Chapter 13, Problem 48E

(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.

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

  $\Delta \text{H}$ of reaction can also be calculated with the help of standard heat of formation of reactant and product.

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

Answer to Problem 48E

  $\text{ΔH}$ is approximate same as calculated by the bond energy values.

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 $\text{ΔH}$ from bond energy $\text{ΔH= -183kJ/mol}$

Calculate $\text{ΔH}$ from standard enthalpy of formation:

  $\begin{array}{l}\text{ΔH =}\sum {\text{ΔH}}_{\text{product }}\text{- }\sum {\text{ΔH}}_{\text{reactant }}\\ \text{ΔH =}\sum {\text{2×ΔH}}_{\text{HCl }}\text{- }\sum {\text{ΔH}}_{\text{H2 }}{\text{+ΔH}}_{\text{Cl2 }}\\ \text{ΔH =}\sum \text{2×(-92kJ/mol)- }\sum \text{0+0}\\ \text{ΔH = -184kJ/mol}\end{array}$

Hence $\text{ΔH}$ is approximate same as calculated by the bond energy values.

(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.

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

  $\Delta \text{H}$ of reaction can also calculate with the help standard heat of formation of reactant and product.

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

Answer to Problem 48E

  $\text{ΔH}$ is not matched with the bond energy values.

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 from the bond energy $\text{ΔH= -109 kJ/mol}$

Calculate $\text{ΔH}$ from standard enthalpy of formation:

  $\begin{array}{l}\text{ΔH =}\sum {\text{ΔH}}_{\text{product }}\text{- }\sum {\text{ΔH}}_{\text{reactant }}\\ \text{ΔH =}\sum {\text{2×ΔH}}_{\text{NH3 }}\text{- }\sum {\text{ΔH}}_{\text{N2 }}\text{+3}\times {\text{ΔH}}_{\text{H2 }}\\ \text{ΔH =}\sum \text{2×(-46kJ/mol)- }\sum \text{0+0}\\ \text{ΔH = - 92 kJ/mol}\end{array}$

Hence, $\text{ΔH}$ is not matched with the bond energy values.

(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.

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

  $\Delta \text{H}$ of reaction can also calculate with the help standard heat of formation of reactant and product.

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

Answer to Problem 48E

Hence $\text{ΔH}$ is approximate same as calculated by the bond energy values.

Explanation of Solution

Given:

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

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

For the given reaction $\text{ΔH= -158 kJ/mol}$ calculated by bond energy values.

Calculate $\text{ΔH}$ from standard enthalpy of formation:

  $\begin{array}{l}\text{ΔH =}\sum {\text{ΔH}}_{\text{product }}\text{- }\sum {\text{ΔH}}_{\text{reactant }}\\ \text{ΔH =}\sum {\text{ΔH}}_{\text{CH3NH2 }}\text{- }\sum {\text{ΔH}}_{\text{HCN }}\text{+2}\times {\text{ΔH}}_{\text{H2 }}\\ \text{ΔH =}\sum \text{(-44}\text{.68 kJ/mol)- }\sum \text{129}\text{.6 kJ/mol +0}\\ \text{ΔH = - 174}\text{.4 kJ/mol}\end{array}$

Hence $\text{ΔH}$ is approximate same as calculated by the bond energy values.

(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.

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

  $\Delta \text{H}$ of reaction can also calculate with the help standard heat of formation of reactant and product.

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

Answer to Problem 48E

  $\text{ΔH}$ is approximate same as calculated by the bond energy values.

Explanation of Solution

  

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

For the given reaction $\text{ΔH= -1169 kJ/mol}$ from the bond energy.

Calculate $\text{ΔH}$ from standard enthalpy of formation:

  $\begin{array}{l}\text{ΔH =}\sum {\text{ΔH}}_{\text{product }}\text{- }\sum {\text{ΔH}}_{\text{reactant }}\\ \text{ΔH =}\sum {\text{2×ΔH}}_{\text{F2 }}{\text{+ΔH}}_{\text{N2H4 }}\text{- }\sum {\text{ΔH}}_{\text{N2 }}\text{+4}\times {\text{ΔH}}_{\text{HF }}\\ \text{ΔH =}\sum \text{2×0 +10 - }\sum \text{0 +4}\times \text{(-271kJ/mol)}\\ \text{ΔH =1094kJ/mol}\end{array}$

Hence $\text{ΔH}$ is approximately same as calculated by the bond energy values.