Chapter 18, Problem 18.22QP

(a)

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Explanation of Solution

To explain: The change in standard free energy for the given reaction

Given reaction

${\text{A}}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{B}}_{\left(\text{g}\right)}$

The relationship between ${\text{ΔG}}^{\text{o}}\text{and}\text{K}$ is given by

${\text{ΔG}}^{\text{o}}\text{=}\text{-RTlnK}$

Above equation shows that, if the temperature remains constant then the free energy change will remains constant.

(b)

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Explanation of Solution

To explain: The change in free energy for the given reaction

Given reaction

${\text{A}}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{B}}_{\left(\text{g}\right)}$

The relationship between $\text{ΔG}\text{and}\text{Q}$ is given by

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{o}}\text{+}\text{RTlnQ}\\ \text{where,}\\ \text{ΔG}\text{-}\text{free}\text{energy}\text{change}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change}\\ \text{R}\text{-}\text{Gas}\text{constant}\\ \text{T}\text{-}\text{temperature}\text{and}\text{Q}\text{-}\text{reaction}\text{quotient}\text{.}\end{array}$

Above equation shows that, if the value of reaction quotient changes then the value of free energy also change.

(c)

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Explanation of Solution

To explain: The spontaneity of the given reaction

Given reaction and information

$\begin{array}{l}{\text{A}}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{B}}_{\left(\text{g}\right)}\\ \text{positive}\text{value}\text{of}\text{K}\text{is}\text{very}\text{large}\text{.}\end{array}$

Since, the value of equilibrium constant is very large, the given reaction is spontaneous.

(d)

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Explanation of Solution

To check: The given statement

Given statement

${\text{ΔG}}^{\text{o}}\text{=}\text{ΔG}\text{=}\text{0}$

The given statement is not correct.  At equilibrium the value of free energy is zero, but the value of standard free energy change is constant.  Hence, the given statement is not true.

(e)

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Explanation of Solution

To check: The composition of the mixture at equilibrium

Given reaction

${\text{A}}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{B}}_{\left(\text{g}\right)}$

The value of equilibrium constant is very large, so at equilibrium the composition will mostly the product side $\left(\text{i}\text{.e}\text{.}{\text{B}}_{\left(\text{g}\right)}\right)$.

(f)

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Explanation of Solution

To give: The values of ${\text{ΔG}}^{\text{o}}\text{,}\text{ΔG}\text{and}\text{K}$ and also the composition of the mixture at equilibrium, spontaneity of the reverse reaction

Given reaction

${\text{B}}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{A}}_{\left(\text{g}\right)}$

For the reverse reaction, ${\text{K}}_{\text{r}}\text{=}{\text{1/K}}_{\text{f}}$ so the value of standard free energy change is same but opposite sign and will be a constant.

At equilibrium, the value of free energy change is zero.  The reaction mixture will mostly reactant side $\left(\text{i}\text{.e}\text{.}{\text{B}}_{\left(\text{g}\right)}\right)\text{.}$

The given reverse reaction is non-spontaneous.