The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined. Concept Introduction: The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure. A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation. The relation to determine Gibbs free energy of a reaction is shown below. Δ G ° = ∑ n Δ G f ∘ ( p r o d u c t s ) − ∑ n Δ G f ∘ ( reactants )

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Answer 1

Question

Chapter 4, Problem 4J.7E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure. A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation. The relation to determine Gibbs free energy of a reaction is shown below.

ΔG° =∑nΔGf∘(products)− ∑nΔGf∘(reactants)

(a)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Explanation of Solution

The given chemical equation for the decomposition of hydrogen peroxide is shown below.

2H2O2(l)→2H2O(l)+O2(g)

The relation for the calculation of standard change in the enthalpy is shown below.

ΔH° =∑nΔHf∘(products)− ∑nΔHf∘(reactants)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(2 mol)×ΔHf∘(H2O, l)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(H2O2, l)} (1)

Where,

ΔHf∘(H2O2, l) is the standard enthalpy of formation of H2O2(l).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).

The value of ΔHf∘(H2O2, l) is −187.78 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔHf∘(H2O2, l), ΔHf∘(H2O, l) and ΔHf∘(O2, g) in equation (1).

ΔH°={(2 mol)×(−285.83 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−187.78 kJ⋅mol−1)}={−571.66 kJ}−{−375.56 kJ}=−196.1 kJ

The relation for the calculation of standard change in the entropy is shown below.

ΔS° =∑nΔSf∘(products)− ∑nΔSf∘(reactants)

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(2 mol)×ΔSf∘(H2O, l)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(H2O2, l)} (2)

Where,

ΔSf∘(H2O2, l) is the standard entropy of formation of H2O2(l).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).

The value of ΔSf∘(H2O2, l) is +109.6 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is +69.91 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(H2O2, l), ΔSf∘(H2O, l) and ΔSf∘(O2, g) in equation (2).

ΔS°={(2 mol)×(69.91 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(109.6 J⋅K−1⋅mol−1)}={139.82 J⋅K−1+205.138 J⋅K−1}−{219.2 J⋅K−1}=125.758 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(2 mol)×ΔGf∘(H2O, l)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(H2O2, l)} (3)

Where,

ΔGf∘(H2O2, l) is the standard Gibbs free energy of formation of H2O2(l).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).

The value of ΔGf∘(H2O2, l) is −120.35 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔGf∘(H2O2, l), ΔGf∘(H2O, l) and ΔGf∘(O2, g) in equation (3).

ΔG°={(2 mol)×(−237.13 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−120.35 kJ⋅mol−1)}={−474.26 kJ}−{−240.7 kJ}=−233.56 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Explanation of Solution

The given chemical equation for the production of hydrofluoric acid is shown below.

2F2(g)+2H2O(l)→4HF(g)+O2(g)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(4 mol)×ΔHf∘(HF, g)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(F2, g)+(2 mol)×ΔHf∘(H2O, l)} (4)

Where,

ΔHf∘(HF, g) is the standard enthalpy of formation of HF(g).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).
ΔHf∘(F2, g) is the standard enthalpy of formation of F2(g).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).

The value of ΔHf∘(HF, g) is −332.63 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

Substitute the value of ΔHf∘(HF, g), ΔHf∘(O2, g), ΔHf∘(F2, g) and ΔHf∘(H2O, l) in equation (4).

ΔH°={(4 mol)×(−332.63 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−285.83 kJ⋅mol−1)}={−1330.52 kJ}−{−571.66 kJ}=−758.86 kJ

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(4 mol)×ΔSf∘(HF, g)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(F2, g)+(2 mol)×ΔSf∘(H2O, l)} (5)

Where,

ΔSf∘(HF, g) is the standard entropy of formation of HF(g).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).
ΔSf∘(F2, g) is the standard entropy of formation of F2(g).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).

The value of ΔSf∘(HF, g) is −13.8 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

The value of ΔHf∘(F2, g) is +202.78 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is 69.91 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(HF, g), ΔSf∘(F2, g), ΔSf∘(O2, g) and ΔSf∘(H2O, l) in equation (5).

ΔS°={(4 mol)×(−13.8 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(202.78 J⋅K−1⋅mol−1)+(2 mol)×(69.91 J⋅K−1⋅mol−1)}={−55.2 J⋅K−1+205.138 J⋅K−1}−{405.56 J⋅K−1+139.82 J⋅K−1}=−395.442 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(4 mol)×ΔGf∘(HF, g)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(F2, g)+(2 mol)×ΔGf∘(H2O, l)} (6)

Where,

ΔGf∘(HF, g) is the standard Gibbs free energy of formation of HF(g).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).
ΔGf∘(F2, g) is the standard Gibbs free energy of formation of F2(g).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).

The value of ΔGf∘(HF, g) is −278.79 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

Substitute the value of ΔGf∘(HF, g), ΔGf∘(O2, g), ΔGf∘(H2O, l) and ΔGf∘(F2, g) in equation (6).

ΔG°={(4 mol)×(−278.79 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−237.13 kJ⋅mol−1)}={−1115.16 kJ}−{−474.26 kJ}=−640.9 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

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Answer 2

Question

Chapter 4, Problem 4J.7E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure. A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation. The relation to determine Gibbs free energy of a reaction is shown below.

ΔG° =∑nΔGf∘(products)− ∑nΔGf∘(reactants)

(a)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Explanation of Solution

The given chemical equation for the decomposition of hydrogen peroxide is shown below.

2H2O2(l)→2H2O(l)+O2(g)

The relation for the calculation of standard change in the enthalpy is shown below.

ΔH° =∑nΔHf∘(products)− ∑nΔHf∘(reactants)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(2 mol)×ΔHf∘(H2O, l)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(H2O2, l)} (1)

Where,

ΔHf∘(H2O2, l) is the standard enthalpy of formation of H2O2(l).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).

The value of ΔHf∘(H2O2, l) is −187.78 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔHf∘(H2O2, l), ΔHf∘(H2O, l) and ΔHf∘(O2, g) in equation (1).

ΔH°={(2 mol)×(−285.83 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−187.78 kJ⋅mol−1)}={−571.66 kJ}−{−375.56 kJ}=−196.1 kJ

The relation for the calculation of standard change in the entropy is shown below.

ΔS° =∑nΔSf∘(products)− ∑nΔSf∘(reactants)

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(2 mol)×ΔSf∘(H2O, l)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(H2O2, l)} (2)

Where,

ΔSf∘(H2O2, l) is the standard entropy of formation of H2O2(l).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).

The value of ΔSf∘(H2O2, l) is +109.6 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is +69.91 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(H2O2, l), ΔSf∘(H2O, l) and ΔSf∘(O2, g) in equation (2).

ΔS°={(2 mol)×(69.91 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(109.6 J⋅K−1⋅mol−1)}={139.82 J⋅K−1+205.138 J⋅K−1}−{219.2 J⋅K−1}=125.758 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(2 mol)×ΔGf∘(H2O, l)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(H2O2, l)} (3)

Where,

ΔGf∘(H2O2, l) is the standard Gibbs free energy of formation of H2O2(l).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).

The value of ΔGf∘(H2O2, l) is −120.35 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔGf∘(H2O2, l), ΔGf∘(H2O, l) and ΔGf∘(O2, g) in equation (3).

ΔG°={(2 mol)×(−237.13 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−120.35 kJ⋅mol−1)}={−474.26 kJ}−{−240.7 kJ}=−233.56 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Explanation of Solution

The given chemical equation for the production of hydrofluoric acid is shown below.

2F2(g)+2H2O(l)→4HF(g)+O2(g)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(4 mol)×ΔHf∘(HF, g)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(F2, g)+(2 mol)×ΔHf∘(H2O, l)} (4)

Where,

ΔHf∘(HF, g) is the standard enthalpy of formation of HF(g).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).
ΔHf∘(F2, g) is the standard enthalpy of formation of F2(g).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).

The value of ΔHf∘(HF, g) is −332.63 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

Substitute the value of ΔHf∘(HF, g), ΔHf∘(O2, g), ΔHf∘(F2, g) and ΔHf∘(H2O, l) in equation (4).

ΔH°={(4 mol)×(−332.63 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−285.83 kJ⋅mol−1)}={−1330.52 kJ}−{−571.66 kJ}=−758.86 kJ

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(4 mol)×ΔSf∘(HF, g)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(F2, g)+(2 mol)×ΔSf∘(H2O, l)} (5)

Where,

ΔSf∘(HF, g) is the standard entropy of formation of HF(g).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).
ΔSf∘(F2, g) is the standard entropy of formation of F2(g).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).

The value of ΔSf∘(HF, g) is −13.8 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

The value of ΔHf∘(F2, g) is +202.78 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is 69.91 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(HF, g), ΔSf∘(F2, g), ΔSf∘(O2, g) and ΔSf∘(H2O, l) in equation (5).

ΔS°={(4 mol)×(−13.8 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(202.78 J⋅K−1⋅mol−1)+(2 mol)×(69.91 J⋅K−1⋅mol−1)}={−55.2 J⋅K−1+205.138 J⋅K−1}−{405.56 J⋅K−1+139.82 J⋅K−1}=−395.442 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(4 mol)×ΔGf∘(HF, g)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(F2, g)+(2 mol)×ΔGf∘(H2O, l)} (6)

Where,

ΔGf∘(HF, g) is the standard Gibbs free energy of formation of HF(g).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).
ΔGf∘(F2, g) is the standard Gibbs free energy of formation of F2(g).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).

The value of ΔGf∘(HF, g) is −278.79 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

Substitute the value of ΔGf∘(HF, g), ΔGf∘(O2, g), ΔGf∘(H2O, l) and ΔGf∘(F2, g) in equation (6).

ΔG°={(4 mol)×(−278.79 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−237.13 kJ⋅mol−1)}={−1115.16 kJ}−{−474.26 kJ}=−640.9 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

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Answer 3

Question

Chapter 4, Problem 4J.7E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure. A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation. The relation to determine Gibbs free energy of a reaction is shown below.

ΔG° =∑nΔGf∘(products)− ∑nΔGf∘(reactants)

(a)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Explanation of Solution

The given chemical equation for the decomposition of hydrogen peroxide is shown below.

2H2O2(l)→2H2O(l)+O2(g)

The relation for the calculation of standard change in the enthalpy is shown below.

ΔH° =∑nΔHf∘(products)− ∑nΔHf∘(reactants)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(2 mol)×ΔHf∘(H2O, l)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(H2O2, l)} (1)

Where,

ΔHf∘(H2O2, l) is the standard enthalpy of formation of H2O2(l).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).

The value of ΔHf∘(H2O2, l) is −187.78 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔHf∘(H2O2, l), ΔHf∘(H2O, l) and ΔHf∘(O2, g) in equation (1).

ΔH°={(2 mol)×(−285.83 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−187.78 kJ⋅mol−1)}={−571.66 kJ}−{−375.56 kJ}=−196.1 kJ

The relation for the calculation of standard change in the entropy is shown below.

ΔS° =∑nΔSf∘(products)− ∑nΔSf∘(reactants)

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(2 mol)×ΔSf∘(H2O, l)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(H2O2, l)} (2)

Where,

ΔSf∘(H2O2, l) is the standard entropy of formation of H2O2(l).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).

The value of ΔSf∘(H2O2, l) is +109.6 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is +69.91 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(H2O2, l), ΔSf∘(H2O, l) and ΔSf∘(O2, g) in equation (2).

ΔS°={(2 mol)×(69.91 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(109.6 J⋅K−1⋅mol−1)}={139.82 J⋅K−1+205.138 J⋅K−1}−{219.2 J⋅K−1}=125.758 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(2 mol)×ΔGf∘(H2O, l)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(H2O2, l)} (3)

Where,

ΔGf∘(H2O2, l) is the standard Gibbs free energy of formation of H2O2(l).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).

The value of ΔGf∘(H2O2, l) is −120.35 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔGf∘(H2O2, l), ΔGf∘(H2O, l) and ΔGf∘(O2, g) in equation (3).

ΔG°={(2 mol)×(−237.13 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−120.35 kJ⋅mol−1)}={−474.26 kJ}−{−240.7 kJ}=−233.56 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Explanation of Solution

The given chemical equation for the production of hydrofluoric acid is shown below.

2F2(g)+2H2O(l)→4HF(g)+O2(g)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(4 mol)×ΔHf∘(HF, g)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(F2, g)+(2 mol)×ΔHf∘(H2O, l)} (4)

Where,

ΔHf∘(HF, g) is the standard enthalpy of formation of HF(g).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).
ΔHf∘(F2, g) is the standard enthalpy of formation of F2(g).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).

The value of ΔHf∘(HF, g) is −332.63 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

Substitute the value of ΔHf∘(HF, g), ΔHf∘(O2, g), ΔHf∘(F2, g) and ΔHf∘(H2O, l) in equation (4).

ΔH°={(4 mol)×(−332.63 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−285.83 kJ⋅mol−1)}={−1330.52 kJ}−{−571.66 kJ}=−758.86 kJ

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(4 mol)×ΔSf∘(HF, g)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(F2, g)+(2 mol)×ΔSf∘(H2O, l)} (5)

Where,

ΔSf∘(HF, g) is the standard entropy of formation of HF(g).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).
ΔSf∘(F2, g) is the standard entropy of formation of F2(g).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).

The value of ΔSf∘(HF, g) is −13.8 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

The value of ΔHf∘(F2, g) is +202.78 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is 69.91 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(HF, g), ΔSf∘(F2, g), ΔSf∘(O2, g) and ΔSf∘(H2O, l) in equation (5).

ΔS°={(4 mol)×(−13.8 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(202.78 J⋅K−1⋅mol−1)+(2 mol)×(69.91 J⋅K−1⋅mol−1)}={−55.2 J⋅K−1+205.138 J⋅K−1}−{405.56 J⋅K−1+139.82 J⋅K−1}=−395.442 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(4 mol)×ΔGf∘(HF, g)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(F2, g)+(2 mol)×ΔGf∘(H2O, l)} (6)

Where,

ΔGf∘(HF, g) is the standard Gibbs free energy of formation of HF(g).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).
ΔGf∘(F2, g) is the standard Gibbs free energy of formation of F2(g).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).

The value of ΔGf∘(HF, g) is −278.79 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

Substitute the value of ΔGf∘(HF, g), ΔGf∘(O2, g), ΔGf∘(H2O, l) and ΔGf∘(F2, g) in equation (6).

ΔG°={(4 mol)×(−278.79 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−237.13 kJ⋅mol−1)}={−1115.16 kJ}−{−474.26 kJ}=−640.9 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

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Answer 4

Question

Chapter 4, Problem 4J.7E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure. A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation. The relation to determine Gibbs free energy of a reaction is shown below.

ΔG° =∑nΔGf∘(products)− ∑nΔGf∘(reactants)

(a)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Explanation of Solution

The given chemical equation for the decomposition of hydrogen peroxide is shown below.

2H2O2(l)→2H2O(l)+O2(g)

The relation for the calculation of standard change in the enthalpy is shown below.

ΔH° =∑nΔHf∘(products)− ∑nΔHf∘(reactants)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(2 mol)×ΔHf∘(H2O, l)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(H2O2, l)} (1)

Where,

ΔHf∘(H2O2, l) is the standard enthalpy of formation of H2O2(l).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).

The value of ΔHf∘(H2O2, l) is −187.78 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔHf∘(H2O2, l), ΔHf∘(H2O, l) and ΔHf∘(O2, g) in equation (1).

ΔH°={(2 mol)×(−285.83 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−187.78 kJ⋅mol−1)}={−571.66 kJ}−{−375.56 kJ}=−196.1 kJ

The relation for the calculation of standard change in the entropy is shown below.

ΔS° =∑nΔSf∘(products)− ∑nΔSf∘(reactants)

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(2 mol)×ΔSf∘(H2O, l)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(H2O2, l)} (2)

Where,

ΔSf∘(H2O2, l) is the standard entropy of formation of H2O2(l).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).

The value of ΔSf∘(H2O2, l) is +109.6 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is +69.91 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(H2O2, l), ΔSf∘(H2O, l) and ΔSf∘(O2, g) in equation (2).

ΔS°={(2 mol)×(69.91 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(109.6 J⋅K−1⋅mol−1)}={139.82 J⋅K−1+205.138 J⋅K−1}−{219.2 J⋅K−1}=125.758 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(2 mol)×ΔGf∘(H2O, l)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(H2O2, l)} (3)

Where,

ΔGf∘(H2O2, l) is the standard Gibbs free energy of formation of H2O2(l).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).

The value of ΔGf∘(H2O2, l) is −120.35 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔGf∘(H2O2, l), ΔGf∘(H2O, l) and ΔGf∘(O2, g) in equation (3).

ΔG°={(2 mol)×(−237.13 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−120.35 kJ⋅mol−1)}={−474.26 kJ}−{−240.7 kJ}=−233.56 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Explanation of Solution

The given chemical equation for the production of hydrofluoric acid is shown below.

2F2(g)+2H2O(l)→4HF(g)+O2(g)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(4 mol)×ΔHf∘(HF, g)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(F2, g)+(2 mol)×ΔHf∘(H2O, l)} (4)

Where,

ΔHf∘(HF, g) is the standard enthalpy of formation of HF(g).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).
ΔHf∘(F2, g) is the standard enthalpy of formation of F2(g).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).

The value of ΔHf∘(HF, g) is −332.63 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

Substitute the value of ΔHf∘(HF, g), ΔHf∘(O2, g), ΔHf∘(F2, g) and ΔHf∘(H2O, l) in equation (4).

ΔH°={(4 mol)×(−332.63 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−285.83 kJ⋅mol−1)}={−1330.52 kJ}−{−571.66 kJ}=−758.86 kJ

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(4 mol)×ΔSf∘(HF, g)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(F2, g)+(2 mol)×ΔSf∘(H2O, l)} (5)

Where,

ΔSf∘(HF, g) is the standard entropy of formation of HF(g).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).
ΔSf∘(F2, g) is the standard entropy of formation of F2(g).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).

The value of ΔSf∘(HF, g) is −13.8 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

The value of ΔHf∘(F2, g) is +202.78 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is 69.91 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(HF, g), ΔSf∘(F2, g), ΔSf∘(O2, g) and ΔSf∘(H2O, l) in equation (5).

ΔS°={(4 mol)×(−13.8 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(202.78 J⋅K−1⋅mol−1)+(2 mol)×(69.91 J⋅K−1⋅mol−1)}={−55.2 J⋅K−1+205.138 J⋅K−1}−{405.56 J⋅K−1+139.82 J⋅K−1}=−395.442 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(4 mol)×ΔGf∘(HF, g)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(F2, g)+(2 mol)×ΔGf∘(H2O, l)} (6)

Where,

ΔGf∘(HF, g) is the standard Gibbs free energy of formation of HF(g).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).
ΔGf∘(F2, g) is the standard Gibbs free energy of formation of F2(g).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).

The value of ΔGf∘(HF, g) is −278.79 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

Substitute the value of ΔGf∘(HF, g), ΔGf∘(O2, g), ΔGf∘(H2O, l) and ΔGf∘(F2, g) in equation (6).

ΔG°={(4 mol)×(−278.79 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−237.13 kJ⋅mol−1)}={−1115.16 kJ}−{−474.26 kJ}=−640.9 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

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Answer 5

Chapter 4, Problem 4J.7E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure. A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation. The relation to determine Gibbs free energy of a reaction is shown below.

ΔG° =∑nΔGf∘(products)− ∑nΔGf∘(reactants)

(a)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Explanation of Solution

The given chemical equation for the decomposition of hydrogen peroxide is shown below.

2H2O2(l)→2H2O(l)+O2(g)

The relation for the calculation of standard change in the enthalpy is shown below.

ΔH° =∑nΔHf∘(products)− ∑nΔHf∘(reactants)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(2 mol)×ΔHf∘(H2O, l)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(H2O2, l)} (1)

Where,

ΔHf∘(H2O2, l) is the standard enthalpy of formation of H2O2(l).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).

The value of ΔHf∘(H2O2, l) is −187.78 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔHf∘(H2O2, l), ΔHf∘(H2O, l) and ΔHf∘(O2, g) in equation (1).

ΔH°={(2 mol)×(−285.83 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−187.78 kJ⋅mol−1)}={−571.66 kJ}−{−375.56 kJ}=−196.1 kJ

The relation for the calculation of standard change in the entropy is shown below.

ΔS° =∑nΔSf∘(products)− ∑nΔSf∘(reactants)

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(2 mol)×ΔSf∘(H2O, l)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(H2O2, l)} (2)

Where,

ΔSf∘(H2O2, l) is the standard entropy of formation of H2O2(l).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).

The value of ΔSf∘(H2O2, l) is +109.6 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is +69.91 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(H2O2, l), ΔSf∘(H2O, l) and ΔSf∘(O2, g) in equation (2).

ΔS°={(2 mol)×(69.91 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(109.6 J⋅K−1⋅mol−1)}={139.82 J⋅K−1+205.138 J⋅K−1}−{219.2 J⋅K−1}=125.758 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(2 mol)×ΔGf∘(H2O, l)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(H2O2, l)} (3)

Where,

ΔGf∘(H2O2, l) is the standard Gibbs free energy of formation of H2O2(l).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).

The value of ΔGf∘(H2O2, l) is −120.35 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔGf∘(H2O2, l), ΔGf∘(H2O, l) and ΔGf∘(O2, g) in equation (3).

ΔG°={(2 mol)×(−237.13 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−120.35 kJ⋅mol−1)}={−474.26 kJ}−{−240.7 kJ}=−233.56 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Explanation of Solution

The given chemical equation for the production of hydrofluoric acid is shown below.

2F2(g)+2H2O(l)→4HF(g)+O2(g)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(4 mol)×ΔHf∘(HF, g)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(F2, g)+(2 mol)×ΔHf∘(H2O, l)} (4)

Where,

ΔHf∘(HF, g) is the standard enthalpy of formation of HF(g).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).
ΔHf∘(F2, g) is the standard enthalpy of formation of F2(g).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).

The value of ΔHf∘(HF, g) is −332.63 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

Substitute the value of ΔHf∘(HF, g), ΔHf∘(O2, g), ΔHf∘(F2, g) and ΔHf∘(H2O, l) in equation (4).

ΔH°={(4 mol)×(−332.63 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−285.83 kJ⋅mol−1)}={−1330.52 kJ}−{−571.66 kJ}=−758.86 kJ

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(4 mol)×ΔSf∘(HF, g)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(F2, g)+(2 mol)×ΔSf∘(H2O, l)} (5)

Where,

ΔSf∘(HF, g) is the standard entropy of formation of HF(g).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).
ΔSf∘(F2, g) is the standard entropy of formation of F2(g).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).

The value of ΔSf∘(HF, g) is −13.8 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

The value of ΔHf∘(F2, g) is +202.78 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is 69.91 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(HF, g), ΔSf∘(F2, g), ΔSf∘(O2, g) and ΔSf∘(H2O, l) in equation (5).

ΔS°={(4 mol)×(−13.8 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(202.78 J⋅K−1⋅mol−1)+(2 mol)×(69.91 J⋅K−1⋅mol−1)}={−55.2 J⋅K−1+205.138 J⋅K−1}−{405.56 J⋅K−1+139.82 J⋅K−1}=−395.442 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(4 mol)×ΔGf∘(HF, g)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(F2, g)+(2 mol)×ΔGf∘(H2O, l)} (6)

Where,

ΔGf∘(HF, g) is the standard Gibbs free energy of formation of HF(g).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).
ΔGf∘(F2, g) is the standard Gibbs free energy of formation of F2(g).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).

The value of ΔGf∘(HF, g) is −278.79 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

Substitute the value of ΔGf∘(HF, g), ΔGf∘(O2, g), ΔGf∘(H2O, l) and ΔGf∘(F2, g) in equation (6).

ΔG°={(4 mol)×(−278.79 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−237.13 kJ⋅mol−1)}={−1115.16 kJ}−{−474.26 kJ}=−640.9 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Answer 6

Chapter 4, Problem 4J.7E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure. A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation. The relation to determine Gibbs free energy of a reaction is shown below.

ΔG° =∑nΔGf∘(products)− ∑nΔGf∘(reactants)

(a)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Explanation of Solution

The given chemical equation for the decomposition of hydrogen peroxide is shown below.

2H2O2(l)→2H2O(l)+O2(g)

The relation for the calculation of standard change in the enthalpy is shown below.

ΔH° =∑nΔHf∘(products)− ∑nΔHf∘(reactants)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(2 mol)×ΔHf∘(H2O, l)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(H2O2, l)} (1)

Where,

ΔHf∘(H2O2, l) is the standard enthalpy of formation of H2O2(l).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).

The value of ΔHf∘(H2O2, l) is −187.78 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔHf∘(H2O2, l), ΔHf∘(H2O, l) and ΔHf∘(O2, g) in equation (1).

ΔH°={(2 mol)×(−285.83 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−187.78 kJ⋅mol−1)}={−571.66 kJ}−{−375.56 kJ}=−196.1 kJ

The relation for the calculation of standard change in the entropy is shown below.

ΔS° =∑nΔSf∘(products)− ∑nΔSf∘(reactants)

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(2 mol)×ΔSf∘(H2O, l)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(H2O2, l)} (2)

Where,

ΔSf∘(H2O2, l) is the standard entropy of formation of H2O2(l).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).

The value of ΔSf∘(H2O2, l) is +109.6 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is +69.91 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(H2O2, l), ΔSf∘(H2O, l) and ΔSf∘(O2, g) in equation (2).

ΔS°={(2 mol)×(69.91 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(109.6 J⋅K−1⋅mol−1)}={139.82 J⋅K−1+205.138 J⋅K−1}−{219.2 J⋅K−1}=125.758 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(2 mol)×ΔGf∘(H2O, l)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(H2O2, l)} (3)

Where,

ΔGf∘(H2O2, l) is the standard Gibbs free energy of formation of H2O2(l).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).

The value of ΔGf∘(H2O2, l) is −120.35 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔGf∘(H2O2, l), ΔGf∘(H2O, l) and ΔGf∘(O2, g) in equation (3).

ΔG°={(2 mol)×(−237.13 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−120.35 kJ⋅mol−1)}={−474.26 kJ}−{−240.7 kJ}=−233.56 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Answer 7

Chapter 4, Problem 4J.7E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure. A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation. The relation to determine Gibbs free energy of a reaction is shown below.

ΔG° =∑nΔGf∘(products)− ∑nΔGf∘(reactants)

Answer 8

(a)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The given chemical equation for the decomposition of hydrogen peroxide is shown below.

2H2O2(l)→2H2O(l)+O2(g)

The relation for the calculation of standard change in the enthalpy is shown below.

ΔH° =∑nΔHf∘(products)− ∑nΔHf∘(reactants)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(2 mol)×ΔHf∘(H2O, l)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(H2O2, l)} (1)

Where,

ΔHf∘(H2O2, l) is the standard enthalpy of formation of H2O2(l).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).

The value of ΔHf∘(H2O2, l) is −187.78 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔHf∘(H2O2, l), ΔHf∘(H2O, l) and ΔHf∘(O2, g) in equation (1).

ΔH°={(2 mol)×(−285.83 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−187.78 kJ⋅mol−1)}={−571.66 kJ}−{−375.56 kJ}=−196.1 kJ

The relation for the calculation of standard change in the entropy is shown below.

ΔS° =∑nΔSf∘(products)− ∑nΔSf∘(reactants)

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(2 mol)×ΔSf∘(H2O, l)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(H2O2, l)} (2)

Where,

ΔSf∘(H2O2, l) is the standard entropy of formation of H2O2(l).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).

The value of ΔSf∘(H2O2, l) is +109.6 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is +69.91 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(H2O2, l), ΔSf∘(H2O, l) and ΔSf∘(O2, g) in equation (2).

ΔS°={(2 mol)×(69.91 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(109.6 J⋅K−1⋅mol−1)}={139.82 J⋅K−1+205.138 J⋅K−1}−{219.2 J⋅K−1}=125.758 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(2 mol)×ΔGf∘(H2O, l)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(H2O2, l)} (3)

Where,

ΔGf∘(H2O2, l) is the standard Gibbs free energy of formation of H2O2(l).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).

The value of ΔGf∘(H2O2, l) is −120.35 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

Substitute the value of ΔGf∘(H2O2, l), ΔGf∘(H2O, l) and ΔGf∘(O2, g) in equation (3).

ΔG°={(2 mol)×(−237.13 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(−120.35 kJ⋅mol−1)}={−474.26 kJ}−{−240.7 kJ}=−233.56 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -196.1 kJ_, 125.758 J⋅K-1_ and -233.56 kJ_ respectively.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Explanation of Solution

The given chemical equation for the production of hydrofluoric acid is shown below.

2F2(g)+2H2O(l)→4HF(g)+O2(g)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(4 mol)×ΔHf∘(HF, g)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(F2, g)+(2 mol)×ΔHf∘(H2O, l)} (4)

Where,

ΔHf∘(HF, g) is the standard enthalpy of formation of HF(g).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).
ΔHf∘(F2, g) is the standard enthalpy of formation of F2(g).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).

The value of ΔHf∘(HF, g) is −332.63 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

Substitute the value of ΔHf∘(HF, g), ΔHf∘(O2, g), ΔHf∘(F2, g) and ΔHf∘(H2O, l) in equation (4).

ΔH°={(4 mol)×(−332.63 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−285.83 kJ⋅mol−1)}={−1330.52 kJ}−{−571.66 kJ}=−758.86 kJ

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(4 mol)×ΔSf∘(HF, g)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(F2, g)+(2 mol)×ΔSf∘(H2O, l)} (5)

Where,

ΔSf∘(HF, g) is the standard entropy of formation of HF(g).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).
ΔSf∘(F2, g) is the standard entropy of formation of F2(g).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).

The value of ΔSf∘(HF, g) is −13.8 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

The value of ΔHf∘(F2, g) is +202.78 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is 69.91 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(HF, g), ΔSf∘(F2, g), ΔSf∘(O2, g) and ΔSf∘(H2O, l) in equation (5).

ΔS°={(4 mol)×(−13.8 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(202.78 J⋅K−1⋅mol−1)+(2 mol)×(69.91 J⋅K−1⋅mol−1)}={−55.2 J⋅K−1+205.138 J⋅K−1}−{405.56 J⋅K−1+139.82 J⋅K−1}=−395.442 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(4 mol)×ΔGf∘(HF, g)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(F2, g)+(2 mol)×ΔGf∘(H2O, l)} (6)

Where,

ΔGf∘(HF, g) is the standard Gibbs free energy of formation of HF(g).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).
ΔGf∘(F2, g) is the standard Gibbs free energy of formation of F2(g).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).

The value of ΔGf∘(HF, g) is −278.79 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

Substitute the value of ΔGf∘(HF, g), ΔGf∘(O2, g), ΔGf∘(H2O, l) and ΔGf∘(F2, g) in equation (6).

ΔG°={(4 mol)×(−278.79 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−237.13 kJ⋅mol−1)}={−1115.16 kJ}−{−474.26 kJ}=−640.9 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The given chemical equation for the production of hydrofluoric acid is shown below.

2F2(g)+2H2O(l)→4HF(g)+O2(g)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

ΔH°={(4 mol)×ΔHf∘(HF, g)+(1 mol)×ΔHf∘(O2, g)}−{(2 mol)×ΔHf∘(F2, g)+(2 mol)×ΔHf∘(H2O, l)} (4)

Where,

ΔHf∘(HF, g) is the standard enthalpy of formation of HF(g).
ΔHf∘(O2, g) is the standard enthalpy of formation of O2(g).
ΔHf∘(F2, g) is the standard enthalpy of formation of F2(g).
ΔHf∘(H2O, l) is the standard enthalpy of formation of H2O(l).

The value of ΔHf∘(HF, g) is −332.63 kJ⋅mol−1.

The value of ΔHf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔHf∘(H2O, l) is −285.83 kJ⋅mol−1.

Substitute the value of ΔHf∘(HF, g), ΔHf∘(O2, g), ΔHf∘(F2, g) and ΔHf∘(H2O, l) in equation (4).

ΔH°={(4 mol)×(−332.63 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−285.83 kJ⋅mol−1)}={−1330.52 kJ}−{−571.66 kJ}=−758.86 kJ

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

ΔS°={(4 mol)×ΔSf∘(HF, g)+(1 mol)×ΔSf∘(O2, g)}−{(2 mol)×ΔSf∘(F2, g)+(2 mol)×ΔSf∘(H2O, l)} (5)

Where,

ΔSf∘(HF, g) is the standard entropy of formation of HF(g).
ΔSf∘(O2, g) is the standard entropy of formation of O2(g).
ΔSf∘(F2, g) is the standard entropy of formation of F2(g).
ΔSf∘(H2O, l) is the standard entropy of formation of H2O(l).

The value of ΔSf∘(HF, g) is −13.8 J⋅K−1⋅mol−1.

The value of ΔSf∘(O2, g) is +205.138 J⋅K−1⋅mol−1.

The value of ΔHf∘(F2, g) is +202.78 J⋅K−1⋅mol−1.

The value of ΔSf∘(H2O, l) is 69.91 J⋅K−1⋅mol−1.

Substitute the value of ΔSf∘(HF, g), ΔSf∘(F2, g), ΔSf∘(O2, g) and ΔSf∘(H2O, l) in equation (5).

ΔS°={(4 mol)×(−13.8 J⋅K−1⋅mol−1)+(1 mol)×(205.138 J⋅K−1⋅mol−1)}−{(2 mol)×(202.78 J⋅K−1⋅mol−1)+(2 mol)×(69.91 J⋅K−1⋅mol−1)}={−55.2 J⋅K−1+205.138 J⋅K−1}−{405.56 J⋅K−1+139.82 J⋅K−1}=−395.442 J⋅K−1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

ΔG°={(4 mol)×ΔGf∘(HF, g)+(1 mol)×ΔGf∘(O2, g)}−{(2 mol)×ΔGf∘(F2, g)+(2 mol)×ΔGf∘(H2O, l)} (6)

Where,

ΔGf∘(HF, g) is the standard Gibbs free energy of formation of HF(g).
ΔGf∘(O2, g) is the standard Gibbs free energy of formation of O2(g).
ΔGf∘(F2, g) is the standard Gibbs free energy of formation of F2(g).
ΔGf∘(H2O, l) is the standard Gibbs free energy of formation of H2O(l).

The value of ΔGf∘(HF, g) is −278.79 kJ⋅mol−1.

The value of ΔGf∘(O2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(F2, g) is 0.0 kJ⋅mol−1.

The value of ΔGf∘(H2O, l) is −237.13 kJ⋅mol−1.

Substitute the value of ΔGf∘(HF, g), ΔGf∘(O2, g), ΔGf∘(H2O, l) and ΔGf∘(F2, g) in equation (6).

ΔG°={(4 mol)×(−278.79 kJ⋅mol−1)+(1 mol)×(0.0 kJ⋅mol−1)}−{(2 mol)×(0.0 kJ⋅mol−1)+(2 mol)×(−237.13 kJ⋅mol−1)}={−1115.16 kJ}−{−474.26 kJ}=−640.9 kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction -758.86 kJ_, -395.442 J⋅K-1_ and -640.9 kJ_ respectively.