Chapter 18, Problem 84SCQ

(a)

Interpretation Introduction

Interpretation:

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ for reaction between $\text{NO}\left(\text{g}\right)$ and ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ should be determined.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

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

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}$ is the enthalpy change for the reaction.

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,

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

${\text{ΔG}}^{\text{o}}$ is also related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

Answer to Problem 84SCQ

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ for the reaction is $-60\text{.}5J\text{/}K\cdot mol\text{-}rxn$.

Explanation of Solution

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ for the reaction is calculated below.

Given:

Refer to Appendix L for the values of standard entropies.

The standard entropy of $\text{NOCl}\left(\text{g}\right)$ is $\text{261}\text{.8 J/K}\cdot \text{mol}$.

The standard entropy of $\text{NO}\left(\text{g}\right)$ is $\text{210}\text{.76 J/K}\cdot \text{mol}$.

The standard entropy of ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ is $\text{223}\text{.08 J/K}\cdot \text{mol}$.

The balanced chemical equation is:

  $\text{NO}\left(\text{g}\right)+\frac{1}{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to \text{NOCl}\left(\text{g}\right)$

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

  $\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{system}\right)={\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left(\text{1 mol NOCl}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[\text{NOCl}\left(\text{g}\right)\right]\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[\text{NO}\left(\text{g}\right)\right]\text{+}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{Cl}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the values,

  $\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{system}\right)\text{= }\left[\begin{array}{l}\left(\text{1 mol NOCl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{261}\text{.8 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{210}\text{.76 J/K×mol}\right)\text{+}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{223}\text{.08 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ {\text{ΔS}}^{\text{o}}\text{= -60}\text{.5 J/K×mol-rxn}\end{array}$

(b)

Interpretation Introduction

Interpretation:

It should be identified that ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ changes with temperature or not.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

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

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}$ is the enthalpy change for the reaction.

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,

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

${\text{ΔG}}^{\text{o}}$ is also related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

Answer to Problem 84SCQ

The value of ${\text{ΔS}}^{\text{o}}\left(\text{system}\right)$ changes with temperature.

Explanation of Solution

The entropy of the system is dependent upon temperature. Entropy of any system increases with increase in the temperature due to the heat which is added to the system at higher temperatures.

(c)

Interpretation Introduction

Interpretation:

It should be identified that ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ changes with temperature.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

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

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}$ is the enthalpy change for the reaction.

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,

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

${\text{ΔG}}^{\text{o}}$ is also related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

Answer to Problem 84SCQ

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ for any reaction changes with temperature.

Explanation of Solution

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{= }\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ is the enthalpy change for the reaction.

Thus, ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ is inversely related to temperature.

(d)

Interpretation Introduction

Interpretation:

It should be identified that ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ changes with increase in temperature or not.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

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

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}$ is the enthalpy change for the reaction.

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,

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

${\text{ΔG}}^{\text{o}}$ is also related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

Answer to Problem 84SCQ

The value of ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ changes with an increase in temperature.

Explanation of Solution

The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

Both ${\text{ΔS}}^{\text{o}}\left(\text{system}\right)$ and ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ are dependent upon temperatures. Thus, the value of ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ changes with an increase in temperature.

(e)

Interpretation Introduction

Interpretation:

It should be identified that does exothermic reaction will always results in positive ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ value.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

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

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}$ is the enthalpy change for the reaction.

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,

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

${\text{ΔG}}^{\text{o}}$ is also related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

Answer to Problem 84SCQ

The exothermic reaction does not necessarily leads to a positive value of ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$.

Explanation of Solution

The exothermic reaction have negative value of free energy change which means that the ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ will be positive. However, if the value of ${\text{ΔS}}^{\text{o}}\left(\text{system}\right)$ is negative and has greater magnitude than the ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ than the sign of ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ is negative. The reaction in such a case is non-spontaneous.

(f)

Interpretation Introduction

Interpretation:

It should be identified that reaction of $\text{NO}\left(\text{g}\right)$ and ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ is spontaneous or not.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

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

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}$ is the enthalpy change for the reaction.

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,

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

${\text{ΔG}}^{\text{o}}$ is also related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

Answer to Problem 84SCQ

The reaction is spontaneous at $\text{298 K}$.

The reaction is not spontaneous at $\text{700 K}$.

Explanation of Solution

The free energy change for the given reaction is calculated below.

Given:

Refer to Appendix L for the values of standard entropies.

The standard enthalpy of $\text{NOCl}\left(\text{g}\right)$ is $\text{51}\text{.71 kJ/mol}$.

The standard enthalpy of $\text{NO}\left(\text{g}\right)$ is $\text{90}\text{.29 kJ/mol}$.

The standard enthalpy of ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The balanced chemical equation is:

  $\text{NO}\left(\text{g}\right)+\frac{1}{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to \text{NOCl}\left(\text{g}\right)$

The ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ can be calculated by the following expression,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left(\text{1 mol NOCl}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[\text{NOCl}\left(\text{g}\right)\right]\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[\text{NO}\left(\text{g}\right)\right]\text{+}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{Cl}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left(\text{1 mol NOCl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{51}\text{.71 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left(\text{1 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{90}\text{.29 kJ/mol}\right)\text{+}\\ \left(\text{0}{\text{.5 mol Cl}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ {\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= -38}\text{.58 J/K×mol-rxn}\end{array}$

Now,

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

Substitute the values at temperature $\text{298 K}$,

  $\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{=-38}\text{.58 kJ/mol-rxn-}\left[\left(\text{298K}\right)\left(\text{-60}\text{.5 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{=-20}\text{.551 kJ/mol-rxn}\end{array}$

Thus, the reaction is spontaneous at $\text{298 K}$.

Substitute the values at temperature $\text{700 K}$,

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= -38}\text{.58 kJ/mol-rxn-}\left[\left(\text{700K}\right)\left(\text{-60}\text{.5 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= +3}\text{.77 kJ/mol-rxn}\end{array}$

Thus, the reaction is not spontaneous at $\text{700 K}$.