Chapter 16, Problem 78QRT

(a)

Interpretation Introduction

Interpretation:

The minimum mass of hydrogen gas that would have to be burned to form water vapor 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 78QRT

The minimum mass of hydrogen gas that would have to be burned to form water vapor to provide the necessary work is $\underset{\_}{13.844g}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\to \text{2Al}\left(\text{s}\right)+\frac{\text{3}}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)$

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 $\text{Al}\left(\text{s}\right)$, ${\text{O}}_{\text{2}}\left(\text{g}\right)$, and ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$ is $0\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$, and $-1582.3\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\begin{array}{c}\left(\begin{array}{l}{\text{n}}_{\text{Al}\left(\text{s}\right)}\times {\Delta }_{\text{f}}G°\left(\text{Al}\left(\text{s}\right)\right)+\\ {\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\end{array}\right)-\\ \left({\text{n}}_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)}\times {\Delta }_{\text{f}}G°\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\begin{array}{l}2\times \left(0\text{kJ}/\text{mol}\right)+\\ \frac{3}{2}\times \left(0\text{kJ}/\text{mol}\right)\end{array}\right)\\ -\left(1\times \left(-1582.3\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =1582.3\text{kJ}/\text{mol}\end{array}$

The value of Gibbs free energy is positive.  Therefore, the reaction is reactant favored and the reaction requires $1582.3\text{kJ}/\text{mol}$ work.

The reaction for the burning of hydrogen gas is shown below.

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

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

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)-\left(\begin{array}{l}{\text{n}}_{{\text{H}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_2\left(\text{g}\right)\right)+\\ {\text{n}}_{{\text{O}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right)\right)\\ =\left(\left(\text{1}\times -228.572\text{kJ}/\text{mol}\right)-\left(\begin{array}{l}\text{1}\times 0\text{kJ}/\text{mol}+\\ \text{1}\times 0\text{kJ}/\text{mol}\end{array}\right)\right)\\ =-228.572\text{kJ}/\text{mol}\end{array}$

The Gibbs energy required for the burning of hydrogen gas is $228.572\text{kJ}/\text{mol}$.

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

    $\text{Number of moles}=\frac{{\text{Gibbs energy for decomposition of Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)}{\text{Gibbs energy for burning of hydrogen gas}}$

Substitute the value of Gibbs energy for decomposition of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$ and combustion of graphite in the above equation.

    $\begin{array}{c}\text{Number of moles}=\frac{{\text{Gibbs energy for decomposition of Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)}{\text{Gibbs energy for burning of hydrogen gas}}\\ =\frac{1582.3\text{kJ}/\text{mol}}{228.572\text{kJ}/\text{mol}}\\ =6.922\end{array}$

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 ${\text{H}}_2$ is $\text{2 g/mol}$.

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

    $\begin{array}{c}\text{Mass}=\text{2 g/mol}\times 6.922\text{ mol}\\ =\underset{\_}{13.844g}\end{array}$

Thus, the minimum mass of hydrogen gas that would have to be burned to form water vapor to provide the necessary work is $\underset{\_}{13.844g}$.

(b)

Interpretation Introduction

Interpretation:

The minimum mass of hydrogen gas that would have to be burned to form water vapor to provide the necessary work has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 78QRT

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

Explanation of Solution

The given reaction is shown below.

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

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 ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{CO}\left(\text{g}\right)$, and ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $-394.359\text{kJ}/\text{mol}$, $-137.168\text{kJ}/\text{mol}$, and $0\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\begin{array}{c}\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)-\\ \left({\text{n}}_{\text{CO}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left(\text{CO}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(-394.359\text{kJ}/\text{mol}\right)\right)\\ -\left(2\times \left(-137.168\text{kJ}/\text{mol}\right)+1\times \left(0\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =-514.382\text{kJ}/\text{mol}\end{array}$

The value of Gibbs free energy is negative.  Therefore, the reaction is product favored and the reaction is capable of being harnessed to do $514.382\text{kJ}/\text{mol}$ work.

(c)

Interpretation Introduction

Interpretation:

The minimum mass of hydrogen gas that would have to be burned to form water vapor to provide the necessary work has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 78QRT

The minimum mass of hydrogen gas that would have to be burned to form water vapor to provide the necessary work is $\underset{\_}{0.8834g}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

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 ${\text{C}}_{\text{2}}{\text{H}}_4\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$, and ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ is $68.15\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$, and  $-32.82\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\begin{array}{c}\left(\begin{array}{l}{\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_4\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{C}}_{\text{2}}{\text{H}}_4\left(\text{g}\right)\right)+\\ {\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\end{array}\right)-\\ \left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\begin{array}{l}1\times \left(68.15\text{kJ}/\text{mol}\right)+\\ 1\times \left(0\text{kJ}/\text{mol}\right)\end{array}\right)\\ -\left(1\times \left(-32.82\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =100.97\text{kJ}/\text{mol}\end{array}$

The value of Gibbs free energy is positive.  Therefore, the reaction is reactant favored and the reaction requires $100.97\text{kJ}/\text{mol}$ work.

The reaction for the burning of hydrogen gas is shown below.

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

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

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)-\left(\begin{array}{l}{\text{n}}_{{\text{H}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_2\left(\text{g}\right)\right)+\\ {\text{n}}_{{\text{O}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right)\right)\\ =\left(\left(\text{1}\times -228.572\text{kJ}/\text{mol}\right)-\left(\begin{array}{l}\text{1}\times 0\text{kJ}/\text{mol}+\\ \text{1}\times 0\text{kJ}/\text{mol}\end{array}\right)\right)\\ =-228.572\text{kJ}/\text{mol}\end{array}$

The Gibbs energy required for the burning of hydrogen gas is $228.572\text{kJ}/\text{mol}$.

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

    $\text{Number of moles}=\frac{{\text{Gibbs energy for decomposition of C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)}{\text{Gibbs energy for burning of hydrogen gas}}$

Substitute the value of Gibbs energy for decomposition of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ and combustion of graphite in the above equation.

    $\begin{array}{c}\text{Number of moles}=\frac{{\text{Gibbs energy for decomposition of C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)}{\text{Gibbs energy for burning of hydrogen gas}}\\ =\frac{100.97\text{kJ}/\text{mol}}{228.572\text{kJ}/\text{mol}}\\ =0.4417\end{array}$

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 ${\text{H}}_2$ is $\text{2 g/mol}$.

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

    $\begin{array}{c}\text{Mass}=\text{2 g/mol}\times 0.4417\text{ mol}\\ =\underset{\_}{0.8834g}\end{array}$

Thus, the minimum mass of hydrogen gas that would have to be burned to form water vapor to provide the necessary work is $\underset{\_}{0.8834g}$.