Chapter 18, Problem 34PS

Using values of Δ_fH° and S°, calculate the standard molar free energy of formation, Δ_fG°, for each of the following:

1. (a) Ca(OH)₂(s)
2. (b) Cl(g)
3. (c) Na₂CO₃(s)

Compare your calculated values of Δ_fG° with those listed in Appendix L. Which of these formation reactions are predicted to be product-favored at equilibrium at 25 °C?

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{ΔG}}^{\text{o}}$. It is related to entropy and enthalpy by the following expression,

  ${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}$

The sign of ${\text{ΔG}}^{\text{o}}$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 34PS

The standard molar energy of formation for $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)$ is $-898.486kJ\text{/}mol\text{-}rxn$.

Explanation of Solution

The standard molar energy of formation for $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$\begin{array}{cccccccc}& \text{Ca}\left(\text{s}\right)& \text{+}& {\text{H}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)& \to & \text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{0}& & \text{0}& & \text{-986}\text{.09}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{41}\text{.59}& & \text{130}\text{.7}& & \text{205}\text{.07}& & \text{83}\text{.39}\end{array}$

The enthalpy change is expressed as,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ \text{= }\left[\begin{array}{l}\left(\text{1 mol Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\right]\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{H}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left(\text{1 mol Ca}\left(\text{s}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[\text{Ca}\left(\text{s}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{1 mol Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{-986}\text{.09 kJ/mol}\right)\\ \left[\begin{array}{l}\left({\text{1 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{1 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{=-986}\text{.09 kJ/mol-rxn}\end{array}$

The entropy change is expressed as,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ \text{= }\left[\begin{array}{l}\left(\text{1 mol Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\right]\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{H}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left(\text{1 mol Ca}\left(\text{s}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[\text{Ca}\left(\text{s}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{1 mol Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{83}\text{.39 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{130}\text{.7 J/K×mol}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{205}\text{.07 J/K×mol}\right)\text{+}\\ \left(\text{1 mol Ca}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{41}\text{.59 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{=-293}\text{.97 J/K×mol-rxn}\end{array}$

Now,

${\text{ΔG}}^{\text{o}}{\text{= ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}$

Substitute the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ and ${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}$.

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= - 986}\text{.09 kJ/mol-}\left[\left(\text{298 K}\right)\left(\text{-293}\text{.97 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= - 898}\text{.486 kJ/mol-rxn}\end{array}$

The value in Appendix L is $-898.43\text{ kJ/mol-rxn}$.

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $\text{Cl}\left(\text{g}\right)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{ΔG}}^{\text{o}}$. It is related to entropy and enthalpy by the following expression,

  ${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}$

The sign of ${\text{ΔG}}^{\text{o}}$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 34PS

The standard molar energy of formation for $\text{Cl}\left(\text{g}\right)$ is $105.3kJ\text{/}mol\text{-}rxn$.

Explanation of Solution

The standard molar energy of formation for $\text{Cl}\left(\text{g}\right)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$\begin{array}{cccc}& \frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)& \to & \text{Cl}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{121}\text{.3}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{223}\text{.08}& & \text{165}\text{.19}\end{array}$

The enthalpy change is expressed as,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ \text{= }\left[\begin{array}{l}\left(\text{1 mol Cl}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[\text{Cl}\left(\text{g}\right)\right]\text{-}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{Cl}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}$

Substituting the respective values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{1 mol Cl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{121}\text{.3 kJ/mol}\right)\text{-}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\\ \text{= 121}\text{.3 kJ/mol-rxn}\end{array}$

The entropy change is expressed as,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left(\text{1 mol Cl}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[\text{Cl}\left(\text{g}\right)\right]\text{-}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{Cl}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}$

Substituting the respective values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{1 mol Cl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{165}\text{.19 J/K×mol}\right)\text{-}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{223}\text{.08 J/K×mol}\right)\end{array}\right]\\ \text{= 53}\text{.65 J/K×mol-rxn}\end{array}$

Now,

${\text{Δ}}_{\text{f}}{\text{G}}^{\text{o}}{\text{= Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}$

Substitute the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ and ${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}$.

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= 121}\text{.3 kJ/mol-}\left[\left(\text{298 K}\right)\left(\text{53}\text{.65 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= 105}\text{.31 kJ/mol-rxn}\end{array}$

The value in Appendix L is $105.3\text{ kJ/mol-rxn}$.

The value of free energy change is positive. Thus, the reaction is reactant-favored at equilibrium.

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{ΔG}}^{\text{o}}$. It is related to entropy and enthalpy by the following expression,

  ${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}$

The sign of ${\text{ΔG}}^{\text{o}}$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 34PS

The standard molar energy of formation for ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)$ is $-1047.08kJ\text{/}mol\text{-}rxn$.

Explanation of Solution

The standard molar energy of formation for ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$\begin{array}{cccccccc}& \text{2Na}\left(\text{s}\right)& +& \text{C}\left(\text{s}\right)& +& \frac{3}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)& \to & {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\\ {\Delta }_f{H}^{\text{°}}\left(\text{kJ/mol}\right)& 0& & 0& & 0& & -1130.77\\ S^{\circ }\left(\text{J/K}\cdot \text{mol}\right)& 51.21& & 5.6& & 205.07& & 134.79\end{array}$

The enthalpy change is expressed as,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left({\text{1 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right]\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol C}\left(\text{s}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[\text{ C}\left(\text{s}\right)\right]\text{+}\\ \left(\text{1}{\text{.5 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left(\text{2 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[\text{Na}\left(\text{s}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{1 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{-1130}\text{.77 kJ/mol}\right)\\ \left[\begin{array}{l}\left(\text{1 mol C}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{1}{\text{.5 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{2 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{=-1130}\text{.77 kJ/mol-rxn}\end{array}$

The entropy change is expressed as,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left({\text{1 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right]\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol C}\left(\text{s}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[\text{ C}\left(\text{s}\right)\right]\text{+}\\ \left(\text{1}{\text{.5 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left(\text{2 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[\text{Na}\left(\text{s}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{S}^{\text{°}}=\left[\begin{array}{l}\left({\text{1 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{134}\text{.79 J/K}\cdot \text{mol}\right)-\\ \left[\begin{array}{l}\left(\text{1 mol C}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{5}\text{.6 J/K}\cdot \text{mol}\right)+\\ \left(\text{1}{\text{.5 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{205}\text{.07 J/K}\cdot \text{mol}\right)+\\ \left(\text{2 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{51}\text{.21 J/K}\cdot \text{mol}\right)\end{array}\right]\end{array}\right]\\ \text{=}-\text{280}\text{.83 J/K}\cdot \text{mol-rxn}\end{array}$

Now,

${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}$

Substituting the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ and ${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}$.

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= -1130}\text{.77 kJ/mol-}\left[\left(\text{298 K}\right)\left(\text{-280}\text{.83 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= -1047}\text{.08 kJ/mol-rxn}\end{array}$

The value in Appendix L is $-1047.08\text{ kJ/mol-rxn}$.

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.