The ΔG ° has to be calculated at 25 ° C for the reaction of SO 2 to form SO 3 at standard conditions. The reaction is spontaneous or not has to be given. Concept Introduction: Standard state free energy of formation ( ΔG ° ): 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: ΔG ° = ∑ ΔG f products ° - ∑ ΔG f reactants ° If ΔG ° > 0 , the reaction is non-spontaneous. If ΔG ° < 0 , the reaction is spontaneous.

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Answer 1

Question

Chapter 22, Problem 22.52P

(a)

Interpretation Introduction

Interpretation:

The $ΔG°$ has to be calculated at $25°C$ for the reaction of $SO2$ to form $SO3$ at standard conditions. The reaction is spontaneous or not has to be given.

Concept Introduction:

Standard state free energy of formation ( $ΔG°$ ):

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:

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(a)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔG°$ can be calculated as

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°ΔGr×n°= [(2 mol SO3)(ΔGf° of SO3)]-[(2 mol of SO2)(ΔGf° of SO2) + (1 mol O2)(ΔGf° of O2)]ΔGr×n°= [(2mol)(-371 kJ/mol)]-[(2 mol)(-300.2 kJ/mol)+(1 mol O2)(0 kJ/mol)]ΔGr×n°= -141.6 = -142 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous.

(b)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The reason has to be given for the reaction not performed at $25°C$ .

(b)

Expert Solution

Explanation of Solution

At $25°C$ , the rate of the reaction is very slow. So, the formation of $SO3$ is not significant at this temperature.

(c)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

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

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(c)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔH°$ can be calculated as

$ΔH°=∑nΔHfproducts°-∑mΔHfreactants°ΔHr×n°= [(2 mol SO3)(ΔHf° of SO3)]-[(2 mol of SO2)(ΔHf° of SO2)+(1 mol O2)(ΔHf° of O2)]ΔHr×n°= [(2 mol)(-396 kJ/mol)]-[(2 mol)(-296.8 kJ/mol)+(1 mol)(0 kJ/mol)]ΔHr×n°= -198.4 = -198 kJ$

The $ΔS°$ can be calculated as

$ΔS°=∑mSproducts°-∑nSreactants°ΔSr×n°= [(2 mol SO3)(S° of SO3)]-[(2 mol of SO2)(S° of SO2)+(1 mol O2)(S° of O2)]ΔSr×n°= [(2 mol)(256.66 J/mol.K)]-[(2 mol)(248.1 J/mol.K)+(1 mol)(205.0 J/mol.K)]ΔSr×n°= -187.88 = -187.9 J/K$

From the above values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated as

$ΔG°= ΔH°- TΔS°ΔG°= -198.4 kJ- ((273+500)K)(-187.88 J/K)(1 kJ/103 J)ΔG°= -53.16876 = -53 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous at $500°C$ .

(d)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The $K$ values has to be compared at $500°C$ and at $25°C$ .

Concept Introduction:

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

$ΔG = ΔG°+ RT ln Kwhere, ΔG = free energy ΔG°=standard-state free energy R = ideal gas constant T = temperature$

If a reaction is at equilibrium, $ΔG=0$ .

$0 = ΔG°+ RT ln KΔG° = -RT ln K$

(d)

Expert Solution

Explanation of Solution

Equilibrium constant, $K$ at $25°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-142 kJ)(103J1 kJ)(8.314 J/molgK)((273+25)K)ln K = 57.31417694$

The value of $K$ is

$K25= e57.31417694 = 7.8×1024$

Equilibrium constant, $K$ at $500°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-53 kJ)(103J1 kJ)(8.314 J/molgK)((273+500)K)ln K = 8.246817$

The value of $K$ is

$K500= e8.246817 = 3.8×103$

(e)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The highest $T$ at which the reaction is spontaneous has to be given.

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

(e)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

$ΔGr×n°= 0 = ΔHr×n°-TΔSr×n°ΔHr×n°= TΔSr×n°T = ΔH°ΔS°T = -198 kJ-187.9 J/K(1 kJ103J)T = 1.05×103 K$

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Answer 2

Question

Chapter 22, Problem 22.52P

(a)

Interpretation Introduction

Interpretation:

The $ΔG°$ has to be calculated at $25°C$ for the reaction of $SO2$ to form $SO3$ at standard conditions. The reaction is spontaneous or not has to be given.

Concept Introduction:

Standard state free energy of formation ( $ΔG°$ ):

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:

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(a)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔG°$ can be calculated as

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°ΔGr×n°= [(2 mol SO3)(ΔGf° of SO3)]-[(2 mol of SO2)(ΔGf° of SO2) + (1 mol O2)(ΔGf° of O2)]ΔGr×n°= [(2mol)(-371 kJ/mol)]-[(2 mol)(-300.2 kJ/mol)+(1 mol O2)(0 kJ/mol)]ΔGr×n°= -141.6 = -142 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous.

(b)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The reason has to be given for the reaction not performed at $25°C$ .

(b)

Expert Solution

Explanation of Solution

At $25°C$ , the rate of the reaction is very slow. So, the formation of $SO3$ is not significant at this temperature.

(c)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

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

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(c)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔH°$ can be calculated as

$ΔH°=∑nΔHfproducts°-∑mΔHfreactants°ΔHr×n°= [(2 mol SO3)(ΔHf° of SO3)]-[(2 mol of SO2)(ΔHf° of SO2)+(1 mol O2)(ΔHf° of O2)]ΔHr×n°= [(2 mol)(-396 kJ/mol)]-[(2 mol)(-296.8 kJ/mol)+(1 mol)(0 kJ/mol)]ΔHr×n°= -198.4 = -198 kJ$

The $ΔS°$ can be calculated as

$ΔS°=∑mSproducts°-∑nSreactants°ΔSr×n°= [(2 mol SO3)(S° of SO3)]-[(2 mol of SO2)(S° of SO2)+(1 mol O2)(S° of O2)]ΔSr×n°= [(2 mol)(256.66 J/mol.K)]-[(2 mol)(248.1 J/mol.K)+(1 mol)(205.0 J/mol.K)]ΔSr×n°= -187.88 = -187.9 J/K$

From the above values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated as

$ΔG°= ΔH°- TΔS°ΔG°= -198.4 kJ- ((273+500)K)(-187.88 J/K)(1 kJ/103 J)ΔG°= -53.16876 = -53 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous at $500°C$ .

(d)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The $K$ values has to be compared at $500°C$ and at $25°C$ .

Concept Introduction:

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

$ΔG = ΔG°+ RT ln Kwhere, ΔG = free energy ΔG°=standard-state free energy R = ideal gas constant T = temperature$

If a reaction is at equilibrium, $ΔG=0$ .

$0 = ΔG°+ RT ln KΔG° = -RT ln K$

(d)

Expert Solution

Explanation of Solution

Equilibrium constant, $K$ at $25°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-142 kJ)(103J1 kJ)(8.314 J/molgK)((273+25)K)ln K = 57.31417694$

The value of $K$ is

$K25= e57.31417694 = 7.8×1024$

Equilibrium constant, $K$ at $500°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-53 kJ)(103J1 kJ)(8.314 J/molgK)((273+500)K)ln K = 8.246817$

The value of $K$ is

$K500= e8.246817 = 3.8×103$

(e)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The highest $T$ at which the reaction is spontaneous has to be given.

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

(e)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

$ΔGr×n°= 0 = ΔHr×n°-TΔSr×n°ΔHr×n°= TΔSr×n°T = ΔH°ΔS°T = -198 kJ-187.9 J/K(1 kJ103J)T = 1.05×103 K$

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Answer 3

Question

Chapter 22, Problem 22.52P

(a)

Interpretation Introduction

Interpretation:

The $ΔG°$ has to be calculated at $25°C$ for the reaction of $SO2$ to form $SO3$ at standard conditions. The reaction is spontaneous or not has to be given.

Concept Introduction:

Standard state free energy of formation ( $ΔG°$ ):

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:

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(a)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔG°$ can be calculated as

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°ΔGr×n°= [(2 mol SO3)(ΔGf° of SO3)]-[(2 mol of SO2)(ΔGf° of SO2) + (1 mol O2)(ΔGf° of O2)]ΔGr×n°= [(2mol)(-371 kJ/mol)]-[(2 mol)(-300.2 kJ/mol)+(1 mol O2)(0 kJ/mol)]ΔGr×n°= -141.6 = -142 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous.

(b)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The reason has to be given for the reaction not performed at $25°C$ .

(b)

Expert Solution

Explanation of Solution

At $25°C$ , the rate of the reaction is very slow. So, the formation of $SO3$ is not significant at this temperature.

(c)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

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

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(c)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔH°$ can be calculated as

$ΔH°=∑nΔHfproducts°-∑mΔHfreactants°ΔHr×n°= [(2 mol SO3)(ΔHf° of SO3)]-[(2 mol of SO2)(ΔHf° of SO2)+(1 mol O2)(ΔHf° of O2)]ΔHr×n°= [(2 mol)(-396 kJ/mol)]-[(2 mol)(-296.8 kJ/mol)+(1 mol)(0 kJ/mol)]ΔHr×n°= -198.4 = -198 kJ$

The $ΔS°$ can be calculated as

$ΔS°=∑mSproducts°-∑nSreactants°ΔSr×n°= [(2 mol SO3)(S° of SO3)]-[(2 mol of SO2)(S° of SO2)+(1 mol O2)(S° of O2)]ΔSr×n°= [(2 mol)(256.66 J/mol.K)]-[(2 mol)(248.1 J/mol.K)+(1 mol)(205.0 J/mol.K)]ΔSr×n°= -187.88 = -187.9 J/K$

From the above values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated as

$ΔG°= ΔH°- TΔS°ΔG°= -198.4 kJ- ((273+500)K)(-187.88 J/K)(1 kJ/103 J)ΔG°= -53.16876 = -53 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous at $500°C$ .

(d)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The $K$ values has to be compared at $500°C$ and at $25°C$ .

Concept Introduction:

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

$ΔG = ΔG°+ RT ln Kwhere, ΔG = free energy ΔG°=standard-state free energy R = ideal gas constant T = temperature$

If a reaction is at equilibrium, $ΔG=0$ .

$0 = ΔG°+ RT ln KΔG° = -RT ln K$

(d)

Expert Solution

Explanation of Solution

Equilibrium constant, $K$ at $25°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-142 kJ)(103J1 kJ)(8.314 J/molgK)((273+25)K)ln K = 57.31417694$

The value of $K$ is

$K25= e57.31417694 = 7.8×1024$

Equilibrium constant, $K$ at $500°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-53 kJ)(103J1 kJ)(8.314 J/molgK)((273+500)K)ln K = 8.246817$

The value of $K$ is

$K500= e8.246817 = 3.8×103$

(e)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The highest $T$ at which the reaction is spontaneous has to be given.

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

(e)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

$ΔGr×n°= 0 = ΔHr×n°-TΔSr×n°ΔHr×n°= TΔSr×n°T = ΔH°ΔS°T = -198 kJ-187.9 J/K(1 kJ103J)T = 1.05×103 K$

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Answer 4

Question

Chapter 22, Problem 22.52P

(a)

Interpretation Introduction

Interpretation:

The $ΔG°$ has to be calculated at $25°C$ for the reaction of $SO2$ to form $SO3$ at standard conditions. The reaction is spontaneous or not has to be given.

Concept Introduction:

Standard state free energy of formation ( $ΔG°$ ):

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:

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(a)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔG°$ can be calculated as

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°ΔGr×n°= [(2 mol SO3)(ΔGf° of SO3)]-[(2 mol of SO2)(ΔGf° of SO2) + (1 mol O2)(ΔGf° of O2)]ΔGr×n°= [(2mol)(-371 kJ/mol)]-[(2 mol)(-300.2 kJ/mol)+(1 mol O2)(0 kJ/mol)]ΔGr×n°= -141.6 = -142 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous.

(b)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The reason has to be given for the reaction not performed at $25°C$ .

(b)

Expert Solution

Explanation of Solution

At $25°C$ , the rate of the reaction is very slow. So, the formation of $SO3$ is not significant at this temperature.

(c)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

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

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(c)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔH°$ can be calculated as

$ΔH°=∑nΔHfproducts°-∑mΔHfreactants°ΔHr×n°= [(2 mol SO3)(ΔHf° of SO3)]-[(2 mol of SO2)(ΔHf° of SO2)+(1 mol O2)(ΔHf° of O2)]ΔHr×n°= [(2 mol)(-396 kJ/mol)]-[(2 mol)(-296.8 kJ/mol)+(1 mol)(0 kJ/mol)]ΔHr×n°= -198.4 = -198 kJ$

The $ΔS°$ can be calculated as

$ΔS°=∑mSproducts°-∑nSreactants°ΔSr×n°= [(2 mol SO3)(S° of SO3)]-[(2 mol of SO2)(S° of SO2)+(1 mol O2)(S° of O2)]ΔSr×n°= [(2 mol)(256.66 J/mol.K)]-[(2 mol)(248.1 J/mol.K)+(1 mol)(205.0 J/mol.K)]ΔSr×n°= -187.88 = -187.9 J/K$

From the above values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated as

$ΔG°= ΔH°- TΔS°ΔG°= -198.4 kJ- ((273+500)K)(-187.88 J/K)(1 kJ/103 J)ΔG°= -53.16876 = -53 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous at $500°C$ .

(d)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The $K$ values has to be compared at $500°C$ and at $25°C$ .

Concept Introduction:

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

$ΔG = ΔG°+ RT ln Kwhere, ΔG = free energy ΔG°=standard-state free energy R = ideal gas constant T = temperature$

If a reaction is at equilibrium, $ΔG=0$ .

$0 = ΔG°+ RT ln KΔG° = -RT ln K$

(d)

Expert Solution

Explanation of Solution

Equilibrium constant, $K$ at $25°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-142 kJ)(103J1 kJ)(8.314 J/molgK)((273+25)K)ln K = 57.31417694$

The value of $K$ is

$K25= e57.31417694 = 7.8×1024$

Equilibrium constant, $K$ at $500°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-53 kJ)(103J1 kJ)(8.314 J/molgK)((273+500)K)ln K = 8.246817$

The value of $K$ is

$K500= e8.246817 = 3.8×103$

(e)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The highest $T$ at which the reaction is spontaneous has to be given.

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

(e)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

$ΔGr×n°= 0 = ΔHr×n°-TΔSr×n°ΔHr×n°= TΔSr×n°T = ΔH°ΔS°T = -198 kJ-187.9 J/K(1 kJ103J)T = 1.05×103 K$

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Answer 5

Chapter 22, Problem 22.52P

(a)

Interpretation Introduction

Interpretation:

The $ΔG°$ has to be calculated at $25°C$ for the reaction of $SO2$ to form $SO3$ at standard conditions. The reaction is spontaneous or not has to be given.

Concept Introduction:

Standard state free energy of formation ( $ΔG°$ ):

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:

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(a)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔG°$ can be calculated as

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°ΔGr×n°= [(2 mol SO3)(ΔGf° of SO3)]-[(2 mol of SO2)(ΔGf° of SO2) + (1 mol O2)(ΔGf° of O2)]ΔGr×n°= [(2mol)(-371 kJ/mol)]-[(2 mol)(-300.2 kJ/mol)+(1 mol O2)(0 kJ/mol)]ΔGr×n°= -141.6 = -142 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous.

(b)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The reason has to be given for the reaction not performed at $25°C$ .

(b)

Expert Solution

Explanation of Solution

At $25°C$ , the rate of the reaction is very slow. So, the formation of $SO3$ is not significant at this temperature.

(c)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

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

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(c)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔH°$ can be calculated as

$ΔH°=∑nΔHfproducts°-∑mΔHfreactants°ΔHr×n°= [(2 mol SO3)(ΔHf° of SO3)]-[(2 mol of SO2)(ΔHf° of SO2)+(1 mol O2)(ΔHf° of O2)]ΔHr×n°= [(2 mol)(-396 kJ/mol)]-[(2 mol)(-296.8 kJ/mol)+(1 mol)(0 kJ/mol)]ΔHr×n°= -198.4 = -198 kJ$

The $ΔS°$ can be calculated as

$ΔS°=∑mSproducts°-∑nSreactants°ΔSr×n°= [(2 mol SO3)(S° of SO3)]-[(2 mol of SO2)(S° of SO2)+(1 mol O2)(S° of O2)]ΔSr×n°= [(2 mol)(256.66 J/mol.K)]-[(2 mol)(248.1 J/mol.K)+(1 mol)(205.0 J/mol.K)]ΔSr×n°= -187.88 = -187.9 J/K$

From the above values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated as

$ΔG°= ΔH°- TΔS°ΔG°= -198.4 kJ- ((273+500)K)(-187.88 J/K)(1 kJ/103 J)ΔG°= -53.16876 = -53 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous at $500°C$ .

(d)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The $K$ values has to be compared at $500°C$ and at $25°C$ .

Concept Introduction:

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

$ΔG = ΔG°+ RT ln Kwhere, ΔG = free energy ΔG°=standard-state free energy R = ideal gas constant T = temperature$

If a reaction is at equilibrium, $ΔG=0$ .

$0 = ΔG°+ RT ln KΔG° = -RT ln K$

(d)

Expert Solution

Explanation of Solution

Equilibrium constant, $K$ at $25°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-142 kJ)(103J1 kJ)(8.314 J/molgK)((273+25)K)ln K = 57.31417694$

The value of $K$ is

$K25= e57.31417694 = 7.8×1024$

Equilibrium constant, $K$ at $500°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-53 kJ)(103J1 kJ)(8.314 J/molgK)((273+500)K)ln K = 8.246817$

The value of $K$ is

$K500= e8.246817 = 3.8×103$

(e)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The highest $T$ at which the reaction is spontaneous has to be given.

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

(e)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

$ΔGr×n°= 0 = ΔHr×n°-TΔSr×n°ΔHr×n°= TΔSr×n°T = ΔH°ΔS°T = -198 kJ-187.9 J/K(1 kJ103J)T = 1.05×103 K$

Answer 6

Chapter 22, Problem 22.52P

(a)

Interpretation Introduction

Interpretation:

The $ΔG°$ has to be calculated at $25°C$ for the reaction of $SO2$ to form $SO3$ at standard conditions. The reaction is spontaneous or not has to be given.

Concept Introduction:

Standard state free energy of formation ( $ΔG°$ ):

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:

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(a)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔG°$ can be calculated as

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°ΔGr×n°= [(2 mol SO3)(ΔGf° of SO3)]-[(2 mol of SO2)(ΔGf° of SO2) + (1 mol O2)(ΔGf° of O2)]ΔGr×n°= [(2mol)(-371 kJ/mol)]-[(2 mol)(-300.2 kJ/mol)+(1 mol O2)(0 kJ/mol)]ΔGr×n°= -141.6 = -142 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous.

Answer 7

Chapter 22, Problem 22.52P

(a)

Interpretation Introduction

Interpretation:

The $ΔG°$ has to be calculated at $25°C$ for the reaction of $SO2$ to form $SO3$ at standard conditions. The reaction is spontaneous or not has to be given.

Concept Introduction:

Standard state free energy of formation ( $ΔG°$ ):

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:

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

Answer 8

(a)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔG°$ can be calculated as

$ΔG°=∑ΔGfproducts°-∑ΔGfreactants°ΔGr×n°= [(2 mol SO3)(ΔGf° of SO3)]-[(2 mol of SO2)(ΔGf° of SO2) + (1 mol O2)(ΔGf° of O2)]ΔGr×n°= [(2mol)(-371 kJ/mol)]-[(2 mol)(-300.2 kJ/mol)+(1 mol O2)(0 kJ/mol)]ΔGr×n°= -141.6 = -142 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The reason has to be given for the reaction not performed at $25°C$ .

(b)

Expert Solution

Explanation of Solution

At $25°C$ , the rate of the reaction is very slow. So, the formation of $SO3$ is not significant at this temperature.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The reason has to be given for the reaction not performed at $25°C$ .

Answer 14

(b)

Expert Solution

Explanation of Solution

At $25°C$ , the rate of the reaction is very slow. So, the formation of $SO3$ is not significant at this temperature.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

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

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

(c)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔH°$ can be calculated as

$ΔH°=∑nΔHfproducts°-∑mΔHfreactants°ΔHr×n°= [(2 mol SO3)(ΔHf° of SO3)]-[(2 mol of SO2)(ΔHf° of SO2)+(1 mol O2)(ΔHf° of O2)]ΔHr×n°= [(2 mol)(-396 kJ/mol)]-[(2 mol)(-296.8 kJ/mol)+(1 mol)(0 kJ/mol)]ΔHr×n°= -198.4 = -198 kJ$

The $ΔS°$ can be calculated as

$ΔS°=∑mSproducts°-∑nSreactants°ΔSr×n°= [(2 mol SO3)(S° of SO3)]-[(2 mol of SO2)(S° of SO2)+(1 mol O2)(S° of O2)]ΔSr×n°= [(2 mol)(256.66 J/mol.K)]-[(2 mol)(248.1 J/mol.K)+(1 mol)(205.0 J/mol.K)]ΔSr×n°= -187.88 = -187.9 J/K$

From the above values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated as

$ΔG°= ΔH°- TΔS°ΔG°= -198.4 kJ- ((273+500)K)(-187.88 J/K)(1 kJ/103 J)ΔG°= -53.16876 = -53 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous at $500°C$ .

Answer 19

(c)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

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

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

If $ΔG°> 0$ , the reaction is non-spontaneous.

If $ΔG°< 0$ , the reaction is spontaneous.

Answer 20

(c)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

The $ΔH°$ can be calculated as

$ΔH°=∑nΔHfproducts°-∑mΔHfreactants°ΔHr×n°= [(2 mol SO3)(ΔHf° of SO3)]-[(2 mol of SO2)(ΔHf° of SO2)+(1 mol O2)(ΔHf° of O2)]ΔHr×n°= [(2 mol)(-396 kJ/mol)]-[(2 mol)(-296.8 kJ/mol)+(1 mol)(0 kJ/mol)]ΔHr×n°= -198.4 = -198 kJ$

The $ΔS°$ can be calculated as

$ΔS°=∑mSproducts°-∑nSreactants°ΔSr×n°= [(2 mol SO3)(S° of SO3)]-[(2 mol of SO2)(S° of SO2)+(1 mol O2)(S° of O2)]ΔSr×n°= [(2 mol)(256.66 J/mol.K)]-[(2 mol)(248.1 J/mol.K)+(1 mol)(205.0 J/mol.K)]ΔSr×n°= -187.88 = -187.9 J/K$

From the above values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated as

$ΔG°= ΔH°- TΔS°ΔG°= -198.4 kJ- ((273+500)K)(-187.88 J/K)(1 kJ/103 J)ΔG°= -53.16876 = -53 kJ$

As the $ΔG°$ is negative ( $ΔG°< 0$ ), the reaction is spontaneous at $500°C$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The $K$ values has to be compared at $500°C$ and at $25°C$ .

Concept Introduction:

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

$ΔG = ΔG°+ RT ln Kwhere, ΔG = free energy ΔG°=standard-state free energy R = ideal gas constant T = temperature$

If a reaction is at equilibrium, $ΔG=0$ .

$0 = ΔG°+ RT ln KΔG° = -RT ln K$

(d)

Expert Solution

Explanation of Solution

Equilibrium constant, $K$ at $25°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-142 kJ)(103J1 kJ)(8.314 J/molgK)((273+25)K)ln K = 57.31417694$

The value of $K$ is

$K25= e57.31417694 = 7.8×1024$

Equilibrium constant, $K$ at $500°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-53 kJ)(103J1 kJ)(8.314 J/molgK)((273+500)K)ln K = 8.246817$

The value of $K$ is

$K500= e8.246817 = 3.8×103$

Answer 25

(d)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The $K$ values has to be compared at $500°C$ and at $25°C$ .

Concept Introduction:

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

$ΔG = ΔG°+ RT ln Kwhere, ΔG = free energy ΔG°=standard-state free energy R = ideal gas constant T = temperature$

If a reaction is at equilibrium, $ΔG=0$ .

$0 = ΔG°+ RT ln KΔG° = -RT ln K$

Answer 26

(d)

Expert Solution

Explanation of Solution

Equilibrium constant, $K$ at $25°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-142 kJ)(103J1 kJ)(8.314 J/molgK)((273+25)K)ln K = 57.31417694$

The value of $K$ is

$K25= e57.31417694 = 7.8×1024$

Equilibrium constant, $K$ at $500°C$ :

$ΔG° = -RT ln Kln K = ΔG°-RTln K = (-53 kJ)(103J1 kJ)(8.314 J/molgK)((273+500)K)ln K = 8.246817$

The value of $K$ is

$K500= e8.246817 = 3.8×103$

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

(e)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The highest $T$ at which the reaction is spontaneous has to be given.

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

(e)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

$ΔGr×n°= 0 = ΔHr×n°-TΔSr×n°ΔHr×n°= TΔSr×n°T = ΔH°ΔS°T = -198 kJ-187.9 J/K(1 kJ103J)T = 1.05×103 K$

Answer 31

(e)

Interpretation Introduction

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The highest $T$ at which the reaction is spontaneous has to be given.

Concept Introduction:

$ΔH°$ can be calculated by the expression

$ΔH°=∑nHfproducts°-∑mHfreactants°$

$ΔS°$ can be calculated by the expression

$ΔS°=∑nSproducts°-∑mSreactants°$

From the values of $ΔH° and ΔS°$ , $ΔG°$ can be calculated by the expression

$ΔG°= ΔH°- TΔS°$

Answer 32

(e)

Expert Solution

Explanation of Solution

The balanced reaction of formation of $SO3$ from $SO2$ is given as

$2SO2(g)+ O2(g)→2SO3(g)$

$ΔGr×n°= 0 = ΔHr×n°-TΔSr×n°ΔHr×n°= TΔSr×n°T = ΔH°ΔS°T = -198 kJ-187.9 J/K(1 kJ103J)T = 1.05×103 K$

Answer 33

(e)

Expert Solution

Answer 34

(e)

Expert Solution

Answer 35

Expert Solution

Answer 36

Ch. 22 - Prob. 22.1P Ch. 22 - Prob. 22.2P Ch. 22 - Prob. 22.3P Ch. 22 - Prob. 22.4P Ch. 22 - Prob. 22.5P Ch. 22 - Prob. 22.6P Ch. 22 - Prob. 22.7P Ch. 22 - Prob. 22.8P Ch. 22 - Prob. 22.9P Ch. 22 - Prob. 22.10P

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Explanation of Solution

Interpretation:

Consider the reaction of $SO2$ to form $SO3$ at standard conditions.

The reason has to be given for the reaction not performed at $25°C$ .

Explanation of Solution

Explanation of Solution

Explanation of Solution

Explanation of Solution

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