Chapter 7, Problem 7B.4E

(a)

Interpretation Introduction

Interpretation:

The rate constant for the given first order reaction has to be determined.

Concept Introduction:

According to the integrated rate law for the first order reaction, the concentration of reactant is the exponential function of time.  The two equations that represent the integrated rate law for the first order kinetics is shown below.

    $\begin{array}{c}\ln \frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_0}=-k_{\text{r}}t\\ {\left[\text{A}\right]}_{\text{t}}={\left[\text{A}\right]}_0{\text{e}}^{-k_{\text{r}}t}\end{array}$

Answer to Problem 7B.4E

The rate constant of the given first order reaction is $\underset{\_}{1.10\times 10^{-2}mi{n}^{-1}}$.

Explanation of Solution

The given first order reaction is shown below.

  $\text{A}\to \text{B}+\text{C}$

As per the data given in the question, the time taken for the decomposition of reactant to one-fourth of its initial value is $125\min$.

The final concentration of $\text{A}$ after $125\min$ is represented as shown below.

    ${\left[\text{A}\right]}_t=\frac{1}{4}{\left[\text{A}\right]}_0$

The relation between the changes in the concentration of reactant after time $t$ for the first order reaction is shown below.

    $\ln \frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_0}=-k_{\text{r}}t$        (1)

Where,

• ${\left[\text{A}\right]}_t$ is the concentration of $\text{A}$ at time $t$.
• ${\left[\text{A}\right]}_0$ is the initial concentration of $\text{A}$.
• $t$ is the time taken.
• $k_{\text{r}}$ is the rate constant of the reaction.

The value of ${\left[\text{A}\right]}_t$ is $\frac{1}{4}{\left[\text{A}\right]}_0$.

The value of $t$ is $125\min$.

Substitute the value of $t$ and ${\left[\text{A}\right]}_t$ in equation (1).

  $\begin{array}{c}\ln \frac{\frac{1}{4}{\left[\text{A}\right]}_0}{{\left[\text{A}\right]}_0}=-k_{\text{r}}\times \left(125\min \right)\\ \ln \frac{1}{4}=-k_{\text{r}}\times \left(125\min \right)\\ -1.3862=-k_{\text{r}}\times \left(125\min \right)\\ k_{\text{r}}=\frac{1.3862}{125\min }\end{array}$

On further calculation the rate constant of the give reaction is calculated as shown below.

    $k_{\text{r}}=1.10\times {10}^{-2}{\text{min}}^{-1}$

Thus, the rate constant of the given first order reaction is $\underset{\_}{1.10\times 10^{-2}mi{n}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The rate constant for the given first order reaction has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 7B.4E

The rate constant of the given first order reaction is $\underset{\_}{1.028\times 10^{-3}{s}^{-1}}$.

Explanation of Solution

The given first order reaction is shown below.

  $\text{2A}\to \text{D}+\text{E}$

According to the given chemical reaction, $2.0\text{mol}\cdot {\text{L}}^{-1}$ of $\text{A}$ gets decomposed to form $1.0\text{mol}\cdot {\text{L}}^{-1}$ of $\text{D}$.

As per the data given in the question, the initial concentration of $\text{A}$ is $0.0421\text{mol}\cdot {\text{L}}^{-1}$ and concentration of rises to $0.00132\text{mol}\cdot {\text{L}}^{-1}$ in $63\text{s}$.  Therefore the left over concentration of $\text{A}$ after $63\text{s}$ is calculated by the expression  shown below.

    ${\left[\text{A}\right]}_t={\left[\text{A}\right]}_0-\text{2}\left[\text{D}\right]$        (2)

Where,

• ${\left[\text{A}\right]}_t$ is the concentration of $\text{A}$ at time $t$.
• ${\left[\text{A}\right]}_0$ is the initial concentration of $\text{A}$.
• $\left[\text{D}\right]$ is the concentration of $\text{D}$ after time $t$.

The value of ${\left[\text{A}\right]}_0$ is $0.0421\text{mol}\cdot {\text{L}}^{-1}$.

The value of $\left[\text{D}\right]$ is $0.00132\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of ${\left[\text{A}\right]}_0$ and $\left[\text{D}\right]$ in equation (2).

  $\begin{array}{c}{\left[\text{A}\right]}_t=0.0421\text{mol}\cdot {\text{L}}^{-1}-\text{2}\times 0.00132\text{mol}\cdot {\text{L}}^{-1}\\ =\text{0}\text{.03946}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Therefore the final concentration of $\text{A}$ after $63\text{s}$ is $\text{0}\text{.03946}\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[\text{A}\right]}_t$ is $\text{0}\text{.03946}\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[\text{A}\right]}_0$ is $0.0421\text{mol}\cdot {\text{L}}^{-1}$.

The value of $t$ is $63\text{s}$.

Substitute the value of ${\left[\text{A}\right]}_0$, ${\left[\text{A}\right]}_t$ and $t$ in equation (1).

  $\begin{array}{c}\ln \frac{0.03946\text{mol}\cdot {\text{L}}^{-1}}{0.0421\text{mol}\cdot {\text{L}}^{-1}}=-k_{\text{r}}\times \left(63\text{s}\right)\\ \ln 0.9372=-k_{\text{r}}\times \left(63\text{s}\right)\\ -0.0648=-k_{\text{r}}\times \left(63\text{s}\right)\\ k_{\text{r}}=\frac{0.0648}{63\text{s}}\end{array}$

On further calculation the rate constant of the give reaction is calculated as shown below.

    $k_{\text{r}}=1.028\times {10}^{-3}{\text{s}}^{-1}$

Thus, the rate constant of the given first order reaction is $\underset{\_}{1.028\times 10^{-3}{s}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The rate constant for the given first order reaction has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 7B.4E

The rate constant of the given first order reaction is $\underset{\_}{7.25\times 10^{-2}mi{n}^{-1}}$.

Explanation of Solution

The given first order reaction is shown below.

  $\text{3A}\to \text{F}+\text{G}$

According to the given chemical reaction, $3.0\text{mol}\cdot {\text{L}}^{-1}$ of $\text{A}$ gets decomposed to form $1.0\text{mol}\cdot {\text{L}}^{-1}$ of $\text{F}$.

As per the data given in the question, the initial concentration of $\text{A}$ is $0.080\text{mol}\cdot {\text{L}}^{-1}$ and concentration of $\text{F}$ rises to $0.015\text{mol}\cdot {\text{L}}^{-1}$ in $11.4\min$.  Therefore the left over concentration of $\text{A}$ after $11.4\min$ is calculated by the expression shown below.

    ${\left[\text{A}\right]}_t={\left[\text{A}\right]}_0-\text{3}\left[\text{F}\right]$        (3)

Where,

• ${\left[\text{A}\right]}_t$ is the concentration of $\text{A}$ at time $t$.
• ${\left[\text{A}\right]}_0$ is the initial concentration of $\text{A}$.
• $\left[\text{F}\right]$ is the concentration of $\text{F}$ after time $t$.

The value of ${\left[\text{A}\right]}_0$ is $0.080\text{mol}\cdot {\text{L}}^{-1}$.

The value of $\left[\text{F}\right]$ is $0.015\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of ${\left[\text{A}\right]}_0$ and $\left[\text{F}\right]$ in equation (3).

  $\begin{array}{c}{\left[\text{A}\right]}_t=0.080\text{mol}\cdot {\text{L}}^{-1}-\text{3}\times 0.015\text{mol}\cdot {\text{L}}^{-1}\\ =\text{0}\text{.035}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Therefore the final concentration of $\text{A}$ after $11.4\min$ is $\text{0}\text{.035}\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[\text{A}\right]}_t$ is $\text{0}\text{.035}\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[\text{A}\right]}_0$ is $0.080\text{mol}\cdot {\text{L}}^{-1}$.

The value of $t$ is $11.4\min$.

Substitute the value of ${\left[\text{A}\right]}_0$, ${\left[\text{A}\right]}_t$ and $t$ in equation (1).

  $\begin{array}{c}\ln \frac{0.035\text{mol}\cdot {\text{L}}^{-1}}{0.080\text{mol}\cdot {\text{L}}^{-1}}=-k_{\text{r}}\times \left(11.4\min \right)\\ \ln 0.4375=-k_{\text{r}}\times \left(11.4\min \right)\\ -0.8266=-k_{\text{r}}\times \left(11.4\min \right)\\ k_{\text{r}}=\frac{0.8266}{11.4\min }\end{array}$

On further calculation the rate constant of the give reaction is calculated as shown below.

    $k_{\text{r}}=7.25\times {10}^{-2}{\text{min}}^{-1}$

Thus, the rate constant of the given first order reaction is $\underset{\_}{7.25\times 10^{-2}mi{n}^{-1}}$.