Chapter 4, Problem 4J.14E

(a)

Interpretation Introduction

Interpretation:

The stability of $C_3{H}_6\left(g\right)$ with respect to the decomposition into its elements at $25°C$ has to be predicted.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure.  A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation.  The relation to determine Gibbs free energy of a reaction is shown below.

$\Delta \text{G}°={\displaystyle \sum \text{n}\Delta {\text{G}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{G}}_f^{\circ }}\left(\text{reactants}\right)$

Answer to Problem 4J.14E

$C_3{H}_6\left(g\right)$ is an unstable compound with respect todecomposition into its elements.

Explanation of Solution

The chemical equation for the decomposition of $C_3{H}_6\left(g\right)$ into its elements is shown below.

    $C_3{H}_6\left(g\right)\to 3C\left(s\right)+6H\left(g\right)$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(3mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C,s\right)\\ +\left(6mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C_3{H}_6,g\right)\right\}$        (1)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(C,s\right)$ is the standard Gibbs free energy of formation of $C\left(s\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(H,g\right)$ is the standard Gibbs free energy of formation of $H\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(C_3{H}_6,g\right)$ is the standard Gibbs free energy of formation of $C_3{H}_6\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C,s\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H,g\right)$ is $+203.25\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C_3{H}_6,g\right)$ is $+\text{104}\text{.45}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(H,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(C,s\right)$ and $\Delta {\text{G}}_f^{\circ }\left(C_3{H}_6,g\right)$ in equation (1).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(3mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(6mol\right)\times \left(+203.25\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(+\text{104}\text{.45}\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{1219.5kJ\right\}-\left\{+\text{104}\text{.45}\text{kJ}\right\}\\ =+1115.05kJ\end{array}$

The calculated value of $\Delta \text{G}°$ is positive, it indicates that the decomposition of $C_3{H}_6\left(g\right)$ into its elements is non-spontaneous at $25°C$.

Thus, with respect to decomposition into its elements $C_3{H}_6\left(g\right)$ is an unstable compound.

(b)

Interpretation Introduction

Interpretation:

The stability of $CaO\left(s\right)$ with respect to the decomposition into its elements at $25°C$ has to be predicted.

Concept Introduction:

Same as part (a).

Answer to Problem 4J.14E

$CaO\left(s\right)$ is an unstable compound with respect to decomposition into its elements.

Explanation of Solution

The chemical equation for the decomposition of $CaO\left(s\right)$ into its elements is shown below.

    $CaO\left(s\right)\to Ca\left(s\right)+O\left(g\right)$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(Ca,s\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(O,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(CaO,s\right)\right\}$        (2)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(Ca,s\right)$ is the standard Gibbs free energy of formation of $Ca\left(s\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ is the standard Gibbs free energy of formation of $O\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(CaO,s\right)$ is the standard Gibbs free energy of formation of $CaO\left(s\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(Ca,s\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ is $+231.73\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(CaO,s\right)$ is $-\text{604}\text{.03}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(Ca,s\right)$, $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(CaO,s\right)$ in equation (2).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(+231.73\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(-\text{604}\text{.03}\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{+231.73\text{kJ}\right\}-\left\{-\text{604}\text{.03}\text{kJ}\right\}\\ =+835.76kJ\end{array}$

The calculated value of $\Delta \text{G}°$ is positive, it indicates that the decomposition of $CaO\left(s\right)$ into its elements is non-spontaneous at $25°C$.

Thus, with respect to decomposition into its elements $CaO\left(s\right)$ is an unstable compound.

(c)

Interpretation Introduction

Interpretation:

The stability of $N_{2}O\left(g\right)$ with respect to the decomposition into its elements at $25°C$ has to be predicted.

Concept Introduction:

Same as part (a).

Answer to Problem 4J.14E

$N_{2}O\left(g\right)$ is an unstable compound with respect to decomposition into its elements.

Explanation of Solution

The chemical equation for the decomposition of $N_{2}O\left(g\right)$ into its elements is shown below.

    $N_{2}O\left(g\right)\to 2N\left(g\right)+O\left(g\right)$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(2mol\right)\times \Delta {\text{G}}_f^{\circ }\left(N,g\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(O,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(N_{2}O,g\right)\right\}$        (3)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(N,g\right)$ is the standard Gibbs free energy of formation of $N\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ is the standard Gibbs free energy of formation of $O\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(N_{2}O,g\right)$ is the standard Gibbs free energy of formation of $N_{2}O\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(N,g\right)$ is $+\text{455}\text{.56}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ is $+231.73\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(N_{2}O,g\right)$ is $+\text{104}\text{.20}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(N,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(N_{2}O,g\right)$ in equation (3).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(+\text{455}\text{.56}\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(+231.73\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(+\text{104}\text{.20}\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{+\text{455}\text{.56}\text{kJ}+231.73\text{kJ}\right\}-\left\{+\text{104}\text{.20}\text{kJ}\right\}\\ =+583.09kJ\end{array}$

The calculated value of $\Delta \text{G}°$ is positive, it indicates that the decomposition of $N_{2}O\left(g\right)$ into its elements is non-spontaneous at $25°C$.

Thus, with respect to decomposition into its elements $N_{2}O\left(g\right)$ is an unstable compound.

(d)

Interpretation Introduction

Interpretation:

The stability of $H{N}_3\left(g\right)$ with respect to the decomposition into its elements at $25°C$ has to be predicted.

Concept Introduction:

Same as part (a).

Answer to Problem 4J.14E

$H{N}_3\left(g\right)$ is an unstable compound with respect to decomposition into its elements.

Explanation of Solution

The chemical equation for the decomposition of $H{N}_3\left(g\right)$ into its elements is shown below.

    $H{N}_3\left(g\right)\to H\left(g\right)+3N\left(g\right)$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H,g\right)\\ +\left(3mol\right)\times \Delta {\text{G}}_f^{\circ }\left(N,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H{N}_3,g\right)\right\}$        (4)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(H,g\right)$ is the standard Gibbs free energy of formation of $H\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(N,g\right)$ is the standard Gibbs free energy of formation of $N\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(H{N}_3,g\right)$ is the standard Gibbs free energy of formation of $H{N}_3\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H,g\right)$ is $+203.25\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(N,g\right)$ is $+\text{455}\text{.56}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H{N}_3,g\right)$ is $+\text{328}\text{.1}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(N,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(H,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(H{N}_3,g\right)$ in equation (4).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(+203.25\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(3mol\right)\times \left(+\text{455}\text{.56}\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(+\text{328}\text{.1}\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{203.25\text{kJ}+1366.68\text{kJ}\right\}-\left\{+\text{328}\text{.1}\text{kJ}\right\}\\ =+1241.83kJ\end{array}$

The calculated value of $\Delta \text{G}°$ is positive, it indicates that the decomposition of $H{N}_3\left(g\right)$ into its elements is non-spontaneous at $25°C$.

Thus, with respect to decomposition into its elements $H{N}_3\left(g\right)$ is an unstable compound.