Chapter 17, Problem 17.65QE

(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.

  ${\text{CH}}_3\text{COOH}\left(\text{l}\right)\text{+}\text{NaOH}\left(\text{s}\right)\to {\text{Na}}^{+}\left(\text{aq}\right)\text{+}{\text{CH}}_3{\text{COO}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Explanation of Solution

The given reaction as follows,

  ${\text{CH}}_3\text{COOH}\left(\text{l}\right)\text{+}\text{NaOH}\left(\text{s}\right)\to {\text{Na}}^{+}\left(\text{aq}\right)\text{+}{\text{CH}}_3{\text{COO}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

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

  $\begin{array}{l}{\text{CH}}_3\text{COOH}\left(\text{l}\right)\text{+}\text{NaOH}\left(\text{s}\right)\to {\text{Na}}^{+}\left(\text{aq}\right)\text{+}{\text{CH}}_3{\text{COO}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)-484.5-425.6-240.12-486.01-285.83\end{array}$

  $\begin{array}{c}{\text{ΔH}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-240.12\text{kJ/mol}\text{.K}\right)+\left(1\text{×}-486.01\text{kJ/mol}\text{.K}\right)+\left(1\text{×}-285.83\text{kJ/mol}\text{.K}\right)\\ -\left(1\text{×}-484.5\text{kJ/mol}\text{.K}\right)-\left(1\text{×}-425.6\text{kJ/mol}\text{.K}\right)\\ \\ =-101.86kJ/mol.\end{array}$

Therefore, value of ${\text{ΔH}}^{\text{o}}$ for the process is $-101.86kJ/mol.$

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

  $\begin{array}{l}{\text{CH}}_3\text{COOH}\left(\text{l}\right)\text{+}\text{NaOH}\left(\text{s}\right)\to {\text{Na}}^{+}\left(\text{aq}\right)\text{+}{\text{CH}}_3{\text{COO}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ {\text{ΔS}}_{rxn}^{\text{o}}\left(\text{J/mol}\text{.K}\right)159.864.4659.086.669.91\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}59\text{J/mol}\text{.K}\right)+\left(1\text{×}86.6\text{J/mol}\text{.K}\right)+\left(1\text{×}69.91\text{J/mol}\text{.K}\right)\\ -\left(1\text{×}159.8\text{J/mol}\text{.K}\right)-\left(1\text{×}64.46\text{J/mol}\text{.K}\right)\\ \\ =-8.8J/mol.K.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $-8.8J/mol.K.$

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

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}-{\text{TΔS}}_{\text{rxn}}^{\text{o}}\\ \\ =-101.86\text{kJ/mol}-\left(298\text{K}\right)\left(-8.8\times {\text{10}}^{-3}\frac{\text{kJ}}{\text{mol}\text{.K}}\right)\\ \\ =-99.24kJ.\end{array}$

Therefore, value of ${\text{ΔG}}^{\text{o}}$ for the process is $-99.24kJ.$

From the obtained values the sign of enthalpy is negative, entropy is negative and the Gibbs free energy is negative. For the reaction to be spontaneous, here both ${\text{ΔS}}^{\text{o}}$ and ${\text{ΔH}}^{\text{o}}$ are negative, then the sign of ${\text{ΔH}}^{\text{o}}$ is consistent but not the sign of ${\text{ΔS}}^{\text{o}}.$

(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.

  ${\text{AgNO}}_{\text{3}}\left(\text{s}\right)\text{+}{\text{Cl}}^{-}\left(\text{aq}\right)\to \text{AgCl}\left(\text{s}\right)\text{+}{\text{NO}}_{\text{3}}^{-}\left(\text{aq}\right)$

Explanation of Solution

The given reaction as follows,

  ${\text{AgNO}}_{\text{3}}\left(\text{s}\right)\text{+}{\text{Cl}}^{-}\left(\text{aq}\right)\to \text{AgCl}\left(\text{s}\right)\text{+}{\text{NO}}_{\text{3}}^{-}\left(\text{aq}\right)$

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

  $\begin{array}{l}{\text{AgNO}}_{\text{3}}\left(\text{s}\right)\text{+}{\text{Cl}}^{-}\left(\text{aq}\right)\to \text{AgCl}\left(\text{s}\right)\text{+}{\text{NO}}_{\text{3}}^{-}\left(\text{aq}\right)\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)-124.39-167.16-127.07-205.0\end{array}$

  $\begin{array}{c}{\text{ΔH}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-127.07\text{J/mol}\text{.K}\right)+\left(\text{1×}-205.0\text{J/mol}\text{.K}\right)\\ -\left(1\text{×}-124.39\text{J/mol}\text{.K}\right)-\left(1\text{×}-167.16\text{J/mol}\text{.K}\right)\\ \\ =-40.52kJ/mol.\end{array}$

Therefore, value of ${\text{ΔH}}^{\text{o}}$ for the process is $-40.52kJ/mol.$

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

  $\begin{array}{l}{\text{AgNO}}_{\text{3}}\left(\text{s}\right)\text{+}{\text{Cl}}^{-}\left(\text{aq}\right)\to \text{AgCl}\left(\text{s}\right)\text{+}{\text{NO}}_{\text{3}}^{-}\left(\text{aq}\right)\\ {\text{ΔS}}_{rxn}^{\text{o}}\left(\text{J/mol}\text{.K}\right)140.9256.596.2146.4\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}96.2\text{J/mol}\text{.K}\right)+\left(1\text{×}146.4\text{J/mol}\text{.K}\right)-\left(1\text{×}140.92\text{J/mol}\text{.K}\right)-\left(1\text{×}56.5\text{J/mol}\text{.K}\right)\\ \\ =0.0452kJ/mol.K.\end{array}$

Therefore, value of ${\text{ΔS}}^{\text{o}}$ for the process is $0.0452kJ/mol.K.$

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

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}-{\text{TΔS}}_{\text{rxn}}^{\text{o}}\\ \\ =-40.52\text{kJ}-\left(298\text{K}\right)\left(\text{0}\text{.0452}\frac{\text{kJ}}{\text{mol}\text{.K}}\right)\\ \\ =-54.0kJ.\end{array}$

Therefore, value of ${\text{ΔG}}^{\text{o}}$ for the process is $-54.0kJ.$

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