Chapter 7, Problem 7B.5E

(a)

Interpretation Introduction

Interpretation:

The half-life for the decomposition of ${\text{N}}_2{\text{O}}_5$ 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.5E

The half-life for the decomposition of ${\text{N}}_2{\text{O}}_5$ is $\underset{\_}{4.62s}$.

Explanation of Solution

The decomposition of ${\text{N}}_2{\text{O}}_5$ follows first order kinetics with rate constant $3.7\times {10}^{-5}{\text{s}}^{-1}$.

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 $k_{\text{r}}$ is $3.7\times {10}^{-5}{\text{s}}^{-1}$.

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

    $\begin{array}{c}{t}_{1/2}=\frac{\ln 2}{3.7\times {10}^{-5}{\text{s}}^{-1}}\\ =1.87\times {10}^4\text{s}\times \frac{1\min }{60\text{s}}\times \frac{1\text{h}}{60\text{min}}\\ =5.19\text{h}\end{array}$

Thus, the half-life for the decomposition of ${\text{N}}_2{\text{O}}_5$ is $\underset{\_}{5.19h}$.

(b)

Interpretation Introduction

Interpretation:

The concentration of ${\text{N}}_2{\text{O}}_5$ after $3.5\text{h}$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.5E

The final concentration of ${\text{N}}_2{\text{O}}_5$ after $3.5\text{h}$ is $\underset{\_}{0.035mol\cdot L^{-1}}$.

Explanation of Solution

According to the data given in the question, the initial concentration of ${\text{N}}_2{\text{O}}_5$ is $0.0567\text{mol}\cdot {\text{L}}^{-1}$.  The decomposition of ${\text{N}}_2{\text{O}}_5$ proceeds for $3.5\text{h}$.

The unit conversion of time from $\text{h}$ to $\text{s}$ is shown below.

  $3.5\text{h}\times \frac{60\min }{1\text{h}}\times \frac{60\text{s}}{1\min }=12600\text{s}$

The relation between the changes in the concentration of ${\text{N}}_2{\text{O}}_5$ after time $t$ for the first order reaction is shown below.

    ${\left[{\text{N}}_2{\text{O}}_5\right]}_t={\left[{\text{N}}_2{\text{O}}_5\right]}_0{e}^{-k_{\text{r}}t}$        (2)

Where,

• ${\left[{\text{N}}_2{\text{O}}_5\right]}_t$ is the concentration of ${\text{N}}_2{\text{O}}_5$ at time $t$.
• ${\left[{\text{N}}_2{\text{O}}_5\right]}_0$ is the initial concentration of ${\text{N}}_2{\text{O}}_5$.
• $t$ is the time taken.
• $k_{\text{r}}$ is the rate constant of the reaction.

The value of ${\left[{\text{N}}_2{\text{O}}_5\right]}_0$ is $0.0567\text{mol}\cdot {\text{L}}^{-1}$.

The value of $k_{\text{r}}$ is $3.7\times {10}^{-5}{\text{s}}^{-1}$.

The value of $t$ is $12600\text{s}$.

Substitute the value of $k_{\text{r}}$, $t$ and ${\left[{\text{N}}_2{\text{O}}_5\right]}_0$ in equation (2).

  $\begin{array}{c}{\left[{\text{N}}_2{\text{O}}_5\right]}_t=\left(0.0567\text{mol}\cdot {\text{L}}^{-1}\right)e^{-\left(3.7\times {10}^{-5}{\text{s}}^{-1}\right)\times \left(12600\text{s}\right)}\\ =\left(0.0567\text{mol}\cdot {\text{L}}^{-1}\right)e^{-0.4662}\\ =\left(0.0567\text{mol}\cdot {\text{L}}^{-1}\right)\times 0.6273\\ =0.035\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the final concentration of ${\text{N}}_2{\text{O}}_5$ after $3.5\text{h}$ is $\underset{\_}{0.035mol\cdot L^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The time taken for the decomposition of ${\text{N}}_2{\text{O}}_5$ from $0.0567\text{mol}\cdot {\text{L}}^{-1}$ to $0.0135\text{mol}\cdot {\text{L}}^{-1}$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.5E

The time taken for the decomposition of ${\text{N}}_2{\text{O}}_5$ from $0.0567\text{mol}\cdot {\text{L}}^{-1}$ to $0.0135\text{mol}\cdot {\text{L}}^{-1}$ is $\underset{\_}{645min}$.

Explanation of Solution

According to the data given in the question, the initial concentration and the final concentration of ${\text{N}}_2{\text{O}}_5$ is $0.0567\text{mol}\cdot {\text{L}}^{-1}$ and $0.0135\text{mol}\cdot {\text{L}}^{-1}$ respectively.

The value of ${\left[{\text{N}}_2{\text{O}}_5\right]}_0$ is $0.0567\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[{\text{N}}_2{\text{O}}_5\right]}_t$ is $0.0135\text{mol}\cdot {\text{L}}^{-1}$.

The value of $k_{\text{r}}$ is $3.7\times {10}^{-5}{\text{s}}^{-1}$.

Substitute the value of $k_{\text{r}}$, ${\left[{\text{N}}_2{\text{O}}_5\right]}_t$ and ${\left[{\text{N}}_2{\text{O}}_5\right]}_0$ in equation (2).

  $\begin{array}{c}0.0135\text{mol}\cdot {\text{L}}^{-1}=\left(0.0567\text{mol}\cdot {\text{L}}^{-1}\right)e^{-\left(3.7\times {10}^{-5}{\text{s}}^{-1}\right)t}\\ e^{-\left(3.7\times {10}^{-5}{\text{s}}^{-1}\right)t}=\frac{0.0135\text{mol}\cdot {\text{L}}^{-1}}{0.0567\text{mol}\cdot {\text{L}}^{-1}}\\ -\left(3.7\times {10}^{-5}{\text{s}}^{-1}\right)t=\ln 0.2380\\ -\left(3.7\times {10}^{-5}{\text{s}}^{-1}\right)t=-1.4354\end{array}$

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

  $\begin{array}{c}t=\frac{1.4354}{3.7\times {10}^{-5}{\text{s}}^{-1}}\\ =3.87\times {10}^4\text{s}\times \frac{1\min }{60\text{s}}\\ =645\min \end{array}$

Thus, the time taken for the decomposition of ${\text{N}}_2{\text{O}}_5$ from $0.0567\text{mol}\cdot {\text{L}}^{-1}$ to $0.0135\text{mol}\cdot {\text{L}}^{-1}$ is $\underset{\_}{645min}$.