [HSO3

-] Rate

1 4.2 s 0.071M 0.0072M 0.0017 mol L-1 s-1

2 8.5 s 0.036M 0.0072M 0.00085 mol L-1 s-1

3 16.8 s 0.018M 0.0072M 0.00043 mol L-1 s-1

4 8.9 s 0.071M 0.0036M 0.00041 mol L-1 s-1

Rate Law

Between runs 1 + 2, and 2 + 3 the initial [IO3-] halves. The rate also halves when

comparing these runs. Therefore, the order of reaction with respect to [IO3-] is 1st order.

While the [HSO3-] is halved, (in comparing runs 1 + 4) the rate changes by a factor of 1/4.

Therefore, the order of reaction with respect to [HSO3-] is 2nd order.

Calculating K

With the rate law of: Rate = k [IO3-][HSO3

-]2

We can calculate the value of the rate constant.

Rate = 0.0017 mol L-1 s-1

Morrow 4

[IO3-] = 0.071M

[HSO3-] = 0.0072M

0.0017 mol L-1 s-1 = k(0.071M)( 0.0072M)2

K = (0.0017 mol L-1 s-1)/(3.7E-6 mol3 L-3)

K = 460 L2 mol-2 s-1

Discussion

Although the rate calculated is fairly accurate, there is some error produced by the procedure. It was assumed that the initial rate is found by the slope of the starting concentration to the ending concentration (because the [HSO3

-] cannot be found until it is gone). This is not true because the initial slope is steeper than the calculated slope.

[HSO3-]

Calculated rate

Initial rate time

Therefore, the calculated initial rate is lower than the real initial rate.

Conclusion

Therefore, the calculated rate law of this reaction is:

Rate = (460 L2 mol-2 s-1) [IO3-] [HSO3

-]2