Chapter 18, Problem 40PS

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) I₂(g) → 2 I(g)
2. (b) 2 SO₂(g) + O₂(g) → 2 SO₃(g)
3. (c) SiCl₄(g) + 2 H₂O(ℓ) → SiO₂(s) + 4 HCl(g)
4. (d) P₄(s, white) + 6 H₂(g) → 4 PH₃(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 ${\text{ΔG}}^{\text{o}}$. 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.

Entropy for any reaction is expressed as,

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

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

Answer to Problem 40PS

The formation of $\text{I}\left(\text{g}\right)$ is entropy favourable, the reaction will become product-favoured at higher temperatures.

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 the values of standard entropies and enthalpies.

$\begin{array}{cccc}& {\text{I}}_{\text{2}}\left(\text{g}\right)& \to & \text{2I}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{62}\text{.438}& & \text{106}\text{.838}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{260}\text{.69}& & \text{180}\text{.791}\end{array}$

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

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left(\text{2 mol I}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{106}\text{.838 kJ/mol}\right)\text{-}\\ \left({\text{1 mol I}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{62}\text{.438 kJ/mol}\right)\end{array}\right]\\ \text{=151}\text{.2 kJ/mol-rxn}\end{array}$

Also,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left(\text{2 mol I}\left(\text{g}\right)\text{/mol-rxn}\right)S^{°}\left[\text{I}\left(\text{g}\right)\right]-\\ \left({\text{1 mol I}}_2\left(\text{g}\right)\text{/mol-rxn}\right)S^{°}\left[{\text{I}}_2\left(\text{g}\right)\right]\end{array}\right]\end{array}$

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left(\text{2 mol I}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{180}\text{.791 J/K×mol}\right)\text{-}\\ \left({\text{1 mol I}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{260}\text{.69 J/K×mol}\right)\end{array}\right]\\ \text{=100}\text{.892 J/K×mol-rxn}\end{array}$

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{=151}\text{.2 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{100}\text{.892 J/K×mol- rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{=121}\text{.14 kJ/mol-rxn}\end{array}$

The formation of $\text{I}\left(\text{g}\right)$ is entropy favourable as the value of entropy change is positive.

The reaction will become product-favored at higher temperature.

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 ${\text{ΔG}}^{\text{o}}$. 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.

Entropy for any reaction is expressed as,

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

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

Answer to Problem 40PS

The formation of ${\text{SO}}_3\left(\text{g}\right)$ is entropy unfavourable as the value of entropy change is negative. The reaction is spontaneous at lower temperatures.

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 the values of standard entropies and enthalpies.

$\begin{array}{cccccc}& {\text{2SO}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)& \to & {\text{2SO}}_{\text{3}}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{-296}\text{.84}& & \text{0}& & \text{-395}\text{.77}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{248}\text{.21}& & \text{205}\text{.07}& & \text{256}\text{.77}\end{array}$

The standard enthalpy change is expressed as,

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

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{2 mol SO}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{-395}\text{.77 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{2 mol SO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{-296}\text{.84 kJ/mol}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{= -197}\text{.86 kJ/mol-rxn}\end{array}$

Also,

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

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{2 mol SO}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{256}\text{.77 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{2 mol SO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{248}\text{.21 J/K×mol}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{205}\text{.07 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{=-187}\text{.95 J/K×mol-rxn}\end{array}$

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= -197}\text{.86 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{-187}\text{.95 J/K×mol- rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= -141}\text{.86 kJ/mol-rxn}\end{array}$

The formation of ${\text{SO}}_3\left(\text{g}\right)$ is entropy unfavourable as the value of entropy change is negative. The reaction is spontaneous at lower temperatures.

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 ${\text{ΔG}}^{\text{o}}$. 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.

Entropy for any reaction is expressed as,

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

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

Answer to Problem 40PS

The reaction of ${\text{SiCl}}_4\left(\text{g}\right)$ and water is entropy favorable as the value of entropy change is negative. The reaction is spontaneous at all temperatures.

Explanation of Solution

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

$\begin{array}{cccccccc}& {\text{SiCl}}_{\text{4}}\left(\text{g}\right)& \text{+}& {\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)& \to & {\text{SiO}}_{\text{2}}\left(\text{s}\right)& \text{+}& \text{4HCl}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{-662}\text{.75}& & \text{-285}\text{.83}& & \text{-910}\text{.86}& & \text{-92}\text{.31}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{330}\text{.86}& & \text{69}\text{.95}& & \text{41}\text{.46}& & \text{186}\text{.2}\end{array}$

The standard enthalpy change is expressed as,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{1 mol SiO}}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{SiO}}_{\text{2}}\left(\text{s}\right)\right]\text{+}\\ \left(\text{4 mol HCl}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[\text{HCl}\left(\text{g}\right)\right]\end{array}\right]\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol SiCl}}_{\text{4}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{SiCl}}_{\text{4}}\left(\text{g}\right)\right]\text{+}\\ \left({\text{2 mol H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{1 mol SiO}}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{-910}\text{.86 kJ/mol}\right)\text{+}\\ \left(\text{4 mol HCl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{-92}\text{.31 kJ/mol}\right)\end{array}\right]\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol SiCl}}_{\text{4}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{-662}\text{.75 kJ/mol}\right)\text{+}\\ \left({\text{2 mol H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{/mol-rxn}\right)\left(\text{-285}\text{.83 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{=-45}\text{.7 kJ/mol-rxn}\end{array}$

Also,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{1 mol SiO}}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right)S^{°}\left[{\text{SiO}}_{\text{2}}\left(\text{s}\right)\right]+\\ \left(\text{4 mol HCl}\left(\text{g}\right)\text{/mol-rxn}\right)S^{°}\left[\text{HCl}\left(\text{g}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{1 mol SiCl}}_4\left(\text{g}\right)\text{/mol-rxn}\right)S^{°}\left[{\text{SiCl}}_4\left(\text{g}\right)\right]+\\ \left({\text{2 mol H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{/mol-rxn}\right)S^{°}\left[{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{1 mol SiO}}_{\text{2}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{41}\text{.46 J/K×mol}\right)\text{+}\\ \left(\text{4 mol HCl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{186}\text{.2 J/K×mol}\right)\end{array}\right]\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol SiCl}}_{\text{4}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{330}\text{.86 J/K×mol}\right)\text{+}\\ \left({\text{2 mol H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{/mol-rxn}\right)\left(\text{69}\text{.95 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{=315}\text{.5 J/K×mol-rxn}\end{array}$

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{=-45}\text{.7 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{315}\text{.5 J/K×mol- rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{=-139}\text{.7 kJ/mol-rxn}\end{array}$

The reaction of ${\text{SiCl}}_4\left(\text{g}\right)$ and water is entropy favorable as the value of entropy change is negative. The reaction is spontaneous at all temperatures.

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 ${\text{ΔG}}^{\text{o}}$. 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.

Entropy for any reaction is expressed as,

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

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

Answer to Problem 40PS

The reaction of ${\text{P}}_4\left(\text{s}\right)$ and hydrogen is entropy favorable as the value of entropy change is negative. The reaction is spontaneous at higher temperatures.

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.

$\begin{array}{cccccc}& {\text{P}}_{\text{4}}\left(\text{s}\right)& \text{+}& {\text{6H}}_{\text{2}}\left(\text{g}\right)& \to & {\text{4PH}}_{\text{3}}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{0}& & \text{5}\text{.47}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{41}\text{.1}& & \text{130}\text{.7}& & \text{210}\text{.24}\end{array}$

The standard enthalpy change is expressed as,

 $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left({\text{4 mol PH}}_3\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{PH}}_3\left(\text{g}\right)\right]-\\ \left[\begin{array}{l}\left({\text{6 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{H}}_{\text{2}}\left(\text{g}\right)\right]+\\ \left({\text{1 mol P}}_{\text{4}}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[{\text{P}}_{\text{4}}\left(\text{s}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{4 mol PH}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{5}\text{.47 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{6 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left({\text{1 mol P}}_{\text{4}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{=21}\text{.88 kJ/mol-rxn}\end{array}$

Also,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left({\text{4 mol PH}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{PH}}_{\text{3}}\left(\text{g}\right)\right]\text{-}\\ \left[\begin{array}{l}\left({\text{6 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{H}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left({\text{1 mol P}}_{\text{4}}\left(\text{s}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{P}}_{\text{4}}\left(\text{s}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{4 mol PH}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{210}\text{.24 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{6 mol H}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{130}\text{.7 J/K×mol}\right)\text{+}\\ \left({\text{1 mol P}}_{\text{4}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{41}\text{.1 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{=15}\text{.66 J/K×mol-rxn}\end{array}$

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= 21}\text{.88 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{15}\text{.66 J/K×mol- rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= 17}\text{.21 kJ/mol-rxn}\end{array}$

The reaction of ${\text{P}}_4\left(\text{s}\right)$ and hydrogen is entropy favorable as the value of entropy change is negative. The reaction is spontaneous at higher temperatures.