Chapter 7, Problem 7B.15E

(a)

Interpretation Introduction

Interpretation:

The half-life for the decomposition of sulfuryl chloride 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.15E

The half-life for the decomposition of sulfuryl chloride is $\underset{\_}{246.6min}$.

Explanation of Solution

As per the given data the rate constant for the decomposition of sulfuryl chloride is $2.81\times {10}^{-3}{\min }^{-1}$.

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 $2.81\times {10}^{-3}{\min }^{-1}$.

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

    $\begin{array}{c}{t}_{1/2}=\frac{\ln 2}{2.81\times {10}^{-3}{\min }^{-1}}\\ =\frac{0.693}{2.81\times {10}^{-3}{\min }^{-1}}\\ =246.6\min \end{array}$

Thus, the half-life for the decomposition of sulfuryl chloride is $\underset{\_}{246.6min}$.

(b)

Interpretation Introduction

Interpretation:

The time required to reduce the concentration of sulfuryl chloride by $10\%$ of the initial value has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.15E

The time required to reduce the concentration of sulfuryl chloride by $10\%$ of the initial value is $\underset{\_}{819.57s}$.

Explanation of Solution

The final concentration of sulfuryl chloride is $10\%$ of the initial value.  Mathematically the final value, ${\left[{\text{SO}}_2{\text{Cl}}_2\right]}_{\text{t}}$ in terms of initial value, ${\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0$ is shown below.

    $\begin{array}{c}{\left[{\text{SO}}_2{\text{Cl}}_2\right]}_{\text{t}}=10\%{\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0\\ =\frac{10}{100}{\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0\\ =0.10{\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0\end{array}$

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

  ${\left[{\text{SO}}_2{\text{Cl}}_2\right]}_{\text{t}}={\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0{\text{e}}^{-k_{\text{r}}t}$        (2)

Where,

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

The value $k_{\text{r}}$ is $2.81\times {10}^{-3}{\min }^{-1}$.

The value of ${\left[{\text{SO}}_2{\text{Cl}}_2\right]}_{\text{t}}$ is ${\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0$.

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

    $\begin{array}{c}-\left(2.81\times {10}^{-3}{\min }^{-1}\right)\times t=\ln \frac{0.10{\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0}{{\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0}\\ -\left(2.81\times {10}^{-3}{\min }^{-1}\right)\times t=\ln 0.10\\ t=\frac{\ln 0.10}{-\left(2.81\times {10}^{-3}{\min }^{-1}\right)}\\ =819.57\text{s}\end{array}$

The time required to reduce the concentration of sulfuryl chloride by $10\%$ of the initial value is $\underset{\_}{819.57s}$.

(c)

Interpretation Introduction

Interpretation:

The mass of sulfuryl chloride remained after $1.5\text{h}$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7B.15E

The mass of sulfuryl chloride remained after $1.5\text{h}$ is $\underset{\_}{10.86g}$.

Explanation of Solution

The given mass and volume of sulfuryl chloride is $14\text{g}$ and $2500\text{L}$ respectively.

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

  $n=\frac{m}{M_{\text{w}}}$        (3)

Where,

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

The value of $m$ for ${\text{SO}}_2{\text{Cl}}_2$ is $14\text{g}$.

The value of $M_{\text{w}}$ for ${\text{SO}}_2{\text{Cl}}_2$ is $134.965\text{g/mol}$.

Substitute the values of $m$ and $M_{\text{w}}$ for ${\text{SO}}_2{\text{Cl}}_2$ in the equation (3).

  $\begin{array}{c}n=\frac{14\text{g}}{134.965\text{g/mol}}\\ =0.1037\text{mol}\end{array}$

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

  $M=\frac{n}{V}$        (4)

Where,

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

The value of $n$ for ${\text{SO}}_2{\text{Cl}}_2$ is $0.1037\text{mol}$.

The value of $V$ for ${\text{SO}}_2{\text{Cl}}_2$ is $2500\text{L}$.

Substitute the values of $n$ and $V$ for ${\text{SO}}_2{\text{Cl}}_2$ in the equation (4).

  $\begin{array}{c}M=\frac{0.1037\text{mol}}{2500\text{L}}\\ =4.148\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the initial concentration of sulfuryl chloride is $4.148\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$.

The time taken by the reaction is $1.5\text{h}$.  The unit conversion of time from $\text{h}$ to $\min$ is shown below.

  $1.5\text{h}\times \frac{60\min }{1\text{h}}=90\min$

The value $k_{\text{r}}$ is $2.81\times {10}^{-3}{\min }^{-1}$.

The value of ${\left[{\text{SO}}_2{\text{Cl}}_2\right]}_0$ is $4.148\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$.

The value of $t$ is $90\min$.

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

    $\begin{array}{c}{\left[{\text{SO}}_2{\text{Cl}}_2\right]}_t=\left(4.148\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\right)\times {\text{e}}^{-\left(2.81\times {10}^{-3}{\min }^{-1}\right)\times \left(90\min \right)}\\ =3.22\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the final concentration of sulfuryl chloride is $3.22\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$.

The number of moles of sulfuryl chloride contained in $3.22\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$ of sulfuryl chloride is calculated using the relation shown below.

  $n=M\times V$        (5)

Where,

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

The value of $M$ for ${\text{SO}}_2{\text{Cl}}_2$ is $3.22\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$.

The value of $V$ for ${\text{SO}}_2{\text{Cl}}_2$ is $2500\text{L}$.

Substitute the values of $n$ and $V$ for ${\text{SO}}_2{\text{Cl}}_2$ in the equation (5).

  $\begin{array}{c}n=3.22\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\times 2500\text{L}\\ =0.0805\text{mol}\end{array}$

The mass of $0.0805\text{mol}$ of sulfuryl chloride which was left in the vessel is calculated by the relation shown below.

  $m=n\times M_{\text{w}}$        (6)

Where,

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

The value of $n$ for ${\text{SO}}_2{\text{Cl}}_2$ is $0.0805\text{mol}$.

The value of $M_{\text{w}}$ for ${\text{SO}}_2{\text{Cl}}_2$ is $134.965\text{g/mol}$.

Substitute the values of $n$ and $M_{\text{w}}$ for ${\text{SO}}_2{\text{Cl}}_2$ in the equation (6).

  $\begin{array}{c}m=0.0805\text{mol}\times 134.965\text{g/mol}\\ =10.86\text{g}\end{array}$

Thus, the mass of sulfuryl chloride remained after $1.5\text{h}$ is $\underset{\_}{10.86g}$.