Chapter 22, Problem 22.52P

(a)

Interpretation Introduction

Interpretation:

The ${\text{ΔG}}^{\text{°}}$ has to be calculated at ${\text{25}}^{\text{°}}\text{C}$ for the reaction of ${\text{SO}}_{\text{2}}$ to form ${\text{SO}}_{\text{3}}$ at standard conditions. The reaction is spontaneous or not has to be given.

Concept Introduction:

Standard state free energy of formation (${\text{ΔG}}^{\text{°}}$):

It is the difference between the free energy of a substance and the free energy of its elements in thermodynamically stable states at standard conditions.

It can be calculated by the following expression:

  ${\text{ΔG}}^{\text{°}}\text{=}{\displaystyle \sum _{}^{}{\text{ΔG}}_{{\text{f}}_{\text{products}}}^{\text{°}}\text{-}}{\displaystyle \sum _{}^{}{\text{ΔG}}_{{\text{f}}_{\text{reactants}}}^{\text{°}}}$

If ${\text{ΔG}}^{\text{°}}\text{> 0}$, the reaction is non-spontaneous.

If ${\text{ΔG}}^{\text{°}}\text{< 0}$, the reaction is spontaneous.

Explanation of Solution

The balanced reaction of formation of ${\text{SO}}_{\text{3}}$ from ${\text{SO}}_{\text{2}}$ is given as

  ${\text{2SO}}_{\text{2(g)}}{\text{+ O}}_{\text{2(g)}}\to {\text{2SO}}_{\text{3(g)}}$

The ${\text{ΔG}}^{\text{°}}$ can be calculated as

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}\text{=}{\displaystyle \sum _{}^{}{\text{ΔG}}_{{\text{f}}_{\text{products}}}^{\text{°}}\text{-}}{\displaystyle \sum _{}^{}{\text{ΔG}}_{{\text{f}}_{\text{reactants}}}^{\text{°}}}\\ \\ {\text{ΔG}}_{\text{r×n}}^{\text{°}}{\text{= [(2 mol SO}}_{\text{3}}{\text{)(ΔG}}_{\text{f}}^{\text{°}}\text{of}{\text{SO}}_{\text{3}}\text{)]-[(2}\text{mol}\text{of}{\text{SO}}_{\text{2}}{\text{)(ΔG}}_{\text{f}}^{\text{°}}\text{of}{\text{SO}}_{\text{2}}\text{) + (1}\text{mol}{\text{O}}_{\text{2}}{\text{)(ΔG}}_{\text{f}}^{\text{°}}\text{of}{\text{O}}_{\text{2}}\text{)]}\\ \\ {\text{ΔG}}_{\text{r×n}}^{\text{°}}\text{= [(2mol)(-371}\text{kJ/mol)]-[(2}\text{mol)(-300}\text{.2}\text{kJ/mol)+(1}\text{mol}{\text{O}}_{\text{2}}\text{)(0}\text{kJ/mol)]}\\ \\ {\text{ΔG}}_{\text{r×n}}^{\text{°}}\text{= -141}\text{.6 = -142}\text{kJ}\end{array}$

As the ${\text{ΔG}}^{\text{°}}$ is negative (${\text{ΔG}}^{\text{°}}\text{< 0}$), the reaction is spontaneous.

(b)

Interpretation Introduction

Interpretation:

Consider the reaction of ${\text{SO}}_{\text{2}}$ to form ${\text{SO}}_{\text{3}}$ at standard conditions.

The reason has to be given for the reaction not performed at ${\text{25}}^{\text{°}}\text{C}$.

Explanation of Solution

At ${\text{25}}^{\text{°}}\text{C}$, the rate of the reaction is very slow. So, the formation of ${\text{SO}}_{\text{3}}$ is not significant at this temperature.

(c)

Interpretation Introduction

Interpretation:

Consider the reaction of ${\text{SO}}_{\text{2}}$ to form ${\text{SO}}_{\text{3}}$ at standard conditions.

The reaction is spontaneous or not at ${\text{500}}^{\text{°}}\text{C}$ has to be given (assume that ${\text{ΔH}}^{\text{°}}\text{and}{\text{ΔS}}^{\text{°}}$ are constant with changing $\text{T}$).

Concept Introduction:

${\text{ΔH}}^{\text{°}}$ can be calculated by the expression

  ${\text{ΔH}}^{\text{°}}\text{=}{\displaystyle \sum _{}^{}{\text{nH}}_{{\text{f}}_{\text{products}}}^{\text{°}}\text{-}}{\displaystyle \sum _{}^{}{\text{mH}}_{{\text{f}}_{\text{reactants}}}^{\text{°}}}$

${\text{ΔS}}^{\text{°}}$ can be calculated by the expression

  ${\text{ΔS}}^{\text{°}}\text{=}{\displaystyle \sum _{}^{}{\text{nS}}_{{}_{\text{products}}}^{\text{°}}\text{-}}{\displaystyle \sum _{}^{}{\text{mS}}_{{}_{\text{reactants}}}^{\text{°}}}$

From the values of ${\text{ΔH}}^{\text{°}}\text{and}{\text{ΔS}}^{\text{°}}$, ${\text{ΔG}}^{\text{°}}$ can be calculated by the expression

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

If ${\text{ΔG}}^{\text{°}}\text{> 0}$, the reaction is non-spontaneous.

If ${\text{ΔG}}^{\text{°}}\text{< 0}$, the reaction is spontaneous.

Explanation of Solution

The balanced reaction of formation of ${\text{SO}}_{\text{3}}$ from ${\text{SO}}_{\text{2}}$ is given as

  ${\text{2SO}}_{\text{2(g)}}{\text{+ O}}_{\text{2(g)}}\to {\text{2SO}}_{\text{3(g)}}$

The ${\text{ΔH}}^{\text{°}}$ can be calculated as

  $\begin{array}{l}{\text{ΔH}}^{\text{°}}\text{=}{\displaystyle \sum _{}^{}{\text{nΔH}}_{{\text{f}}_{\text{products}}}^{\text{°}}\text{-}}{\displaystyle \sum _{}^{}{\text{mΔH}}_{{\text{f}}_{\text{reactants}}}^{\text{°}}}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}{\text{= [(2 mol SO}}_{\text{3}}{\text{)(ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{SO}}_{\text{3}}\text{)]-[(2}\text{mol}\text{of}{\text{SO}}_{\text{2}}{\text{)(ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{SO}}_{\text{2}}\text{)+(1}\text{mol}{\text{O}}_{\text{2}}{\text{)(ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{O}}_{\text{2}}\text{)]}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= [(2 mol)(-396}\text{kJ/mol)]-[(2}\text{mol)(-296}\text{.8}\text{kJ/mol)+(1}\text{mol)(0}\text{kJ/mol)]}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= -198}\text{.4 = -198}\text{kJ}\end{array}$

The ${\text{ΔS}}^{\text{°}}$ can be calculated as

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}\text{=}{\displaystyle \sum _{}^{}{\text{mS}}_{{}_{\text{products}}}^{\text{°}}\text{-}}{\displaystyle \sum _{}^{}{\text{nS}}_{{}_{\text{reactants}}}^{\text{°}}}\\ \\ {\text{ΔS}}_{\text{r×n}}^{\text{°}}{\text{= [(2 mol SO}}_{\text{3}}{\text{)(S}}_{}^{\text{°}}\text{of}{\text{SO}}_{\text{3}}\text{)]-[(2}\text{mol}\text{of}{\text{SO}}_{\text{2}}{\text{)(S}}_{}^{\text{°}}\text{of}{\text{SO}}_{\text{2}}\text{)+(1}\text{mol}{\text{O}}_{\text{2}}{\text{)(S}}_{}^{\text{°}}\text{of}{\text{O}}_{\text{2}}\text{)]}\\ \\ {\text{ΔS}}_{\text{r×n}}^{\text{°}}\text{= [(2 mol)(256}\text{.66}\text{J/mol}\text{.K)]-[(2}\text{mol)(248}\text{.1}\text{J/mol}\text{.K)+(1}\text{mol)(205}\text{.0}\text{J/mol}\text{.K)]}\\ \\ {\text{ΔS}}_{\text{r×n}}^{\text{°}}\text{= -187}\text{.88 = -187}\text{.9}\text{J/K}\end{array}$

From the above values of ${\text{ΔH}}^{\text{°}}\text{and}{\text{ΔS}}^{\text{°}}$, ${\text{ΔG}}^{\text{°}}$ can be calculated as

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{\text{= ΔH}}^{\text{°}}{\text{- TΔS}}^{\text{°}}\\ \\ {\text{ΔG}}^{\text{°}}\text{= -198}\text{.4}\text{kJ- ((273+500)K)(-187}\text{.88}\text{J/K)(1}{\text{kJ/10}}^{\text{3}}\text{J)}\\ \\ {\text{ΔG}}^{\text{°}}\text{= -53}\text{.16876}\text{=}\text{-53}\text{kJ}\end{array}$

As the ${\text{ΔG}}^{\text{°}}$ is negative (${\text{ΔG}}^{\text{°}}\text{< 0}$), the reaction is spontaneous at ${\text{500}}^{\text{°}}\text{C}$.

(d)

Interpretation Introduction

Interpretation:

Consider the reaction of ${\text{SO}}_{\text{2}}$ to form ${\text{SO}}_{\text{3}}$ at standard conditions.

The $\text{K}$ values has to be compared at ${\text{500}}^{\text{°}}\text{C}$ and at ${\text{25}}^{\text{°}}\text{C}$.

Concept Introduction:

The value of $\text{K}$ can be calculated by the expression of free energy and equilibrium constant.

  $\begin{array}{l}{\text{ΔG = ΔG}}^{\text{°}}\text{+ RT ln K}\\ \\ \text{where,}\text{ΔG = free}\text{energy}\\ \\ {\text{ΔG}}^{\text{°}}\text{=standard-state}\text{free}\text{energy}\\ \\ \text{R = ideal}\text{gas}\text{constant}\\ \\ \text{T = temperature}\end{array}$

If a reaction is at equilibrium, $\text{ΔG=0}$.

  $\begin{array}{l}{\text{0 = ΔG}}^{\text{°}}\text{+ RT ln K}\\ \\ {\text{ΔG}}^{\text{°}}\text{ = -RT ln K}\end{array}$

Explanation of Solution

Equilibrium constant, $\text{K}$ at ${\text{25}}^{\text{°}}\text{C}$:

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}\text{ = -RT ln K}\\ \\ \text{ln}\text{K = }\frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\\ \\ \text{ln}\text{K}\text{=}\frac{\text{(-142}\text{kJ)}\left(\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\right)}{\text{(8}\text{.314}\text{J/molgK)((273+25)K)}}\\ \\ \text{ln}\text{K}\text{=}\text{57}\text{.31417694}\end{array}$

The value of $\text{K}$ is

  ${\text{K}}_{\text{25}}{\text{= e}}^{\text{57}\text{.31417694}}\text{=}\text{7}{\text{.8×10}}^{24}$

Equilibrium constant, $\text{K}$ at ${\text{500}}^{\text{°}}\text{C}$:

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}\text{ = -RT ln K}\\ \\ \text{ln}\text{K = }\frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\\ \\ \text{ln}\text{K}\text{=}\frac{\text{(-53}\text{kJ)}\left(\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\right)}{\text{(8}\text{.314}\text{J/molgK)((273+500)K)}}\\ \\ \text{ln}\text{K}\text{=}\text{8}\text{.246817}\end{array}$

The value of $\text{K}$ is

  ${\text{K}}_{\text{500}}{\text{= e}}^{\text{8}\text{.246817}}\text{=}\text{3}{\text{.8×10}}^{\text{3}}$

(e)

Interpretation Introduction

Interpretation:

Consider the reaction of ${\text{SO}}_{\text{2}}$ to form ${\text{SO}}_{\text{3}}$ at standard conditions.

The highest $\text{T}$ at which the reaction is spontaneous has to be given.

Concept Introduction:

${\text{ΔH}}^{\text{°}}$ can be calculated by the expression

  ${\text{ΔH}}^{\text{°}}\text{=}{\displaystyle \sum _{}^{}{\text{nH}}_{{\text{f}}_{\text{products}}}^{\text{°}}\text{-}}{\displaystyle \sum _{}^{}{\text{mH}}_{{\text{f}}_{\text{reactants}}}^{\text{°}}}$

${\text{ΔS}}^{\text{°}}$ can be calculated by the expression

  ${\text{ΔS}}^{\text{°}}\text{=}{\displaystyle \sum _{}^{}{\text{nS}}_{{}_{\text{products}}}^{\text{°}}\text{-}}{\displaystyle \sum _{}^{}{\text{mS}}_{{}_{\text{reactants}}}^{\text{°}}}$

From the values of ${\text{ΔH}}^{\text{°}}\text{and}{\text{ΔS}}^{\text{°}}$, ${\text{ΔG}}^{\text{°}}$ can be calculated by the expression

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

Explanation of Solution

The balanced reaction of formation of ${\text{SO}}_{\text{3}}$ from ${\text{SO}}_{\text{2}}$ is given as

  ${\text{2SO}}_{\text{2(g)}}{\text{+ O}}_{\text{2(g)}}\to {\text{2SO}}_{\text{3(g)}}$

  $\begin{array}{l}{\text{ΔG}}_{\text{r×n}}^{\text{°}}{\text{= 0 = ΔH}}_{\text{r×n}}^{\text{°}}{\text{-TΔS}}_{\text{r×n}}^{\text{°}}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}{\text{= TΔS}}_{\text{r×n}}^{\text{°}}\\ \\ \text{T = }\frac{{\text{ΔH}}_{}^{\text{°}}}{{\text{ΔS}}_{}^{\text{°}}}\\ \\ \text{T = }\frac{\text{-198}\text{kJ}}{\text{-187}\text{.9}\text{J/K}\left(\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)}\\ \\ \text{T = 1}{\text{.05×10}}^{\text{3}}\text{K}\end{array}$