Chapter 13, Problem 13.79QP

The bromination of acetone is acid-catalyzed:

${\text{CH}}_3{\text{COCH}}_3+{\text{Br}}_2\underset{\text{catalyst}}{\overset{{\text{H}}^{+}}{\to }}{\text{CH}}_3{\text{COCH}}_2\text{Br}+{\text{H}}^{+}+{\text{Br}}^{-}$

The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and H⁺ ions at a certain temperature:

1. (a) What is the rate law for the reaction? (b) Determine the rate constant. (c) The following mechanism has been proposed for the reaction:

Show that the rate law deduced from the mechanism is consistent with that shown in (a).

Interpretation Introduction

Interpretation:

The following from the given information has to be determined.

The rate law for the given reaction

Concept introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of reactants consumed and how much concentration of product formed in a unit of time is said to be rate of reaction.

• Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

To determine the rate law for the given reaction

Answer to Problem 13.79QP

The following from the given information has to be determined.

The rate law for the given reaction is ${\text{rate = k[CH}}_{\text{3}}{\text{COCH}}_{\text{3}}{\text{][H}}^{\text{+}}\text{]}$

Explanation of Solution

The given reaction is

${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}{\text{+Br}}_{\text{2}}\stackrel{\begin{array}{l}\text{Catlyst}\\ \\ {\text{ H}}^{\text{+}}\end{array}}{\to }{\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{Br + H}}^{\text{+}}{\text{+Br}}^{\text{-}}$

The rate law for the above reaction in general is

${\text{rate = k[CH}}_{\text{3}}{\text{COCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}^{\text{+}}\text{]}}^{\text{z}}$

In the above equation we need to find out the exponents of the reactant concentrations, which can be determined by comparing the information given in the table.

The exponents for ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}$, let’s compare the experiments 1 and 5 then find out as first order in ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}$

The exponents for ${\text{Br}}_{\text{2}}$, let’s compare the experiments 1 and 2 then find out as zero order in ${\text{Br}}_{\text{2}}$

The exponents for ${\text{H}}^{\text{+}}$, let’s compare the experiments 1 and 3 then find out as first order in ${\text{H}}^{\text{+}}$

Substituting these orders in the above reaction and we get rate law for the given reaction as

${\text{rate = k[CH}}_{\text{3}}{\text{COCH}}_{\text{3}}{\text{][H}}^{\text{+}}\text{]}$

Interpretation Introduction

Interpretation:

The following from the given information has to be determined.

The rate constant for the given reaction

Concept introduction:

Rateof a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of reactants consumed and how much concentration of product formed in a unit of time is said to be rate of reaction.

• Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Rate law: It is an equation that related to the rate of reaction to the concentrationsor pressures of substrates (reactants).  It is also said to be as rate equation.

To determine the rate constant for the given reaction

Answer to Problem 13.79QP

The following from the given information has to be determined.

The rate constant for the given reaction is $\text{k}=3.8\times {10}^{-3}/M.s$

Explanation of Solution

In order to calculate the rate constant for the given reaction, modify the rate law as follows

$\text{ k= }\frac{\text{rate}}{{\text{[CH}}_{\text{3}}{\text{COCH}}_{\text{3}}{\text{][H}}^{\text{+}}\text{]}}$

$\begin{array}{l}\text{ k = }\frac{5.7\times {10}^{-5}M/s}{\left(0.30M\right)\left(0.050M\right)}=3.8\times {10}^{-3}/M.s\\ \\ \text{k}=3.8\times {10}^{-3}/M.s\end{array}$

Interpretation Introduction

Interpretation:

The following from the given information has to be determined.

 The rate law deduced from the mechanism is consistent with that shown in part (a) has to be shown.

Concept introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of reactants consumed and how much concentration of product formed in a unit of time is said to be rate of reaction.

• Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Rate law: It is an equation that related to the rate of reaction to the concentrationsor pressures of substrates (reactants).  It is also said to be as rate equation.

Answer to Problem 13.79QP

The following from the given information has to be determined.

The rate law deduced from the mechanism is consistent with that shown in part (a) is shown.

Explanation of Solution

Assume that $k_2$ as rate constant for the slow step

${\text{rate = k}}_{\text{2}}\text{[CH3COCH3][H3O+]}$

Now, $k_{1}and{k}_{-1}$ are the rate constants for the forward and reverse steps in the equilibrium

${\text{k}}_{\text{1}}{\text{[CH}}_{\text{3}}{\text{COCH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = k}}_{\text{-1}}\text{[CH3COCH3][H3O+]}$

$\text{rate =}\frac{{\text{ k}}_{\text{1}}{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-1}}}{\text{[CH}}_{\text{3}}{\text{COCH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }$

And which is same as $\text{rate =}\frac{{\text{ k}}_{\text{1}}{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-1}}}$