Chapter 7, Problem 7B.13E

(a)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

The equation that represents the integrated rate law for the second order kinetics is shown below.  The unit for the rate constant for the second order reaction is $\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

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

Answer to Problem 7B.13E

The time taken by $\text{A}$ to get reduced into one-sixteenth of its initial concentration is $\underset{\_}{759.87s}$.

Explanation of Solution

As per the given data the initial concentration of $\text{A}$ in the second order reaction is $0.84\text{mol}\cdot {\text{L}}^{-1}$.

The expression for the half-life of $\text{A}$ in the second order reaction in terms of $k_{\text{r}}$ is shown below.

  $k_{\text{r}}=\frac{1}{t_{1/2}{\left[\text{A}\right]}_{\text{0}}}$        (1)

Where,

• $t_{1/2}$ is the half-life.
• $k_{\text{r}}$ is the order rate constant.
• ${\left[\text{A}\right]}_0$ is the initial concentration.

The value of $t_{1/2}$ is $50.5\text{s}$.

The value of ${\left[\text{A}\right]}_0$ is $0.84\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $t_{1/2}$ and ${\left[\text{A}\right]}_0$ in equation (1).

    $\begin{array}{c}{k}_{\text{r}}=\frac{1}{\left(50.5\text{s}\right)\times \left(0.84\text{mol}\cdot {\text{L}}^{-1}\right)}\\ =0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\end{array}$

Therefore, the rate constant is $0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The concentration of $\text{A}$ after time $t$ is one-sixteenth 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.

    $\begin{array}{c}{\left[\text{A}\right]}_{\text{t}}=\frac{1}{16}{\left[\text{A}\right]}_0\\ =\frac{1}{16}\times 0.84\text{mol}\cdot {\text{L}}^{-1}\\ =0.0525\text{mol}\cdot {\text{L}}^{-1}\end{array}$

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

  $\frac{1}{{\left[\text{A}\right]}_{\text{t}}}=k_{\text{r}}t+\frac{1}{{\left[\text{A}\right]}_0}$        (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 $0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The value of ${\left[\text{A}\right]}_{\text{t}}$ is $0.0525\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[\text{A}\right]}_0$ is $0.84\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $k_{\text{r}}$, ${\left[\text{A}\right]}_0$ and ${\left[\text{A}\right]}_{\text{t}}$ in equation (2).

    $\begin{array}{c}\frac{1}{0.0525\text{mol}\cdot {\text{L}}^{-1}}=\left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t+\frac{1}{0.84\text{mol}\cdot {\text{L}}^{-1}}\\ \left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t=\frac{1}{0.0525\text{mol}\cdot {\text{L}}^{-1}}-\frac{1}{0.84\text{mol}\cdot {\text{L}}^{-1}}\\ t=\frac{1}{\left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times 0.20\text{mol}\cdot {\text{L}}^{-1}}\\ =759.87\text{s}\end{array}$

Thus, the time taken by $\text{A}$ to get reduced into one-sixteenth of its initial concentration is $\underset{\_}{759.87s}$.

(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.13E

The time taken by $\text{A}$ to get reduced into one-fourth of its initial concentration is $\underset{\_}{151.9s}$.

Explanation of Solution

As per the given data the initial concentration of $\text{A}$ in the second order reaction is $0.84\text{mol}\cdot {\text{L}}^{-1}$.

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.

    $\begin{array}{c}{\left[\text{A}\right]}_{\text{t}}=\frac{1}{4}{\left[\text{A}\right]}_0\\ =\frac{1}{4}\times 0.84\text{mol}\cdot {\text{L}}^{-1}\\ =0.21\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The value $k_{\text{r}}$ is $0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The value of ${\left[\text{A}\right]}_{\text{t}}$ is $0.21\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[\text{A}\right]}_0$ is $0.84\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $k_{\text{r}}$, ${\left[\text{A}\right]}_0$ and ${\left[\text{A}\right]}_{\text{t}}$ in equation (2).

    $\begin{array}{c}\frac{1}{0.21\text{mol}\cdot {\text{L}}^{-1}}=\left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t+\frac{1}{0.84\text{mol}\cdot {\text{L}}^{-1}}\\ \left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t=\frac{1}{0.21\text{mol}\cdot {\text{L}}^{-1}}-\frac{1}{0.84\text{mol}\cdot {\text{L}}^{-1}}\\ t=\frac{3.57\text{L}\cdot {\text{mol}}^{-1}}{\left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)}\\ =151.9\text{s}\end{array}$

Thus, the time taken by $\text{A}$ to get reduced into one-fourth of its initial concentration is $\underset{\_}{151.9s}$.

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Same as part (a).

Answer to Problem 7B.13E

The time taken by $\text{A}$ to get reduced into one-fifth of its initial concentration is $\underset{\_}{202.55s}$.

Explanation of Solution

As per the given data the initial concentration of $\text{A}$ in the second order reaction is $0.84\text{mol}\cdot {\text{L}}^{-1}$.

The concentration of $\text{A}$ after time $t$ is one-fifth 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.

    $\begin{array}{c}{\left[\text{A}\right]}_{\text{t}}=\frac{1}{5}{\left[\text{A}\right]}_0\\ =\frac{1}{5}\times 0.84\text{mol}\cdot {\text{L}}^{-1}\\ =0.168\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The value $k_{\text{r}}$ is $0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The value of ${\left[\text{A}\right]}_{\text{t}}$ is $0.168\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[\text{A}\right]}_0$ is $0.84\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $k_{\text{r}}$, ${\left[\text{A}\right]}_0$ and ${\left[\text{A}\right]}_{\text{t}}$ in equation (2).

    $\begin{array}{c}\frac{1}{0.168\text{mol}\cdot {\text{L}}^{-1}}=\left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t+\frac{1}{0.84\text{mol}\cdot {\text{L}}^{-1}}\\ \left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t=\frac{1}{0.168\text{mol}\cdot {\text{L}}^{-1}}-\frac{1}{0.84\text{mol}\cdot {\text{L}}^{-1}}\\ t=\frac{4.76\text{L}\cdot {\text{mol}}^{-1}}{\left(0.0235\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)}\\ =202.55\text{s}\end{array}$

Thus, the time taken by $\text{A}$ to get reduced into one-fifth of its initial concentration is $\underset{\_}{202.55s}$.