Chapter 17, Problem 17.53QE

(a)

Interpretation Introduction

Interpretation:

The standard entropy change has to be calculated.

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

Answer to Problem 17.53QE

The value of ${\text{ΔS}}^{\text{o}}$ for the process is $-332kJ/mol.$

Explanation of Solution

The balanced chemical equation as follows,

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

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

  $\begin{array}{l}\text{CO}\left(\text{g}\right)\text{+}{\text{2H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{. K}\right)197.56130.57126.8\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}126.8\text{J/mol}\text{.K}\right)-\left(1\text{×}197.56\text{J/mol}\text{.K}\right)\text{-}\left(2\text{×}130.57\text{J/mol}\text{.K}\right)\\ \\ \text{=}\left(126.8\text{J/mol}\text{.K}\right)-\left(197.56\text{J/mol}\text{.K}\right)\text{-}\left(261.14\text{J/mol}\text{.K}\right)\\ \\ =-332kJ/mol.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $-332kJ/mol.$

(b)

Interpretation Introduction

Interpretation:

The standard entropy change has to be calculated.

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

Answer to Problem 17.53QE

The value of ${\text{ΔS}}^{\text{o}}$ for the process is $-198.53kJ/mol.$

Explanation of Solution

The balanced chemical equation as follows,

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

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

  $\begin{array}{l}{\text{3H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{N}}_{\text{2}}\left(\text{g}\right)\to 2{\text{NH}}_{\text{3}}\left(g\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)130.57191.50192.34\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(2\text{×}192.34\text{J/mol}\text{.K}\right)-\left(3\text{×}130.57\text{J/mol}\text{.K}\right)\text{-}\left(1\text{×}191.50\text{J/mol}\text{.K}\right)\\ \\ \text{=}\left(384.68\text{J/mol}\text{.K}\right)-\left(391.71\text{J/mol}\text{.K}\right)\text{-}\left(191.50\text{J/mol}\text{.K}\right)\\ \\ =-198.53kJ/mol.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $-198.53kJ/mol.$

(c)

Interpretation Introduction

Interpretation:

The standard entropy change has to be calculated.

  ${\text{2C}}_2{H}_2\left(\text{g}\right)\text{+}5O_{\text{2}}\left(\text{g}\right)\to 4{\text{CO}}_2\left(g\right)\text{+}2{\text{H}}_{2}O\left(\text{l}\right)$

Answer to Problem 17.53QE

The value of ${\text{ΔS}}^{\text{o}}$ for the process is $-432.47kJ/mol.$

Explanation of Solution

The balanced chemical equation as follows,

  ${\text{2C}}_2{H}_2\left(\text{g}\right)\text{+}5O_{\text{2}}\left(\text{g}\right)\to 4{\text{CO}}_2\left(g\right)\text{+}2{\text{H}}_{2}O\left(\text{l}\right)$

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

  $\begin{array}{l}{\text{2C}}_2{H}_2\left(\text{g}\right)\text{+}5O_{\text{2}}\left(\text{g}\right)\to 4{\text{CO}}_2\left(g\right)\text{+}2{\text{H}}_{2}O\left(\text{l}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)200.83205.03213.6369.91\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(2\text{×}69.91\text{J/mol}\text{.K}\right)+\left(4\text{×}\text{313}\text{.63}\text{J/mol}\text{.K}\right)\text{-}\left(2\text{×}200.83\text{J/mol}\text{.K}\right)-\left(5\text{×}205.03\text{J/mol}\text{.K}\right)\\ \\ \text{=}\left(139.82\text{J/mol}\text{.K}\right)+\left(854.52\text{J/mol}\text{.K}\right)\text{-}\left(401.66\text{J/mol}\text{.K}\right)-\left(1025.15\text{J/mol}\text{.K}\right)\\ \\ =-432.47kJ/mol.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $-432.47kJ/mol.$

(d)

Interpretation Introduction

Interpretation:

The standard entropy change has to be calculated.

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

Answer to Problem 17.53QE

The value of ${\text{ΔS}}^{\text{o}}$ for the process is $178.61kJ/mol.$

Explanation of Solution

The balanced chemical equation as follows,

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

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

  $\begin{array}{l}\text{2C}\left(s\right)\text{+}{O}_{\text{2}}\left(\text{g}\right)\to 2\text{CO}\left(g\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)5.74205.03197.56\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(2\text{×}197.56\text{J/mol}\text{.K}\right)-\left(2\text{×}\text{5}\text{.74}\text{J/mol}\text{.K}\right)-\left(1\text{×}205.03\text{J/mol}\text{.K}\right)\\ \\ \text{=}\left(395.12\text{J/mol}\text{.K}\right)-\left(11.48\text{J/mol}\text{.K}\right)-\left(205.03\text{J/mol}\text{.K}\right)\\ \\ =178.61kJ/mol.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $178.61kJ/mol.$