Chapter 16, Problem 59QRT

(a)

Interpretation Introduction

Interpretation:

For the given reaction, the temperature at which it shifts from being reactant-favored to being product favored has to be calculated.

Concept Introduction:

The entropy of a system is the randomness created by the molecules.  The order of entropy is $\text{gas}>\text{liquid}>\text{solid}$.  Entropy change of a reaction will be positive when the total number of product molecules is greater than the total number of reactants molecules and entropy change will be negative when total reactants molecules are greater than product molecules.

Answer to Problem 59QRT

For the given reaction, the temperature at which it shifts from being reactant-favored to being product favored is $\underset{\_}{385.67K}$.

Explanation of Solution

The given reaction is shown below.

  $\text{CO}\left(\text{g}\right)+{\text{2H}}_2\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(l\right)$

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 ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$, $\text{CO}\left(\text{g}\right)$ and ${\text{H}}_2\left(\text{g}\right)$ is $-238.66\text{kJ}/\text{mol}$, $-110.525\text{kJ}/\text{mol}$ and $0\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{l}{\text{n}}_{{\text{CH}}_{\text{3}}\text{OH}\left(l\right)}\times {\Delta }_{\text{f}}H°\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)-\\ \left({\text{n}}_{\text{CO}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left(\text{CO}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_2\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}1\times -238.66\text{kJ}/\text{mol}-\\ \left(\text{1}\times -110.525\text{kJ}/\text{mol}+\left(2\times 0\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =-128.135\text{kJ}/\text{mol}\\ =-128135\text{ J}/\text{mol}\end{array}$

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

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (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 $S°$ for ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$, $\text{CO}\left(\text{g}\right)$ and ${\text{H}}_2\left(\text{g}\right)$ is $126.8{\text{J mol}}^{-1}{\text{ K}}^{-1}$, $197.674{\text{J mol}}^{-1}{\text{ K}}^{-1}$ and $130.684{\text{J mol}}^{-1}{\text{ K}}^{-1}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}{\text{n}}_{{\text{CH}}_{\text{3}}\text{OH}\left(l\right)}\times S°\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)-\\ \left({\text{n}}_{\text{CO}\left(\text{g}\right)}\times S°\left(\text{CO}\left(\text{g}\right)\right)+\left({\text{n}}_{{\text{H}}_2\left(\text{g}\right)}\times S°\left({\text{H}}_2\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\text{1}\times 126.8{\text{J mol}}^{-1}{\text{ K}}^{-1}\right)-\\ \left(\text{1}\times \left(197.674{\text{J mol}}^{-1}{\text{ K}}^{-1}\right)+\left(\text{2}\times 130.684{\text{J mol}}^{-1}{\text{ K}}^{-1}\right)\right)\end{array}\right)\\ =-332.242{\text{J mol}}^{-1}{\text{ K}}^{-1}\end{array}$

The value of ${\Delta }_{\text{r}}G°$ is assumed to be zero to calculate the value of $T$.

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ 0=-128135\text{ J}/\text{mol}-\left(T\times \left(-332.242{\text{J mol}}^{-1}{\text{ K}}^{-1}\right)\right)\\ T=\frac{128135}{332.242}\text{ K}\\ =\underset{\_}{385.67K}\end{array}$

For the given reaction, the temperature at which it shifts from being reactant-favored to being product favored is $\underset{\_}{385.67K}$.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, the temperature at which it shifts from being reactant-favored to being product favored has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 59QRT

For the given reaction, the temperature at which it shifts from being reactant-favored to being product favored is $\underset{\_}{836.79K}$.

Explanation of Solution

The given reaction is shown below.

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

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 $\text{Fe}\left(\text{s}\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$ and $\text{C}\left(\text{s, graphite}\right)$ is $0\text{kJ}/\text{mol}$, $-393.509\text{kJ}/\text{mol}$, $-824.2\text{kJ}/\text{mol}$ and $0\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{l}\left({\text{n}}_{\text{Ag}\left(\text{s}\right)}\times {\Delta }_{\text{f}}H°\left(\text{Fe}\left(\text{s}\right)\right)+\left({\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)}\times {\Delta }_{\text{f}}H°\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)+{\text{n}}_{\text{C}\left(\text{s, graphite}\right)}\times {\Delta }_{\text{f}}H°\left(\text{C}\left(\text{s, graphite}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\text{4}\times 0\text{kJ}/\text{mol}+\left(\text{3}\times -393.509\text{kJ}/\text{mol}\right)\right)-\\ \left(\text{2}\times -824.2\text{kJ}/\text{mol}+\text{3}\times 0\text{kJ}/\text{mol}\right)\end{array}\right)\\ =467.873\text{kJ}/\text{mol}\\ =467873\text{ J}/\text{mol}\end{array}$

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

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (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 $S°$ for $\text{Fe}\left(\text{s}\right)$, ${\text{CO}}_{\text{2}}\left(\text{g}\right)$, ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$ and $\text{C}\left(\text{s, graphite}\right)$ is $27.78{\text{J mol}}^{-1}{\text{ K}}^{-1}$, $213.74{\text{J mol}}^{-1}{\text{ K}}^{-1}$, $87.4{\text{J mol}}^{-1}{\text{ K}}^{-1}$ and $5.74{\text{J mol}}^{-1}{\text{ K}}^{-1}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}\left({\text{n}}_{\text{Ag}\left(\text{s}\right)}\times S°\left(\text{Fe}\left(\text{s}\right)\right)+\left({\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times S°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)-\\ \left({\text{n}}_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)}\times S°\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)+{\text{n}}_{\text{C}\left(\text{s, graphite}\right)}\times S°\left(\text{C}\left(\text{s, graphite}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(\text{4}\times 27.78{\text{J mol}}^{-1}{\text{ K}}^{-1}+\left(\text{3}\times 213.74{\text{J mol}}^{-1}{\text{ K}}^{-1}\right)\right)-\\ \left(\text{2}\times 87.4{\text{J mol}}^{-1}{\text{ K}}^{-1}+\text{3}\times 5.74{\text{J mol}}^{-1}{\text{ K}}^{-1}\right)\end{array}\right)\\ =\text{560}\text{.32}{\text{J mol}}^{-1}{\text{ K}}^{-1}\end{array}$

The value of ${\Delta }_{\text{r}}G°$ is assumed to be zero to calculate the value of $T$.

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ -1000\text{ J}/\text{mol}=467873\text{ J}/\text{mol}-\left(T\times \left(\text{560}\text{.32}{\text{J mol}}^{-1}{\text{ K}}^{-1}\right)\right)\\ T=-\frac{\left(-1000\text{ J}/\text{mol}-467873\text{ J}/\text{mol}\right)}{\text{560}\text{.32}{\text{J mol}}^{-1}}\text{ K}\\ =\underset{\_}{836.79K}\end{array}$

For the given reaction, the temperature at which it shifts from being reactant-favored to being product favored is $\underset{\_}{836.79K}$.