Chapter 20, Problem 54E

Interpretation Introduction

Interpretation: Reaction of commercial production of $\text{NO}$ is given. The value of $\Delta H°,\Delta G°$ and $\Delta S°$ is to be calculated for the given reaction. The explanation of the fact that $\text{NO}$ formed in an automobile engine but does not readily decompose back to ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ in the atmosphere is to be stated.

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 $\Delta H°,\Delta G°$ and $\Delta S°$ for the given reaction; the explanation of the fact that $\text{NO}$ is formed in an automobile engine but does not readily decompose back to ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ in the atmosphere.

Answer to Problem 54E

Answer

The value of $\Delta H°$ for the given reaction is $\underset{\_}{180kJ}$.

The value of $\Delta S°$ for the given reaction is $\underset{\_}{25J/mol}$.

The value of $\Delta G°$ for the given reaction is $\underset{\_}{87kJ}$.

The molecule $\text{NO}$ does not readily decompose back to ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ in the atmosphere because rate of reaction becomes very slow at reduced temperature.

Explanation of Solution

Explanation

The value of $\Delta H°$ for the given reaction is $\underset{\_}{180kJ}$.

The stated reaction is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\stackrel{}{\to }\text{2NO}\left(g\right)$

Refer to Appendix $4$.

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

Molecules	$\Delta H°\left(\text{kJ/mol}\right)$
$\text{NO}(g)$	$\text{90}$
${\text{N}}_{\text{2}}(g)$	$\text{0}$
${\text{O}}_{\text{2}}(g)$	$\text{0}$

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 values from the table in the above equation.

$\begin{array}{c}\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(\text{90}\right)-\left\{\left(\text{0}\right)+\left(\text{0}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{180kJ}\end{array}$

The value of $\Delta S°$ for the given reaction is $\underset{\_}{25J/mol}$.

Refer to Appendix $4$

The stated reaction is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\stackrel{}{\to }\text{2NO}\left(g\right)$

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

Molecules	$\Delta S°\left(\text{J/K}\cdot \text{mol}\right)$
$\text{NO}(g)$	$\text{211}$
${\text{N}}_{\text{2}}(g)$	$\text{192}$
${\text{O}}_{\text{2}}(g)$	$\text{205}$

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

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(\text{211}\right)-\left\{\left(\text{192}\right)+\left(\text{205}\right)\right\}\right]\text{J/mol}\\ =\underset{\_}{25J/mol}\end{array}$

The value of $\Delta G°$ for the given reaction is $\underset{\_}{87kJ}$.

The stated reaction is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\stackrel{}{\to }\text{2NO}\left(g\right)$

Refer to Appendix $\text{4}$.

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

Molecules	$\Delta G°\left(\text{kJ/mol}\right)$
$\text{NO}(g)$	$\text{87}$
${\text{N}}_{\text{2}}(g)$	$\text{0}$
${\text{O}}_{\text{2}}(g)$	$\text{0}$

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,

• $n_{\text{p}}$ 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 free energy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta G{°}_{\left(\text{reactant}\right)}$ is the free energy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}\\ =\left[\left(\text{87}\right)-\left\{\frac{\text{1}}{\text{2}}\left(\text{0}\right)+\frac{\text{1}}{\text{2}}\left(\text{0}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{87kJ}\end{array}$

The molecule $\text{NO}$ does not readily decompose back to ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ in the atmosphere because rate of reaction becomes very slow at reduced temperature.

The reaction between ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ to form $\text{NO}$ at room temperature is unfavorable. It is formed in the combustion chamber of an automobile engine at very high temperature. When it escape into the atmosphere, the temperature decreases so rapidly that $\text{NO}$ does not readily decompose back to ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ in the atmosphere because rate of reaction becomes very slow at reduced temperature.

Conclusion

Conclusion

The required value of $\Delta S°$ for the given reaction is $\underset{\_}{25J/mol}$ , $\Delta G°$ is $\underset{\_}{87kJ}$ and $\Delta H°$ is $\underset{\_}{180kJ}$. The molecule $\text{NO}$ does not readily decompose back to ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ in the atmosphere because rate of reaction becomes very slow at reduced temperature