Chapter 17, Problem 85QRT

(a)

Interpretation Introduction

Interpretation:

The structure formula of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ has to be drawn.  The geometry around the $\text{S}$ atom has to be stated.  The bond angles $\text{O}-\text{S}-\text{O}$ and $\text{O}-\text{S}-\text{F}$ has to be stated.

Concept Introduction:

The molecular shape is determined primarily by the repulsions between pairs of electrons in the molecule or molecular ion, be they bonding pairs or lone pairs by the theory of the valence-shell-electron-pair repulsion (VSEPR).  The specific sets of bond pairs and lone pairs of electrons give different molecular shape to molecules.

Explanation of Solution

The given compound is ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$.  In ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$, the sulfur is the center atom.  The sulfur atom is attached to the two oxygen, a fluorine atom, and methyl group.

Therefore, the structure formula of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is shown below.

Figure 1

The number of ligands around sulfur atom of the molecule is $4$.

The number of lone pairs on sulfur atom of the molecule is $0$.

The formula to calculate number of sets of electron pairs is shown below.

    $N=N_{\text{ligand}}+N_{\text{lone}\text{pair}}$

Where,

• $N$ is number of sets of electron pairs.
• $N_{\text{ligand}}$ is the number of ligands around the chromium atom of the molecule.
• $N_{\text{lone}\text{pair}}$ is number of lone pairs on the chromium atom of the molecule.

Substitute the values of $N_{\text{ligand}}$ and $N_{\text{lone}\text{pair}}$ in the above equation.

    $\begin{array}{c}N=4+0\\ =4\end{array}$

When number of sets of electron pairs is $4$, then the four orbital are required.  Therefore, the hybridization of inner atom of the molecule is $s{p}^3$.  The corresponding geometery will be tetrahedral.

The bond angle for tetrahedral geometry is $109.5°$.  Therefore, the bond angles $\text{O}-\text{S}-\text{O}$ and $\text{O}-\text{S}-\text{F}$ are apporximatly $\underset{\_}{109.5°}$.

(b)

Interpretation Introduction

Interpretation:

The mass of $\text{HF}$ required to electrolyze $150\text{g}{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ has to be calculated.  The mass of each product has to be calculated.

Concept Introduction:

The stoichiometry of a chemical species involved in a chemical reaction represents the number of chemical species involved in the chemical reaction.  The stoichiometry of a chemical species helps in calculating the expected mass of reactant and product.  The stoichiometry of a chemical species is also represented in a number of moles.

Explanation of Solution

The given mass of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is $150\text{g}$.

The molar mass of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is $98.09\text{g}/\text{mol}$.

The molar mass of $\text{HF}$ is $20.01\text{g}/\text{mol}$.

The number of moles a substance is given by the formula as shown below.

  $n=\frac{m}{M}$

Where,

• $m$ is the mass of the substance.
• $M$ is the molar mass of the substance.

Substitute the value of mass and molar mass of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ in the above equation.

    $\begin{array}{c}n=\frac{150\text{g}}{98.09\text{g}/\text{mol}}\\ =1.53\text{mol}\end{array}$

The given reaction is shown below.

    ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}+3\text{HF}\to {\text{CF}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}+3{\text{H}}_{\text{2}}$

Three moles of $\text{HF}$ reacts with one mole of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$.  Therefore, the relation between the number of moles of $\text{HF}$ and ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is given by the expression as shown below.

  $n_{\text{HF}}=3n_{{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}}$

Where,

• $n_{\text{HF}}$ is the number of moles of $\text{HF}$.
• $n_{{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}}$ is the number of moles of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$.

Substitute the value of $n_{{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}}$ in the above equation.

  $\begin{array}{c}{n}_{\text{HF}}=3\left(1.53\text{mol}\right)\\ =4.59\text{mol}\end{array}$

The relation between number of moles and mass of a substance is given by the expression as shown below.

  $m=n\times M$        (1)

Where,

• $m$ is the mass of the substance.
• $n$ is the number of moles of the substance.

Substitute the value of number of moles and molar mass of $\text{HF}$ in the above equation.

  $\begin{array}{c}m=\left(4.59\text{mol}\right)\times \left(20.01\text{g}/\text{mol}\right)\\ =\underset{\_}{91.84g}\end{array}$

Therefore, the mass of $\text{HF}$ required to electrolyze $150\text{g}{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is $\underset{\_}{91.84g}$.

The molar mass of ${\text{H}}_2$ is $2.016\text{g}/\text{mol}$.

Three moles of ${\text{H}}_2$ produced by one mole of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$.   Therefore, the relation between the number of moles of ${\text{H}}_2$ and ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is given by the expression as shown below.

  $n_{{\text{H}}_2}=3n_{{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}}$

Where,

• $n_{{\text{H}}_2}$ is the number of moles of ${\text{H}}_2$.
• $n_{{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}}$ is the number of moles of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$.

Substitute the value of $n_{{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}}$ in the above equation.

  $\begin{array}{c}{n}_{\text{HF}}=3\left(1.53\text{mol}\right)\\ =4.59\text{mol}\end{array}$

Substitute the value of number of moles and molar mass of ${\text{H}}_2$ in the equation (1).

  $\begin{array}{c}m=\left(4.59\text{mol}\right)\times \left(2.016\text{g}/\text{mol}\right)\\ =\underset{\_}{9.25g}\end{array}$

Therefore, the mass of ${\text{H}}_2$ produced by $150\text{g}{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is $\underset{\_}{9.25g}$.

One

The molar mass of ${\text{CF}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is $282.13\text{g}/\text{mol}$.

One mole of ${\text{CF}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ produced by one mole of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$.  Therefore, $1.53\text{mol}$ of ${\text{CF}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ produced by $1.53\text{mol}$ of ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$.

Substitute the value of number of moles and molar mass of ${\text{CF}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ in the equation (1).

  $\begin{array}{c}m=\left(1.53\text{mol}\right)\times \left(282.13\text{g}/\text{mol}\right)\\ =\underset{\_}{431.66g}\end{array}$

Therefore, the mass of ${\text{CF}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ produced by $150\text{g}{\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}$ is $\underset{\_}{431.66g}$.

(c)

Interpretation Introduction

Interpretation:

Whether the hydrogen gas is produced at anode or cathode of the electrolysis cell has to be stated.

Concept Introduction:

A reducing agent is a reacting species in the reaction that reduces another reacting species.  A reducing agent itself gets oxidized when reacts with another species.  The reducing agent donates electrons.  An oxidizing agent is a reacting species in the reaction that oxidize another reacting species.  An oxidizing agent itself gets reduced when it reacts with another species.  The oxidizing agent accepts electrons.

Explanation of Solution

The given reaction is shown below.

    ${\text{CH}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}+3\text{HF}\to {\text{CF}}_{\text{3}}{\text{SO}}_{\text{2}}\text{F}+3{\text{H}}_{\text{2}}$

The reduction of hydrogen ion is shown below.

    ${\text{H}}^{+}\left(\text{aq}\right)+{\text{e}}^{-}\to {\text{H}}_2\left(\text{g}\right)$

Hydrogen has been reduced in the reaction.  The reduction takes place at cathode.  Therefore, ${\text{H}}_{\text{2}}$ is produced at the cathode.

(d)

Interpretation Introduction

Interpretation:

The amount of energy transfer in the cell at $250\text{A}$ and $8\text{V}$ for $24\text{h}$ has to be calculated.

Concept Introduction:

During a redox reaction, the amount of electrons transferred is determined by measuring the current flowing in an external electric circuit for a particular period of time.  The electric charge flowing through an electric circuit is calculated by multiplying the current and the time interval.

Explanation of Solution

The cell potential is $8\text{V}$.

The unit conversion of cell potential from $\text{V}$ to $\text{J}/\text{C}$ is shown below.

  $\begin{array}{c}\text{Cell potential}=\left(8\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{V}}\right)\\ =8\text{J}/\text{C}\end{array}$

The current pass through the cell is $250\text{A}$.

The unit conversion of current from $\text{A}$ to $\text{C}/\text{s}$ is shown below.

  $\begin{array}{c}\text{Current}=\left(250\text{A}\right)\left(\frac{1\text{C}/\text{s}}{1\text{A}}\right)\\ =250\text{C}/\text{s}\end{array}$

The time of current flow is $24\text{h}$.

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

  $\begin{array}{c}\text{Time}=\left(24\text{h}\right)\left(\frac{60\text{s}}{1\text{h}}\right)\\ =1440\text{s}\end{array}$

The energy of the cell is calculated by the formula shown below.

    $\text{Energy}=\text{Cell potential}\times \text{Current}\times \text{Time}$

Substitute the value of cell potential, current, and time in the above equation.

    $\begin{array}{c}\text{Energy}=\left(8\text{J}/\text{C}\right)\times \left(250\text{C}/\text{s}\right)\times \left(1440\text{s}\right)\\ =2.88\times {10}^6\text{J}\end{array}$

The unit conversion of energy from $\text{J}$ to $\text{kWh}$ is shown below.

  $\begin{array}{c}\text{Energy}=\left(2.88\times {10}^6\text{J}\right)\left(\frac{1\text{kWh}}{3.60\times {10}^6\text{J}}\right)\\ =\underset{\_}{0.8kWh}\end{array}$

Therefore, the amount of energy transfer in the cell at $250\text{A}$ and $8\text{V}$ for $24\text{h}$ is $\underset{\_}{0.8kWh}$.