The ∆ H r x n 0 value of the following combustion reactions of methylamine: a) 4 C H 3 N H 2 g + 13 O 2 g → 4 C O 2 g + 4 N O 2 g + 10 H 2 O ( l ) b) 4 C H 3 N H 2 g + 6 O 2 g → 4 C O 2 g + 4 N H 3 g + 4 H 2 O ( l )

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Answer 1

Question

Chapter 19, Problem 19.76QA

Interpretation Introduction

To determine:

The ∆Hrxn0 value of the following combustion reactions of methylamine:

a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l)

b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)

Expert Solution & Answer

Answer to Problem 19.76QA

Solution:

a) For the reaction 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l), the ∆Hrxn0=-4207.2 kJ.

b) For the reaction 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O l, the ∆Hrxn0=-2809.6 kJ.

Explanation of Solution

1) Concept:

To calculate the standard enthalpy of the reaction, we need to use the heat of formation of reactant and product. We will calculate the ∆Hrxn0 from the thermodynamic value, ∆Hf0, of the compounds present in the reaction.

2) Formula:

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

3) Given:

i) CH3NH2 (g), ∆Hf0 =-23.0 kJ/mol

ii) CO2 (g), ∆Hf0 = -393.5 kJ/mol

iii) H2O (l), ∆Hf0 = -285.8 kJ/mol

iv) O2 (g), ∆Hf0 = -0.0 kJ/mol

v) NH3 (g), ∆Hf0 = -46.1 kJ/mol

vi) NO2 (g), ∆Hf0 = 33.2 kJ/mol

4) Calculations:

a) The combustion reaction of methylamine is

4 CH3NH2 g+ 13 O2g→4 CO2g+4 NO2g+10 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× 33.2 kJmol+10 × -285.8 kJmol-4 × -23.0 kJmol+13 × 0.0 kJmol

∆Hrxn0 =-4299.2 kJ--92.0 kJ= -4207.2 kJ

The heat of the reaction ∆Hrxn0 is -4207.2 kJ

b) The combustion reaction of methylamine is

4 CH3NH2 g+ 6 O2g→4 CO2g+4 NH3g+4 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× -46.1 kJmol+4 × -285.8 kJmol-4 × -23.0 kJmol+6 × 0.0 kJmol

∆Hrxn0 =-2901.6 kJ--92.0 kJ= -2809.6 kJ

The heat of the reaction ∆Hrxn0 is -2809.6 kJ

Conclusion:

The heat of reaction is calculated from the thermodynamic data and combustion reaction.

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Answer 2

Question

Chapter 19, Problem 19.76QA

Interpretation Introduction

To determine:

The ∆Hrxn0 value of the following combustion reactions of methylamine:

a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l)

b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)

Expert Solution & Answer

Answer to Problem 19.76QA

Solution:

a) For the reaction 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l), the ∆Hrxn0=-4207.2 kJ.

b) For the reaction 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O l, the ∆Hrxn0=-2809.6 kJ.

Explanation of Solution

1) Concept:

To calculate the standard enthalpy of the reaction, we need to use the heat of formation of reactant and product. We will calculate the ∆Hrxn0 from the thermodynamic value, ∆Hf0, of the compounds present in the reaction.

2) Formula:

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

3) Given:

i) CH3NH2 (g), ∆Hf0 =-23.0 kJ/mol

ii) CO2 (g), ∆Hf0 = -393.5 kJ/mol

iii) H2O (l), ∆Hf0 = -285.8 kJ/mol

iv) O2 (g), ∆Hf0 = -0.0 kJ/mol

v) NH3 (g), ∆Hf0 = -46.1 kJ/mol

vi) NO2 (g), ∆Hf0 = 33.2 kJ/mol

4) Calculations:

a) The combustion reaction of methylamine is

4 CH3NH2 g+ 13 O2g→4 CO2g+4 NO2g+10 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× 33.2 kJmol+10 × -285.8 kJmol-4 × -23.0 kJmol+13 × 0.0 kJmol

∆Hrxn0 =-4299.2 kJ--92.0 kJ= -4207.2 kJ

The heat of the reaction ∆Hrxn0 is -4207.2 kJ

b) The combustion reaction of methylamine is

4 CH3NH2 g+ 6 O2g→4 CO2g+4 NH3g+4 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× -46.1 kJmol+4 × -285.8 kJmol-4 × -23.0 kJmol+6 × 0.0 kJmol

∆Hrxn0 =-2901.6 kJ--92.0 kJ= -2809.6 kJ

The heat of the reaction ∆Hrxn0 is -2809.6 kJ

Conclusion:

The heat of reaction is calculated from the thermodynamic data and combustion reaction.

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Answer 3

Question

Chapter 19, Problem 19.76QA

Interpretation Introduction

To determine:

The ∆Hrxn0 value of the following combustion reactions of methylamine:

a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l)

b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)

Expert Solution & Answer

Answer to Problem 19.76QA

Solution:

a) For the reaction 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l), the ∆Hrxn0=-4207.2 kJ.

b) For the reaction 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O l, the ∆Hrxn0=-2809.6 kJ.

Explanation of Solution

1) Concept:

To calculate the standard enthalpy of the reaction, we need to use the heat of formation of reactant and product. We will calculate the ∆Hrxn0 from the thermodynamic value, ∆Hf0, of the compounds present in the reaction.

2) Formula:

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

3) Given:

i) CH3NH2 (g), ∆Hf0 =-23.0 kJ/mol

ii) CO2 (g), ∆Hf0 = -393.5 kJ/mol

iii) H2O (l), ∆Hf0 = -285.8 kJ/mol

iv) O2 (g), ∆Hf0 = -0.0 kJ/mol

v) NH3 (g), ∆Hf0 = -46.1 kJ/mol

vi) NO2 (g), ∆Hf0 = 33.2 kJ/mol

4) Calculations:

a) The combustion reaction of methylamine is

4 CH3NH2 g+ 13 O2g→4 CO2g+4 NO2g+10 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× 33.2 kJmol+10 × -285.8 kJmol-4 × -23.0 kJmol+13 × 0.0 kJmol

∆Hrxn0 =-4299.2 kJ--92.0 kJ= -4207.2 kJ

The heat of the reaction ∆Hrxn0 is -4207.2 kJ

b) The combustion reaction of methylamine is

4 CH3NH2 g+ 6 O2g→4 CO2g+4 NH3g+4 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× -46.1 kJmol+4 × -285.8 kJmol-4 × -23.0 kJmol+6 × 0.0 kJmol

∆Hrxn0 =-2901.6 kJ--92.0 kJ= -2809.6 kJ

The heat of the reaction ∆Hrxn0 is -2809.6 kJ

Conclusion:

The heat of reaction is calculated from the thermodynamic data and combustion reaction.

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Answer 4

Question

Chapter 19, Problem 19.76QA

Interpretation Introduction

To determine:

The ∆Hrxn0 value of the following combustion reactions of methylamine:

a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l)

b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)

Expert Solution & Answer

Answer to Problem 19.76QA

Solution:

a) For the reaction 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l), the ∆Hrxn0=-4207.2 kJ.

b) For the reaction 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O l, the ∆Hrxn0=-2809.6 kJ.

Explanation of Solution

1) Concept:

To calculate the standard enthalpy of the reaction, we need to use the heat of formation of reactant and product. We will calculate the ∆Hrxn0 from the thermodynamic value, ∆Hf0, of the compounds present in the reaction.

2) Formula:

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

3) Given:

i) CH3NH2 (g), ∆Hf0 =-23.0 kJ/mol

ii) CO2 (g), ∆Hf0 = -393.5 kJ/mol

iii) H2O (l), ∆Hf0 = -285.8 kJ/mol

iv) O2 (g), ∆Hf0 = -0.0 kJ/mol

v) NH3 (g), ∆Hf0 = -46.1 kJ/mol

vi) NO2 (g), ∆Hf0 = 33.2 kJ/mol

4) Calculations:

a) The combustion reaction of methylamine is

4 CH3NH2 g+ 13 O2g→4 CO2g+4 NO2g+10 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× 33.2 kJmol+10 × -285.8 kJmol-4 × -23.0 kJmol+13 × 0.0 kJmol

∆Hrxn0 =-4299.2 kJ--92.0 kJ= -4207.2 kJ

The heat of the reaction ∆Hrxn0 is -4207.2 kJ

b) The combustion reaction of methylamine is

4 CH3NH2 g+ 6 O2g→4 CO2g+4 NH3g+4 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× -46.1 kJmol+4 × -285.8 kJmol-4 × -23.0 kJmol+6 × 0.0 kJmol

∆Hrxn0 =-2901.6 kJ--92.0 kJ= -2809.6 kJ

The heat of the reaction ∆Hrxn0 is -2809.6 kJ

Conclusion:

The heat of reaction is calculated from the thermodynamic data and combustion reaction.

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Answer 5

Chapter 19, Problem 19.76QA

Interpretation Introduction

To determine:

The ∆Hrxn0 value of the following combustion reactions of methylamine:

a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l)

b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)

Expert Solution & Answer

Answer to Problem 19.76QA

Solution:

a) For the reaction 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l), the ∆Hrxn0=-4207.2 kJ.

b) For the reaction 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O l, the ∆Hrxn0=-2809.6 kJ.

Explanation of Solution

1) Concept:

To calculate the standard enthalpy of the reaction, we need to use the heat of formation of reactant and product. We will calculate the ∆Hrxn0 from the thermodynamic value, ∆Hf0, of the compounds present in the reaction.

2) Formula:

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

3) Given:

i) CH3NH2 (g), ∆Hf0 =-23.0 kJ/mol

ii) CO2 (g), ∆Hf0 = -393.5 kJ/mol

iii) H2O (l), ∆Hf0 = -285.8 kJ/mol

iv) O2 (g), ∆Hf0 = -0.0 kJ/mol

v) NH3 (g), ∆Hf0 = -46.1 kJ/mol

vi) NO2 (g), ∆Hf0 = 33.2 kJ/mol

4) Calculations:

a) The combustion reaction of methylamine is

4 CH3NH2 g+ 13 O2g→4 CO2g+4 NO2g+10 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× 33.2 kJmol+10 × -285.8 kJmol-4 × -23.0 kJmol+13 × 0.0 kJmol

∆Hrxn0 =-4299.2 kJ--92.0 kJ= -4207.2 kJ

The heat of the reaction ∆Hrxn0 is -4207.2 kJ

b) The combustion reaction of methylamine is

4 CH3NH2 g+ 6 O2g→4 CO2g+4 NH3g+4 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× -46.1 kJmol+4 × -285.8 kJmol-4 × -23.0 kJmol+6 × 0.0 kJmol

∆Hrxn0 =-2901.6 kJ--92.0 kJ= -2809.6 kJ

The heat of the reaction ∆Hrxn0 is -2809.6 kJ

Conclusion:

The heat of reaction is calculated from the thermodynamic data and combustion reaction.

Answer 6

Chapter 19, Problem 19.76QA

Interpretation Introduction

To determine:

The ∆Hrxn0 value of the following combustion reactions of methylamine:

a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l)

b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)

Expert Solution & Answer

Answer to Problem 19.76QA

Solution:

a) For the reaction 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l), the ∆Hrxn0=-4207.2 kJ.

b) For the reaction 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O l, the ∆Hrxn0=-2809.6 kJ.

Explanation of Solution

1) Concept:

To calculate the standard enthalpy of the reaction, we need to use the heat of formation of reactant and product. We will calculate the ∆Hrxn0 from the thermodynamic value, ∆Hf0, of the compounds present in the reaction.

2) Formula:

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

3) Given:

i) CH3NH2 (g), ∆Hf0 =-23.0 kJ/mol

ii) CO2 (g), ∆Hf0 = -393.5 kJ/mol

iii) H2O (l), ∆Hf0 = -285.8 kJ/mol

iv) O2 (g), ∆Hf0 = -0.0 kJ/mol

v) NH3 (g), ∆Hf0 = -46.1 kJ/mol

vi) NO2 (g), ∆Hf0 = 33.2 kJ/mol

4) Calculations:

a) The combustion reaction of methylamine is

4 CH3NH2 g+ 13 O2g→4 CO2g+4 NO2g+10 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× 33.2 kJmol+10 × -285.8 kJmol-4 × -23.0 kJmol+13 × 0.0 kJmol

∆Hrxn0 =-4299.2 kJ--92.0 kJ= -4207.2 kJ

The heat of the reaction ∆Hrxn0 is -4207.2 kJ

b) The combustion reaction of methylamine is

4 CH3NH2 g+ 6 O2g→4 CO2g+4 NH3g+4 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× -46.1 kJmol+4 × -285.8 kJmol-4 × -23.0 kJmol+6 × 0.0 kJmol

∆Hrxn0 =-2901.6 kJ--92.0 kJ= -2809.6 kJ

The heat of the reaction ∆Hrxn0 is -2809.6 kJ

Conclusion:

The heat of reaction is calculated from the thermodynamic data and combustion reaction.

Answer 7

Chapter 19, Problem 19.76QA

Interpretation Introduction

To determine:

The ∆Hrxn0 value of the following combustion reactions of methylamine:

a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l)

b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)

Answer 8

Expert Solution & Answer

Answer to Problem 19.76QA

Solution:

a) For the reaction 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l), the ∆Hrxn0=-4207.2 kJ.

b) For the reaction 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O l, the ∆Hrxn0=-2809.6 kJ.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

1) Concept:

To calculate the standard enthalpy of the reaction, we need to use the heat of formation of reactant and product. We will calculate the ∆Hrxn0 from the thermodynamic value, ∆Hf0, of the compounds present in the reaction.

2) Formula:

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

3) Given:

i) CH3NH2 (g), ∆Hf0 =-23.0 kJ/mol

ii) CO2 (g), ∆Hf0 = -393.5 kJ/mol

iii) H2O (l), ∆Hf0 = -285.8 kJ/mol

iv) O2 (g), ∆Hf0 = -0.0 kJ/mol

v) NH3 (g), ∆Hf0 = -46.1 kJ/mol

vi) NO2 (g), ∆Hf0 = 33.2 kJ/mol

4) Calculations:

a) The combustion reaction of methylamine is

4 CH3NH2 g+ 13 O2g→4 CO2g+4 NO2g+10 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× 33.2 kJmol+10 × -285.8 kJmol-4 × -23.0 kJmol+13 × 0.0 kJmol

∆Hrxn0 =-4299.2 kJ--92.0 kJ= -4207.2 kJ

The heat of the reaction ∆Hrxn0 is -4207.2 kJ

b) The combustion reaction of methylamine is

4 CH3NH2 g+ 6 O2g→4 CO2g+4 NH3g+4 H2O l

Calculate the ∆Hrxn0 of the above combustion reaction by using the thermodynamic value given from Appendix 4.

∆Hrxn0 = ∑m ×∆Hf, product0-∑n ×∆Hf, reactant0

∆Hrxn0 = 4 × -393.5 kJmol+ 4× -46.1 kJmol+4 × -285.8 kJmol-4 × -23.0 kJmol+6 × 0.0 kJmol

∆Hrxn0 =-2901.6 kJ--92.0 kJ= -2809.6 kJ

The heat of the reaction ∆Hrxn0 is -2809.6 kJ

Conclusion:

The heat of reaction is calculated from the thermodynamic data and combustion reaction.

Answer 12

Ch. 19 - Prob. 19.1VP Ch. 19 - Prob. 19.2VP Ch. 19 - Prob. 19.3VP Ch. 19 - Prob. 19.4VP Ch. 19 - Prob. 19.5VP Ch. 19 - Prob. 19.6VP Ch. 19 - Prob. 19.7VP Ch. 19 - Prob. 19.8VP Ch. 19 - Prob. 19.9VP Ch. 19 - Prob. 19.10VP

The ∆ H r x n 0 value of the following combustion reactions of methylamine: a) 4 C H 3 N H 2 g + 13 O 2 g → 4 C O 2 g + 4 N O 2 g + 10 H 2 O ( l ) b) 4 C H 3 N H 2 g + 6 O 2 g → 4 C O 2 g + 4 N H 3 g + 4 H 2 O ( l )

To determine: The ∆Hrxn0 value of the following combustion reactions of methylamine: a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l) b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)

Answer to Problem 19.76QA

Explanation of Solution

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Chapter 19 Solutions

To determine:

The ∆Hrxn0 value of the following combustion reactions of methylamine:

a) 4 CH3NH2 g+13 O2g →4 CO2g+4 NO2g+10 H2O (l)

b) 4 CH3NH2 g+6 O2g →4 CO2g+4 NH3g+4 H2O (l)