Chapter 13, Problem 47E

(a)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the given reaction in gas phase needs to be determined, with the help of bond energy value:

  ${\text{H}}_{\text{2(g)}}{\text{+Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{2HCl}}_{\text{(g)}}$

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 47E

  $\text{ΔH= -183kJ/mol}$

Explanation of Solution

Given:

  ${\text{H}}_{\text{2(g)}}{\text{+Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{2HCl}}_{\text{(g)}}$

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

For the given reaction:

  $\begin{array}{l}\text{ΔH=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}\\ \text{ΔH=}\sum {\text{BE}}_{\text{H-H }}{\text{+BE}}_{\text{Cl-Cl }}\text{- }\sum {\text{2×BE}}_{\text{H-Cl }}\\ \text{ΔH=}\sum \text{432kJ/mol+ 239 kJ/mol - }\sum \text{2× 427 kJ/mol }\\ \text{ΔH= -183kJ/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the given reaction in gas phase needs to be determined with the help of bond energy value:

  $\text{N}\equiv {\text{N}}_{\text{(g)}}{\text{+ 3H}}_{\text{2(g)}}\stackrel{}{\to }{\text{2NH}}_{\text{3}}{}_{\text{(g)}}$

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 47E

  $\text{ΔH= -109 kJ/mol}$

Explanation of Solution

Given:

  $\text{N}\equiv {\text{N}}_{\text{(g)}}{\text{+ 3H}}_{\text{2(g)}}\stackrel{}{\to }{\text{2NH}}_{\text{3}}{}_{\text{(g)}}$

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

For the given reaction:

  $\begin{array}{l}\text{ΔH=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}\\ \text{ΔH=}\sum {\text{BE}}_{\text{N}\equiv \text{N }}\text{+ 3 }\times {\text{BE}}_{\text{H-H }}\text{- }\sum {\text{2× 3 BE}}_{\text{N-H }}\\ \text{ΔH=}\sum \text{941 kJ/mol+ 3 }\times \text{432 kJ/mol - }\sum \text{6× 391 kJ/mol }\\ \text{ΔH= -109 kJ/mol}\end{array}$

(c)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the given reaction in gas phase needs to be determined with the help of bond energy value:

  $\text{H-C}\equiv {\text{N}}_{\text{(g)}}{\text{+ 2H}}_{\text{2(g)}}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{-NH}}_{\text{2}}{}_{\text{(g)}}$

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 47E

  $\text{ΔH= -158 kJ/mol}$

Explanation of Solution

  $\text{H-C}\equiv {\text{N}}_{\text{(g)}}{\text{+ 2H}}_{\text{2(g)}}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{-NH}}_{\text{2}}{}_{\text{(g)}}$

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

For the given reaction:

  $\begin{array}{l}\text{ΔH=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}\\ \text{ΔH=}\sum {\text{BE}}_{\text{C-H }}{\text{+BE}}_{\text{C}\equiv \text{N }}{\text{+ 2 ×BE}}_{\text{H-H }}\text{- }\sum {\text{3 ×BE}}_{\text{C-H }}{\text{+BE}}_{\text{C-N }}{\text{+ 2 ×BE}}_{\text{N-H }}\\ \text{ΔH=}\sum \text{413 kJ/mol+ 891 kJ/mol + 2 × 432kJ/mol- }\sum \text{3 ×413kJ/mol+ 305kJ/mol + 2×391 kJ/mol }\\ \text{ΔH= 2168 kJ/mol - 2326 kJ/mol}\\ \text{ΔH= -158 kJ/mol}\end{array}$

(d)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$

for the given reaction in gas phase needs to be determined with the help of bond energy value:

  

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 47E

  $\text{ΔH= -1169 kJ/mol}$

Explanation of Solution

  

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

For the given reaction:

  $\begin{array}{l}\text{ΔH=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}\\ \text{ΔH=}\sum 4\times {\text{BE}}_{\text{N-H }}{\text{+ BE}}_{N-\text{N }}{\text{+ 2 ×BE}}_{\text{F-F }}\text{- }\sum {\text{BE}}_{\text{N}\equiv \text{N }}{\text{+ 4 ×BE}}_{\text{H-F }}\\ \text{ΔH=}\sum 4\times \text{ 391 kJ/mol+ 160 kJ/mol + 2 × 154 kJ/mol- }\sum \text{ 941 kJ/mol + 4 × 565 kJ/mol }\\ \text{ΔH= 2032 kJ/mol - 3201 kJ/mol}\\ \text{ΔH= -1169 kJ/mol}\end{array}$