Chapter 14, Problem 85AP

The bromination of acetone is acid-catalyzed:

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

The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and ${\text{H}}^{\text{+}}$ ions at a certain temperature:

	$\begin{array}{l}\left[C{H}_{3}COC{H}_3\right]\left(M\right)\hfill \end{array}$	$\left[B{r}_2\right]\left(M\right)$	$\left[{\text{H}}^{+}\right]\left(M\right)$	Rate of Disappearance of ${\text{Br}}_2\left(M/s\right)$
(1)	0.30	0.050	0.050	$5.7\times {10}^{-5}$
(2)	0.30	0.10	0.050	$5.7\times {10}^{-5}$
(3)	0.30	0.050	0.10	$1.2\times {10}^{-}{}^4$
(4)	0.40	0.050	0.20	$3.1\times {10}^{-4}$
(5)	0.40	0.050	0.050	$7.6\times {10}^{-5}$

(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 part (a).

Interpretation Introduction

Interpretation:

The rate law and rate constant for the reaction are to be determined. The rate law is consistent with the mechanism shown in part (a), is to be determined.

Concept introduction:

The rate of reaction in terms of concentration is called the rate law.

The step-by-step reaction is called the reaction mechanism. The reaction that takes place in a single step is called an elementary reaction.

Answer to Problem 85AP

Solution:

(a) $\text{rate}=k\left[{\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}\right]\left[{\text{H}}^{\text{+}}\right]$

(b) $3.8\times {10}^{-3}{M}^{-1}{\text{s}}^{-1}$

(c) The rate law deduced from the mechanism is consistent.

Explanation of Solution

a)The rate law for the reaction

For the determination of rate law, calculate the exponents as follows:

Based on $1$ st and $5$ th experiments, the rate law for acetone is:

$\begin{array}{c}\frac{{\left[C{H}_{3}COC{H}_3\right]}_5}{{\left[C{H}_{3}COC{H}_3\right]}_1}=\frac{0.40}{0.30}\\ =1.333\\ \frac{{\left[rate\right]}_5}{{\left[rate\right]}_1}=\frac{7.6\times {10}^{-5}M}{5.7\times {10}^{-5}M}\\ =1.333\end{array}$

Based on $1$ st and5th experiments, $\left[B{r}_2\right]$ and $\left[H^{+}\right]$ concentrations are constant. The concentration of acetone is increased by $1.333$. So, the reactionis a first order reaction.

Based on $1$ st and $2$ nd experiments, $\left(B{r}_2\right)$ concentration doubles but the reaction rate does not change. So, the reaction is zero order.

Based on $1$ st and $3$ rd experiments, $\left(H^{+}\right)$ concentration is doubled and the reaction rate is also doubled. So, the reaction is first-order.

Hence, the rate law for the reaction is:

$\text{rate}=k\left[{\text{CH}}_3{\text{COCH}}_3\right]\left[{\text{H}}^{+}\right]$

b)The rate constant

According to rate law, the rate constant for the reaction is:

$\text{rate}=k\left[{\text{CH}}_3{\text{COCH}}_3\right]\left[{\text{H}}^{+}\right]$

Rearrange the above equation for $k$ as follows:

$k=\frac{\text{rate}}{\left[{\text{CH}}_3{\text{COCH}}_3\right]\left[{\text{H}}^{+}\right]}$

Here, $k$ is the rate constant, $\left[{\text{CH}}_3{\text{COCH}}_3\right]$ is the concentration of ${\text{CH}}_3{\text{COCH}}_3$, $\left[{\text{H}}^{+}\right]$ is the concentration of ${\text{H}}^{+}$.

From experiment 1,

The rate law is $5.7\times {10}^{-5}M/\text{s}$.

The concentration of $\left[{\text{CH}}_3{\text{COCH}}_3\right]$ is $0.30M$ and the concentration of $\left[{\text{H}}^{+}\right]$ is $0.050M$.

Substitute, $5.7\times {10}^{-5}M/\text{s}$ for rate, $0.30M$ for $\left[{\text{CH}}_3{\text{COCH}}_3\right]$, and $0.050M$ for $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}k=\frac{5.7\times {10}^{-5}M/\text{s}}{\left[0.30M\right]\left[0.050M\right]}\\ =3.8\times {10}^{-3}{M}^{-1}{\text{s}}^{-1}\end{array}$

The rate constant for the reaction is $3.8\times {10}^{-3}{M}^{-1}{\text{s}}^{-1}$.

c) The rate law deduced from the given following mechanism is consistent or not with $\text{rate}=k\left[{\text{CH}}_3{\text{COCH}}_3\right]\left[{\text{H}}^{+}\right]$.

The proposed mechanism for the reaction is:

Let the rate constant for the slow step be $k_2$.

…… (1)

The rate constants for the forward and reverse steps in the fast equilibrium are $k_1$ and $k_{-1}$.

…… (2)

So, equation (1) becomes

$\text{rate}=\frac{k_1{k}_2}{k_{-1}}\left[{\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}\right]\left[{\text{H}}_3{\text{O}}^{+}\right]$ …… (3)

The overall equation from part (a) is:

$\text{rate}=k\left[{\text{CH}}_3{\text{COCH}}_3\right]\left[{\text{H}}^{+}\right]$ …… (4)

From equation (3) and (4),

$\begin{array}{c}k\cancel{\left[{\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}\right]}\cancel{\left[{\text{H}}_3{\text{O}}^{+}\right]}=\frac{k_1{k}_2}{k_{-1}}\cancel{\left[{\text{CH}}_3{\text{COCH}}_3\right]}\cancel{\left[{\text{H}}^{+}\right]}\\ k=k_1{k}_2/k_{-1}\end{array}$

So, the rate law is consistent with the mechanism shown in part (a).