Chapter 3, Problem 3.56E

Determine the entropy of formation, ${\Delta }_{\text{f}}S$, of the following compounds. Assume $25°\text{C}$. (a) ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ (b) ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ (c) ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ (d) ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

Interpretation Introduction

(a)

Interpretation:

The entropy of formation of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 3.56E

The entropy of formation of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ is $-163.34\text{J}/\text{K}$.

Explanation of Solution

The formation reaction of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ is shown below.

${\text{H}}_{\text{2}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)$

The entropy of formation of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ is calculated by the formula as shown below.

${\Delta }_{\text{f}}S=\Delta S°\left({\text{H}}_{\text{2}}\text{O}\left(l\right)\right)-\left(\Delta S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)+\frac{1}{2}\Delta S°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)$ …(1)

Where,

• $\Delta S°\left({\text{H}}_{\text{2}}\text{O}\left(l\right)\right)$ is the standard entropy of ${\text{H}}_{\text{2}}\text{O}$ at $298\text{K}$.

• $\Delta S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)$ is the standard entropy of ${\text{H}}_{\text{2}}$ at $298\text{K}$.

• $\Delta S°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)$ is the standard entropy of ${\text{O}}_{\text{2}}$ at $298\text{K}$.

The standard entropy of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ at $298\text{K}$ is $69.91\text{J}/\text{mol}\cdot \text{K}$.

The standard entropy of ${\text{H}}_{\text{2}}$ at $298\text{K}$ is $130.68\text{J}/\text{mol}\cdot \text{K}$.

The standard entropy of ${\text{O}}_{\text{2}}$ at $298\text{K}$ is $205.14\text{J}/\text{mol}\cdot \text{K}$.

Substitute the standard entropy of ${\text{H}}_{\text{2}}\text{O}$, ${\text{H}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ in equation (1).

$\begin{array}{rll}{\Delta }_{\text{f}}S& =& 69.91\text{J}/\text{mol}\cdot \text{K}-\left(130.68\text{J}/\text{mol}\cdot \text{K}+\frac{1}{2}\times 205.14\text{J}/\text{mol}\cdot \text{K}\right)\\ & =& 69.91\text{J}/\text{K}-130.68\text{J}/\text{K}-102.57\text{J}/\text{K}\\ & =& -163.34\text{J}/\text{K}\end{array}$

Conclusion

The entropy of formation of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ is $-163.34\text{J}/\text{K}$.

Interpretation Introduction

(b)

Interpretation:

The entropy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 3.56E

The entropy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-44.42\text{J}/\text{K}$.

Explanation of Solution

The formation reaction of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is shown below,

${\text{H}}_{\text{2}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The entropy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is calculated by the formula as shown below.

${\Delta }_{\text{f}}S=\Delta S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)-\left(\Delta S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)+\frac{1}{2}\Delta S°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)$ …(2)

Where,

• $\Delta S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)$ is the standard entropy of ${\text{H}}_{\text{2}}\text{O}$ at $298\text{K}$.

• $\Delta S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)$ is the standard entropy of ${\text{H}}_{\text{2}}$ at $298\text{K}$.

• $\Delta S°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)$ is the standard entropy of ${\text{O}}_{\text{2}}$ at $298\text{K}$.

The standard entropy of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ at $298\text{K}$ is $188.83\text{J}/\text{mol}\cdot \text{K}$.

The standard entropy of ${\text{H}}_{\text{2}}$ at $298\text{K}$ is $130.68\text{J}/\text{mol}\cdot \text{K}$.

The standard entropy of ${\text{O}}_{\text{2}}$ at $298\text{K}$ is $205.14\text{J}/\text{mol}\cdot \text{K}$.

Substitute the entropy enthalpy of ${\text{H}}_{\text{2}}\text{O}$, ${\text{H}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ in equation (2).

$\begin{array}{rll}{\Delta }_{\text{f}}S& =& 188.83\text{J}/\text{mol}\cdot \text{K}-\left(130.68\text{J}/\text{mol}\cdot \text{K}+\frac{1}{2}\times 205.14\text{J}/\text{mol}\cdot \text{K}\right)\\ & =& 188.83\text{J}/\text{K}-130.68\text{J}/\text{K}-102.57\text{J}/\text{K}\\ & =& -44.42\text{J}/\text{K}\end{array}$

Conclusion

The entropy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-44.42\text{J}/\text{K}$.

Interpretation Introduction

(c)

Interpretation:

The entropy of formation of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 3.56E

The entropy of formation of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ is $-550.25\text{J}/\text{K}$.

Explanation of Solution

The formation reaction of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ is shown below.

${\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)+3{\text{SO}}_{\text{3}}\left(\text{g}\right)\to {\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{s}\right)$

The entropy of formation of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ is calculated by the formula as shown below.

${\Delta }_{\text{f}}S=\Delta S°\left({\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{s}\right)\right)-\left(\Delta S°\left({\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)\right)+3\Delta S°\left({\text{SO}}_{\text{3}}\left(\text{g}\right)\right)\right)$ … (3)

Where,

• $\Delta S°\left({\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{s}\right)\right)$ is the standard entropy of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{s}\right)$ at $298\text{K}$.

• $\Delta S°\left({\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)\right)$ is the standard entropy of ${\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$ at $298\text{K}$.

• $\Delta S°\left({\text{SO}}_{\text{3}}\left(\text{g}\right)\right)$ is the standard entropy of ${\text{SO}}_{\text{3}}\left(\text{g}\right)$ at $298\text{K}$.

The standard entropy of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{s}\right)$ at $298\text{K}$ is $307.46\text{J}/\text{mol}\cdot \text{K}$.

The standard entropy of ${\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$ at $298\text{K}$ is $87.4\text{J}/\text{mol}\cdot \text{K}$.

The standard entropy of ${\text{SO}}_{\text{3}}\left(\text{g}\right)$ at $298\text{K}$ is $256.77\text{J}/\text{mol}\cdot \text{K}$.

Substitute the standard entropy of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{s}\right)$, ${\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$, ${\text{SO}}_{\text{3}}\left(\text{g}\right)$ in equation (3).

$\begin{array}{rll}{\Delta }_{\text{f}}S& =& 307.46\text{J}/\text{mol}\cdot \text{K}-\left(87.4\text{J}/\text{mol}\cdot \text{K}+3\times 256.77\text{J}/\text{mol}\cdot \text{K}\right)\\ & =& 307.46\text{J}/\text{K}-87.4\text{J}/\text{K}-770.31\text{J}/\text{K}\\ & =& -550.25\text{J}/\text{K}\end{array}$

Conclusion

The entropy of formation of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ is $-550.25\text{J}/\text{K}$.

Interpretation Introduction

(d)

Interpretation:

The entropy of formation of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 3.56E

The entropy of formation of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is $-313.39\text{J}/\text{K}$.

Explanation of Solution

The formation reaction of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is shown below.

$2\text{Al}\left(\text{s}\right)+\frac{3}{2}{\text{O}}_2\left(\text{g}\right)\to {\text{Al}}_{\text{2}}{\text{O}}_3\left(\text{s}\right)$

The entropy of formation of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is calculated by the formula as shown below.

${\Delta }_{\text{f}}S=\Delta S°\left({\text{Al}}_{\text{2}}{\text{O}}_3\left(\text{s}\right)\right)-\left(2\Delta S°\left(\text{Al}\left(\text{s}\right)\right)+\frac{3}{2}\Delta S°\left({\text{O}}_2\left(\text{g}\right)\right)\right)$ … (4)

Where,

• $\Delta S°\left({\text{Al}}_{\text{2}}{\text{O}}_3\left(\text{s}\right)\right)$ is the standard entropy of ${\text{Al}}_{\text{2}}{\text{O}}_3\left(\text{s}\right)$ at $298\text{K}$.

• $\Delta S°\left(\text{Al}\left(\text{s}\right)\right)$ is the standard entropy of $\text{Al}\left(\text{s}\right)$ at $298\text{K}$.

• $\Delta S°\left({\text{O}}_2\left(\text{g}\right)\right)$ is the standard entropy of ${\text{O}}_2\left(\text{g}\right)$ at $298\text{K}$.

The standard entropy of ${\text{Al}}_{\text{2}}{\text{O}}_3\left(\text{s}\right)$ at $298\text{K}$ is $50.92\text{J}/\text{mol}\cdot \text{K}$.

The standard entropy of $\text{Al}\left(\text{s}\right)$ at $298\text{K}$ is $28.30\text{J}/\text{mol}\cdot \text{K}$.

The standard entropy of ${\text{O}}_2\left(\text{g}\right)$ at $298\text{K}$ is $205.14\text{J}/\text{mol}\cdot \text{K}$.

Substitute the standard entropy of ${\text{Al}}_{\text{2}}{\text{O}}_3\left(\text{s}\right)$, $\text{Al}\left(\text{s}\right)$, ${\text{O}}_2\left(\text{g}\right)$ in equation (4).

$\begin{array}{rll}{\Delta }_{\text{f}}S& =& 50.92\text{J}/\text{mol}\cdot \text{K}-\left(2\times 28.30\text{J}/\text{mol}\cdot \text{K}+\frac{3}{2}\times 205.14\text{J}/\text{mol}\cdot \text{K}\right)\\ & =& 50.92\text{J}/\text{K}-56.6\text{J}/\text{K}-307.71\text{J}/\text{K}\\ & =& -313.39\text{J}/\text{K}\end{array}$

Conclusion

The entropy of formation of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is $-313.39\text{J}/\text{K}$.