Chapter 14, Problem 90IL

The acid-catalyzed iodination of acetone

CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq)

is a common laboratory experiment used in general chemistry courses to teach the method of initial rates. The reaction is followed spectrophotometrically by the disappearance of the color of iodine in the solution. The following data (J. P. Birk and D. L Walters, Journal of Chemical Education, Vol. 69, p. 585, 1992) were collected at 23 °C for this reaction.

Determine the rate law for this reaction.

Interpretation Introduction

Interpretation:

The rate law should be determined for the given chemical reaction.

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 90IL

The rate law for the given reaction is as follows,

$\text{Rate law = k}\left[C{H}_{3}COC{H}_3\right]\left[H^{+}\right]$

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 $C{H}_{3}COC{H}_3$ is obtained by considering the experiment 2 and 3 from the given data.

  $\begin{array}{l}\frac{2}{\text{3}}\text{is}\text{as}\text{follows,}\\ \\ \frac{{\left[C{H}_{3}COC{H}_3\right]}_2}{{\left[C{H}_{3}COC{H}_3\right]}_3}\frac{{\left[H^{+}\right]}_2}{{\left[H^{+}\right]}_3}\frac{{\left[I_2\right]}_2}{{\left[I_2\right]}_3}=\frac{{\left[Rate\right]}_2}{{\left[Rate\right]}_3}\\ \\ 1.33\cancel{0.323}\cancel{0.00665}\to 1.7\times {\text{10}}^{\text{-5}}\\ \\ 0.667\cancel{0.323}\cancel{0.00665}\to 7.6\times {\text{10}}^{\text{-6}}\\ \\ \underset{\_}{}\\ \\ 2{}^m\to 2^1\end{array}$

The order for $C{H}_{3}COC{H}_3$ is 1.

Similarly, the order for $H^{+}$ is obtained as follow,

  $\begin{array}{l}\frac{1}{2}\text{is}\text{as}\text{follows,}\\ \\ \frac{{\left[C{H}_{3}COC{H}_3\right]}_1}{{\left[C{H}_{3}COC{H}_3\right]}_2}\frac{{\left[H^{+}\right]}_1}{{\left[H^{+}\right]}_2}\frac{{\left[I_2\right]}_1}{{\left[I_2\right]}_2}=\frac{{\left[Rate\right]}_1}{{\left[Rate\right]}_2}\\ \\ \cancel{1.33}0.162\cancel{0.00665}\to 8.1\times {\text{10}}^{\text{-6}}\\ \\ \cancel{1.33}0.323\cancel{0.00665}\to 1.7\times {\text{10}}^{\text{-5}}\\ \underset{\_}{}\\ \\ 0.5{}^n\to {0.5}^1\end{array}$

The order for $H^{+}$ is 1.

Similarly the order for $I_2$ is determined by using experiments 4 and 5 shows that it is zero order.

  $\begin{array}{l}\frac{4}{5}\text{is}\text{as}\text{follows,}\\ \\ \frac{{\left[C{H}_{3}COC{H}_3\right]}_4}{{\left[C{H}_{3}COC{H}_3\right]}_5}\frac{{\left[H^{+}\right]}_4}{{\left[H^{+}\right]}_5}\frac{{\left[I_2\right]}_4}{{\left[I_2\right]}_5}=\frac{{\left[Rate\right]}_4}{{\left[Rate\right]}_5}\\ \\ \cancel{0.333}\cancel{0.323}0.00665\to 3.8\times {\text{10}}^{\text{-6}}\\ \\ \cancel{0.333}\cancel{0.323}0.00332\to 3.6\times {\text{10}}^{\text{-6}}\\ \underset{\_}{}\\ 2\to 1\\ 2^0\to 1\end{array}$

The order for $I_2$ is 0.

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

  $\begin{array}{l}\text{Rate law = k}{\left[C{H}_{3}COC{H}_3\right]}^1{\left[H^{+}\right]}^1{\left[I_2\right]}^0\\ \\ \text{Rate law = k}\left[C{H}_{3}COC{H}_3\right]\left[H^{+}\right]\end{array}$

Conclusion

The rate law for the given reaction was determined.