Chapter 16, Problem 77QRT

(a)

Interpretation Introduction

Interpretation:

The minimum mass of graphite that would have to be oxidized to carbon dioxide to provide the necessary work has to be calculated.

Concept Introduction:

The Gibbs free energy of a system is defined as the Gibbs free energy of product minus Gibbs free energy of reactants.  The Gibbs free energy of the system is a state function as it is defined in terms of thermodynamic properties that are state functions.

Answer to Problem 77QRT

The given reaction is capable of being harnessed to do the useful work.

Explanation of Solution

The given reaction is shown below.

  ${\text{2C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)+15{\text{O}}_{\text{2}}\left(\text{g}\right)\to 12{\text{CO}}_{\text{2}}\left(\text{g}\right)+{\text{6H}}_2\text{O}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}G°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}G°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}G°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}G°\left(\text{Reactants}\right)$        (1)

Where,

• ${\Delta }_{\text{f}}G°$ is the change in standard Gibbs free energy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}G°$ for ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, ${\text{H}}_2\text{O}\left(\text{g}\right)$, ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)$ and ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $-394.359\text{kJ}/\text{mol}$, $-228.572\text{kJ}/\text{mol}$, $124.5\text{kJ}/\text{mol}$ and $0\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (1) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\begin{array}{c}\left(\begin{array}{l}{\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)+\\ {\text{n}}_{{\text{H}}_2\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_2\text{O}\left(\text{g}\right)\right)\end{array}\right)-\\ \left(\begin{array}{l}{\text{n}}_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)}\times {\Delta }_{\text{f}}G°\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)\right)+\\ {\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\end{array}\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\begin{array}{l}12\times \left(-394.359\text{kJ}/\text{mol}\right)+\\ 6\times \left(-228.572\text{kJ}/\text{mol}\right)\end{array}\right)\\ -\left(\begin{array}{l}2\times \left(124.5\text{kJ}/\text{mol}\right)+\\ 15\times \left(0\text{kJ}/\text{mol}\right)\end{array}\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(-4732.308\text{kJ}/\text{mol}-1371.432\text{kJ}/\text{mol}\right)\\ -249\text{kJ}/\text{mol}\end{array}\right)\\ =-6352.74\text{kJ}/\text{mol}\end{array}$

The value of Gibbs free energy is negative.  Therefore, the reaction is product favored and the reaction can do upto $6352.74\text{kJ}/\text{mol}$ work.

(b)

Interpretation Introduction

Interpretation:

The minimum mass of graphite that would have to be oxidized to carbon dioxide to provide the necessary work has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 77QRT

The minimum mass of graphite that would have to be oxidized to carbon dioxide to provide the necessary work is $\underset{\_}{5.063g}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{2NF}}_{\text{3}}\left(\text{g}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)+{\text{3F}}_{\text{2}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}G°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}G°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}G°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}G°\left(\text{Reactants}\right)$        (1)

Where,

• ${\Delta }_{\text{f}}G°$ is the change in standard Gibbs free energy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}G°$ for ${\text{N}}_{\text{2}}\left(\text{g}\right)$, ${\text{F}}_{\text{2}}\left(\text{g}\right)$, and ${\text{NF}}_{\text{3}}\left(\text{g}\right)$ is $0\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$, and $-83.2\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (1) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\left(\begin{array}{l}{\text{n}}_{{\text{N}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{N}}_{\text{2}}\left(\text{g}\right)\right)+\\ {\text{n}}_{{\text{F}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{F}}_{\text{2}}\left(\text{g}\right)\right)\end{array}\right)-\left({\text{n}}_{{\text{NF}}_{\text{3}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{NF}}_{\text{3}}\left(\text{g}\right)\right)\right)\right)\\ =\left(\left(\begin{array}{l}1\times \left(0\text{kJ}/\text{mol}\right)+\\ 3\times \left(0\text{kJ}/\text{mol}\right)\end{array}\right)-\left(2\times \left(-83.2\text{kJ}/\text{mol}\right)\right)\right)\\ =166.4\text{kJ}/\text{mol}\end{array}$

The value of Gibbs free energy is positive.  Therefore, the reaction is reactant favored and the reaction requires $166.4\text{kJ}/\text{mol}$ work.

The reaction for the combustion of carbon is shown below.

    $\text{C}\left(\text{graphite}\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{f}}G°$ for ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{graphite}\right)$, and ${\text{O}}_2\left(\text{g}\right)$ is $-394.359\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$, and $0\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (1) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)-\left(\begin{array}{l}{\text{n}}_{\text{C}\left(\text{graphite}\right)}\times {\Delta }_{\text{f}}G°\left(\text{C}\left(\text{graphite}\right)\right)+\\ {\text{n}}_{{\text{O}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right)\right)\\ =\left(\left(\text{1}\times -394.359\text{kJ}/\text{mol}\right)-\left(\begin{array}{l}\text{1}\times 0\text{kJ}/\text{mol}+\\ \text{1}\times 0\text{kJ}/\text{mol}\end{array}\right)\right)\\ =-394.359\text{kJ}/\text{mol}\end{array}$

The Gibbs energy required for the conversion of graphite to carbon monoxide is $394.359\text{kJ}/\text{mol}$.

The number of moles required to provide the necessary work is calculated by the formula shown below.

    $\text{Number of moles}=\frac{{\text{Gibbs energy for decomposition of NF}}_{\text{3}}\left(\text{g}\right)}{\text{Gibbs energy for combustion of graphite}}$

Substitute the value of Gibbs energy for decomposition of ${\text{NF}}_{\text{3}}\left(\text{g}\right)$ and combustion of graphite in the above equation.

    $\begin{array}{c}\text{Number of moles}=\frac{{\text{Gibbs energy for decomposition of NF}}_{\text{3}}\left(\text{g}\right)}{\text{Gibbs energy for combustion of graphite}}\\ =\frac{166.4\text{kJ}/\text{mol}}{394.359\text{kJ}/\text{mol}}\\ =0.4219\end{array}$

Mass of carbon required is calculated by the formula shown below.

    $\text{Mass}=\text{Molar mass}\times \text{Number of moles}$

The molar mass of carbon is $\text{12 g/mol}$.

Substitute the value of molar mass and number of moles in the above equation.

    $\begin{array}{c}\text{Mass}=\text{12 g/mol}\times 0.421\text{9 mol}\\ =\underset{\_}{5.063g}\end{array}$

Thus, the minimum mass of graphite that would have to be oxidized to carbon dioxide to provide the necessary work is $\underset{\_}{5.063g}$.

(c)

Interpretation Introduction

Interpretation:

The minimum mass of graphite that would have to be oxidized to carbon dioxide to provide the necessary work has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 77QRT

The minimum mass of graphite that would have to be oxidized to carbon dioxide to provide the necessary work is $\underset{\_}{26.88g}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{TiO}}_2\left(\text{s}\right)\to \text{Ti}\left(\text{s}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}G°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}G°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}G°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}G°\left(\text{Reactants}\right)$        (1)

Where,

• ${\Delta }_{\text{f}}G°$ is the change in standard Gibbs free energy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}G°$ for $\text{Ti}\left(\text{s}\right)$, ${\text{O}}_{\text{2}}\left(\text{g}\right)$, and ${\text{TiO}}_2\left(\text{s}\right)$ is $0\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$, and $-884.5\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (1) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\begin{array}{c}\left(\begin{array}{l}{\text{n}}_{\text{Ti}\left(\text{s}\right)}\times {\Delta }_{\text{f}}G°\left(\text{Ti}\left(\text{s}\right)\right)+\\ {\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\end{array}\right)-\\ \left({\text{n}}_{{\text{TiO}}_2\left(\text{s}\right)}\times {\Delta }_{\text{f}}G°\left({\text{TiO}}_2\left(\text{s}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\begin{array}{l}1\times \left(0\text{kJ}/\text{mol}\right)+\\ 1\times \left(0\text{kJ}/\text{mol}\right)\end{array}\right)\\ -\left(1\times \left(-884.5\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =884.5\text{kJ}/\text{mol}\end{array}$

The value of Gibbs free energy is positive.  Therefore, the reaction is reactant favored and the reaction requires $884.5\text{kJ}/\text{mol}$ work.

The reaction for the combustion of carbon is shown below.

    $\text{C}\left(\text{graphite}\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{f}}G°$ for ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{graphite}\right)$, and ${\text{O}}_2\left(\text{g}\right)$ is $-394.359\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$, and $0\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (1) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)-\left(\begin{array}{l}{\text{n}}_{\text{C}\left(\text{graphite}\right)}\times {\Delta }_{\text{f}}G°\left(\text{C}\left(\text{graphite}\right)\right)+\\ {\text{n}}_{{\text{O}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right)\right)\\ =\left(\left(\text{1}\times -394.359\text{kJ}/\text{mol}\right)-\left(\begin{array}{l}\text{1}\times 0\text{kJ}/\text{mol}+\\ \text{1}\times 0\text{kJ}/\text{mol}\end{array}\right)\right)\\ =-394.359\text{kJ}/\text{mol}\end{array}$

The Gibbs energy required for the conversion of graphite to carbon monoxide is $394.359\text{kJ}/\text{mol}$.

The number of moles required to provide the necessary work is calculated by the formula shown below.

    $\text{Number of moles}=\frac{{\text{Gibbs energy for decomposition of TiO}}_2\left(\text{s}\right)}{\text{Gibbs energy for combustion of graphite}}$

Substitute the value of Gibbs energy for decomposition of ${\text{TiO}}_2\left(\text{s}\right)$ and combustion of graphite in the above equation.

    $\begin{array}{c}\text{Number of moles}=\frac{{\text{Gibbs energy for decomposition of NF}}_{\text{3}}\left(\text{g}\right)}{\text{Gibbs energy for combustion of graphite}}\\ =\frac{884.5\text{kJ}/\text{mol}}{394.359\text{kJ}/\text{mol}}\\ =2.24\end{array}$

Mass of carbon required is calculated by the formula shown below.

    $\text{Mass}=\text{Molar mass}\times \text{Number of moles}$

The molar mass of carbon is $\text{12 g/mol}$.

Substitute the value of molar mass and number of moles in the above equation.

    $\begin{array}{c}\text{Mass}=\text{12 g/mol}\times 2.24\text{ mol}\\ =\underset{\_}{26.88g}\end{array}$

Thus, the minimum mass of graphite that would have to be oxidized to carbon dioxide to provide the necessary work is $\underset{\_}{26.88g}$.