Chapter 17, Problem 17.78QP

(a)

Interpretation Introduction

Interpretation: The values of standard Gibbs free energy $\left(\Delta G_{\text{rxn}}^{\text{ο}}\right)$ for the given reactions are to be calculated. The chemical equation of the overall reaction in which methane and steam combine to produce hydrogen gas and carbon dioxide is to be stated. Whether the coupled reaction is spontaneous under standard condition is to be explained.

Concept introduction: Gibbs free energy is a thermodynamic quantity that is used to calculate the maximum work of reversible reaction performed by a system. It is equal to the difference between the enthalpy and the product of entropy at absolute temperature.

To determine: The value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the given reaction.

Answer to Problem 17.78QP

Solution

The value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the reaction is $\underset{\_}{142.2kJ}$.

Explanation of Solution

Explanation

Given

The reaction of steam-methane to produce carbon monoxide and hydrogen gas is,

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

The reaction consumes one mole of each methane and water vapor to produce one mole of carbon monoxide and three moles of hydrogen.

The $\Delta G_{\text{f}}^{\text{ο}}$ value of ${\text{CH}}_{\text{4}}\left(g\right)$ is $-50.8\text{kJ/mol}$.

The $\Delta G_{\text{f}}^{\text{ο}}$ value of ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is $-228.6\text{kJ/mol}$.

The $\Delta G_{\text{f}}^{\text{ο}}$ value of $\text{CO}\left(g\right)$ is $-137.2\text{kJ/mol}$.

The $\Delta G_{\text{f}}^{\text{ο}}$ value of ${\text{H}}_2\left(g\right)$ is $0.0\text{kJ/mol}$.

The change in standard free energy of the reaction, $\Delta G_{\text{rxn}}^{\text{ο}}$, using the standard free energy of formation is calculated from the given formula,

$\Delta G_{\text{rxn}}^{\text{ο}}={\displaystyle \sum n_{\text{products}}\Delta G_{\text{f}\text{.products}}^{\text{ο}}-}{\displaystyle \sum n_{\text{reactants}}\Delta G_{\text{f}\text{.reactants}}^{\text{ο}}}$

Where,

• $n$ is the number of moles of for products and reactants.
• $\Delta G_{\text{f}}^{\text{ο}}$ is the standard free energy of formation of reactants and products.

Substitute the values of $n_{\text{products}}$, $\Delta G_{\text{f}\text{.products}}^{\text{ο}}$, $n_{\text{reactants}}$ and $\Delta G_{\text{f}\text{.reactants}}^{\text{ο}}$ in the above formula to calculate the $\Delta G_{\text{rxn}}^{\text{ο}}$ for the reaction.

$\begin{array}{c}\Delta G_{\text{rxn}}^{\text{ο}}=\left[\left(n_{\text{CO}}\times \Delta G_{\text{CO}}^{\text{ο}}+n_{{\text{H}}_{\text{2}}}\times \Delta G_{{\text{H}}_{\text{2}}}^{\text{ο}}\right)\right]-\left[\left(n_{{\text{CH}}_4}\times \Delta G_{{\text{CH}}_4}^{\text{ο}}+n_{{\text{H}}_{\text{2}}\text{O}}\times \Delta G_{{\text{H}}_{\text{2}}\text{O}}^{\text{ο}}\right)\right]\\ =\left[\left(1\times \left(-137.2\text{kJ/mol}\right)+3\times 0.0\text{kJ/mol}\right)\right]-\\ \left[\left(1\times \left(-50.8\text{kJ/mol}\right)+1\times \left(-228.6\text{kJ/mol}\right)\right)\right]\\ =\left[-137.2\text{kJ}+0.0\text{kJ}\right]-\left[-50.8\text{kJ}-\text{228}\text{.6}\text{kJ}\right]\\ =-137.2\text{kJ}-\left(-279.4\text{kJ}\right)\end{array}$

Simplify the above equation,

$\begin{array}{c}\Delta G_{\text{rxn}}^{\text{ο}}=-137.2\text{kJ}+279.4\text{kJ}\\ =\underset{\_}{142.2kJ}\end{array}$

Hence, the value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the reaction is $\underset{\_}{142.2kJ}$.

(b)

Interpretation Introduction

To determine: The value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the given reaction.

Answer to Problem 17.78QP

Solution

The value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the reaction is $\underset{\_}{-28.6kJ}$.

Explanation of Solution

Explanation

Given

The reaction of carbon monoxide and water to produce carbon dioxide and hydrogen gas is,

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

The reaction consumes one mole of each carbon monoxide and water vapor to produce one mole of each carbon dioxide and hydrogen.

The $\Delta G_{\text{f}}^{\text{ο}}$ value of $\text{CO}\left(g\right)$ is $-137.2\text{kJ/mol}$.

The $\Delta G_{\text{f}}^{\text{ο}}$ value of ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is $-228.6\text{kJ/mol}$.

The $\Delta G_{\text{f}}^{\text{ο}}$ value of ${\text{CO}}_2\left(g\right)$ is $-394.4\text{kJ/mol}$.

The $\Delta G_{\text{f}}^{\text{ο}}$ value of ${\text{H}}_2\left(g\right)$ is $0.0\text{kJ/mol}$.

The change in standard free energy of the reaction, $\Delta G_{\text{rxn}}^{\text{ο}}$, using the standard free energy of formation is calculated from the given formula,

$\Delta G_{\text{rxn}}^{\text{ο}}={\displaystyle \sum n_{\text{products}}\Delta G_{\text{f}\text{.products}}^{\text{ο}}-}{\displaystyle \sum n_{\text{reactants}}\Delta G_{\text{f}\text{.reactants}}^{\text{ο}}}$

Where,

• $n$ is the number of moles of for products and reactants.
• $\Delta G_{\text{f}}^{\text{ο}}$ is the standard free energy of formation of reactants and products.

Substitute the values of $n_{\text{products}}$, $\Delta G_{\text{f}\text{.products}}^{\text{ο}}$, $n_{\text{reactants}}$ and $\Delta G_{\text{f}\text{.reactants}}^{\text{ο}}$ in the above formula to calculate the $\Delta G_{\text{rxn}}^{\text{ο}}$ for the reaction.

$\begin{array}{c}\Delta G_{\text{rxn}}^{\text{ο}}=\left[\left(n_{{\text{CO}}_2}\times \Delta G_{{\text{CO}}_2}^{\text{ο}}+n_{{\text{H}}_{\text{2}}}\times \Delta G_{{\text{H}}_{\text{2}}}^{\text{ο}}\right)\right]-\left[\left(n_{\text{CO}}\times \Delta G_{\text{CO}}^{\text{ο}}+n_{{\text{H}}_{\text{2}}\text{O}}\times \Delta G_{{\text{H}}_{\text{2}}\text{O}}^{\text{ο}}\right)\right]\\ =\left[\left(1\times \left(-394.4\text{kJ/mol}\right)+1\times 0.0\text{kJ/mol}\right)\right]-\\ \left[\left(1\times \left(-137.2\text{kJ/mol}\right)+1\times \left(-228.6\text{kJ/mol}\right)\right)\right]\\ =\left[-394.4\text{kJ}+0.0\text{kJ}\right]-\left[-137.2\text{kJ}-\text{228}\text{.6}\text{kJ}\right]\\ =-394.4\text{kJ}-\left(-365.8\text{kJ}\right)\end{array}$

Simplify the above equation,

$\begin{array}{c}\Delta G_{\text{rxn}}^{\text{ο}}=-394.4\text{kJ}+365.8\text{kJ}\\ =\underset{\_}{-28.6kJ}\end{array}$

Hence, the value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the reaction is $\underset{\_}{-28.6kJ}$.

(c)

Interpretation Introduction

To determine: The chemical equation of the overall reaction in which methane and steam combine to produce hydrogen gas and carbon dioxide.

Answer to Problem 17.78QP

Solution

The overall balanced chemical equation is,

${\text{CH}}_4\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)\stackrel{}{\to }{\text{CO}}_2\left(g\right)+4{\text{H}}_{\text{2}}\left(g\right)$

Explanation of Solution

Explanation

Given

The reaction of steam-methane reforming is,

${\text{CH}}_{\text{4}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)\stackrel{}{\to }\text{CO}\left(g\right)+{\text{3H}}_{\text{2}}\left(g\right)$ (1)

The reaction of water-gas shift is,

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

The two reactions are combined to give overall reaction as,

${\text{CH}}_4\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)\stackrel{}{\to }{\text{CO}}_2\left(g\right)+4{\text{H}}_{\text{2}}\left(g\right)$

Hence, the overall reaction is,

${\text{CH}}_4\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)\stackrel{}{\to }{\text{CO}}_2\left(g\right)+4{\text{H}}_{\text{2}}\left(g\right)$

(d)

Interpretation Introduction

To determine: The value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the given reaction and if the given reaction is spontaneous under standard conditions.

Answer to Problem 17.78QP

Solution

The value of $\Delta G_{\text{overall}}^{\text{ο}}$ is $\underset{\_}{113.6kJ}$. The value of $\Delta G_{\text{overall}}^{\text{ο}}>0$, the reaction is non-spontaneous under standard condition.

Explanation of Solution

Explanation

The $\Delta G_{\text{rxn}}^{\text{ο}}$ values of the coupled reactions are additive. Hence, their values are added to give $\Delta G_{\text{rxn}}^{\text{ο}}$ of the overall reaction. The change in free energy of the overall reaction is considered to be $\Delta G_{\text{overall}}^{\text{ο}}$.

The change in free energy of the steam-methane reforming reaction is considered to $\Delta G_1^{\text{ο}}$ and the change in free energy of the water-gas shift reaction is considered to be $\Delta G_2^{\text{ο}}$. The value of $\Delta G_1^{\text{ο}}$ is $142.2\text{kJ}$ and the value of $\Delta G_2^{\text{ο}}$ is $-28.6\text{kJ}$ as calculated in part (a) and (b) respectively. The overall change in free energy is calculated as,

$\Delta G_{\text{overall}}^{\text{ο}}=\Delta G_{\text{1}}^{\text{ο}}+\Delta G_2^{\text{ο}}$

Substitute the values of $\Delta G_1^{\text{ο}}$ and $\Delta G_2^{\text{ο}}$ in the above expression to calculate the value of $\Delta G_{\text{overall}}^{\text{ο}}$.

$\begin{array}{c}\Delta G_{\text{overall}}^{\text{ο}}=142.2\text{kJ}+\left(-28.6\text{kJ}\right)\\ =142.2\text{kJ}-28.6\text{kJ}\\ =\underset{\_}{113.6kJ}\end{array}$

Hence, the value of $\Delta G_{\text{overall}}^{\text{ο}}$ is $\underset{\_}{113.6kJ}$. The value of $\Delta G_{\text{overall}}^{\text{ο}}>0$, the reaction is non-spontaneous under standard condition.

Conclusion

1. a. The value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the reaction is $\underset{\_}{142.2kJ}$.
2. b. The value of $\Delta G_{\text{rxn}}^{\text{ο}}$ for the reaction is $\underset{\_}{-28.6kJ}$.
3. c. The overall balanced chemical equation is,

${\text{CH}}_4\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)\stackrel{}{\to }{\text{CO}}_2\left(g\right)+4{\text{H}}_{\text{2}}\left(g\right)$

1. d. The value of $\Delta G_{\text{overall}}^{\text{ο}}$ is $\underset{\_}{113.6kJ}$ which is greater than zero shows the reaction is non-spontaneous under standard condition