Chapter 4, Problem 4J.13E

(a)

Interpretation Introduction

Interpretation:

The stability of $PC{l}_5\left(s\right)$ 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.13E

$PC{l}_5\left(s\right)$ is an unstable compound with respect to decomposition into its elements.

Explanation of Solution

The chemical equation for the decomposition of $PC{l}_5\left(s\right)$ into its elements is shown below.

    $PC{l}_5\left(g\right)\to P\left(s\right)+5Cl\left(g\right)$

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

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(P,s\right)\\ +\left(5mol\right)\times \Delta {\text{G}}_f^{\circ }\left(Cl,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(PC{l}_5,g\right)\right\}$        (1)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(P,s\right)$ is the standard Gibbs free energy of formation of $P\left(s\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(Cl,g\right)$ is the standard Gibbs free energy of formation of $Cl\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(PC{l}_5,g\right)$ is the standard Gibbs free energy of formation of $PC{l}_5\left(g\right)$.

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

The value of $\Delta {\text{G}}_f^{\circ }\left(Cl,g\right)$ is $+105.68\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(PC{l}_5,g\right)$ is $-305.0\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(P,s\right)$, $\Delta {\text{G}}_f^{\circ }\left(Cl,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(PC{l}_5,g\right)$ in equation (1).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(5mol\right)\times \left(+105.68\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(-305.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{528.4kJ\right\}-\left\{305.0\text{kJ}\right\}\\ =+223.4kJ\end{array}$

The calculated value of $\Delta \text{G}°$ is positive, it indicates that the decomposition of $PC{l}_5\left(s\right)$ into its elements is non-spontaneous at $25°C$.

Thus, with respect to decomposition into its elements $PC{l}_5\left(s\right)$ is an unstable compound.

(b)

Interpretation Introduction

Interpretation:

The stability of $HCN\left(g\right)$ 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.13E

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

Explanation of Solution

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

    $HCN\left(g\right)\to H\left(g\right)+C\left(s\right)+N\left(g\right)$

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

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

Where,

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

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

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

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 $\Delta {\text{G}}_f^{\circ }\left(HCN,g\right)$ is $+\text{124}\text{.7}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(H,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(C,s\right)$, $\Delta {\text{G}}_f^{\circ }\left(N,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(HCN,g\right)$ in equation (2).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(+203.25\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(+\text{455}\text{.56}\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(+\text{124}\text{.7}\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{+203.25\text{kJ}+0.0\text{kJ}+\text{455}\text{.56}\text{kJ}\right\}-\left\{\text{124}\text{.7}\text{kJ}\right\}\\ =+534.11kJ\end{array}$

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

Thus, with respect to decomposition into its elements $HCN\left(g\right)$ is anunstable compound.

(c)

Interpretation Introduction

Interpretation:

The stability of $NO\left(g\right)$ 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.13E

$NO\left(g\right)$ is an unstable compound with respect to decomposition into its elements.

Explanation of Solution

The chemical equation for the decomposition of $NO\left(g\right)$ into its elements is shown below.

    $NO\left(g\right)\to N\left(g\right)+O\left(g\right)$

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

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(N,g\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(O,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(NO,g\right)\right\}$        (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 $O\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(NO,g\right)$ is the standard Gibbs free energy of formation of $NO\left(g\right)$.

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 $\Delta {\text{G}}_f^{\circ }\left(NO,g\right)$ is $+86.55\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(N,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(NO,g\right)$ in equation (3).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(+\text{455}\text{.56}\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(+231.73\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(+86.55\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{+\text{455}\text{.56}\text{kJ}+231.73\text{kJ}\right\}-\left\{+86.55\text{kJ}\right\}\\ =+600.74kJ\end{array}$

The calculated value of $\Delta \text{G}°$ is positive, it indicates that the decomposition of $NO\left(g\right)$ into its elements is non-spontaneous at $25°C$.

Thus, with respect to decomposition into its elements $NO\left(g\right)$ is an unstable compound.

(d)

Interpretation Introduction

Interpretation:

The stability of $S{O}_2\left(g\right)$ 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.13E

$S{O}_2\left(g\right)$ is an unstable compound with respect to decomposition into its elements.

Explanation of Solution

The chemical equation for the decomposition of $S{O}_2\left(g\right)$ into its elements is shown below.

    $S{O}_2\left(g\right)\to S\left(g\right)+2O\left(g\right)$

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

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(S,g\right)\\ +\left(2mol\right)\times \Delta {\text{G}}_f^{\circ }\left(O,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(S{O}_2,g\right)\right\}$        (4)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(S,g\right)$ is the standard Gibbs free energy of formation of $S\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ is the standard Gibbs free energy of formation of $O\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(S{O}_2,g\right)$ is the standard Gibbs free energy of formation of $S{O}_2\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(S,g\right)$ is $+238.25\text{kJ}\cdot {\text{mol}}^{-1}$.

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

The value of $\Delta {\text{G}}_f^{\circ }\left(S{O}_2,g\right)$ is $-300.19\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(S,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(O,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(S{O}_2,g\right)$ in equation (4).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(+238.25\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(2mol\right)\times \left(+231.73\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(-300.19\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{+238.25\text{kJ}+463.46\text{kJ}\right\}-\left\{-300.19\text{kJ}\right\}\\ =+1001.9kJ\end{array}$

The calculated value of $\Delta \text{G}°$ is positive, it indicates that the decomposition of $S{O}_2\left(g\right)$ into its elements is non-spontaneous at $25°C$.

Thus, with respect to decomposition into its elements $S{O}_2\left(g\right)$ is an unstable compound.