Chapter 17, Problem 69E

Using data from Appendix 4, calculate ∆H°, ∆S° and ∆G° for the following reactions that produce acetic acid:

Which reaction would you choose as a commercial method for producing acetic acid (CH₃CO₂H) at standard conditions? What temperature conditions would you choose for the reaction? Assume ∆H° and ∆S° do not depend on temperature.

Interpretation Introduction

Interpretation: The values of $\Delta H^{\text{ο}},\Delta S^{\text{ο}}$ and $\Delta G^{\text{ο}}$ for the given reactions are to be calculated. The reaction that involves a commercial method for producing acetic acid at standard condition is to be stated. The temperature conditions are to be stated. The independency of $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ on the temperature is to be assumed.

Concept introduction: The term $\Delta G^{\text{ο}}$ is a thermodynamic function. The superscript on this function represents its standard form. The term $\Delta$ represents the change. This function is known as the standard Gibb’s free energy change. It correlates the enthalpy and entropy of the system in a mathematical formula.

Answer to Problem 69E

Answer

• The values of $\Delta H^{\text{ο}},\Delta S^{\text{ο}}$ and $\Delta G^{\text{ο}}$ for the given reactions are,

$\begin{array}{ccc}& \text{Reaction}1& \text{Reaction}2\\ \Delta H^{\text{ο}}& \underset{\_}{-16kJ}& \underset{\_}{-173kJ}\\ \Delta S^{\text{ο}}& \underset{\_}{-240J/K}& \underset{\_}{-278J/K}\\ \Delta G^{\text{ο}}& \underset{\_}{+56kJ}& \underset{\_}{-89kJ}\end{array}$

• The second reaction is suitable as a commercial method for producing acetic acid at standard conditions.
• The temperature for the suitable reaction is $\underset{\_}{622.30K}$.

Explanation of Solution

Explanation

Given

The reactions are given as,

$\begin{array}{c}{\text{2CH}}_4\left(g\right){\text{+CO}}_2\left(g\right)\to {\text{CH}}_3\text{COOH}\left(l\right)\\ {\text{CH}}_3\text{OH}\left(g\right)\text{+CO}\left(g\right)\to {\text{CH}}_3\text{COOH}\left(l\right)\end{array}$

The given values of $\Delta H_{\text{f}}^{\text{ο}},\Delta S_{\text{f}}^{\text{ο}}$ and $\Delta G_{\text{f}}^{\text{ο}}$ are as follows,

$\begin{array}{cccc}\text{Substance}& \Delta H_{\text{f}}^{\text{ο}}\left(\text{kJ/mol}\right)& \Delta S_{\text{f}}^{\text{ο}}\left(\text{J/K}\cdot \text{mol}\right)& \Delta G_{\text{f}}^{\text{ο}}\left(\text{kJ/mol}\right)\\ {\text{CH}}_{\text{3}}\text{COOH}\left(l\right)& -484& 160& -389\\ {\text{CH}}_4\left(g\right)& -75& 186& -51\\ {\text{CH}}_{\text{3}}\text{OH}\left(g\right)& -201& 240& -163\\ {\text{CO}}_2\left(g\right)& -393.5& 214& -394\\ \text{CO}\left(g\right)& -110.5& 198& -137\end{array}$

The calculation of standard enthalpy change for both the reactions is given below.

The value of standard enthalpy change $\Delta H^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta H^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta H_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard free energy of formation for the reactants.
• $\Delta H_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard free energy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the first reaction, the representation in the above form is written as,

$\begin{array}{c}\Delta H^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta H^{\text{ο}}=\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_4\left(g\right)}-\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CO}}_2\left(g\right)}\end{array}$

Substitute the values of standard enthalpy of formations in the above equation.

$\begin{array}{l}\Delta H^{\text{ο}}=\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_4\left(g\right)}-\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CO}}_2\left(g\right)}\\ \Delta H^{\text{ο}}=1\text{mol}\left(-4\text{84}\text{kJ/mol}\right)-1\text{mol}\left(-75\text{kJ/mol}\right)-1\text{mol}\times \left(-393\text{kJ/mol}\right)\\ \Delta H^{\text{ο}}=\left(-484\text{kJ}\right)+\left(75\text{kJ}\right)+\left(393\text{kJ}\right)\\ \Delta H^{\text{ο}}=\underset{\_}{-\text{1}6kJ}\end{array}$

For the second reaction, the representation in the above form is written as,

$\begin{array}{c}\Delta H^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta H^{\text{ο}}=\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_3\text{OH}\left(g\right)}-\Delta H_{\text{f}}^{\text{ο}}{}_{\text{CO}\left(g\right)}\end{array}$

Substitute the values of standard enthalpy of formations in the above equation.

$\begin{array}{l}\Delta H^{\text{ο}}=\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_3\text{OH}\left(g\right)}-\Delta H_{\text{f}}^{\text{ο}}{}_{\text{CO}\left(g\right)}\\ \Delta H^{\text{ο}}=1\text{mol}\left(-4\text{84}\text{kJ/mol}\right)-1\text{mol}\left(-201\text{kJ/mol}\right)-1\text{mol}\times \left(-110.5\text{kJ/mol}\right)\\ \Delta H^{\text{ο}}=\left(-484\text{kJ}\right)+\left(201\text{kJ}\right)+\left(110.5\text{kJ}\right)\\ \Delta H^{\text{ο}}=\underset{\_}{-173kJ}\end{array}$

The calculation of standard entropy change for both the reactions is given below.

The value of standard entropy change $\Delta S^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta S_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard free energy of formation for the reactants.
• $\Delta S_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard free energy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the first reaction, the representation in the above form is written as,

$\begin{array}{c}\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta S^{\text{ο}}=\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_4\left(g\right)}-\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CO}}_2\left(g\right)}\end{array}$

Substitute the values of standard entropy of formations in the above equation.

$\begin{array}{l}\Delta S^{\text{ο}}=\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_4\left(g\right)}-\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CO}}_2\left(g\right)}\\ \Delta S^{\text{ο}}=1\text{mol}\left(160\text{J/K}\cdot \text{mol}\right)-1\text{mol}\left(186\text{J/K}\cdot \text{mol}\right)-1\text{mol}\times \left(214\text{J/K}\cdot \text{mol}\right)\\ \Delta S^{\text{ο}}=\left(160\text{J/K}\cdot \text{mol}\right)-\left(186\text{J/K}\cdot \text{mol}\right)-\left(214\text{J/mol}\cdot \text{K}\right)\\ \Delta S^{\text{ο}}=\underset{\_}{-240J/K}\end{array}$

For the second reaction, the representation in the above form is written as,

$\begin{array}{c}\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta S^{\text{ο}}=\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_3\text{OH}\left(g\right)}-\Delta S_{\text{f}}^{\text{ο}}{}_{\text{CO}\left(g\right)}\end{array}$

Substitute the values of standard entropy of formations in the above equation.

$\begin{array}{l}\Delta S^{\text{ο}}=\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_3\text{OH}\left(g\right)}-\Delta S_{\text{f}}^{\text{ο}}{}_{\text{CO}\left(g\right)}\\ \Delta S^{\text{ο}}=1\text{mol}\left(160\text{J/K}\cdot \text{mol}\right)-1\text{mol}\left(240\text{J/K}\cdot \text{mol}\right)-1\text{mol}\times \left(198\text{J/K}\cdot \text{mol}\right)\\ \Delta S^{\text{ο}}=\left(160\text{J/K}\right)-\left(240\text{J/K}\right)-\left(198\text{J/K}\right)\\ \Delta S^{\text{ο}}=\underset{\_}{-278J/K}\end{array}$

The calculation of standard entropy change for both the reactions is given below.

The value of standard Gibb’s free energy $\Delta G^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta G_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard free energy of formation for the reactants.
• $\Delta G_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard free energy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the first reaction, the representation in the above form is written as,

$\begin{array}{c}\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta G^{\text{ο}}=\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_4\left(g\right)}-\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CO}}_2\left(g\right)}\end{array}$

Substitute the values of standard free energy in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_4\left(g\right)}-\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CO}}_2\left(g\right)}\\ \Delta G^{\text{ο}}=\left(-\text{389}\text{kJ/mol}\right)-1\text{mol}\left(-51\text{kJ/mol}\right)-1\text{mol}\times \left(-394\text{kJ/mol}\right)\\ \Delta G^{\text{ο}}=\left(-389\text{kJ}\right)+\left(51\text{kJ}\right)+\left(394\text{kJ/mol}\right)\\ \Delta G^{\text{ο}}=\underset{\_}{+56kJ}\end{array}$

For the second reaction, the representation in the above form is written as,

$\begin{array}{c}\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta G^{\text{ο}}=\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_3\text{OH}\left(g\right)}-\Delta G_{\text{f}}^{\text{ο}}{}_{\text{CO}\left(g\right)}\end{array}$

Substitute the values of standard free energy in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)}-\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_3\text{OH}\left(g\right)}-\Delta G_{\text{f}}^{\text{ο}}{}_{\text{CO}\left(g\right)}\\ \Delta G^{\text{ο}}=\left(-\text{389}\text{kJ/mol}\right)-1\text{mol}\left(-163\text{kJ/mol}\right)-1\text{mol}\times \left(-317\text{kJ/mol}\right)\\ \Delta G^{\text{ο}}=\left(-389\text{kJ}\right)+\left(163\text{kJ}\right)+\left(137\text{kJ}\right)\\ \Delta G^{\text{ο}}=\underset{\_}{-89kJ}\end{array}$

According to the law of thermodynamics, the spontaneity of the reaction is determined by the calculated value of $\Delta G^{\text{ο}}$. The value of $\Delta G^{\text{ο}}$ correlates the spontaneity conditions as,

$\begin{array}{cc}\Delta G^{\text{ο}}& \text{Criteria}\text{of}\text{Sponteneity}\\ \text{Positive}& \text{Non-spontaneous}\\ \text{Negative}& \text{Spontaneous}\end{array}$

In the above calculated values of $\Delta G^{\text{ο}}$, the value which is negative is for the second reaction. Therefore, it clearly indicates that the second reaction is spontaneous reaction. Hence, the second reaction is suitable as a commercial method for producing acetic acid at standard conditions.

The temperature for the suitable reaction that is second reaction is calculated below.

The Gibb’s free energy in terms of enthalpy change is defined as,

$\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}$

Where,

• $\Delta S^{\text{ο}}$ is the standard entropy change.
• $\Delta H^{\text{ο}}$ is the standard enthalpy change.
• $\Delta G^{\text{ο}}$ is the standard Gibb’s free energy change.
• $T$ is the temperature.

The above reaction clearly indicates that the negative value of Gibb’s free energy is only possible at a very low temperature. The value of enthalpy change is negative for both the reactions.

But in both the reactions, the reactant is more disordered than the products therefore the entropy decreases and more negative value is observed.

To balance both, the enthalpy and entropy conditions equilibrium is considered and at the equilibrium, the Gibb’s free energy is given as,

$\Delta G^{\text{ο}}=0$

Substitute the value of $\Delta G^{\text{ο}}$ in the above equation to calculate the standard enthalpy.

$\begin{array}{l}\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}\\ 0=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}\\ T=\frac{\Delta H^{\text{ο}}}{\Delta S^{\text{ο}}}\end{array}$

Now substitute the standard values of enthalpy and entropy change.

$\begin{array}{c}T=\frac{\Delta H^{\text{ο}}}{\Delta S^{\text{ο}}}\\ T=\frac{\left(-173\text{kJ}\right)}{\left(-278\text{J/K}\right)}\\ T=\frac{-173000\text{J}}{-278\text{J}\cdot {\text{K}}^{-1}}\\ T=\underset{\_}{622.30K}\end{array}$

Conclusion

Conclusion

• The values of $\Delta H^{\text{ο}},\Delta S^{\text{ο}}$ and $\Delta G^{\text{ο}}$ for the given reactions have been stated as,

$\begin{array}{ccc}& \text{Reaction}1& \text{Reaction}2\\ \Delta H^{\text{ο}}& \underset{\_}{-16kJ}& \underset{\_}{-173kJ}\\ \Delta S^{\text{ο}}& \underset{\_}{-240J/K}& \underset{\_}{-278J/K}\\ \Delta G^{\text{ο}}& \underset{\_}{+56kJ}& \underset{\_}{-89kJ}\end{array}$

• The second reaction is suitable as a commercial method for producing acetic acid at standard conditions.
• The temperature for the suitable reaction is $\underset{\_}{622.30K}$.