Chapter 17, Problem 17.76QE

(a)

Interpretation Introduction

Interpretation:

The values of ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ has to be calculate for given two different temperatures.

  ${\text{BaCO}}_3\left(\text{s}\right)\to \text{BaO}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)$

Answer to Problem 17.76QE

The calculated Standard Gibbs free energy at $\text{300}\text{K}$ is $189.24kJ$ and the calculated Standard Gibbs free energy at $\text{390K}$ is $173.78kJ.$

Explanation of Solution

The given reaction as follows,

  ${\text{BaCO}}_3\left(\text{s}\right)\to \text{BaO}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)$

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

  $\begin{array}{l}{\text{BaCO}}_3\left(\text{s}\right)\to \text{BaO}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔH}}_{rxn}^{\text{o}}\left(\text{kJ/mol}\right)-1216.3-582.0-393.51\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-393.51\text{kJ/mol}\right)+\left(1\text{×}-582.0\text{kJ/mol}\right)-\left(1\text{×}-1216.3\text{kJ/mol}\right)\\ \\ =240.79kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $240.79kJ/mol.$

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

  $\begin{array}{l}{\text{BaCO}}_3\left(\text{s}\right)\to \text{BaO}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)112.170.3213.63\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}213.63\text{J/mol}\text{.K}\right)+\left(1\text{×}70.3\text{J/mol}\text{.K}\right)-\left(1\text{×}112.1\text{J/mol}\text{.K}\right)\\ \\ =171.83J/mol.K.\end{array}$

Hence, value of $\Delta {\text{S}}^{\circ }$ for the process is $171.83J/mol.K.$

Using the equation as follows, the Standard Gibbs free energy can be determined as,

  $\begin{array}{c}At,T=300K,\\ \\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\text{o}}\\ \\ \text{=}240.79\text{kJ}-\left(300K\right)\left(171.83\frac{\text{J}}{\text{K}}\text{×}\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \\ \text{=}189.24kJ.\end{array}$

Therefore, the calculated Standard Gibbs free energy at $\text{300}\text{K}$ is $189.24kJ.$

  $\begin{array}{c}At,T=390K,\\ \\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\text{o}}\\ \\ \text{=}240.79\text{kJ}-\left(390K\right)\left(171.83\frac{\text{J}}{\text{K}}\text{×}\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \\ \text{=}173.78kJ.\end{array}$

Therefore, the calculated Standard Gibbs free energy at $\text{390K}$ is $173.78kJ.$

(b)

Interpretation Introduction

Interpretation:

The values of $\Delta G_{rxn}^{\circ }$ has to be calculate for given two different temperatures.

  ${\text{CH}}_{\text{3}}\text{COOH}\left(\text{l}\right)\to {\text{CH}}_{\text{4}}\left(\text{g}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)$

Answer to Problem 17.76QE

The calculated Standard Gibbs free energy at $\text{300}\text{K}$ is $-55.8kJ$ and the calculated Standard Gibbs free energy at $\text{390K}$ is $-77.4kJ.$

Explanation of Solution

The given reaction as follows,

  ${\text{CH}}_{\text{3}}\text{COOH}\left(\text{l}\right)\to {\text{CH}}_{\text{4}}\left(\text{g}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)$

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

  $\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}\left(\text{l}\right)\to {\text{CH}}_{\text{4}}\left(\text{g}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔH}}_{rxn}^{\text{o}}\left(\text{kJ/mol}\right)-484.5-74.81-393.51\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-74.81\text{kJ/mol}\right)+\left(1\text{×}-393.51\text{kJ/mol}\right)-\left(1\text{×}-484.5\text{kJ/mol}\right)\\ \\ =16.18kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $16.18kJ/mol.$

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

  $\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}\left(\text{l}\right)\to {\text{CH}}_{\text{4}}\left(\text{g}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)159.8186.15213.63\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}186.15\text{J/mol}\text{.K}\right)+\left(1\text{×}213.63\text{J/mol}\text{.K}\right)-\left(1\text{×}159.8\text{J/mol}\text{.K}\right)\\ \\ =239.98J/mol.K.\end{array}$

Hence, value of $\Delta {\text{S}}^{\circ }$ for the process is $239.98J/mol.K.$

Using the equation as follows, the Standard Gibbs free energy can be determined as,

  $\begin{array}{c}\text{At}\text{,}\text{T=}\text{300K,}\\ \\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\text{o}}\\ \\ \text{=}16.18\text{kJ}-\left(300K\right)\left(239.98\frac{\text{J}}{\text{K}}\text{×}\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \\ \text{=}-55.8kJ.\end{array}$

Therefore, the calculated Standard Gibbs free energy at $\text{300}\text{K}$ is $-55.8kJ.$

  $\begin{array}{c}\text{At}\text{,}\text{T=}\text{390K,}\\ \\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\text{o}}\\ \\ \text{=}16.18\text{kJ}-\left(390K\right)\left(239.98\frac{\text{J}}{\text{K}}\text{×}\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \\ \text{=}-77.4kJ.\end{array}$

Therefore, the calculated Standard Gibbs free energy at $\text{390K}$ is $-77.4kJ.$