∆ H r x n 0 for the given reactions of methanogenic bacteria.

Question

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Answer 1

Question

Chapter 19, Problem 19.114QA

Interpretation Introduction

To calculate:

∆Hrxn0 for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.114QA

Solution:

a) ∆Hrxn0=-252.9 kJ

b) ∆Hrxn0=-234.7 kJ

Explanation of Solution

1) Concept:

We are given two balanced reactions of methanogenic bacteria and asked to find the ∆H°rxn for each given reaction. We can find the standard enthalpy formation values for CH4g, CO2g,CH3OH l, H2(g)and H2Ol_ can find out the standard enthalpy formation values for from the Appendix and find the standard enthalpy for the reaction using the following equation:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

2) Formula:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

3) Given:

i) 4H2g+CO2g→CH4g+2H2Ol

ii) 4CH3OHl→3CH4g+CO2g+2H2Ol

4) Calculations:

Values of standard enthalpies of formation of H2g, CO2g, CH4g,

H2Og, CH3OH(l) are taken from Appendix 4 are:

	∆Hf0(kJ/mol)
CH3OH l	-238.7
H2g	0.0
CO2g	-393.5
H2O l	-285.8
CH4(g)	-74.8

a) ∆H°rxn for reaction: 4H2g+CO2g→CH4g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=1 mol CH4g×∆HCH4g0+2 mol H2Ol×∆HH2Ol0-4 mol H2×∆HH2g0+1 mol CO2g×∆HCO2g0

∆H°rxn=1 mol CH4g×-74.8kJmol+2 mol H2Ol×-285.8kJmol-4 mol H2× 0.0kJmol+1 mol CO2g×-393.5kJmol

∆H°rxn=-74.8 kJ+(-571.6 kJ)--393.5 kJ=-252.9 kJ

b) ∆H°rxn for reaction: 4CH3OHl→3CH4g+CO2g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=3 mol CH4g×∆HCH4g0+1 mol CO2g×∆HCO2g0+2 mol H2Ol×∆HH2Ol0-(4 mol CH3OH(l)×∆HCH3OHl0)

∆H°rxn=3 mol CH4g×-74.8kJmol+1 mol CO2g×-393.5kJmol+2 mol H2Ol×-285.8kJmol-(4 mol CH3OHl×-238.7kJmol)

∆H°rxn=-1189.5 kJ--954.8 kJ=-234.7 kJ

Conclusion:

Using standard enthalpy values from Appendix 4, we calculated the enthalpy of reactions of methanogenic bacteria.

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Answer 2

Question

Chapter 19, Problem 19.114QA

Interpretation Introduction

To calculate:

∆Hrxn0 for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.114QA

Solution:

a) ∆Hrxn0=-252.9 kJ

b) ∆Hrxn0=-234.7 kJ

Explanation of Solution

1) Concept:

We are given two balanced reactions of methanogenic bacteria and asked to find the ∆H°rxn for each given reaction. We can find the standard enthalpy formation values for CH4g, CO2g,CH3OH l, H2(g)and H2Ol_ can find out the standard enthalpy formation values for from the Appendix and find the standard enthalpy for the reaction using the following equation:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

2) Formula:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

3) Given:

i) 4H2g+CO2g→CH4g+2H2Ol

ii) 4CH3OHl→3CH4g+CO2g+2H2Ol

4) Calculations:

Values of standard enthalpies of formation of H2g, CO2g, CH4g,

H2Og, CH3OH(l) are taken from Appendix 4 are:

	∆Hf0(kJ/mol)
CH3OH l	-238.7
H2g	0.0
CO2g	-393.5
H2O l	-285.8
CH4(g)	-74.8

a) ∆H°rxn for reaction: 4H2g+CO2g→CH4g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=1 mol CH4g×∆HCH4g0+2 mol H2Ol×∆HH2Ol0-4 mol H2×∆HH2g0+1 mol CO2g×∆HCO2g0

∆H°rxn=1 mol CH4g×-74.8kJmol+2 mol H2Ol×-285.8kJmol-4 mol H2× 0.0kJmol+1 mol CO2g×-393.5kJmol

∆H°rxn=-74.8 kJ+(-571.6 kJ)--393.5 kJ=-252.9 kJ

b) ∆H°rxn for reaction: 4CH3OHl→3CH4g+CO2g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=3 mol CH4g×∆HCH4g0+1 mol CO2g×∆HCO2g0+2 mol H2Ol×∆HH2Ol0-(4 mol CH3OH(l)×∆HCH3OHl0)

∆H°rxn=3 mol CH4g×-74.8kJmol+1 mol CO2g×-393.5kJmol+2 mol H2Ol×-285.8kJmol-(4 mol CH3OHl×-238.7kJmol)

∆H°rxn=-1189.5 kJ--954.8 kJ=-234.7 kJ

Conclusion:

Using standard enthalpy values from Appendix 4, we calculated the enthalpy of reactions of methanogenic bacteria.

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Answer 3

Question

Chapter 19, Problem 19.114QA

Interpretation Introduction

To calculate:

∆Hrxn0 for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.114QA

Solution:

a) ∆Hrxn0=-252.9 kJ

b) ∆Hrxn0=-234.7 kJ

Explanation of Solution

1) Concept:

We are given two balanced reactions of methanogenic bacteria and asked to find the ∆H°rxn for each given reaction. We can find the standard enthalpy formation values for CH4g, CO2g,CH3OH l, H2(g)and H2Ol_ can find out the standard enthalpy formation values for from the Appendix and find the standard enthalpy for the reaction using the following equation:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

2) Formula:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

3) Given:

i) 4H2g+CO2g→CH4g+2H2Ol

ii) 4CH3OHl→3CH4g+CO2g+2H2Ol

4) Calculations:

Values of standard enthalpies of formation of H2g, CO2g, CH4g,

H2Og, CH3OH(l) are taken from Appendix 4 are:

	∆Hf0(kJ/mol)
CH3OH l	-238.7
H2g	0.0
CO2g	-393.5
H2O l	-285.8
CH4(g)	-74.8

a) ∆H°rxn for reaction: 4H2g+CO2g→CH4g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=1 mol CH4g×∆HCH4g0+2 mol H2Ol×∆HH2Ol0-4 mol H2×∆HH2g0+1 mol CO2g×∆HCO2g0

∆H°rxn=1 mol CH4g×-74.8kJmol+2 mol H2Ol×-285.8kJmol-4 mol H2× 0.0kJmol+1 mol CO2g×-393.5kJmol

∆H°rxn=-74.8 kJ+(-571.6 kJ)--393.5 kJ=-252.9 kJ

b) ∆H°rxn for reaction: 4CH3OHl→3CH4g+CO2g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=3 mol CH4g×∆HCH4g0+1 mol CO2g×∆HCO2g0+2 mol H2Ol×∆HH2Ol0-(4 mol CH3OH(l)×∆HCH3OHl0)

∆H°rxn=3 mol CH4g×-74.8kJmol+1 mol CO2g×-393.5kJmol+2 mol H2Ol×-285.8kJmol-(4 mol CH3OHl×-238.7kJmol)

∆H°rxn=-1189.5 kJ--954.8 kJ=-234.7 kJ

Conclusion:

Using standard enthalpy values from Appendix 4, we calculated the enthalpy of reactions of methanogenic bacteria.

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Answer 4

Question

Chapter 19, Problem 19.114QA

Interpretation Introduction

To calculate:

∆Hrxn0 for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.114QA

Solution:

a) ∆Hrxn0=-252.9 kJ

b) ∆Hrxn0=-234.7 kJ

Explanation of Solution

1) Concept:

We are given two balanced reactions of methanogenic bacteria and asked to find the ∆H°rxn for each given reaction. We can find the standard enthalpy formation values for CH4g, CO2g,CH3OH l, H2(g)and H2Ol_ can find out the standard enthalpy formation values for from the Appendix and find the standard enthalpy for the reaction using the following equation:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

2) Formula:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

3) Given:

i) 4H2g+CO2g→CH4g+2H2Ol

ii) 4CH3OHl→3CH4g+CO2g+2H2Ol

4) Calculations:

Values of standard enthalpies of formation of H2g, CO2g, CH4g,

H2Og, CH3OH(l) are taken from Appendix 4 are:

	∆Hf0(kJ/mol)
CH3OH l	-238.7
H2g	0.0
CO2g	-393.5
H2O l	-285.8
CH4(g)	-74.8

a) ∆H°rxn for reaction: 4H2g+CO2g→CH4g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=1 mol CH4g×∆HCH4g0+2 mol H2Ol×∆HH2Ol0-4 mol H2×∆HH2g0+1 mol CO2g×∆HCO2g0

∆H°rxn=1 mol CH4g×-74.8kJmol+2 mol H2Ol×-285.8kJmol-4 mol H2× 0.0kJmol+1 mol CO2g×-393.5kJmol

∆H°rxn=-74.8 kJ+(-571.6 kJ)--393.5 kJ=-252.9 kJ

b) ∆H°rxn for reaction: 4CH3OHl→3CH4g+CO2g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=3 mol CH4g×∆HCH4g0+1 mol CO2g×∆HCO2g0+2 mol H2Ol×∆HH2Ol0-(4 mol CH3OH(l)×∆HCH3OHl0)

∆H°rxn=3 mol CH4g×-74.8kJmol+1 mol CO2g×-393.5kJmol+2 mol H2Ol×-285.8kJmol-(4 mol CH3OHl×-238.7kJmol)

∆H°rxn=-1189.5 kJ--954.8 kJ=-234.7 kJ

Conclusion:

Using standard enthalpy values from Appendix 4, we calculated the enthalpy of reactions of methanogenic bacteria.

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Answer 5

Chapter 19, Problem 19.114QA

Interpretation Introduction

To calculate:

∆Hrxn0 for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.114QA

Solution:

a) ∆Hrxn0=-252.9 kJ

b) ∆Hrxn0=-234.7 kJ

Explanation of Solution

1) Concept:

We are given two balanced reactions of methanogenic bacteria and asked to find the ∆H°rxn for each given reaction. We can find the standard enthalpy formation values for CH4g, CO2g,CH3OH l, H2(g)and H2Ol_ can find out the standard enthalpy formation values for from the Appendix and find the standard enthalpy for the reaction using the following equation:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

2) Formula:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

3) Given:

i) 4H2g+CO2g→CH4g+2H2Ol

ii) 4CH3OHl→3CH4g+CO2g+2H2Ol

4) Calculations:

Values of standard enthalpies of formation of H2g, CO2g, CH4g,

H2Og, CH3OH(l) are taken from Appendix 4 are:

	∆Hf0(kJ/mol)
CH3OH l	-238.7
H2g	0.0
CO2g	-393.5
H2O l	-285.8
CH4(g)	-74.8

a) ∆H°rxn for reaction: 4H2g+CO2g→CH4g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=1 mol CH4g×∆HCH4g0+2 mol H2Ol×∆HH2Ol0-4 mol H2×∆HH2g0+1 mol CO2g×∆HCO2g0

∆H°rxn=1 mol CH4g×-74.8kJmol+2 mol H2Ol×-285.8kJmol-4 mol H2× 0.0kJmol+1 mol CO2g×-393.5kJmol

∆H°rxn=-74.8 kJ+(-571.6 kJ)--393.5 kJ=-252.9 kJ

b) ∆H°rxn for reaction: 4CH3OHl→3CH4g+CO2g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=3 mol CH4g×∆HCH4g0+1 mol CO2g×∆HCO2g0+2 mol H2Ol×∆HH2Ol0-(4 mol CH3OH(l)×∆HCH3OHl0)

∆H°rxn=3 mol CH4g×-74.8kJmol+1 mol CO2g×-393.5kJmol+2 mol H2Ol×-285.8kJmol-(4 mol CH3OHl×-238.7kJmol)

∆H°rxn=-1189.5 kJ--954.8 kJ=-234.7 kJ

Conclusion:

Using standard enthalpy values from Appendix 4, we calculated the enthalpy of reactions of methanogenic bacteria.

Answer 6

Chapter 19, Problem 19.114QA

Interpretation Introduction

To calculate:

∆Hrxn0 for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.114QA

Solution:

a) ∆Hrxn0=-252.9 kJ

b) ∆Hrxn0=-234.7 kJ

Explanation of Solution

1) Concept:

We are given two balanced reactions of methanogenic bacteria and asked to find the ∆H°rxn for each given reaction. We can find the standard enthalpy formation values for CH4g, CO2g,CH3OH l, H2(g)and H2Ol_ can find out the standard enthalpy formation values for from the Appendix and find the standard enthalpy for the reaction using the following equation:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

2) Formula:

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

3) Given:

i) 4H2g+CO2g→CH4g+2H2Ol

ii) 4CH3OHl→3CH4g+CO2g+2H2Ol

4) Calculations:

Values of standard enthalpies of formation of H2g, CO2g, CH4g,

H2Og, CH3OH(l) are taken from Appendix 4 are:

	∆Hf0(kJ/mol)
CH3OH l	-238.7
H2g	0.0
CO2g	-393.5
H2O l	-285.8
CH4(g)	-74.8

a) ∆H°rxn for reaction: 4H2g+CO2g→CH4g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=1 mol CH4g×∆HCH4g0+2 mol H2Ol×∆HH2Ol0-4 mol H2×∆HH2g0+1 mol CO2g×∆HCO2g0

∆H°rxn=1 mol CH4g×-74.8kJmol+2 mol H2Ol×-285.8kJmol-4 mol H2× 0.0kJmol+1 mol CO2g×-393.5kJmol

∆H°rxn=-74.8 kJ+(-571.6 kJ)--393.5 kJ=-252.9 kJ

b) ∆H°rxn for reaction: 4CH3OHl→3CH4g+CO2g+2H2Ol

∆H°rxn= ∑nproducts∆Hf, products0- ∑nreactants∆Hf, reactants0

∆H°rxn=3 mol CH4g×∆HCH4g0+1 mol CO2g×∆HCO2g0+2 mol H2Ol×∆HH2Ol0-(4 mol CH3OH(l)×∆HCH3OHl0)

∆H°rxn=3 mol CH4g×-74.8kJmol+1 mol CO2g×-393.5kJmol+2 mol H2Ol×-285.8kJmol-(4 mol CH3OHl×-238.7kJmol)

∆H°rxn=-1189.5 kJ--954.8 kJ=-234.7 kJ

Conclusion:

Using standard enthalpy values from Appendix 4, we calculated the enthalpy of reactions of methanogenic bacteria.

Answer 7

Chapter 19, Problem 19.114QA

Interpretation Introduction

To calculate:

∆Hrxn0 for the given reactions of methanogenic bacteria.

Answer 8

Expert Solution & Answer

Answer to Problem 19.114QA

Solution:

a) ∆Hrxn0=-252.9 kJ

b) ∆Hrxn0=-234.7 kJ

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

1) Concept:

We are given two balanced reactions of methanogenic bacteria and asked to find the ∆H°rxn for each given reaction. We can find the standard enthalpy formation values for CH4g, CO2g,CH3OH l, H2(g)and H2Ol_ can find out the standard enthalpy formation values for from the Appendix and find the standard enthalpy for the reaction using the following equation: