Chapter 16, Problem 16.85P

(a)

Interpretation Introduction

Interpretation:

The reason for water does not appear in the rate law has to be explained.

Concept introduction:

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

Mechanism of a reaction: The representation of step by step process involved in the chemical process is said to be mechanism of a chemical reaction.

Elementary step: The first step in a reaction mechanism is said to be elementary step.

Intermediate: Sometimes, in between the reaction some separable amount and useful amount of substances are formed during the reaction and which are said to be intermediates

Explanation of Solution

The given reaction involves multiple mechanism steps, by adding the entire individual steps gives rise to an overall reaction equation. Hence, the reaction equation as follows,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{\left(C{H}_3\right)}_{3}CBr\to \cancel{{\left(C{H}_3\right)}_3{C}^{+}}+B{r}^{-}\left[slow\right]\\ \cancel{{\left(C{H}_3\right)}_3{C}^{+}}+H_{2}O\to \cancel{{\left(C{H}_3\right)}_{3}C-O{H}_2^{+}}\left[fast\right]\\ \cancel{{\left(C{H}_3\right)}_{3}C-O{H}_2^{+}}\to H^{+}+{\left(C{H}_3\right)}_{3}C-OH\left[fast\right]\end{array}}\\ {\left(C{H}_3\right)}_{3}CBr+H_{2}O\stackrel{}{\to }{\left(C{H}_3\right)}_{3}C-OH+H^{+}+B{r}^{-}Overallreaction\end{array}}$

In the reaction, the slow step is the rate determining step; and its rate law is the overall rate law. The overall equation becomes,

${\left(C{H}_3\right)}_{3}CBr+H_{2}O\stackrel{}{\to }{\left(C{H}_3\right)}_{3}C-OH+H^{+}+B{r}^{-}$

Water does not appear in the rate law, as the water act as solvent in the reaction. The concentration of water is assumed not to change even though some water is used up as a reactant. Thus, the assumption is valid as long as the solute concentration is low.

(b)

Interpretation Introduction

Interpretation:

The rate laws for elementary steps have to be written.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right).\end{array}$

$\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}{\left[\text{C}\right]}^{\text{p}}\text{,where 'm, n and p' are orders of the reactants}\text{.}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

• Step 1: ${\left(C{H}_3\right)}_{3}CBr\to {\left(C{H}_3\right)}_3{C}^{+}+B{r}^{-}\left[slow\right]$

The rate law of the above step is ${\text{Rate}}_{\text{1}}\text{ = k }\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CBr}\right]\text{.}$

• Step 2: ${\left(C{H}_3\right)}_3{C}^{+}+H_{2}O\to {\left(C{H}_3\right)}_{3}C-O{H}_2^{+}\left[fast\right]$

The rate law of the above step is ${\text{Rate}}_{\text{2}}\text{ = k }\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}{\text{C}}^{\text{+}}\right]\text{.}$

• Step 3: ${\left(C{H}_3\right)}_{3}C-O{H}_2^{+}\to H^{+}+{\left(C{H}_3\right)}_{3}C-OH\left[fast\right]$

The rate law of the above step is ${\text{Rate}}_{\text{3}}\text{ = k }\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}{\text{C-OH}}_{\text{2}}^{\text{+}}\right]\text{.}$

(c)

Interpretation Introduction

Interpretation:

The reaction intermediates that appear in the given mechanism have to be identified.

Concept introduction:

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

Mechanism of a reaction: The representation of step by step process involved in the chemical process is said to be mechanism of a chemical reaction.

Elementary step: The first step in a reaction mechanism is said to be elementary step.

Intermediate: Sometimes, in between the reaction some separable amount and useful amount of substances are formed during the reaction and which are said to be intermediates

Explanation of Solution

The given reaction involves multiple mechanism steps, by adding the entire individual steps gives rise to an overall reaction equation. Hence, the reaction equation as follows,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{\left(C{H}_3\right)}_{3}CBr\to \cancel{{\left(C{H}_3\right)}_3{C}^{+}}+B{r}^{-}\left[slow\right]\\ \cancel{{\left(C{H}_3\right)}_3{C}^{+}}+H_{2}O\to \cancel{{\left(C{H}_3\right)}_{3}C-O{H}_2^{+}}\left[fast\right]\\ \cancel{{\left(C{H}_3\right)}_{3}C-O{H}_2^{+}}\to H^{+}+{\left(C{H}_3\right)}_{3}C-OH\left[fast\right]\end{array}}\\ {\left(C{H}_3\right)}_{3}CBr+H_{2}O\stackrel{}{\to }{\left(C{H}_3\right)}_{3}C-OH+H^{+}+B{r}^{-}Overallreaction\end{array}}$

In the reaction, the intermediates are produced in one step and consumed in the following steps; thus, the intermediates are, $\left({\left(C{H}_3\right)}_3{C}^{+}\right)$, and $\left({\left(C{H}_3\right)}_{3}C-O{H}_2^{+}\right).$

(d)

Interpretation Introduction

Interpretation:

The given mechanism whether consistent with the experimental rate law has to be shown.

Concept introduction:

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

Mechanism of a reaction: The representation of step by step process involved in the chemical process is said to be mechanism of a chemical reaction.

Elementary step: The first step in a reaction mechanism is said to be elementary step.

Intermediate: Sometimes, in between the reaction some separable amount and useful amount of substances are formed during the reaction and which are said to be intermediates

Explanation of Solution

The given reaction involves multiple mechanism steps, by adding the entire individual steps gives rise to an overall reaction equation. Hence, the reaction equation as follows,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{\left(C{H}_3\right)}_{3}CBr\to \cancel{{\left(C{H}_3\right)}_3{C}^{+}}+B{r}^{-}\left[slow\right]\\ \cancel{{\left(C{H}_3\right)}_3{C}^{+}}+H_{2}O\to \cancel{{\left(C{H}_3\right)}_{3}C-O{H}_2^{+}}\left[fast\right]\\ \cancel{{\left(C{H}_3\right)}_{3}C-O{H}_2^{+}}\to H^{+}+{\left(C{H}_3\right)}_{3}C-OH\left[fast\right]\end{array}}\\ {\left(C{H}_3\right)}_{3}CBr+H_{2}O\stackrel{}{\to }{\left(C{H}_3\right)}_{3}C-OH+H^{+}+B{r}^{-}Overallreaction\end{array}}$

In the reaction, the intermediates are produced in one step and consumed in the following steps; thus, the intermediates are, $\left({\left(C{H}_3\right)}_3{C}^{+}\right)$, and $\left({\left(C{H}_3\right)}_{3}C-O{H}_2^{+}\right).$

${\left(C{H}_3\right)}_{3}CBr\to {\left(C{H}_3\right)}_3{C}^{+}+B{r}^{-}\left[slow\right]$

The rate law for the above reaction is, $\text{Rate}\text{ = k}\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CBr}\right]$        (1)

Therefore, the overall reaction rate is as same as the rate of slowest reaction step that is proven as shown above.

Therefore, the given rate law is consistent with the rate law of overall reaction $Rate=k\left[{\left(C{H}_3\right)}_{3}CBr\right]$. The proposed mechanism is valid.