Chapter 20, Problem 11Q

(a)

Interpretation Introduction

Interpretation: Reaction of commercial production of hydrogen is given. The value of $\Delta H°$ and $\Delta S°$ is to be calculated for the given reaction. The temperature at which this reaction is favored is to be calculated.

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 favored if the value of $\Delta G°$ is negative.

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

Explanation of Solution

The stated reaction is,

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

Refer to Appendix $4$

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

Molecules	$\Delta H°\left(\text{kJ/mol}\right)$
${\text{CH}}_{\text{4}}(g)$	$-\text{75}$
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	$-\text{242}$
$\text{CO}(g)$	$-\text{110}\text{.5}$
${\text{H}}_{\text{2}}(g)$	$\text{0}$

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,

• $\Delta H°$ 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 the values from the table in the above equation.

$\begin{array}{c}\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}\\ =\left[\text{3}\left(\text{0}\right)+\left(-\text{110}\text{.5}\right)-\left\{\left(-\text{75}\right)+\left(-\text{242}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-206.5kJ}\end{array}$

The value of $\Delta S°$ for the given reaction is $\underset{\_}{216J/K}$.

Refer to Appendix $4$

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

Molecules	$\Delta S°\left(\text{kJ/mol}\right)$
${\text{CH}}_{\text{4}}(g)$	$\text{186}$
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	$\text{189}$
$\text{CO}(g)$	$\text{198}$
${\text{H}}_{\text{2}}(g)$	$\text{131}$

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

Substitute all the values from the table in the above equation.

$\begin{array}{c}\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}\\ =\left[\text{3}\left(\text{131}\right)+\left(\text{198}\right)-\left\{\left(\text{186}\right)+\left(\text{189}\right)\right\}\right]\text{J/K}\\ =\underset{\_}{216J/K}\end{array}$

(b)

Interpretation Introduction

Interpretation: Reaction of commercial production of hydrogen is given. The value of $\Delta H°$ and $\Delta S°$ is to be calculated for the given reaction. The temperature at which this reaction is favored is to be calculated.

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 favored if the value of $\Delta G°$ is negative.

To determine: The temperature at which this reaction is favored.

Explanation of Solution

The value of $\Delta H$ is $-206.5\text{kJ}$.

The value of $\Delta S$ is $216\text{J/K}$.

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 $-206.5\text{kJ}$ into joule is,

$\begin{array}{c}-\text{206}\text{.5}\text{kJ}=\left(-\text{206}\text{.5}\times {\text{10}}^{\text{3}}\right)\text{J}\\ =-\text{206}\text{.5}\times {\text{10}}^{\text{3}}\text{J}\end{array}$

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 entropy 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.

$\begin{array}{c}\Delta G=\Delta H°-T\Delta S°\\ \text{0}=\left(-\text{206}\text{.5}\times {\text{10}}^{\text{3}}\text{J}\right)-T\left(\text{216}\text{J/K}\right)\\ T=\underset{\_}{956K}\end{array}$

The reaction will be favored if the temperature is greater than $956\text{K}$.

Conclusion

The required value of $\Delta S°$ for the given reaction is $\underset{\_}{216J/K}$ and $\Delta H°$ is $\underset{\_}{-206.5kJ}$. The temperature at which this reaction is favored is $\underset{\_}{956K}$