Chapter 4, Problem 4H.11E

(a)

Interpretation Introduction

Interpretation:

Thestandard reaction entropyfor the formation of $1.00\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ at $\text{25}°\text{C}$ has to be determined.

Concept Introduction:

The degree of randomness in a system is the measured in terms of entropy.  Higher the entropy greater will be the disorder in the system.  The reactionentropy of a reaction in terms of the standard molar entropiesis the difference between the standard entropies of products and that of the reactants.  The negative reaction entropyindicates that the products are less disordered as compared to the reactants.

Answer to Problem 4H.11E

The standard reaction entropy for the formation of $1.00\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ at $\text{25}°\text{C}$ is $\underset{\_}{-163.35J\cdot K^{-1}}$.

Explanation of Solution

The balanced chemical reaction representing the formation of $1.00\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ is shown below.

  ${\text{H}}_2\left(\text{g}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The total entropy of reactants of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}=\left(1\text{mol}\right)\Delta S_{\text{m}}\left({\text{H}}_2,\text{g}\right)+\left(\frac{1}{2}\text{mol}\right)\Delta S_{\text{m}}\left({\text{O}}_2,\text{g}\right)$        (1)

Where,

• ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is total entropy of reactants.
• $\Delta S_{\text{m}}\left({\text{H}}_2,\text{g}\right)$ is the molar entropy of ${\text{H}}_2\left(\text{g}\right)$.
• $\Delta S_{\text{m}}\left({\text{O}}_2,\text{g}\right)$ is the molar entropy of ${\text{O}}_2\left(\text{g}\right)$.

The value of $\Delta S_{\text{m}}\left({\text{H}}_2,\text{g}\right)$ is $130.7\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta S_{\text{m}}\left({\text{O}}_2,\text{g}\right)$ is $205.1\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta S_{\text{m}}\left({\text{H}}_2,\text{g}\right)$ and $\Delta S_{\text{m}}\left({\text{O}}_2,\text{g}\right)$ in equation (1).

  $\begin{array}{c}{\displaystyle \sum \Delta S°\left(\text{reactants}\right)}=\left(1\text{mol}\right)\times \left(130.7\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)+\left(\frac{1}{2}\text{mol}\right)\times \left(205.1\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ =130.7\text{J}\cdot {\text{K}}^{-1}+102.55\text{J}\cdot {\text{K}}^{-1}\\ =233.25\text{J}\cdot {\text{K}}^{-1}\end{array}$

The total entropy of products of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta S°\left(\text{products}\right)}=\left(1\text{mol}\right)\Delta S_{\text{m}}\left({\text{H}}_2\text{O},\text{l}\right)$        (2)

Where,

• ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is total entropy of product.
• $\Delta S_{\text{m}}\left({\text{H}}_2\text{O},\text{l}\right)$ is the molar entropy of ${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$.

The value of $\Delta S_{\text{m}}\left({\text{H}}_2\text{O},\text{l}\right)$ is $69.9\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta S_{\text{m}}\left({\text{H}}_2\text{O},\text{l}\right)$ in equation (2).

  $\begin{array}{c}{\displaystyle \sum \Delta S°\left(\text{products}\right)}=\left(1\text{mol}\right)\times \left(69.9\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ =69.9\text{J}\cdot {\text{K}}^{-1}\end{array}$

The standard reaction entropy of the given reaction is calculated by the relation shown below.

  $\Delta S°={\displaystyle \sum \Delta S°\left(\text{products}\right)}-{\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$        (3)

Where,

• $\Delta S°$ is standard reaction entropy.
• ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is total entropy of reactants.
• ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is total entropy of product.

The value of ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is $233.25\text{J}\cdot {\text{K}}^{-1}$.

The value of ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is $69.9\text{J}\cdot {\text{K}}^{-1}$.

Substitute the value of ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ and ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ in equation (3).

    $\begin{array}{c}\Delta S°=69.9\text{J}\cdot {\text{K}}^{-1}-233.25\text{J}\cdot {\text{K}}^{-1}\\ =-163.35\text{J}\cdot {\text{K}}^{-1}\end{array}$

The obtained standard reaction entropy of the given reaction is negative, which indicates that the products are less disordered as compared to the reactants.

Thus, the standard reaction entropy for the formation of $1.00\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ at $\text{25}°\text{C}$ is $\underset{\_}{-163.35J\cdot K^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The standard reaction entropy for the oxidation of $1.00\text{mol}\text{CO}\left(\text{g}\right)$ at $\text{25}°\text{C}$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4H.11E

The standard reaction entropy for the oxidation of $1.00\text{mol}\text{CO}\left(\text{g}\right)$ at $\text{25}°\text{C}$ is $\underset{\_}{-86.48J\cdot K^{-1}}$.

Explanation of Solution

The balanced chemical reaction representing the oxidation of $1.00\text{mol}\text{CO}\left(\text{g}\right)$ is shown below.

  $\text{CO}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right)$

The total entropy of reactants of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}=\left(1\text{mol}\right)\Delta S_{\text{m}}\left(\text{CO},\text{g}\right)+\left(\frac{1}{2}\text{mol}\right)\Delta S_{\text{m}}\left({\text{O}}_2,\text{g}\right)$        (4)

Where,

• ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is total entropy of reactants.
• $\Delta S_{\text{m}}\left(\text{CO},\text{g}\right)$ is the molar entropy of $\text{CO}\left(\text{g}\right)$.
• $\Delta S_{\text{m}}\left({\text{O}}_2,\text{g}\right)$ is the molar entropy of ${\text{O}}_2\left(\text{g}\right)$.

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

The value of $\Delta S_{\text{m}}\left({\text{O}}_2,\text{g}\right)$ is $205.1\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta S_{\text{m}}\left(\text{CO},\text{g}\right)$ and $\Delta S_{\text{m}}\left({\text{O}}_2,\text{g}\right)$ in equation (4).

  $\begin{array}{c}{\displaystyle \sum \Delta S°\left(\text{reactants}\right)}=\left(1\text{mol}\right)\times \left(197.67\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)+\left(\frac{1}{2}\text{mol}\right)\times \left(205.1\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ =197.67\text{J}\cdot {\text{K}}^{-1}+102.55\text{J}\cdot {\text{K}}^{-1}\\ =300.22\text{J}\cdot {\text{K}}^{-1}\end{array}$

The total entropy of products of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta S°\left(\text{products}\right)}=\left(1\text{mol}\right)\Delta S_{\text{m}}\left({\text{CO}}_2,\text{g}\right)$        (5)

Where,

• ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is total entropy of product.
• $\Delta S_{\text{m}}\left({\text{CO}}_2,\text{g}\right)$ is the molar entropy of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$.

The value of $\Delta S_{\text{m}}\left({\text{CO}}_2,\text{g}\right)$ is $213.74\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta S_{\text{m}}\left({\text{CO}}_2,\text{g}\right)$ in equation (5).

  $\begin{array}{c}{\displaystyle \sum \Delta S°\left(\text{products}\right)}=\left(1\text{mol}\right)\times \left(213.74\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ =213.74\text{J}\cdot {\text{K}}^{-1}\end{array}$

The value of ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is $300.22\text{J}\cdot {\text{K}}^{-1}$.

The value of ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is $213.74\text{J}\cdot {\text{K}}^{-1}$.

Substitute the value of ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ and ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ in equation (3).

    $\begin{array}{c}\Delta S°=213.74\text{J}\cdot {\text{K}}^{-1}-300.22\text{J}\cdot {\text{K}}^{-1}\\ =-86.48\text{J}\cdot {\text{K}}^{-1}\end{array}$

The obtained standard reaction entropy of the given reaction is negative, which indicates that the products are less disordered as compared to the reactants.

Thus, the standard reaction entropy for the oxidation of $1.00\text{mol}\text{CO}\left(\text{g}\right)$ at $\text{25}°\text{C}$ is $\underset{\_}{-86.48J\cdot K^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The standard reaction entropy for the decomposition of $1.00\text{mol}$ calcite at $\text{25}°\text{C}$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4H.11E

The standard reaction entropy for the decomposition of $1.00\text{mol}$ calcite at $\text{25}°\text{C}$ is $\underset{\_}{+160.64J\cdot K^{-1}}$.

Explanation of Solution

The balanced chemical reaction representing the decomposition of $1.00\text{mol}$ calcite is shown below.

  ${\text{CaCO}}_3\left(\text{s}\right)\to \text{CaO}\left(\text{s}\right)+{\text{CO}}_2\left(\text{g}\right)$

The total entropy of reactants of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}=\left(1\text{mol}\right)\Delta S_{\text{m}}\left({\text{CaCO}}_3,\text{s}\right)$        (6)

Where,

• ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is total entropy of reactants.
• $\Delta S_{\text{m}}\left({\text{CaCO}}_3,\text{s}\right)$ is the molar entropy of ${\text{CaCO}}_3\left(\text{s}\right)$.

The value of $\Delta S_{\text{m}}\left({\text{CaCO}}_3,\text{s}\right)$ is $92.9\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta S_{\text{m}}\left({\text{CaCO}}_3,\text{s}\right)$ in equation (6).

  $\begin{array}{c}{\displaystyle \sum \Delta S°\left(\text{reactants}\right)}=\left(1\text{mol}\right)\times \left(92.9\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ =92.9\text{J}\cdot {\text{K}}^{-1}\end{array}$

The total entropy of products of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta S°\left(\text{products}\right)}=\left(1\text{mol}\right)\Delta S_{\text{m}}\left(\text{CaO},\text{s}\right)+\left(1\text{mol}\right)\Delta S_{\text{m}}\left({\text{CO}}_2,\text{g}\right)$        (7)

Where,

• ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is total entropy of product.
• $\Delta S_{\text{m}}\left(\text{CaO},\text{s}\right)$ is the molar entropy of $\text{CaO}\left(\text{s}\right)$.
• $\Delta S_{\text{m}}\left({\text{CO}}_2,\text{g}\right)$ is the molar entropy of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$.

The value of $\Delta S_{\text{m}}\left(\text{CaO},\text{s}\right)$ is $39.8\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta S_{\text{m}}\left({\text{CO}}_2,\text{g}\right)$ is $213.74\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta S_{\text{m}}\left(\text{CaO},\text{s}\right)$ and $\Delta S_{\text{m}}\left({\text{CO}}_2,\text{g}\right)$ in equation (7).

  $\begin{array}{c}{\displaystyle \sum \Delta S°\left(\text{products}\right)}=\left(1\text{mol}\right)\times \left(39.8\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)+\left(1\text{mol}\right)\times \left(213.74\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ =39.8\text{J}\cdot {\text{K}}^{-1}+213.74\text{J}\cdot {\text{K}}^{-1}\\ =253.54\text{J}\cdot {\text{K}}^{-1}\end{array}$

The value of ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is $92.9\text{J}\cdot {\text{K}}^{-1}$.

The value of ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is $253.54\text{J}\cdot {\text{K}}^{-1}$.

Substitute the value of ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ and ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ in equation (3).

    $\begin{array}{c}\Delta S°=253.54\text{J}\cdot {\text{K}}^{-1}-92.9\text{J}\cdot {\text{K}}^{-1}\\ =+160.64\text{J}\cdot {\text{K}}^{-1}\end{array}$

The obtained standard reaction entropy of the given reaction is positive, which indicates that the products are more disordered as compared to the reactants.

Thus, the standard reaction entropy for the decomposition of $1.00\text{mol}$ calcite at $\text{25}°\text{C}$ is $\underset{\_}{+160.64J\cdot K^{-1}}$.

(d)

Interpretation Introduction

Interpretation:

The standard reaction entropy for the decomposition of potassium chlorate at $\text{25}°\text{C}$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4H.11E

The standard reaction entropy for the decomposition of potassium chlorate at $\text{25}°\text{C}$ is $\underset{\_}{-36.29J\cdot K^{-1}}$.

Explanation of Solution

The balanced chemical reaction representing the decomposition of potassium chlorate is shown below.

  ${\text{4KClO}}_3\left(\text{s}\right)\to 3{\text{KClO}}_4\left(\text{s}\right)+\text{KCl}\left(\text{s}\right)$

The total entropy of reactants of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}=\left(4\text{mol}\right)\Delta S_{\text{m}}\left({\text{KClO}}_3,\text{s}\right)$        (8)

Where,

• ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is total entropy of reactants.
• $\Delta S_{\text{m}}\left({\text{KClO}}_3,\text{s}\right)$ is the molar entropy of ${\text{KClO}}_3\left(\text{s}\right)$.

The value of $\Delta S_{\text{m}}\left({\text{KClO}}_3,\text{s}\right)$ is $142.97\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta S_{\text{m}}\left({\text{KClO}}_3,\text{s}\right)$ in equation (8).

  $\begin{array}{c}{\displaystyle \sum \Delta S°\left(\text{reactants}\right)}=\left(4\text{mol}\right)\times \left(142.97\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ =571.88\text{J}\cdot {\text{K}}^{-1}\end{array}$

The total entropy of products of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta S°\left(\text{products}\right)}=\left(3\text{mol}\right)\Delta S_{\text{m}}\left({\text{KClO}}_4,\text{s}\right)+\left(1\text{mol}\right)\Delta S_{\text{m}}\left(\text{KCl},\text{s}\right)$        (9)

Where,

• ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is total entropy of product.
• $\Delta S_{\text{m}}\left({\text{KClO}}_4,\text{s}\right)$ is the molar entropy of ${\text{KClO}}_4\left(\text{s}\right)$.
• $\Delta S_{\text{m}}\left(\text{KCl},\text{s}\right)$ is the molar entropy of $\text{KCl}\left(\text{s}\right)$.

The value of $\Delta S_{\text{m}}\left({\text{KClO}}_4,\text{s}\right)$ is $151.0\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta S_{\text{m}}\left(\text{KCl},\text{s}\right)$ is $82.59\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta S_{\text{m}}\left({\text{KClO}}_4,\text{s}\right)$ and $\Delta S_{\text{m}}\left(\text{KCl},\text{s}\right)$ in equation (9).

  $\begin{array}{c}{\displaystyle \sum \Delta S°\left(\text{products}\right)}=\left(3\text{mol}\right)\times \left(151.0\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)+\left(1\text{mol}\right)\times \left(82.59\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ =453.0\text{J}\cdot {\text{K}}^{-1}+82.59\text{J}\cdot {\text{K}}^{-1}\\ =535.59\text{J}\cdot {\text{K}}^{-1}\end{array}$

The value of ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ is $571.88\text{J}\cdot {\text{K}}^{-1}$.

The value of ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ is $535.59\text{J}\cdot {\text{K}}^{-1}$.

Substitute the value of ${\displaystyle \sum \Delta S°\left(\text{reactants}\right)}$ and ${\displaystyle \sum \Delta S°\left(\text{products}\right)}$ in equation (3).

    $\begin{array}{c}\Delta S°=535.59\text{J}\cdot {\text{K}}^{-1}-571.88\text{J}\cdot {\text{K}}^{-1}\\ =-36.29\text{J}\cdot {\text{K}}^{-1}\end{array}$

The obtained standard reaction entropy of the given reaction is negative, which indicates that the products are less disordered as compared to the reactants.

Thus, the standard reaction entropy for the decomposition of potassium chlorate at $\text{25}°\text{C}$ is $\underset{\_}{-36.29J\cdot K^{-1}}$.