Chapter 16, Problem 35QRT

(a)

Interpretation Introduction

Interpretation:

The entropy change for the reaction, ${\text{C}}_2{\text{H}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ has to be calculated.

Concept Introduction:

The entropy of a system is the randomness created by the molecules.  The order of entropy is $\text{gas}>\text{liquid}>\text{solid}$.  Entropy change of a reaction will be positive when the total number of product molecules is greater than the total number of reactants molecules and entropy change will be negative when total reactants molecules are greater than product molecules.

Answer to Problem 35QRT

The entropy change of the reaction ${\text{C}}_2{\text{H}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ is $\underset{\_}{-120.6J/mol\cdot K}$.

Explanation of Solution

The reaction is shown below.

  ${\text{C}}_2{\text{H}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}S°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (1)

Where,

• ${\Delta }_{\text{f}}S°$ is the change in standard entropy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}S°$ for ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$, ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\left(\text{g}\right)$ is $229.6\text{J}/\text{mol}\cdot \text{K},219.56\text{J}/\text{mol}\cdot \text{K}$ and $130.684\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(1\times \left(229.6\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(1\times \left(219.56\text{J}/\text{mol}\cdot \text{K}\right)+1\times \left(130.684\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =\underset{\_}{-120.6J/mol\cdot K}\end{array}$

The value of ${\Delta }_{\text{r}}S°$ is $\underset{\_}{-120.6J/mol\cdot K}$.

(b)

Interpretation Introduction

Interpretation:

The entropy change for the reaction, ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)+\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+2{\text{H}}_2\text{O}\left(\text{g}\right)$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 35QRT

The entropy change of the reaction ${\text{C}}_2{\text{H}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ is $\underset{\_}{156.9J/mol\cdot K}$.

Explanation of Solution

The reaction is shown below.

  ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)+\frac{3}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+2{\text{H}}_2\text{O}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}S°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (1)

Where,

• ${\Delta }_{\text{f}}S°$ is the change in standard entropy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}S°$ for ${\text{CO}}_2\left(\text{g}\right)$, ${\text{H}}_2\text{O}$, ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ and ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $213.74\text{ J}/\text{mol}\cdot \text{K},188.82\text{J}/\text{mol}\cdot \text{K},\text{126}\text{.8}\text{ J}/\text{mol}\cdot \text{K}$ and $205.138\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}\left(\left({\text{n}}_{{\text{CO}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CO}}_2\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{H}}_2\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_2\text{O}\left(\text{g}\right)\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{CH}}_{\text{3}}\text{OH}\left(l\right)}\times {\Delta }_{\text{f}}S°\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)\right)+\left({\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\left(1\times \left(213.74\text{J}/\text{mol}\cdot \text{K}\right)\right)+\left(2\times \left(188.82\text{J}/\text{mol}\cdot \text{K}\right)\right)\right)\\ -\left(1\times \left(126.8\text{J}/\text{mol}\cdot \text{K}\right)+\frac{3}{2}\times \left(205.138\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =\underset{\_}{156.9J/mol\cdot K}\end{array}$

The value of ${\Delta }_{\text{r}}S°$ is $\underset{\_}{156.9J/mol\cdot K}$.

(c)

Interpretation Introduction

Interpretation:

The entropy change for the reaction ${\text{N}}_2\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\to 2{\text{NH}}_{\text{3}}\left(\text{g}\right)$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 35QRT

The entropy change of the reaction ${\text{N}}_2\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\to 2{\text{NH}}_{\text{3}}\left(\text{g}\right)$ is $\underset{\_}{-197.76J/mol\cdot K}$.

Explanation of Solution

The reaction is shown below.

  ${\text{N}}_2\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\to 2{\text{NH}}_{\text{3}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}S°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (1)

Where,

• ${\Delta }_{\text{f}}S°$ is the change in standard entropy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}S°$ for ${\text{NH}}_{\text{3}}\left(\text{g}\right)$, ${\text{N}}_2\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\left(\text{g}\right)$ is $192.45\text{J}/\text{mol}\cdot \text{K},190.61\text{J}/\text{mol}\cdot \text{K}$ and $130.684\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}\left({\text{n}}_{{\text{NH}}_{\text{3}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{NH}}_{\text{3}}\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{N}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{N}}_2\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(192.45\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(1\times \left(190.61\text{J}/\text{mol}\cdot \text{K}\right)+3\times \left(130.684\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =\underset{\_}{-197.76J/mol\cdot K}\end{array}$

The value of ${\Delta }_{\text{r}}S°$ is $\underset{\_}{-197.76J/mol\cdot K}$.

(d)

Interpretation Introduction

Interpretation:

The entropy change for the reaction ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)\to \text{CaO}\left(\text{s}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 35QRT

The entropy change of the reaction ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)\to \text{CaO}\left(\text{s}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$ is $\underset{\_}{160.6J/mol\cdot K}$.

Explanation of Solution

The reaction is shown below.

  ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)\to \text{CaO}\left(\text{s}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}S°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (1)

Where,

• ${\Delta }_{\text{f}}S°$ is the change in standard entropy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}S°$ for $\text{CaO}\left(\text{s}\right)$, ${\text{CO}}_2\left(\text{g}\right)$, and ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)$ is $39.75\text{J}/\text{mol}\cdot \text{K}\text{, }213.74\text{ J}/\text{mol}\cdot \text{K},$ and $92.9\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{l}\left(\left({\text{n}}_{\text{CaO}\left(\text{s}\right)}\times {\Delta }_{\text{f}}S°\left(\text{CaO}\left(\text{s}\right)\right)\right)+\left({\text{n}}_{{\text{CO}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CO}}_2\left(\text{g}\right)\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{CaCO}}_{\text{3}}\left(\text{s}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CaCO}}_{\text{3}}\left(\text{s}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\left(1\times \left(39.75\text{J}/\text{mol}\cdot \text{K}\right)\right)+\left(1\times \left(213.74\text{J}/\text{mol}\cdot \text{K}\right)\right)\right)\\ -\left(1\times \left(92.9\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =\underset{\_}{160.6J/mol\cdot K}\end{array}$

The value of ${\Delta }_{\text{r}}S°$ is $\underset{\_}{160.6J/mol\cdot K}$.