Chapter 20, Problem 66E

(a)

Interpretation Introduction

Interpretation: The reaction between ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and $\text{CO}$ is given. The value of $\Delta H^{\circ },\Delta S^{\circ }$ and $\Delta G^{\circ }$ are to be calculated for the given options.

Concept introduction: The expression for $\Delta H^{\circ }$ is,

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

The expression for $\Delta S^{\circ }$ is,

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

The expression for standard Gibbs free energy, $\Delta G^{\circ }$, is,

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

To determine: The value of $\Delta H^{\circ }$ and $\Delta S^{\circ }$ for the given reaction.

Answer to Problem 66E

Solution

The value of $\Delta S^{\circ }$ for the given reaction is $\underset{\_}{38J/K}$. The value of $\Delta H^{\circ }$ for the given reaction is $\underset{\_}{-39kJ}$.

Explanation of Solution

Explanation

The reaction that takes place is,

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

Refer to Appendix 4.

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

Molecules	${\text{ΔH}}^{\text{0}}\text{(kJ/mol)}$
${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4(s)}}$	$-1117$
${\text{CO}}_{\text{2(g)}}$	$-393.5$
${\text{Fe}}_2{\text{O}}_{\text{3(s)}}$	$-826$
${\text{CO}}_{\text{(g)}}$	$-110.5$

The formula of $\Delta H^{\circ }$ is,

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

Where,

• $\Delta H^{\circ }$ 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.

• $\Delta H^{\circ }{}_{\left(\text{product}\right)}$ is the standard enthalpy of product at a pressure of $\text{1}\text{atm}$.

• $\Delta H^{\circ }{}_{\left(\text{reactant}\right)}$ is the standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(-\text{1117}\right)+\text{2}\left(-\text{393}\text{.5}\right)-\left\{\text{3}\left(-\text{826}\right)+\left(-\text{110}\text{.5}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-39kJ}\end{array}$

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

Molecules	${\text{ΔS}}^{\text{0}}\text{(kJ/mol)}$
${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4(s)}}$	146
${\text{CO}}_{\text{2(g)}}$	214
${\text{Fe}}_2{\text{O}}_{\text{3(s)}}$	90
${\text{CO}}_{\text{(g)}}$	198

The formula of $\Delta S^{\circ }$ is,

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

Where,

• $\Delta S^{\circ }$ standard enthalpy of reaction.

• $n_{\text{p}}$ is number of moles of each product.

• $n_{\text{r}}$ is number of moles each reactant.

• $\Delta S^{\circ }{}_{\left(\text{product}\right)}$ is standard enthalpy of product at a pressure of $\text{1}\text{atm}$.

• $\Delta S^{\circ }{}_{\left(\text{reactant}\right)}$ is standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta S^{\circ }=\sum n_{\text{p}}\Delta S^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(\text{146}\right)+\text{1}\left(\text{214}\right)-\left\{\text{3}\left(\text{90}\right)+\left(\text{198}\right)\right\}\right]\text{J/K}\\ =\underset{\_}{38J/K}\end{array}$

(b)

Interpretation Introduction

Interpretation: The reaction between ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and $\text{CO}$ is given. The value of $\Delta H^{\circ },\Delta S^{\circ }$ and $\Delta G^{\circ }$ are to be calculated for the given options.

Concept introduction: The expression for $\Delta H^{\circ }$ is,

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

The expression for $\Delta S^{\circ }$ is,

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

The expression for standard Gibbs free energy, $\Delta G^{\circ }$, is,

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

To determine: The value of $\Delta G^{\circ }$ at $800°\text{C}$ for the given reaction.

Answer to Problem 66E

Answer

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

Explanation of Solution

Explanation

Given

Temperature is $800°\text{C}$.

The value of $\Delta H^{\circ }$ is $-39\text{kJ}$.

The value of $\Delta S^{\circ }$ is $38\text{J/K}$.

The conversion of degree Celsius $\left(°\text{C}\right)$ into Kelvin $\left(\text{K}\right)$ is done as,

$T\left(\text{K}\right)=T\left(°\text{C}\right)+\text{273}$

Hence,

The conversion of $\text{800}°\text{C}$ into Kelvin is,

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+\text{273}\\ T\left(\text{K}\right)=\left(\text{800}+\text{273}\right)\text{K}\\ =\text{1073}\text{K}\end{array}$

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

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

Hence,

The conversion of $38\text{J/K}$ into kilo-joule is,

$\begin{array}{c}\text{38}\text{J}=\left(\text{38}\times {\text{10}}^{-\text{3}}\right)\text{kJ}\\ =\text{38}\times {\text{10}}^{-\text{3}}\text{kJ}\end{array}$

Formula

The expression for standard Gibbs free energy, $\Delta G^{\circ }$, is,

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

Where,

• $\Delta H^{\circ }$ is the standard enthalpy of reaction.

• $\Delta G^{\circ }$ is the free energy change.

• $T$ is the given temperature.

• $\Delta S^{\circ }$ is the entropy of reaction.

Substitute the values of $\Delta H^{\circ }$ and $\Delta S^{\circ }$ in the above equation.

$\begin{array}{c}\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }\\ =-\text{39}\text{kJ}-\text{1073}\text{K}\left(\text{38}\times {\text{10}}^{-\text{3}}\text{kJ/K}\right)\\ =\underset{\_}{-80kJ}\end{array}$