Chapter 8, Problem 86AP

Interpretation Introduction

Interpretation:

The energy change for the given reaction is to be calculated.

Concept introduction:

Ionization energy is the energy required when an electron is removed from a gaseous atom.

Electron affinity is the energy released when an electron is added to a gaseous atom.

Answer to Problem 86AP

Solution:

(a) $225\text{ kJ}/\text{mol}$

(b) $168\text{ kJ}/\text{mol}$

(c) $70\text{ kJ}/\text{mol}$

(a)

Explanation of Solution

Given information: The reaction is as follows:

$\text{Li}\left(g\right)+\text{I}\left(g\right)\to {\text{Li }}^{+}\left(g\right)+{\text{I}}^{-}\left(g\right)$

The given reaction is as follows:

$\text{Li}\left(g\right)+\text{I}\left(g\right)\to {\text{Li }}^{+}\left(g\right)+{\text{I}}^{-}\left(g\right)$

In this reaction, lithium loses an electron whereas iodine gains an electron.

With reference tofigure $7.8$,

The ionization energy of lithium is $520\text{ kJ}/\text{mol}$

$\text{Li}\left(g\right)\to {\text{Li }}^{+}\left(g\right)+e^{-}\Delta H°=520\text{ kJ}/\text{mol}$…… (1)

With reference tofigure $7.10$,

The electron affinity of iodine is $-295\text{ kJ}/\text{mol}$

$\text{I}\left(g\right)+e^{-}\to {\text{I}}^{-}\left(g\right)\Delta H°=-295\text{ kJ}/\text{mol}$…… (2)

Add equations(1) and (2) as follows:

$\begin{array}{c}\text{Li}\left(g\right)\to {\text{Li }}^{+}\left(g\right)+\cancel{e^{-}}\Delta H°=520\text{ kJ}/\text{mol}\\ \text{I}\left(g\right)+\cancel{e^{-}}\to {\text{I}}^{-}\left(g\right)\Delta H°=-295\text{ kJ}/\text{mol}\\ \overline{\text{Li}\left(g\right)+\text{I}\left(g\right)}\to \overline{{\text{Li }}^{+}\left(g\right)+{\text{I}}^{-}\left(g\right)}\overline{\Delta H°=225\text{ kJ}/\text{mol}}\end{array}$

Therefore, the energy change for the reactionis $225\text{ kJ}/\text{mol}$.

(b)

Given information: The reaction is as follows:

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

The given reaction is as follows:

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

In this reaction, sodium loses an electron whereas fluorine gains an electron.

With referencetofigure $7.8$,

The ionization energy of sodium is $496\text{ kJ}/\text{mol}$

$\text{Na}\left(g\right)\to {\text{Na }}^{+}\left(g\right)+e^{-}\Delta H°=496\text{ kJ}/\text{mol}$…… (3)

With referencetofigure $7.10$,

The electron affinity of fluorine is $-328\text{ kJ}/\text{mol}$

$\text{F}\left(g\right)+e^{-}\to {\text{F}}^{-}\left(g\right)\Delta H°=-328\text{ kJ}/\text{mol}$…… (4)

Add equation (3) and (4) as follows:

$\begin{array}{c}\text{Na}\left(g\right)\to {\text{Na }}^{+}\left(g\right)+\cancel{e^{-}}\Delta H°=496\text{ kJ}/\text{mol}\\ \text{F}\left(g\right)+\cancel{e^{-}}\to {\text{F}}^{-}\left(g\right)\Delta H°=-328\text{ kJ}/\text{mol}\\ \overline{\text{Na}\left(g\right)+\text{F}\left(g\right)}\to \overline{{\text{Na }}^{+}\left(g\right)+{\text{F}}^{-}\left(g\right)}\overline{\Delta H°=168\text{ kJ}/\text{mol}}\end{array}$

Therefore, the energy change for the reaction is $\text{168 kJ}/\text{mol}$.

(c)

Given information: The reaction is as follows:

$\text{K}\left(g\right)+\text{Cl}\left(g\right)\to {\text{K}}^{+}\left(g\right)+{\text{Cl}}^{-}\left(g\right)$

The given reaction is as follows:

$\text{K}\left(g\right)+\text{Cl}\left(g\right)\to {\text{K}}^{+}\left(g\right)+{\text{Cl}}^{-}\left(g\right)$

In the above reaction, potassium loses an electron whereas chlorine gains an electron.

With referencetofigure $7.8$,

The ionization energy of lithium is $419\text{ kJ}/\text{mol}$

$\text{K}\left(g\right)\to {\text{K}}^{+}\left(g\right)+e^{-}\Delta H°=419\text{ kJ}/\text{mol}$…… (5)

With referencetofigure $7.10$,

The electron affinity of iodine is $-349\text{ kJ}/\text{mol}$

$\text{Cl}\left(g\right)+e^{-}\to {\text{Cl}}^{-}\left(g\right)\Delta H°=-349\text{ kJ}/\text{mol}$…… (6)

Add equations(5) and (6) as follows:

$\begin{array}{c}\text{K}\left(g\right)\to {\text{K}}^{+}\left(g\right)+e^{-}\Delta H°=419\text{ kJ}/\text{mol}\\ \text{Cl}\left(g\right)+e^{-}\to {\text{Cl}}^{-}\left(g\right)\Delta H°=-349\text{ kJ}/\text{mol}\\ \overline{\text{K}\left(g\right)+\text{Cl}\left(g\right)}\to \overline{{\text{K}}^{+}\left(g\right)+{\text{Cl}}^{-}\left(g\right)}\overline{\Delta H°=70\text{ kJ}/\text{mol}}\end{array}$

Therefore, the energy change for the reaction is $70\text{ kJ}/\text{mol}$.