Chapter 4, Problem 4.48E

(a)

Interpretation Introduction

Interpretation:

The formation of acetic acid from methanol and carbon monoxide is easiest to accomplish has to be explained.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure.  A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation.  The relation to determine Gibbs free energy of a reaction is shown below.

  $\Delta \text{G}°={\displaystyle \sum \text{n}\Delta {\text{G}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{G}}_f^{\circ }}\left(\text{reactants}\right)$

Answer to Problem 4.48E

The given path formation of acetic acid is easiest to accomplish.

Explanation of Solution

The balanced chemical equation for the formation of acetic acid from methanol and carbon monoxide is shown below.

    $C{H}_{3}OH\left(l\right)+CO\left(g\right)\to C{H}_{3}COOH\left(l\right)$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C{H}_{3}OH,l\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(CO,g\right)\end{array}\right\}$        (1)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$ is the standard Gibbs free energy of formation of $C{H}_{3}COOH\left(l\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}OH,l\right)$ is the standard Gibbs free energy of formation of $C{H}_{3}OH\left(l\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(CO,g\right)$ is the standard Gibbs free energy of formation of $CO\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$ is $-\text{389}\text{.9}k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}OH,l\right)$ is $-\text{166}\text{.27}k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(CO,g\right)$ is $-\text{137}\text{.17}k\text{J}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$, $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}OH,l\right)$ and $\Delta {\text{G}}_f^{\circ }\left(CO,g\right)$ in equation (1).

    $\begin{array}{c}\Delta \text{G}°=\left\{\left(1mol\right)\times \left(-\text{389}\text{.9}k\text{J}\cdot {\text{mol}}^{-1}\right)\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \left(-\text{166}\text{.27}k\text{J}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(-\text{137}\text{.17}k\text{J}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{-\text{389}\text{.9}k\text{J}\right\}-\left\{-\text{166}\text{.27}k\text{J}-\text{137}\text{.17}k\text{J}\right\}\\ =-86.46\text{kJ}\end{array}$

Therefore, the standard change in the Gibbs free energy for the formation of acetic acid from methanol and carbon monoxide is $-86.46\text{kJ}$.

The negative value for the Gibbs free energy indicates that the reaction is spontaneous.

Thus, this path for the formation of acetic acid is easiest to accomplish.

(b)

Interpretation Introduction

Interpretation:

The formation of acetic acid by the oxidation of ethanol is easiest to accomplish has to be explained.

Concept Introduction:

Same as part (a).

Answer to Problem 4.48E

The given path formation of acetic acid is easiest to accomplish.

Explanation of Solution

The balanced chemical equation for the formation of acetic acid by the oxidation of ethanol is shown below.

    $C_2{H}_{5}OH\left(l\right)+O_2\left(g\right)\to C{H}_{3}COOH\left(l\right)+H_{2}O\left(l\right)$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)\end{array}\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C_2{H}_{5}OH,l\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(O_2,g\right)\end{array}\right\}$        (2)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$ is the standard Gibbs free energy of formation of $C{H}_{3}COOH\left(l\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard Gibbs free energy of formation of $H_{2}O\left(l\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(C_2{H}_{5}OH,l\right)$ is the standard Gibbs free energy of formation of $C_2{H}_{5}OH\left(l\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$ is the standard Gibbs free energy of formation of $O_2\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$ is $-\text{389}\text{.9}k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$ is $-\text{237}\text{.13}k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C_2{H}_{5}OH,l\right)$ is $-\text{174}\text{.78}k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$ is $0.0k\text{J}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$, $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$, $\Delta {\text{G}}_f^{\circ }\left(C_2{H}_{5}OH,l\right)$ and $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$ in equation (2).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(-\text{389}\text{.9}k\text{J}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(-237.13k\text{J}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \left(-\text{174}\text{.78}k\text{J}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(0.0k\text{J}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{-\text{389}\text{.9}k\text{J}-237.13k\text{J}\right\}-\left\{-\text{174}\text{.78}k\text{J}\right\}\\ =-452.25\text{kJ}\end{array}$

Therefore, the standard change in the Gibbs free energy for the formation of acetic acid by the oxidation of ethanol is $-452.25\text{kJ}$.

The negative value for the Gibbs free energy indicates that the reaction is spontaneous.

Thus, this path for the formation of acetic acid is easiest to accomplish.

(c)

Interpretation Introduction

Interpretation:

The formation of acetic acid by the reaction of carbon dioxide and methane is easiest to accomplish has to be explained.

Concept Introduction:

Same as part (a).

Answer to Problem 4.48E

The given path formation of acetic acid is not easiest to accomplish.

Explanation of Solution

The balanced chemical equation for the formation of acetic acid by the reaction of carbon dioxide and methane is shown below.

    $C{H}_4\left(g\right)+C{O}_2\left(g\right)\to C{H}_{3}COOH\left(l\right)$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C{H}_4,g\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C{O}_2,g\right)\end{array}\right\}$        (3)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$ is the standard Gibbs free energy of formation of $C{H}_{3}COOH\left(l\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(C{H}_4,g\right)$ is the standard Gibbs free energy of formation of $C{H}_4\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(C{O}_2,g\right)$ is the standard Gibbs free energy of formation of $C{O}_2\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$ is $-\text{389}\text{.9}k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_4,g\right)$ is $-\text{50}\text{.72}k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C{O}_2,g\right)$ is $-\text{394}\text{.36}k\text{J}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_{3}COOH,l\right)$, $\Delta {\text{G}}_f^{\circ }\left(C{H}_4,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(C{O}_2,g\right)$ in equation (3).

    $\begin{array}{c}\Delta \text{G}°=\left\{\left(1mol\right)\times \left(-\text{389}\text{.9}k\text{J}\cdot {\text{mol}}^{-1}\right)\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \left(-\text{50}\text{.72}k\text{J}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(-\text{394}\text{.36}k\text{J}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{-\text{389}\text{.9}k\text{J}\right\}-\left\{-\text{50}\text{.72}k\text{J}-\text{394}\text{.36}k\text{J}\right\}\\ =+55.18\text{kJ}\end{array}$

Therefore, the standard change in the Gibbs free energy for the formation of acetic acid by the reaction of carbon dioxide and methane is $+55.18\text{kJ}$.

The positive value for the Gibbs free energy indicates that the reaction is non-spontaneous.

Thus, this path for the formation of acetic acid is not easiest to accomplish.