Chapter 13, Problem 56E

(a)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the oxidation of methyl hydrazine by dinitrogen tetroxide needs to be determined.

  

Concept Introduction:

A chemical compound can be formed by either ionic bond or covalent bond between bonded atoms. The ionic compound is formed by opposite charge ions; cations and anions. The covalent compound is formed by sharing of electrons between bonded atoms.

The bond energy of a chemical bond can be defined as the energy required to break that chemical bond. The bond energy that is needed to break the bonds in reactant molecule and the energy released to form chemical bonds in product can be used to calculate the $\Delta \text{H}$ of the chemical reaction.

  $\Delta \text{H=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}$

Answer to Problem 56E

  $\text{ΔH= 341 kJ/mol }$

Explanation of Solution

Given:

  m:math display='block'>ΔH=∑BEreactant - ∑BEproduct 

For the given reaction:

  $\begin{array}{l}\text{ΔH=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}\\ \text{ΔH=}\sum 4\times {\text{BE}}_{\text{C-H }}\text{+}2\times {\text{BE}}_{\text{C-O }}\text{+ }1\times {\text{BE}}_{\text{C-C }}\text{+}1\times {\text{BE}}_{\text{C-H }}\text{+}1\times {\text{BE}}_{\text{C}\equiv \text{N }}\text{- }\sum {\text{1×BE}}_{\text{C}\equiv \text{N}}+{\text{1×BE}}_{\text{C=C}}+{\text{3×BE}}_{\text{C-H}}+{\text{2×BE}}_{\text{O-H}}\\ \text{ΔH=}\sum 4\times \text{413+}2\times \text{358+ }1\times \text{347+}1\times \text{413+}1\times 891\text{- }\sum \text{1×}891+\text{1×614}+\text{3×413}+2\text{×467}\\ \text{ΔH= 4019 - 3678 kJ/mol}\\ \text{ΔH= 341 kJ/mol }\end{array}$

(b)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the oxidation of methyl hydrazine by dinitrogen tetroxide for propulsion needs to be determined.

  ${\text{4CH}}_{\text{2}}{\text{=CH-CH}}_{\text{3}}\text{ + 6 NO }\stackrel{\text{Ag/700°C}}{\to }{\text{4CH}}_{\text{2}}{\text{=CH-CN + 6 H}}_{\text{2}}{\text{O +N}}_{\text{2}}$

Concept Introduction:

A chemical compound can be formed by either ionic bond or covalent bond between bonded atoms. The ionic compound is formed by opposite charge ions; cations and anions. The covalent compound is formed by sharing of electrons between bonded atoms.

The bond energy of a chemical bond can be defined as the energy required to break that chemical bond. The bond energy that is needed to break the bonds in reactant molecule and the energy released to form chemical bonds in product can be used to calculate the $\Delta \text{H}$ of the chemical reaction.

  $\Delta \text{H=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}$

Answer to Problem 56E

  $\text{ΔH= -1373 kJ/mol }$

Explanation of Solution

Given:

  ${\text{4CH}}_{\text{2}}{\text{=CH-CH}}_{\text{3}}\text{ + 6 NO }\stackrel{\text{Ag/700°C}}{\to }{\text{4CH}}_{\text{2}}{\text{=CH-CN + 6 H}}_{\text{2}}{\text{O +N}}_{\text{2}}$

  $\Delta \text{H=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}$

For the given reaction:

  $\begin{array}{l}\text{ΔH=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}\\ \text{ΔH=}\sum 24\times {\text{BE}}_{\text{C-H }}\text{+}4\times {\text{BE}}_{\text{C=C }}\text{+ }4\times {\text{BE}}_{\text{C-C }}\text{+}6\times {\text{BE}}_{\text{N}\equiv O}\text{- }\sum {\text{4×BE}}_{\text{C}\equiv \text{N}}+{\text{4×BE}}_{\text{C-C}}+{\text{12×BE}}_{\text{C-H}}+{\text{4×BE}}_{\text{C}=C}+{\text{12×BE}}_{\text{O-H}}+{\text{BE}}_{N\equiv \text{N}}\\ \text{ΔH=}\sum 24\times \text{413+}4\times \text{614+ }4\times \text{347+}6\times 630\text{- }\sum \text{4×}891+\text{4×347}+\text{12×413}+\text{4×}614+\text{12×467+941 kJ/mol}\\ \text{ΔH= 17536 - 18909 kJ/mol}\\ \text{ΔH= -1373 kJ/mol }\end{array}$

(c)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the oxidation of methyl hydrazine by dinitrogen tetroxide for propulsion needs to be determined.

  ${\text{4CH}}_{\text{2}}{\text{=CH-CH}}_{\text{3}}{\text{ + 2 NH}}_{\text{3 }}{\text{+ 3 O}}_{\text{2}}\stackrel{\text{Catalyst/425-510°C}}{\to }{\text{4CH}}_{\text{2}}{\text{=CH-CN + 6 H}}_{\text{2}}\text{O }$

Concept Introduction:

A chemical compound can be formed by either ionic bond or covalent bond between bonded atoms. The ionic compound is formed by opposite charge ions; cations and anions. The covalent compound is formed by sharing of electrons between bonded atoms.

The bond energy of a chemical bond can be defined as the energy required to break that chemical bond. The bond energy that is needed to break the bonds in reactant molecule and the energy released to form chemical bonds in product can be used to calculate the $\Delta \text{H}$ of the chemical reaction.

  $\Delta \text{H=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}$

Answer to Problem 56E

  $\text{ΔH= -381 kJ/mol }$

Explanation of Solution

Given:

  ${\text{4CH}}_{\text{2}}{\text{=CH-CH}}_{\text{3}}{\text{ + 2 NH}}_{\text{3 }}{\text{+ 3 O}}_{\text{2}}\stackrel{\text{Catalyst/425-510°C}}{\to }{\text{4CH}}_{\text{2}}{\text{=CH-CN + 6 H}}_{\text{2}}\text{O }$

  $\Delta \text{H=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}$

For the given reaction:

  $\begin{array}{l}\text{ΔH=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}\\ \text{ΔH=}\sum 24\times {\text{BE}}_{\text{C-H }}\text{+}4\times {\text{BE}}_{\text{C=C }}\text{+ }4\times {\text{BE}}_{\text{C-C }}\text{+}6\times {\text{BE}}_{\text{N}-H}{\text{+3×BE}}_{O=O}\text{- }\sum 4{\text{×BE}}_{\text{C-C}}+{\text{12×BE}}_{\text{C-H}}+{\text{4×BE}}_{\text{C}=C}+{\text{12×BE}}_{\text{O-H}}\\ \text{ΔH=}\sum 24\times \text{413+}4\times \text{614+ }4\times \text{347+}6\times 391+\text{3×}495\text{- }\sum \text{4×}891+\text{4×347}+\text{12×413}+\text{4×}614+\text{12×467 kJ/mol}\\ \text{ΔH= 17587 - 17968 kJ/mol}\\ \text{ΔH= -381 kJ/mol }\end{array}$