Chapter 8, Problem 125AP

Interpretation Introduction

Interpretation:

The bond enthalpy of ${\text{F}}_2^{-}$ is to be calculated and the difference between the bond enthalpies of ${\text{F}}_2^{-}$

and ${\text{F}}_2$ is to be explained.

Concept introduction:

Bond enthalpy is defined as the energy that is required to break a bond of one mole of a substance. It is given in $\text{kJ}/\text{mol}$.

Change of enthalpy or $\Delta H^{\circ }$

is defined as the difference of enthalpy of the products and enthalpy of the reactants, multiplied by their respective stoichiometric coefficients.

Answer to Problem 125AP

Solution:

(a) $114\text{ kJ}/\text{mol}$.

(b) The molecule ${\text{F}}_2^{-}$

has a lesser bond enthalpy than ${\text{F}}_2$, so its bond is weaker than that of ${\text{F}}_2$.

Explanation of Solution

a) The bond enthalpy of the ${\text{F}}_2^{-}$

ion

The reaction for the formation of $2\text{F}$ is as follows:

${\text{F}}_2\left(\text{g}\right)\to 2\text{F}\left(\text{g}\right)\Delta H_{\text{rxn}}^{\circ }=156.9\text{ kJ}/\text{mol }$ …… (1)

The reaction for the formation of $\text{F}$

is as follows:

${\text{F}}^{-}\left(\text{g}\right)\to \text{F}\left(\text{g}\right){\text{ + e}}^{-}\Delta H_{\text{rxn}}^{\circ }=333.0\text{ kJ}/\text{mol }$ …… (2)

The reaction for the formation of ${\text{F}}_2$

is as follows:

${\text{F}}_2^{-}\left(\text{g}\right)\to {\text{F}}_2\left(\text{g}\right){\text{ + e}}^{-}\Delta H_{\text{rxn}}^{\circ }=290.0\text{ kJ}/\text{mol }$ …… (3)

Add equations (1) and (3)

$\begin{array}{r}{\text{F}}_2\left(\text{g}\right)\to 2\text{F}\left(\text{g}\right)\Delta H_{\text{rxn}}^{\circ }=156.9\text{ kJ}/\text{mol }\\ {\text{F}}_2^{-}\left(\text{g}\right)\to {\text{F}}_2\left(\text{g}\right){\text{ + e}}^{-}\Delta H_{\text{rxn}}^{\circ }=290.0\text{ kJ}/\text{mol }\\ \overline{{\text{F}}_2^{-}\left(\text{g}\right)\to 2\text{F}\left(\text{g}\right){\text{ + e}}^{-}\Delta H_{\text{rxn}}^{\circ }=446.9\text{ kJ}/\text{mol}}\end{array}$ …… (4)

Subtract equation (2) from (4)

$\begin{array}{c}{\text{F}}_2^{-}\left(\text{g}\right)\to 2\text{F}\left(\text{g}\right){\text{ + e}}^{-}\Delta H_{\text{rxn}}^{\circ }=446.9\text{ kJ}/\text{mol }\\ \text{F}\left(\text{g}\right){\text{ + e}}^{-}\to {\text{F}}^{-}\left(\text{g}\right)\Delta H_{\text{rxn}}^{\circ }=-333.0\text{ kJ}/\text{mol}\\ \overline{{\text{F}}_2^{-}\left(\text{g}\right)\to \text{F}\left(\text{g}\right){\text{ + F}}^{-}\left(\text{g}\right)\Delta H_{\text{rxn}}^{\circ }=114.0\text{ kJ}/\text{mol }}\end{array}$

Hence, the bond enthalpy of ${\text{F}}_2^{-}$ is $114.0\text{ kJ}/\text{mol }$.

b) The difference between the bond enthalpies of ${\text{F}}_2$

and ${\text{F}}_2^{-}$.

The bond enthalpy of ${\text{F}}_2$ is $156.9\text{ kJ}/\text{mol }$

and the bond enthalpy of ${\text{F}}_2^{-}$ is $114.0\text{ kJ}/\text{mol }$.

The bond enthalpy of ${\text{F}}_2^{-}$ is weaker than the bond enthalpy of ${\text{F}}_2$, because in ${\text{F}}_2^{-}$, an extra electron is present, which increases the repulsion between the fluorine atoms.

Therefore, the bond enthalpy of ${\text{F}}_2^{-}$ is weaker than the bond enthalpy of ${\text{F}}_2$.