Chapter 7, Problem 7B.7E

(a)

Interpretation Introduction

Interpretation:

The time taken by $\text{A}$ to get reduced into one-eighth of its initial concentration 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 equation that represents the integrated rate law for the first order kinetics is shown below.

    ${\left[\text{A}\right]}_{\text{t}}={\left[\text{A}\right]}_0{\text{e}}^{-k_{\text{r}}t}$

The half-life of the particular chemical reaction is the time in which exactly half of the reactant gets consumed.  The mathematical expression for the half-life for a reaction follows first order kinetics is shown below.

  $t_{1/2}=\frac{\ln 2}{k_{\text{r}}}$

Answer to Problem 7B.7E

The time taken by $\text{A}$ to get reduced into one-eighth of its initial concentration is $\underset{\_}{1.44\times 10^{3}min}$.

Explanation of Solution

The decomposition of $\text{A}$ follows the first order kinetics and having the half-life equals to $355\text{s}$.

The relation between the half-life and the rate constant for the first order is shown below.

    $t_{1/2}=\frac{\ln 2}{k_{\text{r}}}$        (1)

Where,

• $t_{1/2}$ is the half-life.
• $k_{\text{r}}$  is the order rate constant.

The value of $t_{1/2}$ is $355\text{s}$.

Substitute the value of $k_{\text{r}}$ in equation (1).

    $\begin{array}{c}{k}_{\text{r}}=\frac{\ln 2}{355\text{s}}\\ =1.95\times {10}^{-3}{\text{s}}^{-1}\end{array}$

Therefore, the rate constant for the decomposition of $\text{A}$ is $1.95\times {10}^{-3}{\text{s}}^{-1}$.

The concentration of $\text{A}$ after time $t$ is one-eighth of the initial concentration.  Mathematically the final concentration, ${\left[\text{A}\right]}_{\text{t}}$ in terms of initial concentration, ${\left[\text{A}\right]}_0$ is shown below.

    ${\left[\text{A}\right]}_{\text{t}}=\frac{1}{8}{\left[\text{A}\right]}_0$

The relation between the changes in the concentration of $\text{A}$ after time $t$ for the first order reaction is shown below.

  ${\left[\text{A}\right]}_{\text{t}}={\left[\text{A}\right]}_0{\text{e}}^{-k_{\text{r}}t}$        (2)

Where,

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

The value $k_{\text{r}}$ is $1.95\times {10}^{-3}{\text{s}}^{-1}$.

The value of ${\left[\text{A}\right]}_{\text{t}}$ is $\frac{1}{8}{\left[\text{A}\right]}_0$.

Substitute the value of $k_{\text{r}}$ and ${\left[\text{A}\right]}_{\text{t}}$ in equation (2).

    $\begin{array}{c}\frac{1}{8}{\left[\text{A}\right]}_0={\left[\text{A}\right]}_0{\text{e}}^{-\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t}\\ \frac{1}{8}={\text{e}}^{-\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t}\\ -\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t=\ln \frac{1}{8}\\ -\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t=-2.0794\end{array}$

On further calculation the required time taken is calculated as shown below.

    $\begin{array}{c}t=\frac{2.0794}{1.95\times {10}^{-3}{\text{s}}^{-1}}\\ =1.06\times {10}^3\text{s}\end{array}$

Thus, the time taken by $\text{A}$ to get reduced into one-eighth of its initial concentration is $\underset{\_}{1.06\times 10^{3}s}$.

(b)

Interpretation Introduction

Interpretation:

The time taken by $\text{A}$ to get reduced into one-fourth of its initial concentration has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.7E

The time taken by $\text{A}$ to get reduced into one-fourth of its initial concentration is $\underset{\_}{710.8s}$.

Explanation of Solution

As per the given data the decomposition of $\text{A}$ follows the first order kinetics and having the rate constant equals to $1.95\times {10}^{-3}{\text{s}}^{-1}$.

The concentration of $\text{A}$ after time $t$ is one-fourth of the initial concentration.  Mathematically the final concentration, ${\left[\text{A}\right]}_{\text{t}}$ in terms of initial concentration, ${\left[\text{A}\right]}_0$ is shown below.

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

The value $k_{\text{r}}$ is $1.95\times {10}^{-3}{\text{s}}^{-1}$.

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

Substitute the value of $k_{\text{r}}$ and ${\left[\text{A}\right]}_{\text{t}}$ in equation (2).

    $\begin{array}{c}\frac{1}{4}{\left[\text{A}\right]}_0={\left[\text{A}\right]}_0{\text{e}}^{-\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t}\\ \frac{1}{4}={\text{e}}^{-\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t}\\ -\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t=\ln \frac{1}{4}\\ -\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t=-1.3862\end{array}$

On further calculation the required time taken is calculated as shown below.

    $\begin{array}{c}t=\frac{1.3862}{1.95\times {10}^{-3}{\text{s}}^{-1}}\\ =710.8\text{s}\end{array}$

Thus, the time taken by $\text{A}$ to get reduced into one-fourth of its initial concentration is $\underset{\_}{710.8s}$.

(c)

Interpretation Introduction

Interpretation:

The time taken by $\text{A}$ to get reduced into $15\%$ of its initial concentration has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.7E

The time taken by $\text{A}$ to get reduced into $15\%$ of its initial concentration is $\underset{\_}{972.8s}$.

Explanation of Solution

As per the given data the decomposition of $\text{A}$ follows the first order kinetics and having the rate constant equals to $1.95\times {10}^{-3}{\text{s}}^{-1}$.

The final concentration of $\text{A}$ is $15\%$ of the initial value.  Mathematically the final value, ${\left[\text{A}\right]}_{\text{t}}$ in terms of initial value, ${\left[\text{A}\right]}_0$ is shown below.

    $\begin{array}{c}{\left[\text{A}\right]}_{\text{t}}=15\%{\left[\text{A}\right]}_0\\ =\frac{15}{100}{\left[\text{A}\right]}_0\\ =0.15{\left[\text{A}\right]}_0\end{array}$

The value $k_{\text{r}}$ is $1.95\times {10}^{-3}{\text{s}}^{-1}$.

The value of ${\left[\text{A}\right]}_{\text{t}}$ is $0.15{\left[\text{A}\right]}_0$.

Substitute the value of $k_{\text{r}}$ and ${\left[\text{A}\right]}_{\text{t}}$ in equation (2).

    $\begin{array}{c}0.15{\left[\text{A}\right]}_0={\left[\text{A}\right]}_0{e}^{-\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)\times t}\\ 0.15=e^{-\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)\times t}\\ -\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t=\ln 0.15\\ -\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t=-1.8971\end{array}$

On further calculation the required time taken is calculated as shown below.

    $\begin{array}{c}t=\frac{1.8971}{1.95\times {10}^{-3}{\text{s}}^{-1}}\\ =972.8\text{s}\end{array}$

Thus, the time taken by $\text{A}$ to get reduced into $15\%$ of its initial concentration is $\underset{\_}{972.8s}$.

(d)

Interpretation Introduction

Interpretation:

The time taken by $\text{A}$ to get reduced into one-ninth of its initial concentration has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.7E

The time taken by $\text{A}$ to get reduced into one-ninth of its initial concentration is $\underset{\_}{1.12\times 10^{3}s}$.

Explanation of Solution

As per the given data the decomposition of $\text{A}$ follows the first order kinetics and having the rate constant equals to $1.95\times {10}^{-3}{\text{s}}^{-1}$.

The concentration of $\text{A}$ after time $t$ is one-ninth of the initial concentration.  Mathematically the final concentration, ${\left[\text{A}\right]}_{\text{t}}$ in terms of initial concentration, ${\left[\text{A}\right]}_0$ is shown below.

    ${\left[\text{A}\right]}_{\text{t}}=\frac{1}{9}{\left[\text{A}\right]}_0$

The value $k_{\text{r}}$ is $1.95\times {10}^{-3}{\text{s}}^{-1}$.

The value of ${\left[\text{A}\right]}_{\text{t}}$ is $\frac{1}{9}{\left[\text{A}\right]}_0$.

Substitute the value of $k_{\text{r}}$ and ${\left[\text{A}\right]}_{\text{t}}$ in equation (2).

    $\begin{array}{c}\frac{1}{9}{\left[\text{A}\right]}_0={\left[\text{A}\right]}_0{\text{e}}^{-\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t}\\ \frac{1}{9}={\text{e}}^{-\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t}\\ -\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t=\ln \frac{1}{9}\\ -\left(1.95\times {10}^{-3}{\text{s}}^{-1}\right)t=-2.1973\end{array}$

On further calculation the required time taken is calculated as shown below.

    $\begin{array}{c}t=\frac{2.1973}{1.95\times {10}^{-3}{\text{s}}^{-1}}\\ =1.12\times {10}^3\text{s}\end{array}$

Thus, the time taken by $\text{A}$ to get reduced into one-ninth of its initial concentration is $\underset{\_}{1.12\times 10^{3}s}$.