Chapter 14, Problem 89IL

The oxidation of iodide ion by the hypochlorite ion in the presence of hydroxide ions

I⁻(aq) + ClO⁻(aq) → IO⁻(aq) + Cl⁻(aq)

was studied at 25 °C, and the following initial rates data (Y. Chia and R. E. Connick, Journal of Physical Chemistry, Vol. 63, p. 1518, 1959) were collected:

1. (a) Determine the rate law for this reaction.
2. (b) One mechanism that has been proposed for this reaction is the following:

Show that the rate law predicted by this mechanism matches the experimentally determined rate law in part a. (Note that when writing the expression for K the equilibrium constant, [H₂O] is not involved. See Chapter 15.)

Interpretation Introduction

Interpretation:

The rate law of the given reaction should be given.

Concept Introduction:

Rate law: It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

Rate constant: The rate constant for a chemical reaction is the proportionality term in the chemical reaction rate law which gives the relationship between the rate and the concentration of the reactant present in the chemical reaction.

Answer to Problem 89IL

The rate law for the given reaction is as follows,

$\text{Rate law = k}\frac{\left[Cl{O}^{-}\right]\left[I-\right]}{\left[O{H}^{-}\right]}$

Explanation of Solution

The rate law is obtained by first determining the order of each reactant present in the given reaction. The order of each reactant is obtained by using the given set of concentration and the rate data as follows,

The order for $Cl{O}^{-}$ is obtained by considering the experiment 1 and 3 from the given data.

  $\begin{array}{l}\frac{\text{1}}{\text{3}}\text{is}\text{as}\text{follows,}\\ \\ \frac{{\left[Cl{O}^{-}\right]}_1}{{\left[Cl{O}^{-}\right]}_3}\frac{{\left[I^{-}\right]}_1}{{\left[I^{-}\right]}_3}\frac{{\left[O{H}^{-}\right]}_1}{{\left[O{H}^{-}\right]}_3}=\frac{{\left[Rate\right]}_1}{{\left[Rate\right]}_3}\\ \\ \text{4}\times {\text{10}}^{\text{-3}}\cancel{2\times {\text{10}}^{\text{-3}}}\cancel{1.0}\to 4.8\times {\text{10}}^{\text{-4}}\\ \\ 2\times {\text{10}}^{\text{-3}}\cancel{2\times {\text{10}}^{\text{-3}}}\cancel{1.0}\to 2.3\times {\text{10}}^{\text{-4}}\\ \\ \underset{\_}{}\\ \\ 2{}^m\to 2^1\end{array}$

The order for $Cl{O}^{-}$ is 1.

Similarly, the order for $I^{-}$ is obtained as follow:

  $\begin{array}{l}\frac{2}{\text{3}}\text{is}\text{as}\text{follows,}\\ \\ \frac{{\left[Cl{O}^{-}\right]}_2}{{\left[Cl{O}^{-}\right]}_3}\frac{{\left[I^{-}\right]}_2}{{\left[I^{-}\right]}_3}\frac{{\left[O{H}^{-}\right]}_2}{{\left[O{H}^{-}\right]}_3}=\frac{{\left[Rate\right]}_2}{{\left[Rate\right]}_3}\\ \\ \cancel{2\times {\text{10}}^{\text{-3}}}4\times {\text{10}}^{\text{-3}}\cancel{1.0}\to 5.0\times {\text{10}}^{\text{-4}}\\ \\ \cancel{2\times {\text{10}}^{\text{-3}}}2\times {\text{10}}^{\text{-3}}\cancel{1.0}\to 2.3\times {\text{10}}^{\text{-4}}\\ \underset{\_}{}\\ \\ 2{}^n\to 2^1\end{array}$

The order for $I^{-}$ is 1.

Similarly the order for $O{H}^{-}$ is determined by using experiments 3 and 4 shows that it is $-1$.

Therefore, the rate law for the given reaction is as follows,

$\text{Rate law = k}\frac{\left[Cl{O}^{-}\right]\left[I-\right]}{\left[O{H}^{-}\right]}$

Interpretation Introduction

Interpretation:

The rate law predicted by the mechanism matches the experimentally determined rate law should be shown.

Concept Introduction:

Rate order: The order of each reactant in a reaction is represented by the exponential term of the respective reactant present in the rate law and the overall order of the reaction is the sum of all the exponents of all reactants present in the chemical reaction. The order of the reaction is directly proportional to the concentration of the reactants.

Rate law: It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

Rate constant: The rate constant for a chemical reaction is the proportionality term in the chemical reaction rate law which gives the relationship between the rate and the concentration of the reactant present in the chemical reaction.

Explanation of Solution

The rate for the rate determining step is $Rate=k\left[I^{-}\right]\left[HOCl\right]$

$K_{eq}$ is obtained from the first step and it is

$\begin{array}{l}{K}_{eq}=\frac{\left[HOCl\right]\left[O{H}^{-}\right]}{\left[Cl{O}^{-}\right]}\\ \left[HOCl\right]=K_{eq}\frac{\left[Cl{O}^{-}\right]}{\left[O{H}^{-}\right]}\end{array}$

Now,

$Rate=k\left[I^{-}\right]K_{eq}\frac{\left[Cl{O}^{-}\right]}{\left[O{H}^{-}\right]}=\text{k}\frac{\left[Cl{O}^{-}\right]\left[I-\right]}{\left[O{H}^{-}\right]}$

Therefore, it clears that the rate determined from experimental data matches with the above obtained expression derived from elementary steps.