Chapter 7, Problem 7B.18E

(a)

Interpretation Introduction

Interpretation:

The time taken by ${\text{NO}}_2$ to get reduced into one-half 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.18E

The time taken by ${\text{NO}}_2$ to get reduced into one-half of its initial concentration is $\underset{\_}{9.25s}$.

Explanation of Solution

The given chemical reaction is shown below.

  ${\text{2NO}}_2\left(\text{g}\right)\to 2\text{NO}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$

As per the given data, the above reaction follows the second order kinetics and having the rate constant equals to $0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The concentration of ${\text{NO}}_2$ after time $t$ is one-half of the initial concentration.  Mathematically the final concentration, ${\left[{\text{NO}}_2\right]}_{\text{t}}$ in terms of initial concentration, ${\left[{\text{NO}}_2\right]}_0$ is shown below.

    $\begin{array}{c}{\left[{\text{NO}}_2\right]}_{\text{t}}=\frac{1}{2}{\left[{\text{NO}}_2\right]}_0\\ =\frac{1}{2}\times 0.20\text{mol}\cdot {\text{L}}^{-1}\\ =0.10\text{mol}\cdot {\text{L}}^{-1}\end{array}$

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

  $\frac{1}{{\left[{\text{NO}}_2\right]}_{\text{t}}}=k_{\text{r}}t+\frac{1}{{\left[{\text{NO}}_2\right]}_0}$        (1)

Where,

• ${\left[{\text{NO}}_2\right]}_{\text{t}}$ is the concentration of reactant at time $t$.
• ${\left[{\text{NO}}_2\right]}_0$ is the initial concentration.
• $k_{\text{r}}$ is the order rate constant.
• $t$ is the time taken.

The value $k_{\text{r}}$ is $0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The value of ${\left[{\text{NO}}_2\right]}_{\text{t}}$ is $0.10\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[{\text{NO}}_2\right]}_0$ is $0.20\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $k_{\text{r}}$, ${\left[{\text{NO}}_2\right]}_0$ and ${\left[{\text{NO}}_2\right]}_{\text{t}}$ in equation (1).

    $\begin{array}{c}\frac{1}{0.10\text{mol}\cdot {\text{L}}^{-1}}=\left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t+\frac{1}{0.20\text{mol}\cdot {\text{L}}^{-1}}\\ \left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t=\frac{1}{0.10\text{mol}\cdot {\text{L}}^{-1}}-\frac{1}{0.20\text{mol}\cdot {\text{L}}^{-1}}\\ t=\frac{1}{\left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times 0.20\text{mol}\cdot {\text{L}}^{-1}}\\ =\underset{\_}{9.25s}\end{array}$

Thus, the time taken by ${\text{NO}}_2$ to get reduced into one-half of its initial concentration is $\underset{\_}{9.25s}$.

(b)

Interpretation Introduction

Interpretation:

The time taken by ${\text{NO}}_2$ to get reduced into one-sixteenth of its initial concentration has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.18E

The time taken by ${\text{NO}}_2$ to get reduced into one-sixteenth of its initial concentration is $\underset{\_}{138.88s}$.

Explanation of Solution

The given chemical reaction is shown below.

  ${\text{2NO}}_2\left(\text{g}\right)\to 2\text{NO}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$

As per the given data the above reaction follows the second order kinetics and having the rate constant equals to $0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The concentration of ${\text{NO}}_2$ after time $t$ is one-sixteenth of the initial concentration.  Mathematically the final concentration, ${\left[{\text{NO}}_2\right]}_{\text{t}}$ in terms of initial concentration, ${\left[{\text{NO}}_2\right]}_0$ is shown below.

    $\begin{array}{c}{\left[{\text{NO}}_2\right]}_{\text{t}}=\frac{1}{16}{\left[{\text{NO}}_2\right]}_0\\ =\frac{1}{16}\times 0.20\text{mol}\cdot {\text{L}}^{-1}\\ =0.0125\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The value $k_{\text{r}}$ is $0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The value of ${\left[{\text{NO}}_2\right]}_{\text{t}}$ is $0.0125\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[{\text{NO}}_2\right]}_0$ is $0.20\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $k_{\text{r}}$, ${\left[{\text{NO}}_2\right]}_0$ and ${\left[{\text{NO}}_2\right]}_{\text{t}}$ in equation (1).

    $\begin{array}{c}\frac{1}{0.0125\text{mol}\cdot {\text{L}}^{-1}}=\left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t+\frac{1}{0.20\text{mol}\cdot {\text{L}}^{-1}}\\ \left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t=\frac{1}{0.0125\text{mol}\cdot {\text{L}}^{-1}}-\frac{1}{0.20\text{mol}\cdot {\text{L}}^{-1}}\\ t=\frac{75\text{L}\cdot {\text{mol}}^{-1}}{\left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)}\\ =\underset{\_}{138.88s}\end{array}$

Thus, the time taken by ${\text{NO}}_2$ to get reduced into one-sixteenth of its initial concentration is $\underset{\_}{138.88s}$.

(c)

Interpretation Introduction

Interpretation:

The time taken by ${\text{NO}}_2$ to get reduced into one-ninth of its initial concentration has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.18E

The time taken by ${\text{NO}}_2$ to get reduced into one-ninth of its initial concentration is $\underset{\_}{74.9s}$.

Explanation of Solution

The given chemical reaction is shown below.

  ${\text{2NO}}_2\left(\text{g}\right)\to 2\text{NO}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$

As per the given data the above reaction follows the second order kinetics and having the rate constant equals to $0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The concentration of ${\text{NO}}_2$ after time $t$ is one-ninth of the initial concentration.  Mathematically the final concentration, ${\left[{\text{NO}}_2\right]}_{\text{t}}$ in terms of initial concentration, ${\left[{\text{NO}}_2\right]}_0$ is shown below.

    $\begin{array}{c}{\left[{\text{NO}}_2\right]}_{\text{t}}=\frac{1}{9}{\left[{\text{NO}}_2\right]}_0\\ =\frac{1}{9}\times 0.20\text{mol}\cdot {\text{L}}^{-1}\\ =0.022\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The value $k_{\text{r}}$ is $0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}$.

The value of ${\left[{\text{NO}}_2\right]}_{\text{t}}$ is $0.022\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[{\text{NO}}_2\right]}_0$ is $0.20\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $k_{\text{r}}$, ${\left[{\text{NO}}_2\right]}_0$ and ${\left[{\text{NO}}_2\right]}_{\text{t}}$ in equation (1).

    $\begin{array}{c}\frac{1}{0.022\text{mol}\cdot {\text{L}}^{-1}}=\left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t+\frac{1}{0.20\text{mol}\cdot {\text{L}}^{-1}}\\ \left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)\times t=\frac{1}{0.022\text{mol}\cdot {\text{L}}^{-1}}-\frac{1}{0.20\text{mol}\cdot {\text{L}}^{-1}}\\ t=\frac{40.45\text{L}\cdot {\text{mol}}^{-1}}{\left(0.54\text{L}\cdot {\text{mol}}^{-1}\cdot {\text{s}}^{-1}\right)}\\ =74.9\text{s}\end{array}$

Thus, the time taken by ${\text{NO}}_2$ to get reduced into one-ninth of its initial concentration is $\underset{\_}{74.9s}$.