Chapter 20, Problem 65E

(a)

Interpretation Introduction

Interpretation: The answers are to be stated for the given options.

Concept introduction: The expression to calculate $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

The expression to calculate $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

A reaction is said to be spontaneous if the value of $\Delta G°$ is negative.

To determine: The value of $\Delta H°$ for the given reaction.

Answer to Problem 65E

Answer

The value of $\Delta H°$ for the given reaction is $\underset{\_}{-11kJ}$.

Explanation of Solution

Explanation

The value of $\Delta H°$ for the given reaction is $\underset{\_}{-11kJ}$.

For the reaction,

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\text{3CO}\left(g\right)\stackrel{}{\to }\text{2Fe}\left(s\right)+{\text{3CO}}_{\text{2}}\left(g\right)$

The value of $\Delta H°$ for the reaction is,

$\Delta H°=-23\text{kJ}$

Multiply the above equation by coefficient $3$ on both side.

${\text{3Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\text{9CO}\left(g\right)\stackrel{}{\to }\text{6Fe}\left(s\right)+{\text{9CO}}_{\text{2}}\left(g\right)$ (1)

The value of $\Delta H°$ for this reaction is,

$\begin{array}{c}\Delta H°=-23\text{kJ}\times \text{3}\\ =-\text{69}\text{kJ}\end{array}$

For the reaction,

${\text{3Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\text{CO}\left(g\right)\stackrel{}{\to }\text{2Fe}\left(s\right)+{\text{9CO}}_{\text{2}}\left(g\right)$ (2)

The value of $\Delta H°$ for the reaction is,

$\Delta H°=-39\text{kJ}$

For the reaction,

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\text{CO}\left(g\right)\stackrel{}{\to }\text{3FeO}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$

The value of $\Delta H°$ for the reaction is,

$\Delta H°=18\text{kJ}$

Multiply the above equation by coefficient $2$ on both side.

${\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\text{2CO}\left(g\right)\stackrel{}{\to }\text{6FeO}\left(s\right)+{\text{2CO}}_{\text{2}}\left(g\right)$ (3)

The value of $\Delta H°$ for this reaction is,

$\begin{array}{c}\Delta H°=18\text{kJ}\times \text{2}\\ =3\text{6}\text{kJ}\end{array}$

Subtract equation (1), (2), (3) and cancel the like terms from both sides to get the reaction,

$\text{6FeO}\left(s\right)+\text{6CO}\left(g\right)\stackrel{}{\to }\text{6Fe}\left(s\right)+{\text{6CO}}_{\text{2}}\left(g\right)$

The resultant value of $\Delta H°$ for this reaction is,

$\begin{array}{c}\Delta H°=\left[-69-(-39)-(36)\right]\text{kJ}\\ =-\text{66}\text{kJ}\end{array}$

Since, the reaction for which $\Delta H°$ is to be calculated is,

$\text{FeO}\left(s\right)+\text{CO}\left(g\right)\stackrel{}{\to }\text{Fe}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Divide the resultant value of $\Delta H°$ by coefficient $6$ to get the required reaction. The value of $\Delta H°$ is,

$\frac{-66\text{kJ}}{6}=\underset{\_}{-11kJ}$

(b)

Interpretation Introduction

Interpretation: The answers are to be stated for the given options.

Concept introduction: The expression to calculate $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

The expression to calculate $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

A reaction is said to be spontaneous if the value of $\Delta G°$ is negative.

Answer to Problem 65E

Answer

The value of $\Delta S°$ for the given reaction is $\underset{\_}{199J/K}$ and $\Delta H°$ is $\underset{\_}{-92kJ}$. The temperature at which this reaction is favored is $\underset{\_}{462K}$.

Explanation of Solution

Explanation

To determine: The value of $\Delta H°$ and $\Delta S°$ for the given reaction; the temperature at which the conversion reaction of ${\text{CO}}_{\text{2}}$ into $\text{CO}$ spontaneous at standard conditions.

The value of $\Delta H°$ for the given reaction is $\underset{\_}{172.5kJ}$.

The reaction that takes place is,

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

Refer to Appendix $\text{4}$.

The value of $\Delta H°\left(\text{kJ/mol}\right)$ for the given reactant and product is,

Molecules	$\Delta H°\left(\text{kJ/mol}\right)$
${\text{CO}}_{\text{2}}(g)$	$-\text{393}\text{.5}$
$\text{C}(g)$	$\text{0}$
$\text{CO}(g)$	$-\text{110}\text{.5}$

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta H{°}_{\left(\text{product}\right)}$ is the standard enthalpy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta H{°}_{\left(\text{reactant}\right)}$ is the standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(-\text{110}\text{.5}\right)-\left\{\text{1}\left(-\text{393}\text{.5}\right)+\left(\text{0}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{172.5kJ}\end{array}$

The value of $\Delta S°$ for the given reaction is $\underset{\_}{182J/K}$.

The value of $\Delta S°\left(\text{J/K}\cdot \text{mol}\right)$ for the given reactant and product is,

Molecules	$\Delta S°\left(\text{J/K}\cdot \text{mol}\right)$
${\text{CO}}_{\text{2}}(g)$	$\text{214}$
$\text{C}(g)$	$\text{0}$
$\text{CO}(g)$	$\text{198}$

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta S°$ is the standard entropy of reaction.
• $n_{\text{p}}$ is number of moles of each product.
• $n_{\text{r}}$ is number of moles each reactant.
• $\Delta S{°}_{\left(\text{product}\right)}$ is standard entropy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta S{°}_{\left(\text{reactant}\right)}$ is standard entropy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(\text{198}\right)-\left\{\left(\text{214}\right)+\left(\text{0}\right)\right\}\right]\text{J/K}\\ =\underset{\_}{182J/K}\end{array}$

The value of $\Delta G°$ for the given reaction is $\underset{\_}{-80kJ}$.

The value of $\Delta G°\left(\text{kJ/mol}\right)$ for the given reactant and product is,

Molecules	$\Delta G°\left(\text{kJ/mol}\right)$
${\text{CO}}_{\text{2}}(g)$	$-\text{394}\text{.4}$
$\text{C}(g)$	$\text{0}$
$\text{CO}(g)$	$-\text{137}\text{.2}$

The formula of $\Delta G°$ is,

$\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta G°$ is the standard enthalpy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta G{°}_{\left(\text{product}\right)}$ is the standard enthalpy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta G{°}_{\left(\text{reactant}\right)}$ is the standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(-\text{137}\text{.2}\right)-\left\{\left(-\text{394}\text{.4}\right)+\left(\text{0}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{120kJ}\end{array}$

The temperature at which the given reaction is spontaneous is $\underset{\_}{462K}$.

The value of $\Delta H$ is $-92\text{kJ}$.

The value of $\Delta S$ is $199\text{J/K}$.

The conversion of kilo-joule $\left(\text{kJ}\right)$ into joule $\left(\text{J}\right)$ is done as,

$\text{1}\text{kJ}={\text{10}}^{\text{3}}\text{J}$

Hence,

The conversion of $-92\text{kJ}$ into joule is,

$\begin{array}{c}-\text{92}\text{kJ}=\left(-\text{92}\times {\text{10}}^{\text{3}}\right)\text{J}\\ =-\text{92}\times {\text{10}}^{\text{3}}\text{J}\end{array}$

Formula

The formula of $\Delta G$ is,

$\Delta G=\Delta H°-T\Delta S°$

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $\Delta G°$ is the free energy change.
• $T$ is the given temperature.
• $\Delta S°$ is the standard enthalpy of reaction.

At equilibrium, the value of $\Delta G$ is zero.

Substitute the values of $\Delta G°,\Delta H°$ and $\Delta S°$ in the above equation.

$\begin{array}{c}\Delta G=\Delta H°-T\Delta S°\\ \text{0}=\left(-\text{92}\times {\text{10}}^{\text{3}}\text{J}\right)-T\left(\text{199}\text{J/K}\right)\\ T\approx \underset{\_}{462K}\end{array}$

The reaction will be spontaneous if the temperature is greater than $\underset{\_}{462K}$.