Chapter 7, Problem 7B.16E

(a)

Interpretation Introduction

Interpretation:

The half-life for the decomposition of ethane 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.16E

The half-life for the formation of methyl radical from ethane is $\underset{\_}{246.6min}$.

Explanation of Solution

As per the given data the rate constant for the formation of methyl radical from ethaneis $1.98{\text{h}}^{-1}$.

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 $1.98{\text{h}}^{-1}$.

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

    $\begin{array}{c}{t}_{1/2}=\frac{\ln 2}{1.98{\text{h}}^{-1}}\\ =\frac{0.693}{1.98{\text{h}}^{-1}}\\ =0.35\text{h}\end{array}$

Thus, the half-life for the formation of methyl radical from ethaneis $\underset{\_}{0.35h}$.

(b)

Interpretation Introduction

Interpretation:

The time required to reduce the amount of ethane from $2.26\times {10}^{-3}\text{mol}$ to $1.45\times {10}^{-4}\text{mol}$ have to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.16E

The time required to reduce the concentration of ethaneis $\underset{\_}{807.17s}$.

Explanation of Solution

As per the given data in the question the initial and final number of moles of ethane are $2.26\times {10}^{-3}\text{mol}$ and $1.45\times {10}^{-4}\text{mol}$ respectively.

The initial concentration of ethane is calculated using the relation shown below.

  $M=\frac{n}{V}$        (2)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $n$ for ethane is $2.26\times {10}^{-3}\text{mol}$.

The value of $V$ for ethane is $500\text{mL}$.

Substitute the values of $n$ and $V$ for initial concentration of ethane in the equation (2).

  $\begin{array}{c}M=\frac{2.26\times {10}^{-3}\text{mol}}{500\text{mL}\times \frac{1\text{L}}{1000\text{mL}}}\\ =4.52\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the initial concentration of ethane is $4.52\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}$.

The final concentration of ethane is calculated using the relation represented in equation (2).

The value of $n$ for ethane is $1.45\times {10}^{-4}\text{mol}$.

The value of $V$ for ethane is $500\text{mL}$.

Substitute the values of $n$ and $V$ for final concentration of ethane in the equation (2).

    $\begin{array}{c}M=\frac{1.45\times {10}^{-4}\text{mol}}{500\text{mL}\times \frac{1\text{L}}{1000\text{mL}}}\\ =2.9\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the final concentration of ethane is $2.9\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}$.

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

  $-kt=\ln \frac{{\left[{\text{C}}_2{\text{H}}_6\right]}_t}{{\left[{\text{C}}_2{\text{H}}_6\right]}_0}$        (3)

Where,

• ${\left[{\text{C}}_2{\text{H}}_6\right]}_{\text{t}}$ is the concentration at time $t$.
• ${\left[{\text{C}}_2{\text{H}}_6\right]}_0$ is the initial concentration.
• $k_{\text{r}}$ is the order rate constant.
• $t$ is the time taken.

The value $k_{\text{r}}$ is $5.5\times {10}^{-4}{\text{s}}^{-1}$.

The value of ${\left[{\text{C}}_2{\text{H}}_6\right]}_{\text{t}}$ is $2.9\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}$.

The value of ${\left[{\text{C}}_2{\text{H}}_6\right]}_0$ is $4.52\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $k_{\text{r}}$, ${\left[{\text{C}}_2{\text{H}}_6\right]}_0$ and ${\left[{\text{C}}_2{\text{H}}_6\right]}_{\text{t}}$ in equation (3).

    $\begin{array}{c}-\left(5.5\times {10}^{-4}{\text{s}}^{-1}\right)\times t=\ln \frac{2.9\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}}{4.52\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}}\\ -\left(5.5\times {10}^{-4}{\text{s}}^{-1}\right)\times t=\ln 0.6415\\ t=\frac{\ln 0.6415}{-\left(5.5\times {10}^{-4}{\text{s}}^{-1}\right)}\\ =807.17\text{s}\end{array}$

The time required to reduce the concentration of ethane is $\underset{\_}{807.17s}$.

(c)

Interpretation Introduction

Interpretation:

The mass of ethane remained after $42\text{min}$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.16E

The mass of ethane remained after $42\min$ is $\underset{\_}{1.35mg}$.

Explanation of Solution

The given mass and volume of ethane is $5.44\text{mg}$ and $500\text{mL}$ respectively.

The initial number of moles of ethane is calculated using the relation shown below.

  $n=\frac{m}{M_{\text{w}}}$        (4)

Where,

• $n$ is the number of moles.
• $m$ is the mass.
• $M_{\text{w}}$ is the molar mass.

The value of $m$ for ethane is $5.44\text{mg}$.

The value of $M_{\text{w}}$ for ethane is $30.07\text{g/mol}$.

Substitute the values of $m$ and $M_{\text{w}}$ for ethane in the equation (4).

  $\begin{array}{c}n=\frac{5.44\text{mg}\times \frac{1\text{g}}{1000\text{mg}}}{30.07\text{g/mol}}\\ =1.8\times {10}^{-4}\text{mol}\end{array}$

The initial concentration of ethane is calculated using the relation shown below.

  $M=\frac{n}{V}$        (5)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $n$ for ethane is $1.8\times {10}^{-4}\text{mol}$.

The value of $V$ for ethane is $500\text{mL}$.

Substitute the values of $n$ and $V$ for ethane in the equation (5).

  $\begin{array}{c}M=\frac{1.8\times {10}^{-4}\text{mol}}{500\text{mL}\times \frac{1\text{L}}{1000\text{mL}}}\\ =3.6\times {10}^{-4}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the initial concentration of ethaneis $3.6\times {10}^{-4}\text{mol}\cdot {\text{L}}^{-1}$.

The time taken by the reaction is $42\text{min}$.  The unit conversion of time from $\min$ to $\text{s}$ is shown below.

  $42\text{min}\times \frac{60\text{s}}{1\text{min}}=2520\text{s}$

The value $k_{\text{r}}$ is $5.5\times {10}^{-4}{\text{s}}^{-1}$.

The value of ${\left[{\text{C}}_2{\text{H}}_6\right]}_0$ is $3.6\times {10}^{-4}\text{mol}\cdot {\text{L}}^{-1}$.

The value of $t$ is $2520\text{s}$.

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

    ${\left[{\text{C}}_2{\text{H}}_6\right]}_{\text{t}}={\left[{\text{C}}_2{\text{H}}_6\right]}_0{\text{e}}^{-k_{\text{r}}t}$        (6)

Where,

• ${\left[{\text{C}}_2{\text{H}}_6\right]}_{\text{t}}$ is the concentration at time $t$.
• ${\left[{\text{C}}_2{\text{H}}_6\right]}_0$ is the initial concentration.
• $k_{\text{r}}$ is the order rate constant.
• $t$ is the time taken.

Substitute the value of $k_{\text{r}}$, $t$ and ${\left[{\text{C}}_2{\text{H}}_6\right]}_0$ in equation (6).

    $\begin{array}{c}{\left[{\text{C}}_2{\text{H}}_6\right]}_t=\left(3.6\times {10}^{-4}\text{mol}\cdot {\text{L}}^{-1}\right)\times {\text{e}}^{-\left(5.5\times {10}^{-4}{\text{s}}^{-1}\right)\times \left(2520\text{s}\right)}\\ =9.00\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the final concentration of ethaneis $9.00\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$.

The number of moles of ethane contained in $9.00\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$ of sulfuryl chloride is calculated using the relation shown below.

  $n=M\times V$        (7)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $M$ for ethane is $9.00\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$.

The value of $V$ for ethane is $500\text{mL}$.

Substitute the values of $n$ and $V$ for ethane in the equation (7).

  $\begin{array}{c}n=9.00\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\times 500\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\\ =4.5\times {10}^{-5}\text{mol}\end{array}$

The mass of $4.5\times {10}^{-5}\text{mol}$ of ethane which was left in the vessel is calculated by the relation shown below.

  $m=n\times M_{\text{w}}$        (8)

Where,

• $n$ is the number of moles.
• $m$ is the mass.
• $M_{\text{w}}$ is the molar mass.

The value of $n$ for ethane is $4.5\times {10}^{-5}\text{mol}$.

The value of $M_{\text{w}}$ for ethane is $30.07\text{g/mol}$.

Substitute the values of $n$ and $M_{\text{w}}$ for ethane in the equation (8).

  $\begin{array}{c}m=4.5\times {10}^{-5}\text{mol}\times 30.07\text{g/mol}\\ =1.35\times {10}^{-3}\text{g}\times \frac{1000\text{mg}}{1\text{g}}\\ =1.35\text{mg}\end{array}$

Thus, the mass of ethane remained after $42\min$ is $\underset{\_}{1.35mg}$.