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

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Answer 1

Question

Chapter 19, Problem 19.113QA

Interpretation Introduction

To calculate:

∆H°rxn for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.113QA

Solution:

a) ∆H°rxn=16.2 kJ

b) ∆H°rxn=-126.9 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, CH3COOH land 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) Standard enthalpy formation value for formic acid ∆H°f=-425.0 kJ/mol

ii) Standard enthalpy formation value for CH4g= -74.8 kJ/mol(from Appendix 4)

iii) Standard enthalpy formation value for CO2g= -393.5 kJ/mol(from Appendix 4)

iv) Standard enthalpy formation value for H2Ol= -285.8 kJ/mol(from Appendix 4)

v) Standard enthalpy formation value for CH3COOHl= -484.5 kJ/mol(from Appendix 4)

vi) CH3COOH l→CH4g+CO2g

vii) 4HCOOHl→CH4g+3CO2g+2H2O(l)

4) Calculations:

(1) Calculating standard enthalpy for the first given reaction:

CH3COOH l→CH4g+CO2g

Calculating ∆H°rxn for this reaction using the formula:

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

Plugging in the values in the equation, we get:

∆H°rxn=1mol CH4×-74.8kJmol CH4+ 1mol CO2×-393.5kJmol CO2-1 mol CH3COOH× -484.5kJmol CH3COOH

∆H°rxn=-468.3 kJ--484.5 kJ

∆H°rxn= -468.3 kJ+484.5 kJ=16.2kJ

2) Calculating standard enthalpy for second reaction:

4HCOOHl→CH4g+3CO2g+2H2O(l)

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

Plugging in the values in the equation, we get:

∆H°rxn=2molH2O×-285.8kJmolH2O+ 3mol CO2×-393.5kJmol CO2+(1mol CH4 ×-74.8kJmol CH4-4 mol HCOOH × -425.0kJmol HCOOH

∆H°rxn=-571.6 kJ -1180.5 kJ-74.8 kJ--1700 kJ

∆H°rxn= -1826.9 kJ+1700kJ=-126.9 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 19.113QA

Interpretation Introduction

To calculate:

∆H°rxn for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.113QA

Solution:

a) ∆H°rxn=16.2 kJ

b) ∆H°rxn=-126.9 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, CH3COOH land 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) Standard enthalpy formation value for formic acid ∆H°f=-425.0 kJ/mol

ii) Standard enthalpy formation value for CH4g= -74.8 kJ/mol(from Appendix 4)

iii) Standard enthalpy formation value for CO2g= -393.5 kJ/mol(from Appendix 4)

iv) Standard enthalpy formation value for H2Ol= -285.8 kJ/mol(from Appendix 4)

v) Standard enthalpy formation value for CH3COOHl= -484.5 kJ/mol(from Appendix 4)

vi) CH3COOH l→CH4g+CO2g

vii) 4HCOOHl→CH4g+3CO2g+2H2O(l)

4) Calculations:

(1) Calculating standard enthalpy for the first given reaction:

CH3COOH l→CH4g+CO2g

Calculating ∆H°rxn for this reaction using the formula:

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

Plugging in the values in the equation, we get:

∆H°rxn=1mol CH4×-74.8kJmol CH4+ 1mol CO2×-393.5kJmol CO2-1 mol CH3COOH× -484.5kJmol CH3COOH

∆H°rxn=-468.3 kJ--484.5 kJ

∆H°rxn= -468.3 kJ+484.5 kJ=16.2kJ

2) Calculating standard enthalpy for second reaction:

4HCOOHl→CH4g+3CO2g+2H2O(l)

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

Plugging in the values in the equation, we get:

∆H°rxn=2molH2O×-285.8kJmolH2O+ 3mol CO2×-393.5kJmol CO2+(1mol CH4 ×-74.8kJmol CH4-4 mol HCOOH × -425.0kJmol HCOOH

∆H°rxn=-571.6 kJ -1180.5 kJ-74.8 kJ--1700 kJ

∆H°rxn= -1826.9 kJ+1700kJ=-126.9 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 19.113QA

Interpretation Introduction

To calculate:

∆H°rxn for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.113QA

Solution:

a) ∆H°rxn=16.2 kJ

b) ∆H°rxn=-126.9 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, CH3COOH land 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) Standard enthalpy formation value for formic acid ∆H°f=-425.0 kJ/mol

ii) Standard enthalpy formation value for CH4g= -74.8 kJ/mol(from Appendix 4)

iii) Standard enthalpy formation value for CO2g= -393.5 kJ/mol(from Appendix 4)

iv) Standard enthalpy formation value for H2Ol= -285.8 kJ/mol(from Appendix 4)

v) Standard enthalpy formation value for CH3COOHl= -484.5 kJ/mol(from Appendix 4)

vi) CH3COOH l→CH4g+CO2g

vii) 4HCOOHl→CH4g+3CO2g+2H2O(l)

4) Calculations:

(1) Calculating standard enthalpy for the first given reaction:

CH3COOH l→CH4g+CO2g

Calculating ∆H°rxn for this reaction using the formula:

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

Plugging in the values in the equation, we get:

∆H°rxn=1mol CH4×-74.8kJmol CH4+ 1mol CO2×-393.5kJmol CO2-1 mol CH3COOH× -484.5kJmol CH3COOH

∆H°rxn=-468.3 kJ--484.5 kJ

∆H°rxn= -468.3 kJ+484.5 kJ=16.2kJ

2) Calculating standard enthalpy for second reaction:

4HCOOHl→CH4g+3CO2g+2H2O(l)

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

Plugging in the values in the equation, we get:

∆H°rxn=2molH2O×-285.8kJmolH2O+ 3mol CO2×-393.5kJmol CO2+(1mol CH4 ×-74.8kJmol CH4-4 mol HCOOH × -425.0kJmol HCOOH

∆H°rxn=-571.6 kJ -1180.5 kJ-74.8 kJ--1700 kJ

∆H°rxn= -1826.9 kJ+1700kJ=-126.9 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 19.113QA

Interpretation Introduction

To calculate:

∆H°rxn for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.113QA

Solution:

a) ∆H°rxn=16.2 kJ

b) ∆H°rxn=-126.9 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, CH3COOH land 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) Standard enthalpy formation value for formic acid ∆H°f=-425.0 kJ/mol

ii) Standard enthalpy formation value for CH4g= -74.8 kJ/mol(from Appendix 4)

iii) Standard enthalpy formation value for CO2g= -393.5 kJ/mol(from Appendix 4)

iv) Standard enthalpy formation value for H2Ol= -285.8 kJ/mol(from Appendix 4)

v) Standard enthalpy formation value for CH3COOHl= -484.5 kJ/mol(from Appendix 4)

vi) CH3COOH l→CH4g+CO2g

vii) 4HCOOHl→CH4g+3CO2g+2H2O(l)

4) Calculations:

(1) Calculating standard enthalpy for the first given reaction:

CH3COOH l→CH4g+CO2g

Calculating ∆H°rxn for this reaction using the formula:

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

Plugging in the values in the equation, we get:

∆H°rxn=1mol CH4×-74.8kJmol CH4+ 1mol CO2×-393.5kJmol CO2-1 mol CH3COOH× -484.5kJmol CH3COOH

∆H°rxn=-468.3 kJ--484.5 kJ

∆H°rxn= -468.3 kJ+484.5 kJ=16.2kJ

2) Calculating standard enthalpy for second reaction:

4HCOOHl→CH4g+3CO2g+2H2O(l)

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

Plugging in the values in the equation, we get:

∆H°rxn=2molH2O×-285.8kJmolH2O+ 3mol CO2×-393.5kJmol CO2+(1mol CH4 ×-74.8kJmol CH4-4 mol HCOOH × -425.0kJmol HCOOH

∆H°rxn=-571.6 kJ -1180.5 kJ-74.8 kJ--1700 kJ

∆H°rxn= -1826.9 kJ+1700kJ=-126.9 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 19.113QA

Interpretation Introduction

To calculate:

∆H°rxn for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.113QA

Solution:

a) ∆H°rxn=16.2 kJ

b) ∆H°rxn=-126.9 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, CH3COOH land 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) Standard enthalpy formation value for formic acid ∆H°f=-425.0 kJ/mol

ii) Standard enthalpy formation value for CH4g= -74.8 kJ/mol(from Appendix 4)

iii) Standard enthalpy formation value for CO2g= -393.5 kJ/mol(from Appendix 4)

iv) Standard enthalpy formation value for H2Ol= -285.8 kJ/mol(from Appendix 4)

v) Standard enthalpy formation value for CH3COOHl= -484.5 kJ/mol(from Appendix 4)

vi) CH3COOH l→CH4g+CO2g

vii) 4HCOOHl→CH4g+3CO2g+2H2O(l)

4) Calculations:

(1) Calculating standard enthalpy for the first given reaction:

CH3COOH l→CH4g+CO2g

Calculating ∆H°rxn for this reaction using the formula:

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

Plugging in the values in the equation, we get:

∆H°rxn=1mol CH4×-74.8kJmol CH4+ 1mol CO2×-393.5kJmol CO2-1 mol CH3COOH× -484.5kJmol CH3COOH

∆H°rxn=-468.3 kJ--484.5 kJ

∆H°rxn= -468.3 kJ+484.5 kJ=16.2kJ

2) Calculating standard enthalpy for second reaction:

4HCOOHl→CH4g+3CO2g+2H2O(l)

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

Plugging in the values in the equation, we get:

∆H°rxn=2molH2O×-285.8kJmolH2O+ 3mol CO2×-393.5kJmol CO2+(1mol CH4 ×-74.8kJmol CH4-4 mol HCOOH × -425.0kJmol HCOOH

∆H°rxn=-571.6 kJ -1180.5 kJ-74.8 kJ--1700 kJ

∆H°rxn= -1826.9 kJ+1700kJ=-126.9 kJ

Conclusion:

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

Answer 6

Chapter 19, Problem 19.113QA

Interpretation Introduction

To calculate:

∆H°rxn for the given reactions of methanogenic bacteria.

Expert Solution & Answer

Answer to Problem 19.113QA

Solution:

a) ∆H°rxn=16.2 kJ

b) ∆H°rxn=-126.9 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, CH3COOH land 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) Standard enthalpy formation value for formic acid ∆H°f=-425.0 kJ/mol

ii) Standard enthalpy formation value for CH4g= -74.8 kJ/mol(from Appendix 4)

iii) Standard enthalpy formation value for CO2g= -393.5 kJ/mol(from Appendix 4)

iv) Standard enthalpy formation value for H2Ol= -285.8 kJ/mol(from Appendix 4)

v) Standard enthalpy formation value for CH3COOHl= -484.5 kJ/mol(from Appendix 4)

vi) CH3COOH l→CH4g+CO2g

vii) 4HCOOHl→CH4g+3CO2g+2H2O(l)

4) Calculations:

(1) Calculating standard enthalpy for the first given reaction:

CH3COOH l→CH4g+CO2g

Calculating ∆H°rxn for this reaction using the formula:

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

Plugging in the values in the equation, we get:

∆H°rxn=1mol CH4×-74.8kJmol CH4+ 1mol CO2×-393.5kJmol CO2-1 mol CH3COOH× -484.5kJmol CH3COOH

∆H°rxn=-468.3 kJ--484.5 kJ

∆H°rxn= -468.3 kJ+484.5 kJ=16.2kJ

2) Calculating standard enthalpy for second reaction:

4HCOOHl→CH4g+3CO2g+2H2O(l)

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

Plugging in the values in the equation, we get:

∆H°rxn=2molH2O×-285.8kJmolH2O+ 3mol CO2×-393.5kJmol CO2+(1mol CH4 ×-74.8kJmol CH4-4 mol HCOOH × -425.0kJmol HCOOH

∆H°rxn=-571.6 kJ -1180.5 kJ-74.8 kJ--1700 kJ

∆H°rxn= -1826.9 kJ+1700kJ=-126.9 kJ

Conclusion:

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

Answer 7

Chapter 19, Problem 19.113QA

Interpretation Introduction

To calculate:

∆H°rxn for the given reactions of methanogenic bacteria.

Answer 8

Expert Solution & Answer

Answer to Problem 19.113QA

Solution:

a) ∆H°rxn=16.2 kJ

b) ∆H°rxn=-126.9 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, CH3COOH land 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: