The major industrial use of hydrogen is in the production of ammonia by the Haber process: 3 H 2 ( g ) + N 2 ( g ) → 2 NH 3 ( g ) a. Using data from Appendix 4, calculate ∆H °, ∆S °, and ∆G ° for the Haber process reaction. b. Is the reaction spontaneous at standard conditions? c. At what temperatures is the reaction spontaneous at standard conditions? Assume ∆H ° and ∆S ° do not depend on temperature.

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Answer 1

Textbook Question

Chapter 16, Problem 54E

The major industrial use of hydrogen is in the production of ammonia by the Haber process:

$3 H 2 ( g ) + N 2 ( g ) → 2 NH 3 ( g )$

a. Using data from Appendix 4, calculate ∆H°, ∆S°, and ∆G° for the Haber process reaction.

b. Is the reaction spontaneous at standard conditions?

c. At what temperatures is the reaction spontaneous at standard conditions? Assume ∆H° and ∆S° do not depend on temperature.

(a)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

The stated reaction is,

$3H2(g)+N2(g)→2NH3(g)$

Refer to Appendix $4$ .

The value of $ΔS∘(J/K⋅mol)$ for the given reactant and product is,

Molecules	$ΔS∘(J/K⋅mol)$
$NH3(g)$	$193$
$N2(g)$	$192$
$H2(g)$	$131$

The formula of $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

Where,

$ΔS∘$ is the standard entropy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS∘(product)$ is the standard entropy of product at a pressure of $1 atm$ .
$ΔS∘(reactant)$ is the standard entropy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)=[2(193)−{3(131)+(192)}] J/K=−199 J/K_$

The value of $ΔH∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔH∘(kJ/mol)$
$NH3(g)$	$−46$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

Where,

$ΔH∘$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH∘(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH∘(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)=[2(−46)−{3(0)+(0)}] kJ=−92 kJ_$

The value of $ΔG∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔG∘(kJ/mol)$
$NH3(g)$	$−17$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Where,

$ΔG∘$ is the standard Gibb’s free energy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔG∘(product)$ is the standard Gibb’s free energy of product at a pressure of $1 atm$ .
$ΔG∘(reactant)$ is the standard Gibb’s free energy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)=[2(−17)−{3(0)+(0)}] kJ=−34 kJ_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

A reaction is said to be spontaneous if the value of $ΔG$ is negative.

Since, the value of $ΔG$ for the given reaction is $−34 kJ$ , therefore, the given reaction is spontaneous.

The formula of $ΔG$ is,

$ΔG=ΔH−TΔS$

Where,

$ΔH$ is the enthalpy of reaction.
$ΔG$ is the free energy change.
$T$ is the temperature.
$ΔS$ is the entropy of reaction.

Since, the value of $ΔH$ is negative, therefore, the reaction is spontaneous at lower temperature.

(c)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer to Problem 54E

Solutions are as follows.

Explanation of Solution

The value of $ΔS∘$ for the given reaction is $−199 J/K$ .

The value of $ΔH∘$ for the given reaction is $−92 kJ$ .

The negative value of $ΔH∘$ suggests that the give reaction is exothermic. But $ΔS∘$ favors opposite process. These opposite tendencies cancel out that means that the reaction has achieved equilibrium. The value of $ΔG∘$ at equilibrium is zero.

The formula of $ΔG$ is,

$ΔG∘=ΔH∘−TΔS∘$

Where,

$ΔH∘$ is the enthalpy of reaction.
$ΔG∘$ is the free energy change.
$T$ is the temperature.
$ΔS∘$ is the entropy of reaction.

Substitute the values of $ΔS∘,ΔG∘$ and $ΔH∘$ in the above equation.

$ΔG∘=ΔH∘−TΔS∘T=ΔH∘ΔS∘−ΔG∘=−92×103 J−199 J/K−0=462.3 K_$

The given reaction will be spontaneous below $462.3 K_$ .

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Answer 2

Textbook Question

Chapter 16, Problem 54E

The major industrial use of hydrogen is in the production of ammonia by the Haber process:

$3 H 2 ( g ) + N 2 ( g ) → 2 NH 3 ( g )$

a. Using data from Appendix 4, calculate ∆H°, ∆S°, and ∆G° for the Haber process reaction.

b. Is the reaction spontaneous at standard conditions?

c. At what temperatures is the reaction spontaneous at standard conditions? Assume ∆H° and ∆S° do not depend on temperature.

(a)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

The stated reaction is,

$3H2(g)+N2(g)→2NH3(g)$

Refer to Appendix $4$ .

The value of $ΔS∘(J/K⋅mol)$ for the given reactant and product is,

Molecules	$ΔS∘(J/K⋅mol)$
$NH3(g)$	$193$
$N2(g)$	$192$
$H2(g)$	$131$

The formula of $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

Where,

$ΔS∘$ is the standard entropy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS∘(product)$ is the standard entropy of product at a pressure of $1 atm$ .
$ΔS∘(reactant)$ is the standard entropy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)=[2(193)−{3(131)+(192)}] J/K=−199 J/K_$

The value of $ΔH∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔH∘(kJ/mol)$
$NH3(g)$	$−46$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

Where,

$ΔH∘$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH∘(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH∘(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)=[2(−46)−{3(0)+(0)}] kJ=−92 kJ_$

The value of $ΔG∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔG∘(kJ/mol)$
$NH3(g)$	$−17$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Where,

$ΔG∘$ is the standard Gibb’s free energy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔG∘(product)$ is the standard Gibb’s free energy of product at a pressure of $1 atm$ .
$ΔG∘(reactant)$ is the standard Gibb’s free energy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)=[2(−17)−{3(0)+(0)}] kJ=−34 kJ_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

A reaction is said to be spontaneous if the value of $ΔG$ is negative.

Since, the value of $ΔG$ for the given reaction is $−34 kJ$ , therefore, the given reaction is spontaneous.

The formula of $ΔG$ is,

$ΔG=ΔH−TΔS$

Where,

$ΔH$ is the enthalpy of reaction.
$ΔG$ is the free energy change.
$T$ is the temperature.
$ΔS$ is the entropy of reaction.

Since, the value of $ΔH$ is negative, therefore, the reaction is spontaneous at lower temperature.

(c)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer to Problem 54E

Solutions are as follows.

Explanation of Solution

The value of $ΔS∘$ for the given reaction is $−199 J/K$ .

The value of $ΔH∘$ for the given reaction is $−92 kJ$ .

The negative value of $ΔH∘$ suggests that the give reaction is exothermic. But $ΔS∘$ favors opposite process. These opposite tendencies cancel out that means that the reaction has achieved equilibrium. The value of $ΔG∘$ at equilibrium is zero.

The formula of $ΔG$ is,

$ΔG∘=ΔH∘−TΔS∘$

Where,

$ΔH∘$ is the enthalpy of reaction.
$ΔG∘$ is the free energy change.
$T$ is the temperature.
$ΔS∘$ is the entropy of reaction.

Substitute the values of $ΔS∘,ΔG∘$ and $ΔH∘$ in the above equation.

$ΔG∘=ΔH∘−TΔS∘T=ΔH∘ΔS∘−ΔG∘=−92×103 J−199 J/K−0=462.3 K_$

The given reaction will be spontaneous below $462.3 K_$ .

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Answer 3

Textbook Question

Chapter 16, Problem 54E

The major industrial use of hydrogen is in the production of ammonia by the Haber process:

$3 H 2 ( g ) + N 2 ( g ) → 2 NH 3 ( g )$

a. Using data from Appendix 4, calculate ∆H°, ∆S°, and ∆G° for the Haber process reaction.

b. Is the reaction spontaneous at standard conditions?

c. At what temperatures is the reaction spontaneous at standard conditions? Assume ∆H° and ∆S° do not depend on temperature.

(a)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

The stated reaction is,

$3H2(g)+N2(g)→2NH3(g)$

Refer to Appendix $4$ .

The value of $ΔS∘(J/K⋅mol)$ for the given reactant and product is,

Molecules	$ΔS∘(J/K⋅mol)$
$NH3(g)$	$193$
$N2(g)$	$192$
$H2(g)$	$131$

The formula of $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

Where,

$ΔS∘$ is the standard entropy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS∘(product)$ is the standard entropy of product at a pressure of $1 atm$ .
$ΔS∘(reactant)$ is the standard entropy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)=[2(193)−{3(131)+(192)}] J/K=−199 J/K_$

The value of $ΔH∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔH∘(kJ/mol)$
$NH3(g)$	$−46$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

Where,

$ΔH∘$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH∘(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH∘(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)=[2(−46)−{3(0)+(0)}] kJ=−92 kJ_$

The value of $ΔG∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔG∘(kJ/mol)$
$NH3(g)$	$−17$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Where,

$ΔG∘$ is the standard Gibb’s free energy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔG∘(product)$ is the standard Gibb’s free energy of product at a pressure of $1 atm$ .
$ΔG∘(reactant)$ is the standard Gibb’s free energy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)=[2(−17)−{3(0)+(0)}] kJ=−34 kJ_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

A reaction is said to be spontaneous if the value of $ΔG$ is negative.

Since, the value of $ΔG$ for the given reaction is $−34 kJ$ , therefore, the given reaction is spontaneous.

The formula of $ΔG$ is,

$ΔG=ΔH−TΔS$

Where,

$ΔH$ is the enthalpy of reaction.
$ΔG$ is the free energy change.
$T$ is the temperature.
$ΔS$ is the entropy of reaction.

Since, the value of $ΔH$ is negative, therefore, the reaction is spontaneous at lower temperature.

(c)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer to Problem 54E

Solutions are as follows.

Explanation of Solution

The value of $ΔS∘$ for the given reaction is $−199 J/K$ .

The value of $ΔH∘$ for the given reaction is $−92 kJ$ .

The negative value of $ΔH∘$ suggests that the give reaction is exothermic. But $ΔS∘$ favors opposite process. These opposite tendencies cancel out that means that the reaction has achieved equilibrium. The value of $ΔG∘$ at equilibrium is zero.

The formula of $ΔG$ is,

$ΔG∘=ΔH∘−TΔS∘$

Where,

$ΔH∘$ is the enthalpy of reaction.
$ΔG∘$ is the free energy change.
$T$ is the temperature.
$ΔS∘$ is the entropy of reaction.

Substitute the values of $ΔS∘,ΔG∘$ and $ΔH∘$ in the above equation.

$ΔG∘=ΔH∘−TΔS∘T=ΔH∘ΔS∘−ΔG∘=−92×103 J−199 J/K−0=462.3 K_$

The given reaction will be spontaneous below $462.3 K_$ .

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Answer 4

Textbook Question

Chapter 16, Problem 54E

The major industrial use of hydrogen is in the production of ammonia by the Haber process:

$3 H 2 ( g ) + N 2 ( g ) → 2 NH 3 ( g )$

a. Using data from Appendix 4, calculate ∆H°, ∆S°, and ∆G° for the Haber process reaction.

b. Is the reaction spontaneous at standard conditions?

c. At what temperatures is the reaction spontaneous at standard conditions? Assume ∆H° and ∆S° do not depend on temperature.

(a)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

The stated reaction is,

$3H2(g)+N2(g)→2NH3(g)$

Refer to Appendix $4$ .

The value of $ΔS∘(J/K⋅mol)$ for the given reactant and product is,

Molecules	$ΔS∘(J/K⋅mol)$
$NH3(g)$	$193$
$N2(g)$	$192$
$H2(g)$	$131$

The formula of $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

Where,

$ΔS∘$ is the standard entropy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS∘(product)$ is the standard entropy of product at a pressure of $1 atm$ .
$ΔS∘(reactant)$ is the standard entropy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)=[2(193)−{3(131)+(192)}] J/K=−199 J/K_$

The value of $ΔH∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔH∘(kJ/mol)$
$NH3(g)$	$−46$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

Where,

$ΔH∘$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH∘(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH∘(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)=[2(−46)−{3(0)+(0)}] kJ=−92 kJ_$

The value of $ΔG∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔG∘(kJ/mol)$
$NH3(g)$	$−17$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Where,

$ΔG∘$ is the standard Gibb’s free energy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔG∘(product)$ is the standard Gibb’s free energy of product at a pressure of $1 atm$ .
$ΔG∘(reactant)$ is the standard Gibb’s free energy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)=[2(−17)−{3(0)+(0)}] kJ=−34 kJ_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

A reaction is said to be spontaneous if the value of $ΔG$ is negative.

Since, the value of $ΔG$ for the given reaction is $−34 kJ$ , therefore, the given reaction is spontaneous.

The formula of $ΔG$ is,

$ΔG=ΔH−TΔS$

Where,

$ΔH$ is the enthalpy of reaction.
$ΔG$ is the free energy change.
$T$ is the temperature.
$ΔS$ is the entropy of reaction.

Since, the value of $ΔH$ is negative, therefore, the reaction is spontaneous at lower temperature.

(c)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer to Problem 54E

Solutions are as follows.

Explanation of Solution

The value of $ΔS∘$ for the given reaction is $−199 J/K$ .

The value of $ΔH∘$ for the given reaction is $−92 kJ$ .

The negative value of $ΔH∘$ suggests that the give reaction is exothermic. But $ΔS∘$ favors opposite process. These opposite tendencies cancel out that means that the reaction has achieved equilibrium. The value of $ΔG∘$ at equilibrium is zero.

The formula of $ΔG$ is,

$ΔG∘=ΔH∘−TΔS∘$

Where,

$ΔH∘$ is the enthalpy of reaction.
$ΔG∘$ is the free energy change.
$T$ is the temperature.
$ΔS∘$ is the entropy of reaction.

Substitute the values of $ΔS∘,ΔG∘$ and $ΔH∘$ in the above equation.

$ΔG∘=ΔH∘−TΔS∘T=ΔH∘ΔS∘−ΔG∘=−92×103 J−199 J/K−0=462.3 K_$

The given reaction will be spontaneous below $462.3 K_$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

The stated reaction is,

$3H2(g)+N2(g)→2NH3(g)$

Refer to Appendix $4$ .

The value of $ΔS∘(J/K⋅mol)$ for the given reactant and product is,

Molecules	$ΔS∘(J/K⋅mol)$
$NH3(g)$	$193$
$N2(g)$	$192$
$H2(g)$	$131$

The formula of $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

Where,

$ΔS∘$ is the standard entropy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS∘(product)$ is the standard entropy of product at a pressure of $1 atm$ .
$ΔS∘(reactant)$ is the standard entropy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)=[2(193)−{3(131)+(192)}] J/K=−199 J/K_$

The value of $ΔH∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔH∘(kJ/mol)$
$NH3(g)$	$−46$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

Where,

$ΔH∘$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH∘(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH∘(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)=[2(−46)−{3(0)+(0)}] kJ=−92 kJ_$

The value of $ΔG∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔG∘(kJ/mol)$
$NH3(g)$	$−17$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Where,

$ΔG∘$ is the standard Gibb’s free energy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔG∘(product)$ is the standard Gibb’s free energy of product at a pressure of $1 atm$ .
$ΔG∘(reactant)$ is the standard Gibb’s free energy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)=[2(−17)−{3(0)+(0)}] kJ=−34 kJ_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

A reaction is said to be spontaneous if the value of $ΔG$ is negative.

Since, the value of $ΔG$ for the given reaction is $−34 kJ$ , therefore, the given reaction is spontaneous.

The formula of $ΔG$ is,

$ΔG=ΔH−TΔS$

Where,

$ΔH$ is the enthalpy of reaction.
$ΔG$ is the free energy change.
$T$ is the temperature.
$ΔS$ is the entropy of reaction.

Since, the value of $ΔH$ is negative, therefore, the reaction is spontaneous at lower temperature.

(c)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer to Problem 54E

Solutions are as follows.

Explanation of Solution

The value of $ΔS∘$ for the given reaction is $−199 J/K$ .

The value of $ΔH∘$ for the given reaction is $−92 kJ$ .

The negative value of $ΔH∘$ suggests that the give reaction is exothermic. But $ΔS∘$ favors opposite process. These opposite tendencies cancel out that means that the reaction has achieved equilibrium. The value of $ΔG∘$ at equilibrium is zero.

The formula of $ΔG$ is,

$ΔG∘=ΔH∘−TΔS∘$

Where,

$ΔH∘$ is the enthalpy of reaction.
$ΔG∘$ is the free energy change.
$T$ is the temperature.
$ΔS∘$ is the entropy of reaction.

Substitute the values of $ΔS∘,ΔG∘$ and $ΔH∘$ in the above equation.

$ΔG∘=ΔH∘−TΔS∘T=ΔH∘ΔS∘−ΔG∘=−92×103 J−199 J/K−0=462.3 K_$

The given reaction will be spontaneous below $462.3 K_$ .

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

The stated reaction is,

$3H2(g)+N2(g)→2NH3(g)$

Refer to Appendix $4$ .

The value of $ΔS∘(J/K⋅mol)$ for the given reactant and product is,

Molecules	$ΔS∘(J/K⋅mol)$
$NH3(g)$	$193$
$N2(g)$	$192$
$H2(g)$	$131$

The formula of $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

Where,

$ΔS∘$ is the standard entropy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS∘(product)$ is the standard entropy of product at a pressure of $1 atm$ .
$ΔS∘(reactant)$ is the standard entropy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)=[2(193)−{3(131)+(192)}] J/K=−199 J/K_$

The value of $ΔH∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔH∘(kJ/mol)$
$NH3(g)$	$−46$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

Where,

$ΔH∘$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH∘(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH∘(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)=[2(−46)−{3(0)+(0)}] kJ=−92 kJ_$

The value of $ΔG∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔG∘(kJ/mol)$
$NH3(g)$	$−17$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Where,

$ΔG∘$ is the standard Gibb’s free energy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔG∘(product)$ is the standard Gibb’s free energy of product at a pressure of $1 atm$ .
$ΔG∘(reactant)$ is the standard Gibb’s free energy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)=[2(−17)−{3(0)+(0)}] kJ=−34 kJ_$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer 13

Explanation of Solution

The stated reaction is,

$3H2(g)+N2(g)→2NH3(g)$

Refer to Appendix $4$ .

The value of $ΔS∘(J/K⋅mol)$ for the given reactant and product is,

Molecules	$ΔS∘(J/K⋅mol)$
$NH3(g)$	$193$
$N2(g)$	$192$
$H2(g)$	$131$

The formula of $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

Where,

$ΔS∘$ is the standard entropy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS∘(product)$ is the standard entropy of product at a pressure of $1 atm$ .
$ΔS∘(reactant)$ is the standard entropy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)=[2(193)−{3(131)+(192)}] J/K=−199 J/K_$

The value of $ΔH∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔH∘(kJ/mol)$
$NH3(g)$	$−46$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

Where,

$ΔH∘$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH∘(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH∘(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)=[2(−46)−{3(0)+(0)}] kJ=−92 kJ_$

The value of $ΔG∘(kJ/mol)$ for the given reactant and product is,

Molecules	$ΔG∘(kJ/mol)$
$NH3(g)$	$−17$
$N2(g)$	$0$
$H2(g)$	$0$

The formula of $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Where,

$ΔG∘$ is the standard Gibb’s free energy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔG∘(product)$ is the standard Gibb’s free energy of product at a pressure of $1 atm$ .
$ΔG∘(reactant)$ is the standard Gibb’s free energy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)=[2(−17)−{3(0)+(0)}] kJ=−34 kJ_$

Answer 14

(b)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Explanation of Solution

A reaction is said to be spontaneous if the value of $ΔG$ is negative.

Since, the value of $ΔG$ for the given reaction is $−34 kJ$ , therefore, the given reaction is spontaneous.

The formula of $ΔG$ is,

$ΔG=ΔH−TΔS$

Where,

$ΔH$ is the enthalpy of reaction.
$ΔG$ is the free energy change.
$T$ is the temperature.
$ΔS$ is the entropy of reaction.

Since, the value of $ΔH$ is negative, therefore, the reaction is spontaneous at lower temperature.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer 19

Explanation of Solution

A reaction is said to be spontaneous if the value of $ΔG$ is negative.

Since, the value of $ΔG$ for the given reaction is $−34 kJ$ , therefore, the given reaction is spontaneous.

The formula of $ΔG$ is,

$ΔG=ΔH−TΔS$

Where,

$ΔH$ is the enthalpy of reaction.
$ΔG$ is the free energy change.
$T$ is the temperature.
$ΔS$ is the entropy of reaction.

Since, the value of $ΔH$ is negative, therefore, the reaction is spontaneous at lower temperature.

Answer 20

(c)

Expert Solution

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer to Problem 54E

Solutions are as follows.

Explanation of Solution

The value of $ΔS∘$ for the given reaction is $−199 J/K$ .

The value of $ΔH∘$ for the given reaction is $−92 kJ$ .

The negative value of $ΔH∘$ suggests that the give reaction is exothermic. But $ΔS∘$ favors opposite process. These opposite tendencies cancel out that means that the reaction has achieved equilibrium. The value of $ΔG∘$ at equilibrium is zero.

The formula of $ΔG$ is,

$ΔG∘=ΔH∘−TΔS∘$

Where,

$ΔH∘$ is the enthalpy of reaction.
$ΔG∘$ is the free energy change.
$T$ is the temperature.
$ΔS∘$ is the entropy of reaction.

Substitute the values of $ΔS∘,ΔG∘$ and $ΔH∘$ in the above equation.

$ΔG∘=ΔH∘−TΔS∘T=ΔH∘ΔS∘−ΔG∘=−92×103 J−199 J/K−0=462.3 K_$

The given reaction will be spontaneous below $462.3 K_$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Interpretation Introduction

Interpretation: The values of $ΔS∘,ΔH∘$ and $ΔG∘$ is to be calculated at $25 °C$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $ΔS∘$ is,

$ΔS∘=∑npΔS∘(product)−∑nfΔS∘(reactant)$

The expression for $ΔH∘$ is,

$ΔH∘=∑npΔH∘(product)−∑nfΔH∘(reactant)$

The expression for $ΔG∘$ is,

$ΔG∘=∑npΔG∘(product)−∑nfΔG∘(reactant)$

Answer 25

Answer to Problem 54E

Solutions are as follows.

Answer 26

Explanation of Solution

The value of $ΔS∘$ for the given reaction is $−199 J/K$ .

The value of $ΔH∘$ for the given reaction is $−92 kJ$ .

The negative value of $ΔH∘$ suggests that the give reaction is exothermic. But $ΔS∘$ favors opposite process. These opposite tendencies cancel out that means that the reaction has achieved equilibrium. The value of $ΔG∘$ at equilibrium is zero.