a. ∆ G o for the given reaction at 298 K. b. If the reaction is spontaneous at 298 K c. ∆ G for the given reaction at 1500 K d. If the reaction is spontaneous at 1500 K e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K )

Question

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Answer 1

Question

Chapter 12, Problem 12.68QA

Interpretation Introduction

To find:

a. ∆Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. ∆G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Expert Solution & Answer

Answer to Problem 12.68QA

Solution:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Explanation of Solution

1) Concept:

We have been asked to calculate the ∆Go at 298 K by using appendix 4. We can calculate ∆Go from the ∆Hrxno and ∆So values. We can determine the spontaneity of the reaction from the sign of the ∆Go values.

2) Given information:

Substance	CaOs	CO2g	CaCO3s
∆HfokJ/mol	-634.9	-393.5	-1206.9
SoJ/mol.K	38.1	213.8	92.9

3) Formulae:

i. ∆Hrxno= ∑nproduct ∆Hf, producto- ∑nreactant ∆Hf,reactanto

ii. ∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

iii. ∆Grxno=∆Hrxno-T∆Srxno

4) Calculations:

We will calculate the ∆Go at 298 K from given data.

a. ∆Go for the given reaction at 298 K:

CaOs+CO2(g)→CaCO3(s)

∆Hrxno(Enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

∆Hrxno= ∑nproduct ∆Hf,producto- ∑nreactant ∆Hf,reactanto

∆Hrxno= nCaCO3s ∆H CaCO3so-n CaOs ∆H CaOso+n CO2g ∆H CO2go

∆Hrxno=1 mol ×-1206.9 kJ/mol-1 mol ×-634.9 kJ/mol+1 mol ×-393.5 kJ/ mol

∆Hrxno=-1206.9 kJ+1028.4 kJ

∆Hrxno=-178.5 kJ

Change in entropy of reaction i.e. ∆Srxno:

∆Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies of nproduct moles of products,

Therefore,

∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

∆Srxno= nCaCO3s S CaCO3so-n CaOs S CaOso+n CO2g S CO2go

∆Srxno=1 mol ×92.9J/mol.K-1 mol ×38.1 J/mol.K+1 mol ×213.8 J/mol.K

∆Srxno=92.9 J/K-251.9J/K

∆Srxno=-159.0 J/K

The Gibb’s free energy change for reaction, ∆Grxno at 298 K:

∆Hrxno=-178.5 kJ

∆Srxno=-159.0 J/K= -0.1590 kJ/K

T = 298 K

Therefore, ∆Grxno=∆Hrxno-T∆Srxno

∆Grxno=(-178.5 kJ- 298 K×-0.1590 kJ/K

∆Grxno=(-178.5 kJ+47.382 kJ)

∆Grxno=-131.1 kJ

b. ∆Grxno=-131.1 kJ at 298K, as the value of ∆Grxno is negative. Thus, the reaction is spontaneous.

c. ∆G for the given reaction at 1500 K:

∆Grxn=∆Hrxn-T∆S rxn

∆Grxn=(-178.5 kJ- 1500 K×-0.1590 kJ/K

∆Grxn=(-178.5 kJ+238.5 kJ)

∆Grxn=60.0 kJ

d. ∆Grxn=60.0 kJ at 1500 K, as the value of ∆Grxn is positive. Thus, the reaction is nonspontaneous.

e. ∆Grxn for reverse reaction at the temperature of kiln (i.e. 1500 K):

For the given reaction,

CaOs+CO2(g)→CaCO3(s); ∆Grxn=60.0 kJ

Therefore, for reverse reaction,

CaCO3s→CaOs+CO2g; ∆Grxn=-60.0 kJ

Since the value of ∆Grxn is negative for reverse reaction, the reaction is spontaneous at 1500 K.

Conclusion:

We can calculate the values of ∆Hrxn0 and ∆Srxn0 using the standard values from the appendix and ∆Grxno can be calculate using this equation, ∆Grxn=∆Hrxn-T∆S rxn. With that, we can conclude the following:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

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Answer 2

Question

Chapter 12, Problem 12.68QA

Interpretation Introduction

To find:

a. ∆Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. ∆G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Expert Solution & Answer

Answer to Problem 12.68QA

Solution:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Explanation of Solution

1) Concept:

We have been asked to calculate the ∆Go at 298 K by using appendix 4. We can calculate ∆Go from the ∆Hrxno and ∆So values. We can determine the spontaneity of the reaction from the sign of the ∆Go values.

2) Given information:

Substance	CaOs	CO2g	CaCO3s
∆HfokJ/mol	-634.9	-393.5	-1206.9
SoJ/mol.K	38.1	213.8	92.9

3) Formulae:

i. ∆Hrxno= ∑nproduct ∆Hf, producto- ∑nreactant ∆Hf,reactanto

ii. ∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

iii. ∆Grxno=∆Hrxno-T∆Srxno

4) Calculations:

We will calculate the ∆Go at 298 K from given data.

a. ∆Go for the given reaction at 298 K:

CaOs+CO2(g)→CaCO3(s)

∆Hrxno(Enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

∆Hrxno= ∑nproduct ∆Hf,producto- ∑nreactant ∆Hf,reactanto

∆Hrxno= nCaCO3s ∆H CaCO3so-n CaOs ∆H CaOso+n CO2g ∆H CO2go

∆Hrxno=1 mol ×-1206.9 kJ/mol-1 mol ×-634.9 kJ/mol+1 mol ×-393.5 kJ/ mol

∆Hrxno=-1206.9 kJ+1028.4 kJ

∆Hrxno=-178.5 kJ

Change in entropy of reaction i.e. ∆Srxno:

∆Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies of nproduct moles of products,

Therefore,

∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

∆Srxno= nCaCO3s S CaCO3so-n CaOs S CaOso+n CO2g S CO2go

∆Srxno=1 mol ×92.9J/mol.K-1 mol ×38.1 J/mol.K+1 mol ×213.8 J/mol.K

∆Srxno=92.9 J/K-251.9J/K

∆Srxno=-159.0 J/K

The Gibb’s free energy change for reaction, ∆Grxno at 298 K:

∆Hrxno=-178.5 kJ

∆Srxno=-159.0 J/K= -0.1590 kJ/K

T = 298 K

Therefore, ∆Grxno=∆Hrxno-T∆Srxno

∆Grxno=(-178.5 kJ- 298 K×-0.1590 kJ/K

∆Grxno=(-178.5 kJ+47.382 kJ)

∆Grxno=-131.1 kJ

b. ∆Grxno=-131.1 kJ at 298K, as the value of ∆Grxno is negative. Thus, the reaction is spontaneous.

c. ∆G for the given reaction at 1500 K:

∆Grxn=∆Hrxn-T∆S rxn

∆Grxn=(-178.5 kJ- 1500 K×-0.1590 kJ/K

∆Grxn=(-178.5 kJ+238.5 kJ)

∆Grxn=60.0 kJ

d. ∆Grxn=60.0 kJ at 1500 K, as the value of ∆Grxn is positive. Thus, the reaction is nonspontaneous.

e. ∆Grxn for reverse reaction at the temperature of kiln (i.e. 1500 K):

For the given reaction,

CaOs+CO2(g)→CaCO3(s); ∆Grxn=60.0 kJ

Therefore, for reverse reaction,

CaCO3s→CaOs+CO2g; ∆Grxn=-60.0 kJ

Since the value of ∆Grxn is negative for reverse reaction, the reaction is spontaneous at 1500 K.

Conclusion:

We can calculate the values of ∆Hrxn0 and ∆Srxn0 using the standard values from the appendix and ∆Grxno can be calculate using this equation, ∆Grxn=∆Hrxn-T∆S rxn. With that, we can conclude the following:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

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Answer 3

Question

Chapter 12, Problem 12.68QA

Interpretation Introduction

To find:

a. ∆Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. ∆G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Expert Solution & Answer

Answer to Problem 12.68QA

Solution:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Explanation of Solution

1) Concept:

We have been asked to calculate the ∆Go at 298 K by using appendix 4. We can calculate ∆Go from the ∆Hrxno and ∆So values. We can determine the spontaneity of the reaction from the sign of the ∆Go values.

2) Given information:

Substance	CaOs	CO2g	CaCO3s
∆HfokJ/mol	-634.9	-393.5	-1206.9
SoJ/mol.K	38.1	213.8	92.9

3) Formulae:

i. ∆Hrxno= ∑nproduct ∆Hf, producto- ∑nreactant ∆Hf,reactanto

ii. ∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

iii. ∆Grxno=∆Hrxno-T∆Srxno

4) Calculations:

We will calculate the ∆Go at 298 K from given data.

a. ∆Go for the given reaction at 298 K:

CaOs+CO2(g)→CaCO3(s)

∆Hrxno(Enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

∆Hrxno= ∑nproduct ∆Hf,producto- ∑nreactant ∆Hf,reactanto

∆Hrxno= nCaCO3s ∆H CaCO3so-n CaOs ∆H CaOso+n CO2g ∆H CO2go

∆Hrxno=1 mol ×-1206.9 kJ/mol-1 mol ×-634.9 kJ/mol+1 mol ×-393.5 kJ/ mol

∆Hrxno=-1206.9 kJ+1028.4 kJ

∆Hrxno=-178.5 kJ

Change in entropy of reaction i.e. ∆Srxno:

∆Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies of nproduct moles of products,

Therefore,

∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

∆Srxno= nCaCO3s S CaCO3so-n CaOs S CaOso+n CO2g S CO2go

∆Srxno=1 mol ×92.9J/mol.K-1 mol ×38.1 J/mol.K+1 mol ×213.8 J/mol.K

∆Srxno=92.9 J/K-251.9J/K

∆Srxno=-159.0 J/K

The Gibb’s free energy change for reaction, ∆Grxno at 298 K:

∆Hrxno=-178.5 kJ

∆Srxno=-159.0 J/K= -0.1590 kJ/K

T = 298 K

Therefore, ∆Grxno=∆Hrxno-T∆Srxno

∆Grxno=(-178.5 kJ- 298 K×-0.1590 kJ/K

∆Grxno=(-178.5 kJ+47.382 kJ)

∆Grxno=-131.1 kJ

b. ∆Grxno=-131.1 kJ at 298K, as the value of ∆Grxno is negative. Thus, the reaction is spontaneous.

c. ∆G for the given reaction at 1500 K:

∆Grxn=∆Hrxn-T∆S rxn

∆Grxn=(-178.5 kJ- 1500 K×-0.1590 kJ/K

∆Grxn=(-178.5 kJ+238.5 kJ)

∆Grxn=60.0 kJ

d. ∆Grxn=60.0 kJ at 1500 K, as the value of ∆Grxn is positive. Thus, the reaction is nonspontaneous.

e. ∆Grxn for reverse reaction at the temperature of kiln (i.e. 1500 K):

For the given reaction,

CaOs+CO2(g)→CaCO3(s); ∆Grxn=60.0 kJ

Therefore, for reverse reaction,

CaCO3s→CaOs+CO2g; ∆Grxn=-60.0 kJ

Since the value of ∆Grxn is negative for reverse reaction, the reaction is spontaneous at 1500 K.

Conclusion:

We can calculate the values of ∆Hrxn0 and ∆Srxn0 using the standard values from the appendix and ∆Grxno can be calculate using this equation, ∆Grxn=∆Hrxn-T∆S rxn. With that, we can conclude the following:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

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Answer 4

Question

Chapter 12, Problem 12.68QA

Interpretation Introduction

To find:

a. ∆Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. ∆G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Expert Solution & Answer

Answer to Problem 12.68QA

Solution:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Explanation of Solution

1) Concept:

We have been asked to calculate the ∆Go at 298 K by using appendix 4. We can calculate ∆Go from the ∆Hrxno and ∆So values. We can determine the spontaneity of the reaction from the sign of the ∆Go values.

2) Given information:

Substance	CaOs	CO2g	CaCO3s
∆HfokJ/mol	-634.9	-393.5	-1206.9
SoJ/mol.K	38.1	213.8	92.9

3) Formulae:

i. ∆Hrxno= ∑nproduct ∆Hf, producto- ∑nreactant ∆Hf,reactanto

ii. ∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

iii. ∆Grxno=∆Hrxno-T∆Srxno

4) Calculations:

We will calculate the ∆Go at 298 K from given data.

a. ∆Go for the given reaction at 298 K:

CaOs+CO2(g)→CaCO3(s)

∆Hrxno(Enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

∆Hrxno= ∑nproduct ∆Hf,producto- ∑nreactant ∆Hf,reactanto

∆Hrxno= nCaCO3s ∆H CaCO3so-n CaOs ∆H CaOso+n CO2g ∆H CO2go

∆Hrxno=1 mol ×-1206.9 kJ/mol-1 mol ×-634.9 kJ/mol+1 mol ×-393.5 kJ/ mol

∆Hrxno=-1206.9 kJ+1028.4 kJ

∆Hrxno=-178.5 kJ

Change in entropy of reaction i.e. ∆Srxno:

∆Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies of nproduct moles of products,

Therefore,

∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

∆Srxno= nCaCO3s S CaCO3so-n CaOs S CaOso+n CO2g S CO2go

∆Srxno=1 mol ×92.9J/mol.K-1 mol ×38.1 J/mol.K+1 mol ×213.8 J/mol.K

∆Srxno=92.9 J/K-251.9J/K

∆Srxno=-159.0 J/K

The Gibb’s free energy change for reaction, ∆Grxno at 298 K:

∆Hrxno=-178.5 kJ

∆Srxno=-159.0 J/K= -0.1590 kJ/K

T = 298 K

Therefore, ∆Grxno=∆Hrxno-T∆Srxno

∆Grxno=(-178.5 kJ- 298 K×-0.1590 kJ/K

∆Grxno=(-178.5 kJ+47.382 kJ)

∆Grxno=-131.1 kJ

b. ∆Grxno=-131.1 kJ at 298K, as the value of ∆Grxno is negative. Thus, the reaction is spontaneous.

c. ∆G for the given reaction at 1500 K:

∆Grxn=∆Hrxn-T∆S rxn

∆Grxn=(-178.5 kJ- 1500 K×-0.1590 kJ/K

∆Grxn=(-178.5 kJ+238.5 kJ)

∆Grxn=60.0 kJ

d. ∆Grxn=60.0 kJ at 1500 K, as the value of ∆Grxn is positive. Thus, the reaction is nonspontaneous.

e. ∆Grxn for reverse reaction at the temperature of kiln (i.e. 1500 K):

For the given reaction,

CaOs+CO2(g)→CaCO3(s); ∆Grxn=60.0 kJ

Therefore, for reverse reaction,

CaCO3s→CaOs+CO2g; ∆Grxn=-60.0 kJ

Since the value of ∆Grxn is negative for reverse reaction, the reaction is spontaneous at 1500 K.

Conclusion:

We can calculate the values of ∆Hrxn0 and ∆Srxn0 using the standard values from the appendix and ∆Grxno can be calculate using this equation, ∆Grxn=∆Hrxn-T∆S rxn. With that, we can conclude the following:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

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Answer 5

Chapter 12, Problem 12.68QA

Interpretation Introduction

To find:

a. ∆Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. ∆G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Expert Solution & Answer

Answer to Problem 12.68QA

Solution:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Explanation of Solution

1) Concept:

We have been asked to calculate the ∆Go at 298 K by using appendix 4. We can calculate ∆Go from the ∆Hrxno and ∆So values. We can determine the spontaneity of the reaction from the sign of the ∆Go values.

2) Given information:

Substance	CaOs	CO2g	CaCO3s
∆HfokJ/mol	-634.9	-393.5	-1206.9
SoJ/mol.K	38.1	213.8	92.9

3) Formulae:

i. ∆Hrxno= ∑nproduct ∆Hf, producto- ∑nreactant ∆Hf,reactanto

ii. ∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

iii. ∆Grxno=∆Hrxno-T∆Srxno

4) Calculations:

We will calculate the ∆Go at 298 K from given data.

a. ∆Go for the given reaction at 298 K:

CaOs+CO2(g)→CaCO3(s)

∆Hrxno(Enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

∆Hrxno= ∑nproduct ∆Hf,producto- ∑nreactant ∆Hf,reactanto

∆Hrxno= nCaCO3s ∆H CaCO3so-n CaOs ∆H CaOso+n CO2g ∆H CO2go

∆Hrxno=1 mol ×-1206.9 kJ/mol-1 mol ×-634.9 kJ/mol+1 mol ×-393.5 kJ/ mol

∆Hrxno=-1206.9 kJ+1028.4 kJ

∆Hrxno=-178.5 kJ

Change in entropy of reaction i.e. ∆Srxno:

∆Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies of nproduct moles of products,

Therefore,

∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

∆Srxno= nCaCO3s S CaCO3so-n CaOs S CaOso+n CO2g S CO2go

∆Srxno=1 mol ×92.9J/mol.K-1 mol ×38.1 J/mol.K+1 mol ×213.8 J/mol.K

∆Srxno=92.9 J/K-251.9J/K

∆Srxno=-159.0 J/K

The Gibb’s free energy change for reaction, ∆Grxno at 298 K:

∆Hrxno=-178.5 kJ

∆Srxno=-159.0 J/K= -0.1590 kJ/K

T = 298 K

Therefore, ∆Grxno=∆Hrxno-T∆Srxno

∆Grxno=(-178.5 kJ- 298 K×-0.1590 kJ/K

∆Grxno=(-178.5 kJ+47.382 kJ)

∆Grxno=-131.1 kJ

b. ∆Grxno=-131.1 kJ at 298K, as the value of ∆Grxno is negative. Thus, the reaction is spontaneous.

c. ∆G for the given reaction at 1500 K:

∆Grxn=∆Hrxn-T∆S rxn

∆Grxn=(-178.5 kJ- 1500 K×-0.1590 kJ/K

∆Grxn=(-178.5 kJ+238.5 kJ)

∆Grxn=60.0 kJ

d. ∆Grxn=60.0 kJ at 1500 K, as the value of ∆Grxn is positive. Thus, the reaction is nonspontaneous.

e. ∆Grxn for reverse reaction at the temperature of kiln (i.e. 1500 K):

For the given reaction,

CaOs+CO2(g)→CaCO3(s); ∆Grxn=60.0 kJ

Therefore, for reverse reaction,

CaCO3s→CaOs+CO2g; ∆Grxn=-60.0 kJ

Since the value of ∆Grxn is negative for reverse reaction, the reaction is spontaneous at 1500 K.

Conclusion:

We can calculate the values of ∆Hrxn0 and ∆Srxn0 using the standard values from the appendix and ∆Grxno can be calculate using this equation, ∆Grxn=∆Hrxn-T∆S rxn. With that, we can conclude the following:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Answer 6

Chapter 12, Problem 12.68QA

Interpretation Introduction

To find:

a. ∆Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. ∆G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Expert Solution & Answer

Answer to Problem 12.68QA

Solution:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Explanation of Solution

1) Concept:

We have been asked to calculate the ∆Go at 298 K by using appendix 4. We can calculate ∆Go from the ∆Hrxno and ∆So values. We can determine the spontaneity of the reaction from the sign of the ∆Go values.

2) Given information:

Substance	CaOs	CO2g	CaCO3s
∆HfokJ/mol	-634.9	-393.5	-1206.9
SoJ/mol.K	38.1	213.8	92.9

3) Formulae:

i. ∆Hrxno= ∑nproduct ∆Hf, producto- ∑nreactant ∆Hf,reactanto

ii. ∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

iii. ∆Grxno=∆Hrxno-T∆Srxno

4) Calculations:

We will calculate the ∆Go at 298 K from given data.

a. ∆Go for the given reaction at 298 K:

CaOs+CO2(g)→CaCO3(s)

∆Hrxno(Enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

∆Hrxno= ∑nproduct ∆Hf,producto- ∑nreactant ∆Hf,reactanto

∆Hrxno= nCaCO3s ∆H CaCO3so-n CaOs ∆H CaOso+n CO2g ∆H CO2go

∆Hrxno=1 mol ×-1206.9 kJ/mol-1 mol ×-634.9 kJ/mol+1 mol ×-393.5 kJ/ mol

∆Hrxno=-1206.9 kJ+1028.4 kJ

∆Hrxno=-178.5 kJ

Change in entropy of reaction i.e. ∆Srxno:

∆Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies of nproduct moles of products,

Therefore,

∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

∆Srxno= nCaCO3s S CaCO3so-n CaOs S CaOso+n CO2g S CO2go

∆Srxno=1 mol ×92.9J/mol.K-1 mol ×38.1 J/mol.K+1 mol ×213.8 J/mol.K

∆Srxno=92.9 J/K-251.9J/K

∆Srxno=-159.0 J/K

The Gibb’s free energy change for reaction, ∆Grxno at 298 K:

∆Hrxno=-178.5 kJ

∆Srxno=-159.0 J/K= -0.1590 kJ/K

T = 298 K

Therefore, ∆Grxno=∆Hrxno-T∆Srxno

∆Grxno=(-178.5 kJ- 298 K×-0.1590 kJ/K

∆Grxno=(-178.5 kJ+47.382 kJ)

∆Grxno=-131.1 kJ

b. ∆Grxno=-131.1 kJ at 298K, as the value of ∆Grxno is negative. Thus, the reaction is spontaneous.

c. ∆G for the given reaction at 1500 K:

∆Grxn=∆Hrxn-T∆S rxn

∆Grxn=(-178.5 kJ- 1500 K×-0.1590 kJ/K

∆Grxn=(-178.5 kJ+238.5 kJ)

∆Grxn=60.0 kJ

d. ∆Grxn=60.0 kJ at 1500 K, as the value of ∆Grxn is positive. Thus, the reaction is nonspontaneous.

e. ∆Grxn for reverse reaction at the temperature of kiln (i.e. 1500 K):

For the given reaction,

CaOs+CO2(g)→CaCO3(s); ∆Grxn=60.0 kJ

Therefore, for reverse reaction,

CaCO3s→CaOs+CO2g; ∆Grxn=-60.0 kJ

Since the value of ∆Grxn is negative for reverse reaction, the reaction is spontaneous at 1500 K.

Conclusion:

We can calculate the values of ∆Hrxn0 and ∆Srxn0 using the standard values from the appendix and ∆Grxno can be calculate using this equation, ∆Grxn=∆Hrxn-T∆S rxn. With that, we can conclude the following:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Answer 7

Chapter 12, Problem 12.68QA

Interpretation Introduction

To find:

a. ∆Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. ∆G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Answer 8

Expert Solution & Answer

Answer to Problem 12.68QA

Solution:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

1) Concept:

We have been asked to calculate the ∆Go at 298 K by using appendix 4. We can calculate ∆Go from the ∆Hrxno and ∆So values. We can determine the spontaneity of the reaction from the sign of the ∆Go values.

2) Given information:

Substance	CaOs	CO2g	CaCO3s
∆HfokJ/mol	-634.9	-393.5	-1206.9
SoJ/mol.K	38.1	213.8	92.9

3) Formulae:

i. ∆Hrxno= ∑nproduct ∆Hf, producto- ∑nreactant ∆Hf,reactanto

ii. ∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

iii. ∆Grxno=∆Hrxno-T∆Srxno

4) Calculations:

We will calculate the ∆Go at 298 K from given data.

a. ∆Go for the given reaction at 298 K:

CaOs+CO2(g)→CaCO3(s)

∆Hrxno(Enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

∆Hrxno= ∑nproduct ∆Hf,producto- ∑nreactant ∆Hf,reactanto

∆Hrxno= nCaCO3s ∆H CaCO3so-n CaOs ∆H CaOso+n CO2g ∆H CO2go

∆Hrxno=1 mol ×-1206.9 kJ/mol-1 mol ×-634.9 kJ/mol+1 mol ×-393.5 kJ/ mol

∆Hrxno=-1206.9 kJ+1028.4 kJ

∆Hrxno=-178.5 kJ

Change in entropy of reaction i.e. ∆Srxno:

∆Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies of nproduct moles of products,

Therefore,

∆Srxno= ∑nproduct Sproducto- ∑nreactant Sreactanto

∆Srxno= nCaCO3s S CaCO3so-n CaOs S CaOso+n CO2g S CO2go

∆Srxno=1 mol ×92.9J/mol.K-1 mol ×38.1 J/mol.K+1 mol ×213.8 J/mol.K

∆Srxno=92.9 J/K-251.9J/K

∆Srxno=-159.0 J/K

The Gibb’s free energy change for reaction, ∆Grxno at 298 K:

∆Hrxno=-178.5 kJ

∆Srxno=-159.0 J/K= -0.1590 kJ/K

T = 298 K

Therefore, ∆Grxno=∆Hrxno-T∆Srxno

∆Grxno=(-178.5 kJ- 298 K×-0.1590 kJ/K

∆Grxno=(-178.5 kJ+47.382 kJ)

∆Grxno=-131.1 kJ

b. ∆Grxno=-131.1 kJ at 298K, as the value of ∆Grxno is negative. Thus, the reaction is spontaneous.

c. ∆G for the given reaction at 1500 K:

∆Grxn=∆Hrxn-T∆S rxn

∆Grxn=(-178.5 kJ- 1500 K×-0.1590 kJ/K

∆Grxn=(-178.5 kJ+238.5 kJ)

∆Grxn=60.0 kJ

d. ∆Grxn=60.0 kJ at 1500 K, as the value of ∆Grxn is positive. Thus, the reaction is nonspontaneous.

e. ∆Grxn for reverse reaction at the temperature of kiln (i.e. 1500 K):

For the given reaction,

CaOs+CO2(g)→CaCO3(s); ∆Grxn=60.0 kJ

Therefore, for reverse reaction,

CaCO3s→CaOs+CO2g; ∆Grxn=-60.0 kJ

Since the value of ∆Grxn is negative for reverse reaction, the reaction is spontaneous at 1500 K.

Conclusion:

We can calculate the values of ∆Hrxn0 and ∆Srxn0 using the standard values from the appendix and ∆Grxno can be calculate using this equation, ∆Grxn=∆Hrxn-T∆S rxn. With that, we can conclude the following:

a. At 298 K, the value of ∆Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of ∆G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Answer 12

Ch. 12 - Prob. 12.1VP Ch. 12 - Prob. 12.2VP Ch. 12 - Prob. 12.3VP Ch. 12 - Prob. 12.4VP Ch. 12 - Prob. 12.5VP Ch. 12 - Prob. 12.6VP Ch. 12 - Prob. 12.7VP Ch. 12 - Prob. 12.8VP Ch. 12 - Prob. 12.9VP Ch. 12 - Prob. 12.10VP

a. ∆ G o for the given reaction at 298 K. b. If the reaction is spontaneous at 298 K c. ∆ G for the given reaction at 1500 K d. If the reaction is spontaneous at 1500 K e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K )

To find: a. ∆Go for the given reaction at 298 K. b. If the reaction is spontaneous at 298 K c. ∆G for the given reaction at 1500 K d. If the reaction is spontaneous at 1500 K e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Answer to Problem 12.68QA

Explanation of Solution

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Chapter 12 Solutions

To find:

a. ∆Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. ∆G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)