Chapter 13, Problem 59E

(a)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the given reaction needs to be determined with the help of bond energy value and ionization energy of hydrogen (1312 kJ/mol) and electron affinity.

  ${\text{HF}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{(g)}^{+}+{\text{F}}_{(g)}^{-}$

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 59E

  $\text{ΔH =1550kJ/mol}$

Explanation of Solution

  ${\text{HF}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{(g)}^{+}+{\text{F}}_{(g)}^{-}$

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

Ionization energy of hydrogen = 1312 kJ/mol

Electron affinity of F = -327 kJ/mol

  $\begin{array}{l}{\text{HF}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{\text{(g)}}^{\text{+}}{\text{+ F}}_{\text{(g)}}^{\text{-}}\text{ ΔH= 565 kJ/mol}\\ {\text{H}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{H}}_{\text{(g)}}^{\text{+}}\text{ ΔH=1312kJ/mol}\\ F_{\text{(g)}}^{}\stackrel{}{\to }{\text{F}}_{\text{(g)}}^{\text{-}}\text{ ΔH=-327 kJ/mol}\end{array}$

Substitute the values:

  $\begin{array}{l}\Delta \text{H =}\sum \Delta {\text{H}}_{\text{product }}\text{- }\sum \Delta {\text{H}}_{\text{reactant }}\\ \Delta \text{H =}\sum \Delta {\text{H}}_{\text{H(g) }}\text{+}\Delta {\text{H}}_{\text{F(g) }}\text{- }\sum \Delta {\text{H}}_{\text{HF(g) }}\\ \Delta \text{H =}\sum 1312\text{+}(-327)\text{- }\sum (-565)\text{kJ/mol}\\ \text{ΔH =1550kJ/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the given reaction needs to be determined with the help of bond energy value and ionization energy of hydrogen (1312 kJ/mol) and electron affinity.

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

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 59E

  $\text{ΔH =1390 kJ/mol}$

Explanation of Solution

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

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

Ionization energy of hydrogen = 1312 kJ/mol

Electron affinity of Cl = -348.7 kJ/mol

  $\begin{array}{l}{\text{HCl}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{\text{(g)}}^{\text{+}}{\text{+ Cl}}_{\text{(g)}}^{\text{-}}\text{ ΔH= 427 kJ/mol}\\ {\text{H}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{H}}_{\text{(g)}}^{\text{+}}\text{ ΔH=1312kJ/mol}\\ C{l}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Cl}}_{\text{(g)}}^{\text{-}}\text{ ΔH=-348}\text{.7 kJ/mol}\end{array}$

Substitute the values:

  $\begin{array}{l}\Delta \text{H =}\sum \Delta {\text{H}}_{\text{product }}\text{- }\sum \Delta {\text{H}}_{\text{reactant }}\\ \Delta \text{H =}\sum \Delta {\text{H}}_{\text{H(g) }}\text{+}\Delta {\text{H}}_{\text{Cl(g) }}\text{- }\sum \Delta {\text{H}}_{\text{HCl(g) }}\\ \Delta \text{H =}\sum 1312\text{+}(-348.7)\text{- }\sum (-427)\text{kJ/mol}\\ \text{ΔH =1390 kJ/mol}\end{array}$

(c)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the given reaction needs to be determined with the help of bond energy value and ionization energy of hydrogen (1312 kJ/mol) and electron affinity.

  ${\text{HI}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{(g)}^{+}+{\text{I}}_{(g)}^{-}$

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 59E

  $\text{ΔH =1312 kJ/mol}$

Explanation of Solution

  ${\text{HI}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{(g)}^{+}+{\text{I}}_{(g)}^{-}$

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

Ionization energy of hydrogen = 1312 kJ/mol

Electron affinity of I = -295.2 kJ/mol

  $\begin{array}{l}{\text{HI}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{\text{(g)}}^{\text{+}}{\text{+ I}}_{\text{(g)}}^{\text{-}}\text{ ΔH= 295 kJ/mol}\\ {\text{H}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{H}}_{\text{(g)}}^{\text{+}}\text{ ΔH=1312kJ/mol}\\ I_{\text{(g)}}^{}\stackrel{}{\to }{\text{I}}_{\text{(g)}}^{\text{-}}\text{ ΔH=-295}\text{.2 kJ/mol}\end{array}$

Substitute the values:

  $\begin{array}{l}\Delta \text{H =}\sum \Delta {\text{H}}_{\text{product }}\text{- }\sum \Delta {\text{H}}_{\text{reactant }}\\ \Delta \text{H =}\sum \Delta {\text{H}}_{\text{H(g) }}\text{+}\Delta {\text{H}}_{\text{I(g) }}\text{- }\sum \Delta {\text{H}}_{\text{HI(g) }}\\ \Delta \text{H =}\sum 1312\text{+}(-295.2)\text{- }\sum (-295)\text{kJ/mol}\\ \text{ΔH =1312 kJ/mol}\end{array}$

(d)

Interpretation Introduction

Interpretation: The value of $\Delta \text{H}$ for the given reaction needs to be determined with the help of bond energy value and ionization energy of hydrogen (1312 kJ/mol) and electron affinity.

  ${\text{HI}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{(g)}^{+}+{\text{I}}_{(g)}^{-}$

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 59E

  $\text{ΔH = 1599 kJ/mol}$

Explanation of Solution

  ${\text{H}}_{\text{2}}^{}{\text{O}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{(g)}^{+}+{\text{OH}}_{(g)}^{-}$

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

Ionization energy of hydrogen = 1312 kJ/mol

Electron affinity of OH^-= -180 kJ/mol

  $\begin{array}{l}{\text{H}}_{\text{2}}^{}{\text{O}}_{\text{(g)}}\stackrel{}{\to }{\text{H}}_{(g)}^{+}+{\text{OH}}_{(g)}^{-}\text{ ΔH= - 467 kJ/mol}\\ {\text{H}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{H}}_{\text{(g)}}^{\text{+}}\text{ ΔH=1312kJ/mol}\\ {\mathrm{OH}}_{(g)}^{}\stackrel{}{\to }O{H}_{(g)}^{-}\text{ ΔH=-180 kJ/mol}\end{array}$

Substitute the values:

  $\begin{array}{l}\Delta \text{H =}\sum \Delta {\text{H}}_{\text{product }}\text{- }\sum \Delta {\text{H}}_{\text{reactant }}\\ \Delta \text{H =}\sum \Delta {\text{H}}_{\text{H(g) }}\text{+}\Delta {\text{H}}_{\text{OH(g) }}\text{- }\sum \Delta {\text{H}}_{\text{HOH(g) }}\\ \Delta \text{H =}\sum 1312\text{+}(-180)\text{- }\sum (-467)\text{kJ/mol}\\ \text{ΔH = 1599 kJ/mol}\end{array}$