Chapter 17, Problem 17.66QE

(a)

Interpretation Introduction

Interpretation:

The value of ${\text{ΔG}}_{\text{f}}^{\text{o}}$ has to be calculated for Cadmium (II) chloride (s).

Explanation of Solution

The given reaction as follows,

  $\text{Cd}\left(\text{s}\right)\text{+}{\text{Cl}}_2\left(\text{g}\right)\to {\text{CdCl}}_{\text{2}}\left(\text{s}\right)$

• The value of $\Delta {\text{H}}_{reaction}^{\circ }$ is calculated by standard entropy change as follows,

$\begin{array}{l}\text{Cd}\left(\text{s}\right)\text{+}{\text{Cl}}_2\left(\text{g}\right)\to {\text{CdCl}}_{\text{2}}\left(\text{s}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)0.00-391.5\end{array}$

$\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-391.5\text{kJ/mol}\right)\\ \\ =-391.5kJ/mol.\end{array}$

Therefore, value of ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ for the process is $-391.5kJ/mol.$

• The value of $\Delta {\text{S}}_f^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}\text{Cd}\left(\text{s}\right)\text{+}{\text{Cl}}_2\left(\text{g}\right)\to {\text{CdCl}}_{\text{2}}\left(\text{s}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)0.05180.22300.1153\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}0.1153\text{kJ/mol}\text{.K}\right)-\left(1\text{×}0.0518\text{kJ/mol}\text{.K}\right)-\left(1\text{×}0.2230\text{kJ/mol}\text{.K}\right)\\ \\ =-0.1595kJ/mol.K.\end{array}$

Therefore, value of $\Delta {\text{G}}_f^{\circ }$ for the process is $-0.1595kJ/mol.K.$

• The value of $\Delta {\text{G}}_f^{\circ }$ is calculated by Gibbs-Helmholtz equation as follows,

  $\begin{array}{c}{\text{ΔG}}_{\text{f}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\circ }\\ \\ \text{=}\left(-391.5\text{kJ/mol}\right)-\left[\left(\text{298K}\right)\left(-0.1595\text{kJ/mol}\text{.K}\right)\right]\\ \\ =-343.97kJ/mol.\end{array}$

Therefore, value of $\Delta {\text{G}}_f^{\circ }$ for the process is $-343.97kJ/mol.$

(b)

Interpretation Introduction

Interpretation:

The value of ${\text{ΔG}}_{\text{f}}^{\text{o}}$ has to be calculated for Methyl alcohol.

Explanation of Solution

The given reaction as follows,

  $\text{C}\left(\text{s}\right)\text{+}2{\text{H}}_2\left(\text{g}\right)\text{+}\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_3\text{OH}\left(\text{l}\right)$

• The value of $\Delta {\text{H}}_{reaction}^{\circ }$ is calculated by standard entropy change as follows,

$\begin{array}{l}\text{C}\left(\text{s}\right)\text{+}2{\text{H}}_2\left(\text{g}\right)\text{+}\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_3\text{OH}\left(\text{s}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)0.000-238.66\end{array}$

$\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-238.66\text{kJ/mol}\right)\\ \\ =-238.66kJ/mol.\end{array}$

Therefore, value of ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ for the process is $-238.66kJ/mol.$

• The value of $\Delta {\text{S}}_f^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}\text{C}\left(\text{s}\right)\text{+}2{\text{H}}_2\left(\text{g}\right)\text{+}\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_3\text{OH}\left(\text{l}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)0.00570.13060.20500.1268\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}0.1268\text{kJ/mol}\text{.K}\right)-\left(1\text{×}0.0057\text{kJ/mol}\text{.K}\right)-\left(2\text{×}0.1306\text{kJ/mol}\text{.K}\right)\\ -\left(\frac{1}{2}\text{×}0.2050\text{kJ/mol}\text{.K}\right)\\ \\ =-0.2426kJ/mol.\end{array}$

Therefore, value of $\Delta {\text{S}}_f^{\circ }$ for the process is $-0.2426kJ/mol.$

• The value of $\Delta {\text{G}}_f^{\circ }$ is calculated by Gibbs-Helmholtz equation as follows,

  $\begin{array}{c}{\text{ΔG}}_{\text{f}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\circ }\\ \\ \text{=}\left(-238.7\text{kJ/mol}\right)-\left[\left(\text{298K}\right)\left(-\text{0}\text{.2426}\text{kJ/mol}\right)\right]\\ \\ =-166.4kJ/mol.\end{array}$

Therefore, value of $\Delta {\text{G}}_f^{\circ }$ for the process is $-166.4kJ/mol.$

(c)

Interpretation Introduction

Interpretation:

The value of ${\text{ΔG}}_{\text{f}}^{\text{o}}$ has to be calculated for Copper (I) Sulfide(s).

Explanation of Solution

The given reaction as follows,

  $\text{Cu}\left(\text{s}\right)\text{+}\text{S}\left(\text{g}\right)\to \text{CuS}\left(\text{s}\right)$

• The value of $\Delta {\text{H}}_{reaction}^{\circ }$ is calculated by standard entropy change as follows,

$\begin{array}{l}\text{Cu}\left(\text{s}\right)\text{+}\text{S}\left(\text{g}\right)\to \text{CuS}\left(\text{s}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)0.00-53.1\end{array}$

$\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-53.1\text{kJ/mol}\right)\\ \\ =-53.1kJ/mol.\end{array}$

Therefore, value of ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ for the process is $-53.1kJ/mol.$

• The value of $\Delta {\text{S}}_{rxn}^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}\text{Cu}\left(\text{s}\right)\text{+}\text{S}\left(\text{g}\right)\to \text{CuS}\left(\text{s}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)0.03320.03180.0665\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}0.0665\text{kJ/mol}\text{.K}\right)-\left(1\text{×}0.0332\text{kJ/mol}\text{.K}\right)-\left(1\text{×}0.0318\text{kJ/mol}\text{.K}\right)\\ \\ =-0.0015kJ/mol.\end{array}$

Therefore, value of $\Delta {\text{S}}_{rxn}^{\circ }$ for the process is $-0.0015kJ/mol.$

• The value of $\Delta {\text{G}}_f^{\circ }$ is calculated by Gibbs-Helmholtz equation as follows,

  $\begin{array}{c}{\text{ΔG}}_{\text{f}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\circ }\\ \\ \text{=}\left(-53.1\text{kJ/mol}\right)-\left[\left(\text{298K}\right)\left(-\text{0}\text{.0015}\text{kJ/mol}\right)\right]\\ \\ =-53.55kJ/mol.\end{array}$

Therefore, value of $\Delta {\text{G}}_f^{\circ }$ for the process is $-53.55kJ/mol.$