Iodine Clock Reaction
"Iodine Clock" refers to a group of reactions which involve the mixing of two colorless solutions to produce a solution which remains colorless for a precise amount of time, then suddenly changes to a deep purple-blue color. The time is controlled by the temperature and/or the concentrations of the reactants.

 The reactions involve the oxidation of iodide ion (I-) to dissolved iodine (I2) or tri-iodide ion (I3-). Either of these combine with starch indicator to produce the characteristic purple-blue color. This color is visible to the eye when the concentration of iodine or tri-iodide ion exceeds 10-5 moles/liter.

 A typical reaction is:
    6 H+ + IO3- + 8 I- -> 3 I3- + 3 H2O

 The rate of this reaction depends on the temperature, and on the concentrations of iodate, iodide, and hydrogen ions. This reaction alone does not give very impressive delays and color changes. The time delay until the appearance of the blue color is inversely related to the rate of the reaction (the faster the reaction, the shorter the delay) but the color development is directly related to the rate (a sharp change in the color requires a moderately fast reaction). Consequently, if the time delay is more than a second, the color development appears relatively slow.

 In order to obtain a time delay of a few seconds to a few minutes with a reasonably sharp color development, a measured amount of a reducing agent (arsenious acid or thiosulfate ion) is included in the mixture. These react very quickly with tri-iodide ion, and very slowly with iodate ion, removing the tri-iodide ion as quickly as it is produced, so that the concentration does not reach the visible level until all of the reducing agent is consumed.
    H3AsO3 + I3- + H2O -> HAsO42- + 3 I- + 4 H+

While the reducing agent is present, the net reaction is:
    IO3- + 3 H3AsO3 -> I- + 6 H+ + 3 HAsO42- .

 This reaction has been used extensively for laboratory experiments involving investigation of the rate expressions in Chemical Kinetics. The concentration of hydrogen ion is maintained by using Acetic Acid/Sodium Acetate buffers in the pH range 4 -5. The initial concentration of arsenious acid is less than 1/3 of the initial concentrations of iodate and iodide ions, so that these concentrations change by less than 10% before the reducing agent is consumed and the sharp color change to blue is observed.

 Solutions are prepared with the iodate ion, starch, arsenious acid, and buffer solution in one flask and a potassium iodide solution in another flask. These solutions are equilibrated at a known temperature in a constant temperature bath. A timer is started as the solutions are mixed, and stopped when the blue color appears. Times are measured for a series of mixtures with the same initial concentration of arsenious acid while varying the initial concentrations of iodate, iodide, and hydrogen ion and possibly the temperature. Typically, one of the initial concentrations will be changed by a factor (2,3, or 4) while the other two initial concentrations are held constant. The process is then repeated for each of the other two components.

Typical Experiment
Run #    [IO3-]o   [I-]o     [H+]o   time to blue color (sec) 
2 x 10-5
2 x 10-5
2 x 10-5
4 x 10-5

 The rate of the reaction:
    6 H+ + IO3- + 8 I- -> 3 I3- + 3 H2O
can be described as the rate of disappearance of iodate ion. In the initial stages of the reaction, this is also equal to 1/3 the rate of disappearance of arsenious acid:
    rate = - d[IO3-]/dt = - (1/3)d[H3AsO3]/dt .

The initial rate is approximated from the initial concentration of arsenious acid and the time (tC) from mixing to the color change:
    initial rate = (1/3) [H3AsO3]o / tC .

The rate of this reaction is expected to be mathematically related to the concentrations of the reactants through a rate constant (kR) which depends only on the temperature:
    rate = kR[IO3-]a[I-]b[H+]c ,

with the exponents (also called "order") a, b, and c expected to be integers (0, 1, 2, or 3) or half-integers (1/2, 3/2, 5/2). The reaction is said to have an "overall order" of a + b + c.

The initial rate is then related to the initial concentrations:
    initial rate = kR[IO3-]oa [I-]ob [H+]oc = (1/3) [H3AsO3]o / tC .

If measurements are performed as suggested above, the results will be:

    Run#1:   kR (0.005)a (0.05)b (2 x 10-5)c = [H3AsO3]o / t1
    Run#2:   kR (0.010)a (0.05)b (2 x 10-5)c = [H3AsO3]o / t2
    Run#3:   kR (0.005)a (0.10)b (2 x 10-5)c = [H3AsO3]o / t3
    Run#4:   kR (0.005)a (0.05)b (4 x 10-5)c = [H3AsO3]o / t4

If the equation for Run #2 is divided by the equation for Run #1, most of the terms cancel, leaving
    (0.010 / 0.005)a = 2a = t1 / t2 .

Similarly, division of the equation for Run #3 by the equation for Run #1 gives
    (0.010 / 0.005)b = 2b = t1 / t3,

and division of the equation for Run #4 by the equation for Run #1 gives
    ((4 x 10-5 / (2 x 10-5)c = 2c = t1 / t4 .

All of these equations involve 2 raised to a power which is an integer or half-integer.  The values of the exponents may be easily estimated by assuming values:

        a     ,      b      ,      c      :     0       1/2       1       3/2       2      5/2       3

    t1 / t2  ,   t1 / t3   , t1 / t4   :    1.0     1.4       2.0     2.8     4.0     5.7       8

After the exponents have been determined, the equations for Runs 1 to 4 may be solved explicitly to obtain a value of kR from each equation, and these values are averaged to obtain the "best" value for kR.

Temperature Effects

    To study the effect of temperature on this reaction, a series of measurements are made at different temperatures with identical initial concentrations of all of the components.  As above,
    initial rate = kR[IO3-]oa [I-]ob [H+]oc = (1/3) [H3AsO3]o / tC
In this case, all of the concentrations remain constant so the changes in kR are inversely related to the changes in the time required for the blue color to appear (tC):
      kR= constant / tC .
The Arrhenius equation gives the mathematical relationship between kR and absolute temperature (T):

    kR= A e - Ea / RT    (A & Ea are constants for the reaction, R is the Gas Law Constant)

or in logarithmic form:

    ln(kR) = ln(A) - Ea / RT  .

From the relationship between kR and tC above,
    ln(tC) = ln(constant / A) + Ea / RT  .

A graph of the logarithm of time, ln(tC), vs  the reciprocal of the absolute temperature (1/T) should be a straight line with slope =  Ea / R .