Chapter 16, Problem 61QRT

(a)

Interpretation Introduction

Interpretation:

The ${\Delta }_{\text{r}}G°$ value for $\text{CO}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ at $2000\text{K}$ has to be calculated.

Concept Introduction:

The Gibbs free energy of a system is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.  The Gibbs free energy of the system is a state function as it is defined in terms of thermodynamic properties that are state functions.

Answer to Problem 61QRT

The ${\Delta }_{\text{r}}G°$ value for $\text{CO}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ at $2000\text{K}$ is $\underset{\_}{536.3kJ}$.

Explanation of Solution

The given reaction is shown below.

  $\text{CO}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{CH}}_{\text{3}}\text{OH}\left(l\right)$

The formula to calculate ${\Delta }_{\text{r}}G°$ is shown below.

    ${\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°$        (1)

Where,

• ${\text{Δ}}_{\text{r}}G°$ is the change in standard Gibbs free energy of reaction.
• ${\Delta }_{\text{r}}H°$ is the change in standard enthalpy of reaction.
• ${\Delta }_{\text{r}}S°$ is the change in standard entropy of reaction.
• $T$ is the temperature.

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

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

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 $\text{CO}\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$ and ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ is $-110.525\text{kJ}/\text{mol},0\text{kJ}/\text{mol}$ and $-238.66\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{c}{\text{n}}_{{\text{CH}}_{\text{3}}\text{OH}\left(l\right)}\times {\Delta }_{\text{f}}H°\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)-\\ \left({\text{n}}_{\text{CO}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left(\text{CO}\left(\text{g}\right)\right)+{\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =1\times \left(-238.66\text{kJ}/\text{mol}\right)-\left(1\times \left(-110.525\text{kJ}/\text{mol}\right)+2\times \left(0\text{kJ}/\text{mol}\right)\right)\\ =-128.135\text{kJ}\end{array}$

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

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

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 $\text{CO}\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$ and ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ is $197.674\text{J}/\text{mol}\cdot \text{K},130.684\text{J}/\text{mol}\cdot \text{K}$ and $126.8\text{J}/\text{mol}\cdot \text{K}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{c}{\text{n}}_{{\text{CH}}_{\text{3}}\text{OH}\left(l\right)}\times {\Delta }_{\text{f}}S°\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)-\\ \left({\text{n}}_{\text{CO}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left(\text{CO}\left(\text{g}\right)\right)+{\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =1\times \left(126.8\text{J}/\text{mol}\cdot \text{K}\right)-\left(1\times \left(197.674\text{J}/\text{mol}\cdot \text{K}\right)+2\times \left(130.684\text{J}/\text{mol}\cdot \text{K}\right)\right)\\ =126.8\text{J}/\text{K}-\left(459.042\text{J}/\text{K}\right)\\ =-332.242\text{J}/\text{K}\end{array}$

Substitute the values of ${\Delta }_{\text{r}}H°$, ${\Delta }_{\text{r}}S°$ and $T$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ =-128.135\text{kJ}-\left(\left(2000\text{K}\right)\times \left(-332.242\text{J}/\text{K}\right)\right)\\ =-128.135\times {\text{10}}^3\text{J}+664484\text{J}\because \text{1}{\text{kJ=10}}^{\text{3}}\text{J}\\ =\text{536349}\text{J}\end{array}$

The value of ${\Delta }_{\text{r}}G°$ in $\text{kJ}$ is written as follows.

    $\begin{array}{c}1\text{J}={10}^{-3}\text{kJ}\\ \text{536349}\text{J}=\text{536349}\times {10}^{-3}\text{kJ}\\ =\underset{\_}{536.3kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The ${\Delta }_{\text{r}}G°$ value for ${\text{2Fe}}_2{\text{O}}_3\left(\text{s}\right)+\text{3C}\left(\text{s,}\text{graphite}\right)\rightleftarrows \text{4Fe}\left(\text{s}\right)+{\text{3CO}}_2\left(\text{g}\right)$ at $2000\text{K}$ has to be calculated

Concept Introduction:

Refer to part (a).

Answer to Problem 61QRT

The ${\Delta }_{\text{r}}G°$ value for ${\text{2Fe}}_2{\text{O}}_3\left(\text{s}\right)+\text{3C}\left(\text{s,}\text{graphite}\right)\rightleftarrows \text{4Fe}\left(\text{s}\right)+{\text{3CO}}_2\left(\text{g}\right)$ at $2000\text{K}$ is $\underset{\_}{-648.8kJ}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{2Fe}}_2{\text{O}}_3\left(\text{s}\right)+\text{3C}\left(\text{s,}\text{graphite}\right)\rightleftarrows \text{4Fe}\left(\text{s}\right)+{\text{3CO}}_2\left(\text{g}\right)$

The value of ${\Delta }_{\text{f}}H°$ for ${\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$, $\text{C}\left(\text{s,}\text{graphite}\right)$, $\text{Fe}\left(\text{s}\right)$ and ${\text{CO}}_2\left(\text{g}\right)$ is $-824.2\text{kJ}/\text{mol},0\text{kJ}/\text{mol},0\text{kJ}/\text{mol}$ and $-393.509\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{c}\left({\text{n}}_{\text{Fe}\left(\text{s}\right)}\times {\Delta }_{\text{f}}H°\left(\text{Fe}\left(\text{s}\right)\right)+{\text{n}}_{{\text{CO}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{CO}}_2\left(\text{g}\right)\right)\right)-\\ \left({\text{n}}_{{\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)}\times {\Delta }_{\text{f}}H°\left({\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)\right)+{\text{n}}_{\text{C}\left(\text{s,}\text{graphite}\right)}\times {\Delta }_{\text{f}}H°\left(\text{C}\left(\text{s,}\text{graphite}\right)\right)\right)\end{array}\right)\\ =\left(4\times \left(0\right)+3\times \left(-393.509\text{kJ}/\text{mol}\right)\right)-\left(2\times \left(-824.2\text{kJ}/\text{mol}\right)+3\times \left(0\text{kJ}/\text{mol}\right)\right)\\ =-1180.527\text{kJ}+\text{1648}\text{.4}\text{kJ}\\ =467.873\text{kJ}\end{array}$

The value of ${\Delta }_{\text{f}}S°$ for ${\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$, $\text{C}\left(\text{s,}\text{graphite}\right)$, $\text{Fe}\left(\text{s}\right)$ and ${\text{CO}}_2\left(\text{g}\right)$ is $87.4\text{J}/\text{mol}\cdot \text{K},5.74\text{J}/\text{mol}\cdot \text{K},27.28\text{J}/\text{mol}\cdot \text{K}$ and $213.74\text{J}/\text{mol}\cdot \text{K}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{c}\left({\text{n}}_{\text{Fe}\left(\text{s}\right)}\times {\Delta }_{\text{f}}S°\left(\text{Fe}\left(\text{s}\right)\right)+{\text{n}}_{{\text{CO}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CO}}_2\left(\text{g}\right)\right)\right)-\\ \left({\text{n}}_{{\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)}\times {\Delta }_{\text{f}}S°\left({\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)\right)+{\text{n}}_{\text{C}\left(\text{s,}\text{graphite}\right)}\times {\Delta }_{\text{f}}S°\left(\text{C}\left(\text{s,}\text{graphite}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(4\times \left(27.28\text{J}/\text{mol}\cdot \text{K}\right)+3\times \left(213.74\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(2\times \left(87.4\text{J}/\text{mol}\cdot \text{K}\right)+3\times \left(5.74\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =\left(109.12\text{J}/\text{K}+\text{641}\text{.22}\text{J}/\text{K}\right)-\left(174.8\text{J}/\text{K}+17.22\text{J}/\text{K}\right)\\ =558.32\text{J}/\text{K}\end{array}$

Substitute the values of ${\Delta }_{\text{r}}H°$, ${\Delta }_{\text{r}}S°$ and $T$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ =467.873\text{kJ}-\left(\left(2000\text{K}\right)\times \left(558.32\text{J}/\text{K}\right)\right)\\ =467.873\times {\text{10}}^3\text{J}-1116640\text{J}\because \text{1}{\text{kJ=10}}^{\text{3}}\text{J}\\ =-\text{648767}\text{J}\end{array}$

The value of ${\Delta }_{\text{r}}G°$ in $\text{kJ}$ is written as follows.

    $\begin{array}{c}1\text{J}={10}^{-3}\text{kJ}\\ -\text{648767}\text{J}=-\text{648767}\text{J}\times {10}^{-3}\text{kJ}\\ =\underset{\_}{-648.8kJ}\end{array}$