Chapter 16, Problem 79QRT

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the decomposition of the oxide has to be stated.  The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide has to be stated. The ${\Delta }_{\text{r}}G°$ for coupling of decomposition of copper oxide at $25°C$ has to be determined.  Whether the metal could be obtain by the corresponding method at $25°C$ or not has to be stated.

Concept Introduction:

The standard Gibbs free energy change of the reaction is the difference of the sum of standard free energy change of products and the sum of standard free energy change of reactants.  The mathematical equation for the calculation of Gibbs free energy for a reaction is shown below.

${\Delta }_r\text{G}°\text{=}{\displaystyle \sum \Delta {\text{G}}_{product}^{\circ }}-{\displaystyle \sum \Delta {\text{G}}_{reac\tan t}^{\circ }}$

Answer to Problem 79QRT

The balanced chemical equation for the decomposition of the oxide is shown below.

  $\text{CuO}\left(\text{s}\right)\to \text{Cu}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide is shown below.

  $\text{CuO}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+\text{Cu}\left(\text{s}\right)$

The ${\Delta }_{\text{r}}G°$ for the coupling reaction of copper oxide at $25°C$ is $\underset{\_}{-7.468kJmo{l}^{-1}}$.

The copper metal can be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide.

Explanation of Solution

The unbalance equation of decomposition of copper (II) oxide is shown below.

  $\text{CuO}\left(\text{s}\right)\to \text{Cu}\left(\text{s}\right)+{\text{O}}_2\left(\text{g}\right)$

A coefficient $\frac{1}{2}$ is added in fornt of ${\text{O}}_2\left(\text{g}\right)$ to balance the chemical equation.

Therefore, the balanced chemical equation for the decomposition of the oxide is shown below.

  $\text{CuO}\left(\text{s}\right)\to \text{Cu}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$ and $\text{Cu}\left(\text{s}\right)$ at $25°C$ is $\text{0}\text{.0}$ whereas the standard Gibbs free energy for the formation of $\text{CuO}\left(\text{s}\right)$ at $25°C$ is $-129.7\text{kJ}{\text{mol}}^{-1}$.

The expression for the Gibbs free energy for the above reaction is shown below.

  $\Delta G°=\left\{\begin{array}{l}\left(1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{Cu}\left(\text{s}\right)\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{CuO}\left(\text{s}\right)\right)\right\}$        (1)

Where,

• $\Delta G_{\text{f}}^{\circ }\left(\text{Cu}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{Cu}\left(\text{s}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is the standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left(\text{CuO}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{CuO}\left(\text{s}\right)$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{Cu}\left(\text{s}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{CuO}\left(\text{s}\right)\right)$ is $-129.7\text{kJ}{\text{mol}}^{-1}$.

Substitute the value $\Delta G_{\text{f}}^{\circ }\left(\text{Cu}\left(\text{s}\right)\right)$, $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ and $\Delta G_{\text{f}}^{\circ }\left(\text{CuO}\left(\text{s}\right)\right)$ of in equation (1).

  $\begin{array}{c}{\Delta }_{\text{r}}G°=\left\{\begin{array}{l}\left(1\text{mol}\times \left(0.0\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \left(0.0\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \left(-129.7\text{kJ}{\text{mol}}^{-1}\right)\right\}\\ =+129.7\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the Gibbs free energy for the decomposition of copper oxide at $25°C$ is $+129.7\text{kJ}{\text{mol}}^{-1}$.

The chemical equation that represents the oxidation of coke is shown below.

  $\text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)$

The standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$ and $\text{C}\left(\text{s}\right)$ at $25°C$ is $\text{0}\text{.0}$ whereas the standard Gibbs free energy for the formation of $\text{CO}\left(\text{g}\right)$ at $25°C$ is $-137.168\text{kJ}{\text{mol}}^{-1}$.

The expression for the Gibbs free energy for the oxidation reaction of coke is shown below.

  $\Delta G°=\left\{\left(1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{CO}\left(g\right)\right)\right)\right\}-\left\{\begin{array}{l}1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{C}\left(\text{s}\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)\right)\end{array}\right\}$        (2)

Where,

• $\Delta G_{\text{f}}^{\circ }\left(\text{C}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{C}\left(\text{s}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is the standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left(\text{CO}\left(\text{g}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{CO}\left(\text{g}\right)$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{C}\left(\text{s}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{CO}\left(\text{g}\right)\right)$ is $-137.168\text{kJ}{\text{mol}}^{-1}$.

Substitute the value $\Delta G_{\text{f}}^{\circ }\left(\text{C}\left(\text{s}\right)\right)$, $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ and $\Delta G_{\text{f}}^{\circ }\left(\text{CO}\left(\text{g}\right)\right)$ of in equation (2).

  $\begin{array}{c}{\Delta }_{\text{r}}G°=\left\{1\text{mol}\times \left(-137.168\text{kJ}{\text{mol}}^{-1}\right)\right\}-\left\{\begin{array}{l}\left(1\text{mol}\times \left(0.0\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \left(0.0\right)\right)\end{array}\right\}\\ =-137.168\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the Gibbs free energy for the oxidation of coke at $25°C$ is $-137.168\text{kJ}{\text{mol}}^{-1}$.

The coupling reaction of the combustion of coke and the decomposition of copper oxide is shown below.

  $\begin{array}{l}\underset{\_}{\begin{array}{c}\text{CuO}\left(\text{s}\right)\to \text{Cu}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\\ \text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)\end{array}}\\ \text{CuO}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+\text{Cu}\left(\text{s}\right)\end{array}$

The resultant Gibbs free energy for the coupling reaction obtained above is calculated as shown below.

  $\begin{array}{c}{\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }=-137.168\text{kJ}{\text{mol}}^{-1}+129.7\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{-7.468kJmo{l}^{-1}}\end{array}$

Thus, the total Gibbs free energy at $25°C$ is $\underset{\_}{-7.468kJmo{l}^{-1}}$.  The negative value of ${\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }$ indicates that the coupling reaction is spontaneous.

Hence, the copper metal can be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide.

(b)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the decomposition of the oxide has to be stated. The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide has to be stated. The ${\Delta }_{\text{r}}G°$ for coupling of decomposition of silver oxide at $25°C$ has to be determined.  Whether the metal could be obtain by the corresponding method at $25°C$ or not has to be stated.

Concept Introduction:

Same as part (a).

Answer to Problem 79QRT

The balanced chemical equation for the decomposition of the oxide is shown below.

  ${\text{Ag}}_2\text{O}\left(\text{s}\right)\to 2\text{Ag}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide is shown below.

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

The ${\Delta }_{\text{r}}G°$ for the coupling reaction of silver oxide at $25°C$ is $\underset{\_}{-125.968kJmo{l}^{-1}}$.

The silver metal can be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide.

Explanation of Solution

The unbalance equation of decomposition of silver oxide is shown below.

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

A coefficient $\frac{1}{2}$ is added in fornt of ${\text{O}}_2\left(\text{g}\right)$ and $2$ in front of $\text{Ag}\left(\text{s}\right)$ to balance the chemical equation.

Therefore, the chemical equation for the coupling of decomposition of silver oxide is shown below.

  ${\text{Ag}}_2\text{O}\left(\text{s}\right)\to 2\text{Ag}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$ and $\text{Ag}\left(\text{s}\right)$ at $25°C$ is $\text{0}\text{.0}$ whereas the standard Gibbs free energy for the formation of ${\text{Ag}}_2\text{O}\left(\text{s}\right)$ at $25°C$ is $-11.2\text{kJ}{\text{mol}}^{-1}$.

The expression for the Gibbs free energy for the above reaction is shown below.

  $\Delta G°=\left\{\begin{array}{l}\left(2\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{Ag}\left(\text{s}\right)\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left({\text{Ag}}_2\text{O}\left(\text{s}\right)\right)\right\}$        (3)

Where,

• $\Delta G_{\text{f}}^{\circ }\left(\text{Ag}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{Ag}\left(\text{s}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is the standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left({\text{Ag}}_2\text{O}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of ${\text{Ag}}_2\text{O}\left(\text{s}\right)$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{Ag}\left(\text{s}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left({\text{Ag}}_2\text{O}\left(\text{s}\right)\right)$ is $-11.2\text{kJ}{\text{mol}}^{-1}$.

Substitute the value $\Delta G_{\text{f}}^{\circ }\left(\text{Ag}\left(\text{s}\right)\right)$, $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ and $\Delta G_{\text{f}}^{\circ }\left({\text{Ag}}_2\text{O}\left(\text{s}\right)\right)$ of in equation (3).

  $\begin{array}{c}{\Delta }_{\text{r}}G°=\left\{\begin{array}{l}\left(2\text{mol}\times \left(0.0\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \left(0.0\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \left(-11.2\text{kJ}{\text{mol}}^{-1}\right)\right\}\\ =+11.2\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the Gibbs free energy for the decomposition of silver oxide at $25°C$ is $+11.2\text{kJ}{\text{mol}}^{-1}$.

The chemical equation that represents the oxidation of coke is shown below.

  $\text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)$

The Gibbs free energy for the oxidation of coke at $25°C$ is $-137.168\text{kJ}{\text{mol}}^{-1}$.

The coupling reaction of the combustion of coke and the decomposition of silver oxide is shown below.

  $\begin{array}{l}\underset{\_}{\begin{array}{c}{\text{Ag}}_2\text{O}\left(\text{s}\right)\to 2\text{Ag}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\\ \text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)\end{array}}\\ {\text{Ag}}_2\text{O}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+2\text{Ag}\left(\text{s}\right)\end{array}$

The resultant Gibbs free energy for the coupling reaction obtained above is calculated as shown below.

  $\begin{array}{c}{\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }=-137.168\text{kJ}{\text{mol}}^{-1}+11.2\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{-125.968kJmo{l}^{-1}}\end{array}$

Thus, the total Gibbs free energy at $25°C$ is $\underset{\_}{-125.968kJmo{l}^{-1}}$.  The negative value of ${\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }$ indicates that the coupling reaction is spontaneous.

Hence, the silver metal can be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide.

(c)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the decomposition of the oxide has to be stated. The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide has to be stated. The ${\Delta }_{\text{r}}G°$ for coupling of decomposition of mercury oxide at $25°C$ has to be determined.  Whether the metal could be obtain by the corresponding method at $25°C$ or not has to be stated.

Concept Introduction:

Same as part (a).

Answer to Problem 79QRT

The balanced chemical equation for the decomposition of the oxide is shown below.

  $\text{HgO}\left(\text{s}\right)\to \text{Hg}\left(l\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide is shown below.

  $\text{HgO}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+\text{Hg}\left(l\right)$

The ${\Delta }_{\text{r}}G°$ for the coupling reaction of mercury oxide at $25°C$ is $\underset{\_}{-78.629kJmo{l}^{-1}}$.

The mercury metal can be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide.

Explanation of Solution

The unbalance equation of decomposition of mercury oxide is shown below.

  $\text{HgO}\left(\text{s}\right)\to \text{Hg}\left(l\right)+{\text{O}}_2\left(\text{g}\right)$

A coefficient $\frac{1}{2}$ is added in fornt of ${\text{O}}_2\left(\text{g}\right)$ to balance the chemical equation.

Therefore, the chemical equation for the coupling of decomposition of mercury oxide is shown below.

  $\text{HgO}\left(\text{s}\right)\to \text{Hg}\left(l\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$ and $\text{Hg}\left(l\right)$ at $25°C$ is $\text{0}\text{.0}$ whereas the standard Gibbs free energy for the formation of $\text{HgO}\left(\text{s}\right)$ at $25°C$ is $-58.539\text{kJ}{\text{mol}}^{-1}$.

The expression for the Gibbs free energy for the above reaction is shown below.

  $\Delta G°=\left\{\begin{array}{l}\left(1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{Hg}\left(l\right)\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{HgO}\left(\text{s}\right)\right)\right\}$        (4)

Where,

• $\Delta G_{\text{f}}^{\circ }\left(\text{Hg}\left(l\right)\right)$ is the standard Gibbs free energy for the formation of $\text{Hg}\left(l\right)$.
• $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is the standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left(\text{HgO}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{HgO}\left(\text{s}\right)$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{Hg}\left(l\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{HgO}\left(\text{s}\right)\right)$ is $-58.539\text{kJ}{\text{mol}}^{-1}$.

Substitute the value $\Delta G_{\text{f}}^{\circ }\left(\text{Hg}\left(l\right)\right)$, $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ and $\Delta G_{\text{f}}^{\circ }\left(\text{HgO}\left(\text{s}\right)\right)$ in equation (4).

  $\begin{array}{c}{\Delta }_{\text{r}}G°=\left\{\begin{array}{l}\left(1\text{mol}\times \left(0.0\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \left(0.0\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \left(-58.539\text{kJ}{\text{mol}}^{-1}\right)\right\}\\ =+58.539\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the Gibbs free energy for the decomposition of mercury oxide at $25°C$ is $+58.539\text{kJ}{\text{mol}}^{-1}$.

The chemical equation that represents the oxidation of coke is shown below.

  $\text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)$

The Gibbs free energy for the oxidation of coke at $25°C$ is $-137.168\text{kJ}{\text{mol}}^{-1}$.

The coupling reaction of the combustion of coke and the decomposition of mercury oxide is shown below.

  $\begin{array}{l}\underset{\_}{\begin{array}{c}\text{HgO}\left(\text{s}\right)\to \text{Hg}\left(l\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\\ \text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)\end{array}}\\ \text{HgO}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+\text{Hg}\left(l\right)\end{array}$

The resultant Gibbs free energy for the coupling reaction obtained above is calculated as shown below.

  $\begin{array}{c}{\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }=-137.168\text{kJ}{\text{mol}}^{-1}+58.539\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{-78.629kJmo{l}^{-1}}\end{array}$

Thus, the total Gibbs free energy at $25°C$ is $\underset{\_}{-78.629kJmo{l}^{-1}}$.  The negative value of ${\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }$ indicates that the coupling reaction is spontaneous.

Hence, the mercury metal can be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide.

(d)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the decomposition of the oxide has to be stated. The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide has to be stated. The ${\Delta }_{\text{r}}G°$ for coupling of decomposition of magnesium oxide at $25°C$ has to be determined.  Whether the metal could be obtain by the corresponding method at $25°C$ or not has to be stated.

Concept Introduction:

Same as part (a).

Answer to Problem 79QRT

The balanced chemical equation for the decomposition of the oxide is shown below.

  $\text{MgO}\left(\text{s}\right)\to \text{Mg}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide is shown below.

  $\text{MgO}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+\text{Mg}\left(\text{s}\right)$

The ${\Delta }_{\text{r}}G°$ for the coupling reaction of magnesium oxide at $25°C$ is $\underset{\_}{432.262kJmo{l}^{-1}}$.

The magnesium metal cannot be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide.

Explanation of Solution

The unbalance equation of decomposition of magnesium oxide is shown below.

  $\text{MgO}\left(\text{s}\right)\to \text{Mg}\left(\text{s}\right)+{\text{O}}_2\left(\text{g}\right)$

A coefficient $\frac{1}{2}$ is added in fornt of ${\text{O}}_2\left(\text{g}\right)$ to balance the chemical equation.

Therefore, thechemical equation for the coupling of decomposition of magnesium oxide is shown below.

  $\text{MgO}\left(\text{s}\right)\to \text{Mg}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$ and $\text{Mg}\left(\text{s}\right)$ at $25°C$ is $\text{0}\text{.0}$ whereas the standard Gibbs free energy for the formation of $\text{MgO}\left(\text{s}\right)$ at $25°C$ is $-569.43\text{kJ}{\text{mol}}^{-1}$.

The expression for the Gibbs free energy for the above reaction is shown below.

  $\Delta G°=\left\{\begin{array}{l}\left(1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{Mg}\left(\text{s}\right)\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{MgO}\left(\text{s}\right)\right)\right\}$        (5)

Where,

• $\Delta G_{\text{f}}^{\circ }\left(\text{Mg}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{Mg}\left(\text{s}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is the standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left(\text{MgO}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{MgO}\left(\text{s}\right)$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{Mg}\left(\text{s}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{MgO}\left(\text{s}\right)\right)$ is $-569.43\text{kJ}{\text{mol}}^{-1}$.

Substitute the value $\Delta G_{\text{f}}^{\circ }\left(\text{Mg}\left(\text{s}\right)\right)$, $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ and $\Delta G_{\text{f}}^{\circ }\left(\text{MgO}\left(\text{s}\right)\right)$ in equation (5).

  $\begin{array}{c}{\Delta }_{\text{r}}G°=\left\{\begin{array}{l}\left(1\text{mol}\times \left(0.0\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \left(0.0\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \left(-569.43\text{kJ}{\text{mol}}^{-1}\right)\right\}\\ =+569.43\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the Gibbs free energy for the decomposition of magnesium oxide at $25°C$ is $+569.43\text{kJ}{\text{mol}}^{-1}$.

The chemical equation that represents the oxidation of coke is shown below.

  $\text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)$

The Gibbs free energy for the oxidation of coke at $25°C$ is $-137.168\text{kJ}{\text{mol}}^{-1}$.

The coupling reaction of the combustion of coke and the decomposition of magnesium oxide is shown below.

  $\begin{array}{l}\underset{\_}{\begin{array}{c}\text{MgO}\left(\text{s}\right)\to \text{Mg}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\\ \text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)\end{array}}\\ \text{MgO}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+\text{Mg}\left(\text{s}\right)\end{array}$

The resultant Gibbs free energy for the coupling reaction obtained above is calculated as shown below.

  $\begin{array}{c}{\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }=-137.168\text{kJ}{\text{mol}}^{-1}+569.43\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{432.262kJmo{l}^{-1}}\end{array}$

Thus, the total Gibbs free energy at $25°C$ is $\underset{\_}{432.262kJmo{l}^{-1}}$.  The positive value of ${\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }$ indicates that the coupling reaction is not feasible.

Hence, the magnesium metal cannot be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide as the value of ${\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }$ is positive.

(e)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the decomposition of the oxide has to be stated. The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide has to be stated. The ${\Delta }_{\text{r}}G°$ for coupling of decomposition of lead oxide at $25°C$ has to be determined.  Whether the metal could be obtain by the corresponding method at $25°C$ or not has to be stated.

Concept Introduction:

Same as part (a).

Answer to Problem 79QRT

The balanced chemical equation for the decomposition of the oxide is shown below.

  $\text{PbO}\left(\text{s}\right)\to \text{Pb}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The overall reaction when the decomposition is coupled to oxidation of coke to carbon monoxide is shown below.

  $\text{PbO}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+\text{Pb}\left(\text{s}\right)$

The ${\Delta }_{\text{r}}G°$ for the coupling reaction of lead oxide at $25°C$ is $\underset{\_}{50.722kJmo{l}^{-1}}$.

The lead metal cannot be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide.

Explanation of Solution

The unbalance equation of decomposition of lead oxide is shown below.

  $\text{PbO}\left(\text{s}\right)\to \text{Pb}\left(\text{s}\right)+{\text{O}}_2\left(\text{g}\right)$

A coefficient $\frac{1}{2}$ is added in fornt of ${\text{O}}_2\left(\text{g}\right)$ to balance the chemical equation.

Therefore, thechemical equation for the coupling of decomposition of lead oxide is shown below.

  $\text{PbO}\left(\text{s}\right)\to \text{Pb}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)$

The standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$ and $\text{Pb}\left(\text{s}\right)$ at $25°C$ is $\text{0}\text{.0}$ whereas the standard Gibbs free energy for the formation of $\text{PbO}\left(\text{s}\right)$ at $25°C$ is $-187.89\text{kJ}{\text{mol}}^{-1}$.

The expression for the Gibbs free energy for the above reaction is shown below.

  $\Delta G°=\left\{\begin{array}{l}\left(1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{Pb}\left(\text{s}\right)\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \Delta G_{\text{f}}^{\circ }\left(\text{PbO}\left(\text{s}\right)\right)\right\}$        (6)

Where,

• $\Delta G_{\text{f}}^{\circ }\left(\text{Pb}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{Pb}\left(\text{s}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is the standard Gibbs free energy for the formation of ${\text{O}}_2\left(\text{g}\right)$.
• $\Delta G_{\text{f}}^{\circ }\left(\text{PbO}\left(\text{s}\right)\right)$ is the standard Gibbs free energy for the formation of $\text{PbO}\left(\text{s}\right)$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{Pb}\left(\text{s}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ is $\text{0}\text{.0}$.

The value of $\Delta G_{\text{f}}^{\circ }\left(\text{PbO}\left(\text{s}\right)\right)$ is $-187.89\text{kJ}{\text{mol}}^{-1}$.

Substitute the value $\Delta G_{\text{f}}^{\circ }\left(\text{Pb}\left(\text{s}\right)\right)$, $\Delta G_{\text{f}}^{\circ }\left({\text{O}}_2\left(\text{g}\right)\right)$ and $\Delta G_{\text{f}}^{\circ }\left(\text{PbO}\left(\text{s}\right)\right)$ in equation (5).

  $\begin{array}{c}{\Delta }_{\text{r}}G°=\left\{\begin{array}{l}\left(1\text{mol}\times \left(0.0\right)\right)+\\ \left(\frac{1}{2}\text{mol}\times \left(0.0\right)\right)\end{array}\right\}-\left\{1\text{mol}\times \left(-187.89\text{kJ}{\text{mol}}^{-1}\right)\right\}\\ =+187.89\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the Gibbs free energy for the decomposition of lead oxide at $25°C$ is $+187.89\text{kJ}{\text{mol}}^{-1}$.

The chemical equation that represents the oxidation of coke is shown below.

  $\text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)$

The Gibbs free energy for the oxidation of coke at $25°C$ is $-137.168\text{kJ}{\text{mol}}^{-1}$.

The coupling reaction of the combustion of coke and the decomposition of lead oxide is shown below.

  $\begin{array}{l}\underset{\_}{\begin{array}{c}\text{PbO}\left(\text{s}\right)\to \text{Pb}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\\ \text{C}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)\end{array}}\\ \text{PbO}\left(\text{s}\right)+\text{C}\left(\text{s}\right)\to \text{CO}\left(\text{g}\right)+\text{Pb}\left(\text{s}\right)\end{array}$

The resultant Gibbs free energy for the coupling reaction obtained above is calculated as shown below.

  $\begin{array}{c}{\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }=-137.168\text{kJ}{\text{mol}}^{-1}+187.89\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{50.722kJmo{l}^{-1}}\end{array}$

Thus, the total Gibbs free energy at $25°C$ is $\underset{\_}{50.722kJmo{l}^{-1}}$.  The positive value of ${\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }$ indicates that the coupling reaction is not feasible.

Hence, the lead metal cannot be obtained from its ore by coupling it with the oxidation of coke to carbon monoxide as the value of ${\Delta }_{\text{r}}{G}_{\text{total}}^{\circ }$ is positive.