Chapter 17, Problem 17.54QE

(a)

Interpretation Introduction

Interpretation:

The standard entropy change has to be calculated.

  $\text{C}\left(s\right)\text{+}{H}_{\text{2}}O\left(\text{g}\right)\to \text{CO}\left(g\right)+H_2\left(g\right)$

Answer to Problem 17.54QE

The value of ${\text{ΔS}}^{\text{o}}$ for the process is $133.67kJ/mol.$

Explanation of Solution

The balanced chemical equation as follows,

  $\text{C}\left(s\right)\text{+}{H}_{\text{2}}O\left(\text{g}\right)\to \text{CO}\left(g\right)+H_2\left(g\right)$

The value of ${\text{ΔS}}^{\text{o}}$ is calculated as follows,

  $\begin{array}{l}\text{C}\left(s\right)\text{+}{H}_{\text{2}}O\left(\text{g}\right)\to \text{CO}\left(g\right)+H_2\left(g\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)5.74188.72197.56130.57\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}130.57\text{J/mol}\text{.K}\right)+\left(1\text{×}197.56\text{J/mol}\text{.K}\right)\text{-}\left(\text{1×}5.74\text{J/mol}\text{.K}\right)-\left(\text{1×}188.72\text{J/mol}\text{.K}\right)\\ \\ =133.67kJ/mol.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $133.67kJ/mol.$

(b)

Interpretation Introduction

Interpretation:

The standard entropy change has to be calculated.

  ${\text{2NO}}_{\text{2}}\left(\text{g}\right)\to 2\text{NO}\left(g\right)\text{+}{O}_{\text{2}}\left(\text{g}\right)$

Answer to Problem 17.54QE

The value of ${\text{ΔS}}^{\text{o}}$ for the process is $146.43kJ/mol.$

Explanation of Solution

The balanced chemical equation as follows,

  ${\text{2NO}}_{\text{2}}\left(\text{g}\right)\to 2\text{NO}\left(g\right)\text{+}{O}_{\text{2}}\left(\text{g}\right)$

The value of $\Delta {\text{H}}_f^{\circ }$ is calculated as follows,

  $\begin{array}{l}{\text{2NO}}_{\text{2}}\left(\text{g}\right)\to 2\text{NO}\left(g\right)\text{+}{O}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)239.95210.65205.03\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}205.03\text{J/mol}\text{.K}\right)+\left(2\text{×}210.65\text{J/mol}\text{.K}\right)\text{-}\left(2\text{×}239.95\text{J/mol}\text{.K}\right)\\ \\ =146.43kJ/mol.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $146.43kJ/mol.$

(c)

Interpretation Introduction

Interpretation:

The standard entropy change has to be calculated.

  $NaCl\left(\text{s}\right)\to N{a}^{+}\left(aq\right)\text{+}C{l}^{-}\left(\text{aq}\right)$

Answer to Problem 17.54QE

The value of ${\text{ΔS}}^{\text{o}}$ for the process is $43.37kJ/mol.$

Explanation of Solution

The balanced chemical equation as follows,

  $NaCl\left(\text{s}\right)\to N{a}^{+}\left(aq\right)\text{+}C{l}^{-}\left(\text{aq}\right)$

The value of $\Delta {\text{H}}_f^{\circ }$ is calculated as follows,

  $\begin{array}{l}NaCl\left(\text{s}\right)\to N{a}^{+}\left(aq\right)\text{+}C{l}^{-}\left(\text{aq}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)72.1359.056.5\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}56.5\text{J/mol}\text{.K}\right)+\left(1\text{×}59.0\text{J/mol}\text{.K}\right)\text{-}\left(1\text{×}72.13\text{J/mol}\text{.K}\right)\\ \\ =43.37kJ/mol.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $43.37kJ/mol.$

(d)

Interpretation Introduction

Interpretation:

The standard entropy change has to be calculated.

  ${\text{C}}_6{H}_{12}{O}_6\left(s\right)\text{+}6O_{\text{2}}\left(\text{g}\right)\to 6{\text{CO}}_2\left(g\right)+6H_{2}O\left(l\right)$

Answer to Problem 17.54QE

The value of ${\text{ΔS}}^{\text{o}}$ for the process is $259.06kJ/mol.$

Explanation of Solution

The balanced chemical equation as follows,

  ${\text{C}}_6{H}_{12}{O}_6\left(s\right)\text{+}6O_{\text{2}}\left(\text{g}\right)\to 6{\text{CO}}_2\left(g\right)+6H_{2}O\left(l\right)$

The value of $\Delta {\text{H}}_f^{\circ }$ is calculated as follows,

  $\begin{array}{l}{\text{C}}_6{H}_{12}{O}_6\left(s\right)\text{+}6O_{\text{2}}\left(\text{g}\right)\to 6{\text{CO}}_2\left(g\right)+6H_{2}O\left(l\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)212205.03213.6369.91\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(6\text{×}69.91\text{J/mol}\text{.K}\right)+\left(6\text{×}213.63\text{J/mol}\text{.K}\right)-\left(1\text{×}212\text{J/mol}\text{.K}\right)-\left(6\text{×}205.03\text{J/mol}\text{.K}\right)\\ \\ \text{=}\left(419.46\text{J/mol}\text{.K}\right)+\left(1281.78\text{J/mol}\text{.K}\right)-\left(212\text{J/mol}\text{.K}\right)-\left(1230.18\text{J/mol}\text{.K}\right)\\ \\ =259.06kJ/mol.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $259.06kJ/mol.$