Chapter 18, Problem 39PS

Determine whether the reactions listed below are entropy-favored or disfavored under standard conditions. Predict how an increase in temperature will affect the value of Δ_rG°.

1. (a) N₂(g) + 2 O₂(g) → 2 NO₂(g)
2. (b) 2 C(s) + O₂(g) → 2 CO(g)
3. (c) CaO(s) + CO₂(g) → CaCO₃(s)
4. (d) 2 NaCl(s) → 2 Na(s) + Cl₂(g)

Interpretation Introduction

Interpretation:

It should be determined that whether the given reaction is entropy favorable and should be identified that how increase in temperature will affect the value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$. 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.

Entropy for any reaction is expressed as,

  ${\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)$

A reaction is said to be entropy-favored if the value of entropy change for reaction is positive.

Answer to Problem 39PS

The formation of ${\text{NO}}_{\text{2}}\left(\text{g}\right)$ is entropy unfavourable. There is no temperature at which reaction will become product-favoured at equilibrium.

Explanation of Solution

The value of ${\text{ΔG}}^{\text{o}}$, ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$ is calculated below.

Given:

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

$\begin{array}{cccccc}& {\text{N}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{2O}}_{\text{2}}\left(\text{g}\right)& \to & {\text{2NO}}_{\text{2}}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{0}& & \text{33}\text{.1}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{191}\text{.56}& & \text{205}\text{.07}& & \text{240}\text{.04}\end{array}$

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left({\text{2 mol NO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{NO}}_{\text{2}}\left(\text{g}\right)\right]\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{N}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left({\text{2 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the respective values

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{2 mol NO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{33}\text{.1 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left({\text{2 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{=66}\text{.2 kJ/mol-rxn}\end{array}$

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left({\text{2 mol NO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{NO}}_{\text{2}}\left(\text{g}\right)\right]\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{N}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left({\text{2 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the respective values

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{2 mol NO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{240}\text{.04 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{191}\text{.56 J/K×mol}\right)\text{+}\\ \left({\text{2 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{205}\text{.07 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{= -121}\text{.62 J/K×mol-rxn}\end{array}$

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

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

  $\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= 66}\text{.2 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{-121}\text{.62 J/K×mol- rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= 102}\text{.5 kJ/mol-rxn}\end{array}$

The formation of ${\text{NO}}_{\text{2}}$ is entropy unfavourable as the value of entropy change is negative.

The Table 18.1 was referred,

Both the enthalpy and entropy change are unfavorable. There is no temperature at which the reaction will become product-favored at equilibrium. It is a Type $4$ reaction and as the temperature is increased, the reaction becomes more reactant-favored.

Interpretation Introduction

Interpretation:

It should be determined that whether the given reaction is entropy favorable and should be identified that how increase in temperature will affect the value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$. 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.

Entropy for any reaction is expressed as,

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)$

A reaction is said to be entropy-favoured if the value of entropy change for reaction is positive.

Answer to Problem 39PS

The formation of $\text{CO}\left(\text{g}\right)$ is entropy favourable. As temperature increases the reaction will become more product-favoured at equilibrium.

Explanation of Solution

The value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$, ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$ is calculated below.

Given:

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

$\begin{array}{cccccc}& \text{2C}\left(\text{s}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)& \to & \text{2CO}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{0}& & \text{-110}\text{.525}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{5}\text{.6}& & \text{205}\text{.07}& & \text{197}\text{.674}\end{array}$

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

Substituting the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{2 mol CO}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{-110}\text{.525 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{2 mol C}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{= -221}\text{.05 kJ/mol-rxn}\end{array}$

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

Substituting the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left(\text{2 mol CO}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{197}\text{.674 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{2 mol C}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{5}\text{.6 J/K×mol}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{205}\text{.07 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{=179}\text{.1 J/K×mol-rxn}\end{array}$

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= -221}\text{.05 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{179}\text{.1 J/K×mol- rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= -274}\text{.45 kJ/mol-rxn}\end{array}$

The formation of $\text{CO}$ is entropy favourable as the value of entropy change is positive.

The Table 18.1 was referred,

Both the enthalpy and entropy change are favorable. As temperature increases the reaction will become more product-favored at equilibrium. It is a Type $1$ reaction and as the temperature is increased, the reaction becomes more product-favored.

Interpretation Introduction

Interpretation:

It should be determined that whether the given reaction is entropy favorable and should be identified that how increase in temperature will affect the value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$. 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.

Entropy for any reaction is expressed as,

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)$

A reaction is said to be entropy-favoured if the value of entropy change for reaction is positive.

Answer to Problem 39PS

The formation of ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)$ is entropy unfavourable. As temperature increases the reaction will become less product-favoured at equilibrium.

Explanation of Solution

The value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$ is calculated below.

Given:

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

$\begin{array}{cccccc}& \text{CaO}\left(\text{s}\right)& \text{+}& {\text{CO}}_{\text{2}}\left(\text{g}\right)& \to & {\text{CaCO}}_{\text{3}}\left(\text{s}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{-635}\text{.09}& & \text{-393}\text{.509}& & \text{-1207}\text{.6}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{38}\text{.2}& & \text{213}\text{.74}& & \text{91}\text{.7}\end{array}$

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left({\text{1mol CaCO}}_3\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{CaCO}}_3\left(\text{s}\right)\right]-\\ \left[\begin{array}{l}\left(\text{1 mol CaO}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[\text{CaO}\left(\text{s}\right)\right]+\\ \left({\text{1 mol CO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{CO}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\left[\begin{array}{l}\left({\text{1mol CaCO}}_3\left(\text{s}\right)\text{/mol-rxn}\right)\left(-\text{1207}\text{.6 kJ/mol}\right)-\\ \left[\begin{array}{l}\left(\text{1 mol CaO}\left(\text{s}\right)\text{/mol-rxn}\right)\left(-\text{635}\text{.09 kJ/mol}\right)\\ \left({\text{1 mol CO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-\text{393}\text{.509 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ =-179\text{ kJ/mol-rxn}\end{array}$

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

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{1 mol CaCO}}_{\text{3}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{91}\text{.7 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol CaO}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{38}\text{.2 J/K×mol}\right)\text{+}\\ \left({\text{1 mol CO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{213}\text{.74 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{=-160}\text{.2 J/K×mol-rxn}\end{array}$

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{=-179 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{-160}\text{.2 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{=-131}\text{.23 kJ/mol-rxn}\end{array}$

The formation of ${\text{CaCO}}_3$ is entropy unfavourable as the value of entropy change is negative. The reaction is enthalpy favored. As the temperature increases, the reaction thus becomes less product-favored.

Interpretation Introduction

Interpretation:

It should be determined that whether the given reaction is entropy favorable and should be identified that how increase in temperature will affect the value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$. 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.

Entropy for any reaction is expressed as,

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)$

A reaction is said to be entropy-favoured if the value of entropy change for reaction is positive.

Answer to Problem 39PS

The decomposition of $\text{NaCl}\left(\text{s}\right)$ is entropy favourable. As temperature increases the reaction will become more product-favored at equilibrium.

Explanation of Solution

The value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$ is calculated below.

Given:

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

$\begin{array}{cccccc}& \text{2NaCl}\left(\text{s}\right)& \to & \text{2Na}\left(\text{s}\right)& \text{+}& {\text{Cl}}_{\text{2}}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{-411}\text{.12}& & \text{0}& & \text{0}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{72}\text{.11}& & \text{51}\text{.21}& & \text{223}\text{.08}\end{array}$

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

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{2 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{-0 kJ/mol}\right)\text{+}\\ \left({\text{1mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{-0 kJ/mol}\right)\end{array}\right]\text{-}\\ \left(\text{2 mol NaCl}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{-411}\text{.12 kJ/mol}\right)\end{array}\right]\\ \text{=822}\text{.2 kJ/mol-rxn}\end{array}$

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

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{2 mol Na}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{51}\text{.21 J/K×mol}\right)\text{+}\\ \left({\text{1mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{223}\text{.08 J/K×mol}\right)\end{array}\right]\text{-}\\ \left(\text{2 mol NaCl}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{72}\text{.11 J/K×mol}\right)\end{array}\right]\\ \text{=181}\text{.28 J/K×mol-rxn}\end{array}$

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= 822}\text{.2 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{181}\text{.28 J/K×mol- rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= 768}\text{.19 kJ/mol-rxn}\end{array}$

The decomposition of $\text{NaCl}$ is entropy favourable as the value of entropy change is positive. The reaction is enthalpy unfavored. As the temperature increases, the reaction thus becomes more product-favored.