Chapter 4, Problem 4J.8E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

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.8E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction $\underset{\_}{+206.1kJ}$, $\underset{\_}{+214.632J\cdot K^{-1}}$ and $\underset{\_}{+142.12kJ}$ respectively.

Explanation of Solution

The given chemical equation for the production of synthesis gas is shown below.

    $C{H}_4\left(g\right)+H_{2}O\left(g\right)\to CO\left(g\right)+3H_2\left(g\right)$

The relation for the calculation of standard change in the enthalpy is shown below.

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

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

    $\Delta \text{H}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(CO,g\right)\\ +\left(3mol\right)\times \Delta {\text{H}}_f^{\circ }\left(H_2,g\right)\end{array}\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(C{H}_4,g\right)\\ +\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)\end{array}\right\}$        (1)

Where,

• $\Delta {\text{H}}_f^{\circ }\left(CO,g\right)$ is the standard enthalpy of formation of $CO\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_2,g\right)$ is the standard enthalpy of formation of $H_2\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(C{H}_4,g\right)$ is the standard enthalpy of formation of $C{H}_4\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ is the standard enthalpy of formation of $H_{2}O\left(g\right)$.

The value of $\Delta {\text{H}}_f^{\circ }\left(CO,g\right)$ is $-110.53\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_2,g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(C{H}_4,g\right)$ is $-74.81\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ is $-241.82\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{H}}_f^{\circ }\left(CO,g\right)$, $\Delta {\text{H}}_f^{\circ }\left(H_2,g\right)$, $\Delta {\text{H}}_f^{\circ }\left(C{H}_4,g\right)$ and $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ in equation (1).

    $\begin{array}{c}\Delta \text{H}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(-110.53\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(3mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \left(-74.81\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(-241.82\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{-110.53\text{kJ}\right\}-\left\{-74.81\text{kJ}-241.82\text{kJ}\right\}\\ =+206.1\text{kJ}\end{array}$

The relation for the calculation of standard change in the entropy is shown below.

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

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

    $\Delta \text{S}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{S}}_f^{\circ }\left(CO,g\right)\\ +\left(3mol\right)\times \Delta {\text{S}}_f^{\circ }\left(H_2,g\right)\end{array}\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{S}}_f^{\circ }\left(C{H}_4,g\right)\\ +\left(1mol\right)\times \Delta {\text{S}}_f^{\circ }\left(H_{2}O,g\right)\end{array}\right\}$        (2)

Where,

• $\Delta {\text{S}}_f^{\circ }\left(CO,g\right)$ is the standard entropy of formation of $CO\left(g\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(H_2,g\right)$ is the standard entropy of formation of $H_2\left(g\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(C{H}_4,g\right)$ is the standard entropy of formation of $C{H}_4\left(g\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,g\right)$ is the standard entropy of formation of $H_{2}O\left(g\right)$.

The value of $\Delta {\text{S}}_f^{\circ }\left(CO,g\right)$ is $197.67\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(H_2,g\right)$ is $130.684\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(C{H}_4,g\right)$ is $186.26\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,g\right)$ is $188.83\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{S}}_f^{\circ }\left(CO,g\right)$, $\Delta {\text{S}}_f^{\circ }\left(H_2,g\right)$, $\Delta {\text{S}}_f^{\circ }\left(C{H}_4,g\right)$ and $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,g\right)$ in equation (2).

    $\begin{array}{c}\Delta \text{S}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(197.67\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ +\left(3mol\right)\times \left(130.684\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \left(186.26\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(188.83\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{197.67\text{J}\cdot {\text{K}}^{-1}+392.052\text{J}\cdot {\text{K}}^{-1}\right\}-\left\{186.26\text{J}\cdot {\text{K}}^{-1}+188.83\text{J}\cdot {\text{K}}^{-1}\right\}\\ =+214.632\text{J}\cdot {\text{K}}^{-1}\end{array}$

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

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(CO,g\right)\\ +\left(3mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H_2,g\right)\end{array}\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(C{H}_4,g\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H_{2}O,g\right)\end{array}\right\}$        (3)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(CO,g\right)$ is the standard Gibbs free energyentropy of formation of $CO\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(H_2,g\right)$ is the standard Gibbs free energy of formation of $H_2\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(C{H}_4,g\right)$ is the standard Gibbs free energy of formation of $C{H}_4\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,g\right)$ is the standard Gibbs free energy of formation of $H_{2}O\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(CO,g\right)$ is $-137.17k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H_2,g\right)$ is $0.0k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(C{H}_4,g\right)$ is $-50.72k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,g\right)$ is $-228.57k\text{J}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(CO,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(H_2,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(C{H}_4,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,g\right)$ in equation (3).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(-137.17k\text{J}\cdot {\text{mol}}^{-1}\right)\\ +\left(3mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\begin{array}{l}\left(1mol\right)\times \left(-50.72k\text{J}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(-228.57k\text{J}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{-137.17k\text{J}\right\}-\left\{-50.72k\text{J}-228.57kJ\right\}\\ =+142.12\text{kJ}\end{array}$

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction $\underset{\_}{+206.1kJ}$, $\underset{\_}{+214.632J\cdot K^{-1}}$ and $\underset{\_}{+142.12kJ}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4J.8E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction $\underset{\_}{+205.79kJ}$, $\underset{\_}{+257.6J\cdot K^{-1}}$ and $\underset{\_}{+59.5kJ}$ respectively.

Explanation of Solution

The given chemical equation for the thermal decomposition of ammonium nitrate is shown below.

    $N{H}_{4}N{O}_3\left(s\right)\to N_{2}O\left(g\right)+H_{2}O\left(g\right)$

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

    $\Delta \text{H}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)\\ +\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(N_{2}O,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)\right\}$        (4)

Where,

• $\Delta {\text{H}}_f^{\circ }\left(N_{2}O,g\right)$ is the standard enthalpy of formation of $N_{2}O\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ is the standard enthalpy of formation of $H_{2}O\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ is the standard enthalpy of formation of $N{H}_{4}N{O}_3\left(s\right)$.

The value of $\Delta {\text{H}}_f^{\circ }\left(N_{2}O,g\right)$ is $82.05\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ is $-241.82\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ is $-365.56\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{H}}_f^{\circ }\left(N_{2}O,g\right)$, $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ and $\Delta {\text{H}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ in equation (4).

    $\begin{array}{c}\Delta \text{H}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(82.05\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(-241.82\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(-365.56\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{82.05\text{kJ}-241.82\text{kJ}\right\}-\left\{-365.56\text{kJ}\right\}\\ =+205.79\text{kJ}\end{array}$

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

    $\Delta \text{S}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{S}}_f^{\circ }\left(H_{2}O,g\right)\\ +\left(1mol\right)\times \Delta {\text{S}}_f^{\circ }\left(N_{2}O,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{S}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)\right\}$        (5)

Where,

• $\Delta {\text{S}}_f^{\circ }\left(N_{2}O,g\right)$ is the standard entropy of formation of $N_{2}O\left(g\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,g\right)$ is the standard entropy of formation of $H_{2}O\left(g\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ is the standard entropy of formation of $N{H}_{4}N{O}_3\left(s\right)$.

The value of $\Delta {\text{S}}_f^{\circ }\left(N_{2}O,g\right)$ is $219.85\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,g\right)$ is $188.83\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ is $151.08\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{S}}_f^{\circ }\left(N_{2}O,g\right)$, $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,g\right)$ and $\Delta {\text{S}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ in equation (5).

    $\begin{array}{c}\Delta \text{S}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(219.85\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(188.83\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(151.08\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{219.85\text{J}\cdot {\text{K}}^{-1}+188.83\text{J}\cdot {\text{K}}^{-1}\right\}-\left\{151.08\text{J}\cdot {\text{K}}^{-1}\right\}\\ =+257.6\text{J}\cdot {\text{K}}^{-1}\end{array}$

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

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H_{2}O,g\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(N_{2}O,g\right)\end{array}\right\}-\left\{\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)\right\}$        (6)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(N_{2}O,g\right)$ is the standard Gibbs free energy of formation of $N_{2}O\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,g\right)$ is the standard Gibbs free energy of formation of $H_{2}O\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ is the standard Gibbs free energy of formation of $N{H}_{4}N{O}_3\left(s\right)$

The value of $\Delta {\text{G}}_f^{\circ }\left(N_{2}O,g\right)$ is $104.20k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,g\right)$ is $-228.57k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ is $-183.87k\text{J}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(N_{2}O,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,g\right)$ and $\Delta {\text{G}}_f^{\circ }\left(N{H}_{4}N{O}_3,s\right)$ in equation (6).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(1mol\right)\times \left(104.20k\text{J}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(-228.57k\text{J}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(1mol\right)\times \left(-183.87\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{104.20k\text{J}-228.57k\text{J}\right\}-\left\{-183.87\text{kJ}\right\}\\ =+59.5\text{kJ}\end{array}$

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction are $\underset{\_}{+205.79kJ}$, $\underset{\_}{+257.6J\cdot K^{-1}}$ and $\underset{\_}{+59.5kJ}$ respectively.