Chapter 17, Problem 17.64QE

(a)

Interpretation Introduction

Interpretation:

For the given reaction, the value of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}$ have to be calculated; the direction of spontaneous is whether consistent with the sign of the enthalpy change, the entropy change or both has to be stated.

  ${\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{+}\text{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to \text{4HCl}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)$

Explanation of Solution

The given reaction as follows,

  ${\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{+}\text{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to \text{4HCl}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)$

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

  $\begin{array}{l}2{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{+}\text{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to \text{4HCl}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)-241.820-92.310\end{array}$

  $\begin{array}{c}{\text{ΔH}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(4\text{×}-92.31\text{kJ/mol}\text{.K}\right)-\left(2\text{×}-241.82\text{kJ/mol}\text{.K}\right)\\ \\ =114.4kJ/mol.\end{array}$

Therefore, value of ${\text{ΔH}}^{\text{o}}$ for the process is $114.4kJ/mol.$

The value of $\Delta {\text{S}}_{rxn}^{\circ }$ is calculated as follows,

  $\begin{array}{l}{\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{+}\text{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to \text{4HCl}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔS}}_{rxn}^{\text{o}}\left(\text{J/mol}\text{.K}\right)188.72222.96186.80205.03\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(4\text{×}186.80\text{J/mol}\text{.K}\right)+\left(1\text{×}205.03\text{J/mol}\text{.K}\right)-\left(2\text{×}188.72\text{J/mol}\text{.K}\right)-\left(2\text{×}222.96\text{J/mol}\text{.K}\right)\\ \\ =128.87J/mol.K.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $0.129kJ/mol.K.$

The value of the Gibbs free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ is calculated as follows,

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}-{\text{TΔS}}_{\text{rxn}}^{\text{o}}\\ \\ =114.4\text{kJ}-\left(298\text{K}\right)\left(0.129\frac{\text{kJ}}{\text{mol}\text{.K}}\right)\\ \\ =75.96kJ.\end{array}$

Therefore, value of ${\text{ΔG}}^{\text{o}}$ for the process is $75.96kJ.$

From the obtained values the sign of enthalpy is positive, entropy is positive and the Gibbs free energy is positive. For the reaction to be spontaneous, here both ${\text{ΔS}}^{\text{o}}$ and ${\text{ΔH}}^{\text{o}}$ are negative, then the sign of both ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}.$ are consistent with the direction of spontaneous.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, the value of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}$ have to be calculated; the direction of spontaneous is whether consistent with the sign of the enthalpy change, the entropy change or both has to be stated.

  ${\text{2CO}}_2\left(\text{g}\right)+4{\text{H}}_{\text{2}}O\left(l\right)\to 2{\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)+3O_2\left(g\right)$

Explanation of Solution

The given reaction as follows,

  ${\text{2CO}}_2\left(\text{g}\right)+4{\text{H}}_{\text{2}}O\left(l\right)\to 2{\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)+3O_2\left(g\right)$

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

  $\begin{array}{l}{\text{2CO}}_2\left(\text{g}\right)+4{\text{H}}_{\text{2}}O\left(l\right)\to 2{\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)+3O_2\left(g\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)-393.51-285.83-238.660\end{array}$

  $\begin{array}{c}{\text{ΔH}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(2\text{×}-238.66\text{J/mol}\text{.K}\right)-\left(\text{2×}-393.51\text{J/mol}\text{.K}\right)-\left(4\text{×}-285.83\text{J/mol}\text{.K}\right)\\ \\ =1453.02kJ/mol.\end{array}$

Therefore, value of ${\text{ΔH}}^{\text{o}}$ for the process is $1453.02kJ/mol.$

The value of $\Delta {\text{S}}_{rxn}^{\circ }$ is calculated as follows,

  $\begin{array}{l}{\text{2CO}}_2\left(\text{g}\right)+4{\text{H}}_{\text{2}}O\left(l\right)\to 2{\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)+3O_2\left(g\right)\\ {\text{ΔS}}_{rxn}^{\text{o}}\left(\text{J/mol}\text{.K}\right)213.6369.91126.8205.03\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(2\text{×}126.8\text{J/mol}\text{.K}\right)+\left(3\text{×}205.03\text{J/mol}\text{.K}\right)-\left(\text{2}\text{×}213.63\text{J/mol}\text{.K}\right)-\left(4\text{×}69.91\text{J/mol}\text{.K}\right)\\ \\ =0.162kJ/mol.K.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $0.162kJ/mol.K.$

The value of the Gibbs free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ is calculated as follows,

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}-{\text{TΔS}}_{\text{rxn}}^{\text{o}}\\ \\ =1453.02\text{kJ}-\left(298\text{K}\right)\left(0.162\frac{\text{kJ}}{\text{mol}\text{.K}}\right)\\ \\ =1404.74kJ.\end{array}$

Therefore, value of ${\text{ΔG}}^{\text{o}}$ for the process is $1404.74kJ.$

From the obtained values the sign of enthalpy is positive, entropy is positive and the Gibbs free energy is positive. For the reaction to be spontaneous, here both ${\text{ΔS}}^{\text{o}}$ and ${\text{ΔH}}^{\text{o}}$ are negative, then the sign of both ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}.$ are consistent with the direction of spontaneous.