Chapter 16, Problem 120QRT

(a)

Interpretation Introduction

Interpretation:

Three balanced equations for the reactions of coal with steam to form ethane gas, propane gas and liquid methanol with carbon dioxide as a byproduct have to be stated.

Concept Introduction:

Reaction of carbon with steam results in the formation of various gases along with carbon dioxide as by product.  The materials whose reaction releases large quantities of Gibbs free energy are useful to society because this Gibbs free energy can be utilized to do some other fruitful works that may not occur by itself.  The burning of coal releases a large amount of Gibbs free energy.  This energy is utilized in doing many useful works.

Answer to Problem 120QRT

Three balanced equations for the reactions of coal with steam to form ethane gas, propane gas and liquid methanol with carbon dioxide as a byproduct are shown below.

    $\begin{array}{l}\text{7C}\left(\text{carbon}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{2C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)+{\text{3CO}}_{\text{2}}\left(\text{g}\right)\\ \text{5C}\left(\text{carbon}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{C}}_3{\text{H}}_8\left(\text{g}\right)+2{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ \text{3C}\left(\text{carbon}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{2CH}}_{\text{3}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

Explanation of Solution

Reaction of carbon with steam results in the formation of various gases along with carbon dioxide as by product.  Three balanced equations for the reactions of coal with steam to form ethane gas, propane gas and liquid methanol with carbon dioxide as a byproduct are shown below.

    $\begin{array}{l}\text{7C}\left(\text{carbon}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{2C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)+{\text{3CO}}_{\text{2}}\left(\text{g}\right)\\ \text{5C}\left(\text{carbon}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{C}}_3{\text{H}}_8\left(\text{g}\right)+2{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ \text{3C}\left(\text{carbon}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{2CH}}_{\text{3}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}H°,{\Delta }_{\text{r}}S°,{\Delta }_{\text{r}}G°$ for each of the reaction has to be calculated.  It has to be commented that whether any of them would be a feasible way to produce the mentioned products.

Concept Introduction:

Refer to part (a).

Answer to Problem 120QRT

For the first reaction, the value of ${\Delta }_{\text{r}}H°$ is $\underset{\_}{101.02kJ/mol}$, the value of ${\Delta }_{\text{r}}G°$ is $\underset{\_}{122.71kJ/mol}$ and the value of ${\Delta }_{\text{r}}S°$ is $\underset{\_}{-72.71J/mol\cdot K}$.

For the second reaction, the value of ${\Delta }_{\text{r}}H°$ is $\underset{\_}{76.5kJ/mol}$, the value of ${\Delta }_{\text{r}}G°$ is $\underset{\_}{102.08kJ/mol}$ and the value of ${\Delta }_{\text{r}}S°$ is $\underset{\_}{-86.6J/mol\cdot K}$.

For the third reaction, the value of ${\Delta }_{\text{r}}H°$ is $\underset{\_}{96.44kJ/mol}$, the value of ${\Delta }_{\text{r}}G°$ is $\underset{\_}{187.39kJ/mol}$ and the value of ${\Delta }_{\text{r}}S°$ is $\underset{\_}{-305.2J/mol\cdot K}$.

All these reactions has a positive value of ${\Delta }_{\text{r}}G°$.  Therefore, none of them is feasible.  The value of ${\Delta }_{\text{r}}H°$ is positive and the value of ${\Delta }_{\text{r}}S°$ is negative.  It means that there is no temperature at which the formation of products is favored.

Explanation of Solution

The given reaction is shown below.

  $\text{7C}\left(\text{g}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{2C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)+{\text{3CO}}_{\text{2}}\left(\text{g}\right)$

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

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

Where,

• ${\Delta }_{\text{f}}H°$ is the change in standard enthalpy 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 ${\Delta }_{\text{f}}H°$ for ${\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-84.68\text{kJ}/\text{mol}$, $-393.509\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$ and $-241.818\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(2\times -84.68\text{kJ}/\text{mol}\right)+\left(\text{3}\times -393.509\text{kJ}/\text{mol}\right)\right)-\\ \left(\text{7}\times 0\text{kJ}/\text{mol}+\left(\text{6}\times -241.818\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =\underset{\_}{101.02kJ/mol}\end{array}$

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

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (2)

Where,

• ${\Delta }_{\text{f}}S°$ is the change in standard entropy 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 ${\Delta }_{\text{f}}S°$ for ${\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $229.6\text{J}/\text{mol}\cdot \text{K}$, $213.74\text{J}/\text{mol}\cdot \text{K}$, $5.74\text{J}/\text{mol}\cdot \text{K}$ and $188.825\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (2) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(2\times 229.6\text{J}/\text{mol}\cdot \text{K}\right)+\left(\text{3}\times 213.74\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(\text{7}\times 5.74\text{J}/\text{mol}\cdot \text{K}+\left(\text{6}\times 188.825\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =\underset{\_}{-72.71J/mol\cdot K}\end{array}$

The value of temperature is given as $25.0°\text{C}$.

Convert the temperature of $25.0°\text{C}$ into Kelvin as follows.

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(25+273.15\right)\text{K}\\ =\text{298}\text{.15}\text{K}\end{array}$

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

    ${\Delta }_{\text{r}}G°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}G°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}G°\left(\text{Reactants}\right)$        (3)

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 ${\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-32.82\text{kJ}/\text{mol}$, $-394.359\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$ and $-228.572\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(2\times -32.82\text{kJ}/\text{mol}\right)+\left(\text{3}\times -394.359\text{kJ}/\text{mol}\right)\right)-\\ \left(\text{7}\times 0\text{kJ}/\text{mol}+\left(\text{6}\times -228.572\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =\underset{\_}{122.71kJ/mol}\end{array}$

The given reaction is shown below.

  $\text{5C}\left(\text{carbon}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{C}}_3{\text{H}}_8\left(\text{g}\right)+2{\text{CO}}_{\text{2}}\left(\text{g}\right)$

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

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

Where,

• ${\Delta }_{\text{f}}H°$ is the change in standard enthalpy 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 ${\Delta }_{\text{f}}H°$ for ${\text{C}}_3{\text{H}}_8\left(\text{g}\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-103.8\text{kJ}/\text{mol}$, $-393.509\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$ and $-241.818\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{C}}_3{\text{H}}_8\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(1\times -103.8\text{kJ}/\text{mol}\right)+\left(2\times -393.509\text{kJ}/\text{mol}\right)\right)-\\ \left(5\times 0\text{kJ}/\text{mol}+\left(\text{4}\times -241.818\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =\underset{\_}{76.5kJ/mol}\end{array}$

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

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (2)

Where,

• ${\Delta }_{\text{f}}S°$ is the change in standard entropy 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 ${\Delta }_{\text{f}}S°$ for ${\text{C}}_3{\text{H}}_8\left(\text{g}\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $269.9\text{J}/\text{mol}\cdot \text{K}$, $213.74\text{J}/\text{mol}\cdot \text{K}$, $5.74\text{J}/\text{mol}\cdot \text{K}$ and $188.825\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (2) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{C}}_3{\text{H}}_8\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(1\times 269.9\text{J}/\text{mol}\cdot \text{K}\right)+\left(2\times 213.74\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(5\times 5.74\text{J}/\text{mol}\cdot \text{K}+\left(4\times 188.825\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =\underset{\_}{-86.6J/mol\cdot K}\end{array}$

The value of temperature is given as $25.0°\text{C}$.

Convert the temperature of $25.0°\text{C}$ into Kelvin as follows.

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(25+273.15\right)\text{K}\\ =\text{298}\text{.15}\text{K}\end{array}$

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

    ${\Delta }_{\text{r}}G°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}G°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}G°\left(\text{Reactants}\right)$        (3)

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 ${\text{C}}_3{\text{H}}_8\left(\text{g}\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-23.49\text{kJ}/\text{mol}$, $-394.359\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$ and $-228.572\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{C}}_3{\text{H}}_8\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(1\times -23.49\text{kJ}/\text{mol}\right)+\left(2\times -394.359\text{kJ}/\text{mol}\right)\right)-\\ \left(\text{5}\times 0\text{kJ}/\text{mol}+\left(\text{4}\times -228.572\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =\underset{\_}{102.08kJ/mol}\end{array}$

The given reaction is shown below.

  $\text{3C}\left(\text{carbon}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{2CH}}_{\text{3}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$

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

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

Where,

• ${\Delta }_{\text{f}}H°$ is the change in standard enthalpy 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 ${\Delta }_{\text{f}}H°$ for ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-238.66\text{kJ}/\text{mol}$, $-393.509\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$ and $-241.818\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(2\times -238.66\text{kJ}/\text{mol}\right)+\left(1\times -393.509\text{kJ}/\text{mol}\right)\right)-\\ \left(3\times 0\text{kJ}/\text{mol}+\left(\text{4}\times -241.818\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =\underset{\_}{96.44kJ/mol}\end{array}$

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

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (2)

Where,

• ${\Delta }_{\text{f}}S°$ is the change in standard entropy 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 ${\Delta }_{\text{f}}S°$ for ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $126.8\text{J}/\text{mol}\cdot \text{K}$, $213.74\text{J}/\text{mol}\cdot \text{K}$, $5.74\text{J}/\text{mol}\cdot \text{K}$ and $188.825\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (2) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(2\times 126.8\text{J}/\text{mol}\cdot \text{K}\right)+\left(1\times 213.74\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(3\times 5.74\text{J}/\text{mol}\cdot \text{K}+\left(4\times 188.825\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =\underset{\_}{-305.2J/mol\cdot K}\end{array}$

The value of temperature is given as $25.0°\text{C}$.

Convert the temperature of $25.0°\text{C}$ into Kelvin as follows.

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(25+273.15\right)\text{K}\\ =\text{298}\text{.15}\text{K}\end{array}$

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

    ${\Delta }_{\text{r}}G°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}G°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}G°\left(\text{Reactants}\right)$        (3)

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 ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, $\text{C}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-166.27\text{kJ}/\text{mol}$, $-394.359\text{kJ}/\text{mol}$, $0\text{kJ}/\text{mol}$ and $-228.572\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_6\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{\text{C}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left(\text{C}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\left(2\times -166.27\text{kJ}/\text{mol}\right)+\left(1\times -394.359\text{kJ}/\text{mol}\right)\right)-\\ \left(\text{3}\times 0\text{kJ}/\text{mol}+\left(\text{4}\times -228.572\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =\underset{\_}{187.39kJ/mol}\end{array}$

All these reactions has a positive value of ${\Delta }_{\text{r}}G°$.  Therefore, none of them is feasible.  The value of ${\Delta }_{\text{r}}H°$ is positive and the value of ${\Delta }_{\text{r}}S°$ is negative.  It means that there is no temperature at which the formation of products is favored.