Chapter 7, Problem 7.19E

(a)

Interpretation Introduction

Interpretation:

The time needed to reduce the concentration of $\text{A}$ to one-half of the initial concentration has to be determined.

Concept Introduction:

The equation that represents the integrated rate law for the third order kinetics is shown below.  The unit for the rate constant for the third order reaction is ${\text{L}}^2\cdot {\text{mol}}^{-2}\cdot {\text{s}}^{-1}$.

    $\frac{1}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}-\frac{1}{2{\left({\left[\text{A}\right]}_t\right)}^2}=-k_{\text{r}}t$

Answer to Problem 7.19E

The timeneeded to reduce the concentration of $\text{A}$ to one-half of the initial concentration is $t_{1/2}=\frac{3}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}$.

Explanation of Solution

The given chemical equation is shown below.

  $\text{A}\to \text{product}$

As per the given data the above reaction follows the third order kinetics.

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

    ${\left[\text{A}\right]}_{\text{t}}=\frac{1}{2}{\left[\text{A}\right]}_0$

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

  $\frac{1}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}-\frac{1}{2{\left({\left[\text{A}\right]}_t\right)}^2}=-k_{\text{r}}t$        (1)

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 of ${\left[\text{A}\right]}_{\text{t}}$ is $\frac{1}{2}{\left[\text{A}\right]}_0$.

The symbol $t$ is replaced by $t_{1/2}$ in equation (1).

Substitute value of ${\left[\text{A}\right]}_{\text{t}}$ for the expression of half-life of $\text{A}$ in equation (1).

  $\begin{array}{c}\frac{1}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}-\frac{1}{2{\left(\frac{1}{2}{\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}t\\ \frac{1}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}-\frac{2}{{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}{t}_{1/2}\\ -\frac{3}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}{t}_{1/2}\\ t_{1/2}=\frac{3}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\end{array}$

Thus, the required expression for the time needed to reduce the concentration of $\text{A}$ to one-half of the initial concentration is shown below.

  $t_{1/2}=\frac{3}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}$

(b)

Interpretation Introduction

Interpretation:

The time needed to reduce the concentration of $\text{A}$ to one-fourth of the initial concentration has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7.19E

The time needed to reduce the concentration of $\text{A}$ to one-half of the initial concentration is $t_{1/4}=5t_{1/2}$.

Explanation of Solution

The given chemical equation is shown below.

  $\text{A}\to \text{product}$

As per the given data the above reaction follows the third order kinetics.

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

The symbol $t$ is replaced by $t_{1/4}$ in equation (1).

Substitute value of ${\left[\text{A}\right]}_{\text{t}}$ for the required expression in equation (1).

  $\begin{array}{c}\frac{1}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}-\frac{1}{2{\left(\frac{1}{4}{\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}{t}_{1/4}\\ \frac{1}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}-\frac{16}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}{t}_{1/4}\\ -\frac{15}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}{t}_{1/4}\\ t_{1/4}=\frac{15}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\end{array}$

The time needed to reduce the concentration of $\text{A}$ to one-fourth of the initial concentration in terms of half-life is shown below.

  $\begin{array}{c}{t}_{1/4}=\frac{15}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\\ =5\left(\frac{3}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\right)\end{array}$

Replace the term $\left(\frac{3}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\right)$ by $t_{1/2}$ in the above expression.  Therefore, the modified expression is shown below.

  $t_{1/4}=5t_{1/2}$

Thus, the required expression for the time needed to reduce the concentration of $\text{A}$ to one-fourth of the initial concentration is shown below.

  $t_{1/4}=5t_{1/2}$

(c)

Interpretation Introduction

Interpretation:

The time needed to reduce the concentration of $\text{A}$ to one-sixteenth of the initial concentration has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7.19E

The time needed to reduce the concentration of $\text{A}$ to one-half of the initial concentration is $t_{1/16}=85t_{1/2}$.

Explanation of Solution

The given chemical equation is shown below.

  $\text{A}\to \text{product}$

As per the given data the above reaction follows the third order kinetics.

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.

    ${\left[\text{A}\right]}_{\text{t}}=\frac{1}{16}{\left[\text{A}\right]}_0$

The value of ${\left[\text{A}\right]}_{\text{t}}$ is $\frac{1}{16}{\left[\text{A}\right]}_0$.

The symbol $t$ is replaced by $t_{1/16}$ in equation (1).

Substitute value of ${\left[\text{A}\right]}_{\text{t}}$ for the required expression in equation (1).

  $\begin{array}{c}\frac{1}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}-\frac{1}{2{\left(\frac{1}{16}{\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}{t}_{1/16}\\ \frac{1}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}-\frac{256}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}{t}_{1/16}\\ -\frac{255}{2{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}=-k_{\text{r}}{t}_{1/16}\\ t_{1/16}=\frac{255}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\end{array}$

The time needed to reduce the concentration of $\text{A}$ to one-sixteenth of the initial concentration in terms of half-life is shown below.

  $\begin{array}{c}{t}_{1/16}=\frac{255}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\\ =85\left(\frac{3}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\right)\end{array}$

Replace the term $\left(\frac{3}{2k_{\text{r}}{\left({\left[\text{A}\right]}_{\text{0}}\right)}^2}\right)$ by $t_{1/2}$ in the above expression.  Therefore, the modified expression is shown below.

  $t_{1/16}=85t_{1/2}$

Thus, the required expression for the time needed to reduce the concentration of $\text{A}$ to one-sixteenth of the initial concentration is shown below.

  $t_{1/16}=85t_{1/2}$