Chapter 10, Problem 10.135QP

(a)

Interpretation Introduction

Interpretation:

The bond enthalpy of ${\text{F}}_{\text{2}}{}^{\text{-}}$ ion and the difference in bond enthalpies of ${\text{F}}_{\text{2}}{}^{\text{-}}$ and ${\text{F}}_{\text{2}}$ has to be explained.

Concept Introduction:

Bond Enthalpy:

The measure of stability of molecule is bond enthalpy.  The change in enthalpy is related in breaking a specific bond of 1 mole of gaseous molecule.  In solids and liquids bond enthalpies are affected by neighboring molecules.  There is possibility to predict the enthalpy of reaction using the average bond enthalpies.  Energy is always needed for the breaking of chemical bonds and release of energy always takes place with the formation of chemical bonds.  The enthalpy of reaction is estimated by counting the total number of broken bonds and bonds formed in a reaction.

The enthalpy of reaction in gas phase is given by,

$\text{ΔH°=}{\displaystyle \sum {\text{BE}}_{\text{(reactants)}}}\text{-}{\displaystyle \sum {\text{BE}}_{\left(\text{products}\right)}}$

Where,

BE= Bond enthalpy and ${\displaystyle \sum \text{=}}$ summation

Answer to Problem 10.135QP

Answer

The bond enthalpy of ${\text{F}}_{\text{2}}{}^{\text{-}}$ is $\text{114}\text{kJ}{\text{mol}}^{\text{-1}}$ .

Explanation of Solution

The energy that is necessary to break up the ${\text{F}}_{\text{2}}{}^{\text{-}}$ into an $\text{F}$ and an ${\text{F}}^{\text{-}}$ atom is the bond enthalpy of ${\text{F}}_{\text{2}}{}^{\text{-}}$.

The equations are summed up based on Hess’s law,

$\begin{array}{l}{\text{F}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{g}\right)}\to {\text{F}}_{\text{2(g)}}{\text{+e}}^{\text{-}}\text{ΔH°=290}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{F}}_{\text{2(g)}}\to {\text{2F}}_{\left(\text{g}\right)}\text{ΔH°=156}\text{.9}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{F}}_{\left(\text{g}\right)}{\text{+e}}^{\text{-}}\to {\text{F}}^{\text{-}}{}_{\left(\text{g}\right)}\text{ΔH°=-333}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

${\text{F}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{g}\right)}\to {\text{F}}_{\left(\text{g}\right)}{\text{+F}}^{\text{-}}{}_{\left(\text{g}\right)}$

The bond enthalpy is given by the sum of enthalpies of reaction

$\text{ΔH°=}{\displaystyle \sum {\text{BE}}_{\text{(reactants)}}}\text{-}{\displaystyle \sum {\text{BE}}_{\left(\text{products}\right)}}$

$\begin{array}{l}{\text{BE(F}}_{\text{2}}{}^{\text{-}}\text{)=}\left[\text{290+156}\text{.9+(-333)}\right]\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{BE(F}}_{\text{2}}{}^{\text{-}}\text{)=114}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The bond enthalpy of ${\text{F}}_{\text{2}}{}^{\text{-}}$= $\text{114}\text{kJ}{\text{mol}}^{\text{-1}}$

(b)

Interpretation Introduction

Interpretation:

The bond enthalpy of ${\text{F}}_{\text{2}}{}^{\text{-}}$ ion and the difference in bond enthalpies of ${\text{F}}_{\text{2}}{}^{\text{-}}$ and ${\text{F}}_{\text{2}}$ has to be explained.

Concept Introduction:

Bond Enthalpy:

The measure of stability of molecule is bond enthalpy.  The change in enthalpy is related in breaking a specific bond of 1 mole of gaseous molecule.  In solids and liquids bond enthalpies are affected by neighboring molecules.  There is possibility to predict the enthalpy of reaction using the average bond enthalpies.  Energy is always needed for the breaking of chemical bonds and release of energy always takes place with the formation of chemical bonds.  The enthalpy of reaction is estimated by counting the total number of broken bonds and bonds formed in a reaction.

The enthalpy of reaction in gas phase is given by,

$\text{ΔH°=}{\displaystyle \sum {\text{BE}}_{\text{(reactants)}}}\text{-}{\displaystyle \sum {\text{BE}}_{\left(\text{products}\right)}}$

Where,

BE= Bond enthalpy and ${\displaystyle \sum \text{=}}$ summation

Answer to Problem 10.135QP

Answer

${\text{F}}_{\text{2}}{}^{\text{-}}$ is said to have weaker bond than ${\text{F}}_{\text{2}}$ due to the presence of additional electron.  The additional electron increases the repulsion between the $\text{F}$ atoms.

Explanation of Solution

${\text{F}}_{\text{2}}{}^{\text{-}}$ is said to have weaker bond than ${\text{F}}_{\text{2}}$ due to the presence of additional electron.  The additional electron increases the repulsion between the $\text{F}$ atoms.

${\text{F}}_{\text{2}}{}^{\text{-}}$ is said to have weaker bond($\text{114}\text{kJ}{\text{mol}}^{\text{-1}}$) than the bond of ${\text{F}}_{\text{2}}$ ($\text{156}\text{.9}\text{kJ}{\text{mol}}^{\text{-1}}$ ) due to the presence of additional electron.  The additional electron increases the repulsion between the $\text{F}$ atoms.