Question

Explain what “k” means, and how it can be calculated for the iodine clock reaction. As part of your explanation, include a sample calculation. Explain in one well-developed paragraph.

Answer 1

"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 -> HAsO₄^2- + 3 I^- + 4 H⁺

While the reducing agent is present, the net reaction is:
    IO₃^- + 3 H₃AsO₃ -> I^- + 6 H⁺ + 3 HAsO₄^2- .

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 (t_C) from mixing to the color change:
    initial rate = (1/3) [H₃AsO₃]_o / t_C .

The rate of this reaction is expected to be mathematically related to the concentrations of the reactants through a rate constant (k_R) which depends only on the temperature:
    rate = k_R[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 = k_R[IO₃^-]_o^a [I^-]_o^b [H⁺]_o^c = (1/3) [H₃AsO₃]_o / t_C .

So K is the rate constant of the reaction. You can also do this by the temperature effect

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 = k_R[IO₃^-]_o^a [I^-]_o^b [H⁺]_o^c = (1/3) [H₃AsO₃]_o / t_C
In this case, all of the concentrations remain constant so the changes in k_R are inversely related to the changes in the time required for the blue color to appear (t_C):
      k_R= constant/ t_C .
The Arrhenius equation gives the mathematical relationship between k_R and absolute temperature (T):

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

or in logarithmic form:

    ln(k_R) = ln(A) - Ea / RT .

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

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

Hope this helps