Chapter 18, Problem 17PS

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

1. (a) CS₂(g)
2. (b) NaOH(s)
3. (c) ICl(g)

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 ${\text{CS}}_{\text{2}}\left(\text{g}\right)$ 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 entropy by the following expression,

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

Here, ${\text{ΔH}}^{\text{o}}$ is the change in enthalpy and ${\text{ΔS}}^{\text{o}}$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for ${\text{CS}}_{\text{2}}\left(\text{g}\right)$ is $+66.6kJ\text{/}mol\text{-}rxn$.

The value of $\Delta G^{\circ }$ is positive, hence the reaction is reactant-favored at equilibrium.

Explanation of Solution

The standard molar free energy of formation for ${\text{CS}}_{\text{2}}\left(\text{g}\right)$ is calculated below.

Given:

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

$\begin{array}{cccccc}& \text{C}\left(\text{s}\right)& \text{+}& \text{2S}\left(\text{s}\right)& \to & {\text{CS}}_{\text{2}}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{0}& & \text{+116}\text{.7}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{+5}\text{.6}& & \text{+32}\text{.1}& & \text{+237}\text{.8}\end{array}$

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

Substituting the respective values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\left[\begin{array}{l}\left({\text{1 mol CS}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{116}\text{.7 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol C}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{2 mol S}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{= 116}\text{.7 kJ/mol-rxn}\end{array}$

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

Substituting the respective values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\left[\begin{array}{l}\left({\text{1 mol CS}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{237}\text{.8 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol C}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{5}\text{.6 J/K×mol}\right)\text{+}\\ \left(\text{2 mol S}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{32}\text{.1 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{= 168 J/K×mol-rxn}\end{array}$

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

Substituting the value of ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$.

  $\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= 116}\text{.7 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{168 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= 66}\text{.6 kJ/mol-rxn}\end{array}$

The value listed in Appendix L is $66.61\text{ kJ/mol-rxn}$. Thus both the calculated value and the value in Appendix L are in agreement with each other.

The value of $\Delta G^{\circ }$ is positive, hence the reaction is reactant-favored at equilibrium.

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $\text{NaOH}\left(\text{s}\right)$ 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 entropy by the following expression,

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

Here, ${\text{ΔH}}^{\text{o}}$ is the change in enthalpy and ${\text{ΔS}}^{\text{o}}$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $\text{NaOH}\left(\text{s}\right)$ is $-379.82kJ\text{/}mol\text{-}rxn$.

The value of $\Delta G^{\circ }$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$\begin{array}{cccccccc}& \text{Na}\left(\text{s}\right)& \text{+}& \frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\left(\text{g}\right)& \text{+}& \frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\left(\text{g}\right)& \to & \text{NaOH}\left(\text{s}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{0}& & \text{0}& & \text{- 425}\text{.93}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{51}\text{.21}& & \text{130}\text{.7}& & \text{205}\text{.07}& & \text{64}\text{.46}\end{array}$

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

Substituting the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{1 mol NaOH}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{-425}\text{.93 kJ/mol}\right)\\ \left[\begin{array}{l}\left(\text{0}{\text{.5 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{0}{\text{.5 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{1 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{= -425}\text{.93 kJ/mol-rxn}\end{array}$

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

Substituting the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{1 mol NaOH}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{64}\text{.46 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{0}{\text{.5 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{130}\text{.7 J/K×mol}\right)\text{+}\\ \left(\text{0}{\text{.5 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{205}\text{.07 J/K×mol}\right)\text{+}\\ \left(\text{1 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{51}\text{.21 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{= -154}\text{.6 J/K×mol-rxn}\end{array}$

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

Substitute the value of ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$.

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= - 425}\text{.93 kJ/mol-rxn- }\left[\left(\text{298K}\right)\left(\text{-154}\text{.6 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= - 379}\text{.82 kJ/mol-rxn}\end{array}$

The value listed in Appendix L is $-\text{379}\text{.75 kJ/mol-rxn}$. Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of ${\text{ΔG}}^{\text{o}}$ is negative, hence the reaction is product-favored at equilibrium.

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $\text{ICl}\left(\text{g}\right)$ 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 entropy by the following expression,

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

Here, ${\text{ΔH}}^{\text{o}}$ is the change in enthalpy and ${\text{ΔS}}^{\text{o}}$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $\text{ICl}\left(\text{g}\right)$ is $-5.73kJ\text{/}mol\text{-}rxn$.

The value of $\Delta G^{\circ }$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$\begin{array}{cccccc}& \frac{\text{1}}{\text{2}}{\text{I}}_{\text{2}}\left(\text{s}\right)& \text{+}& \frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)& \to & \text{ICl}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{0}& & \text{+17}\text{.51}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{+116}\text{.135}& & \text{+223}\text{.08}& & \text{+247}\text{.56}\end{array}$

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

Substituting the respective values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{1 mol ICl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{17}\text{.51 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{0}{\text{.5 mol I}}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{= 17}\text{.51 kJ/mol-rxn}\end{array}$

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

Substituting the respective values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{1 mol ICl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{247}\text{.56 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{0}{\text{.5 mol I}}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{116}\text{.135 J/K×mol}\right)\text{+}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{223}\text{.08 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{= 77}\text{.95 J/K×mol-rxn}\end{array}$

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

Substituting the value of ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$.

  $\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= 17}\text{.51 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{77}\text{.95 J/K×mol- rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= -5}\text{.73 kJ/mol-rxn}\end{array}$

The value listed in Appendix L is $-5.73\text{ kJ/mol-rxn}$. Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $\Delta G^{\circ }$ is negative, hence the reaction is product-favored at equilibrium.