Chapter 8, Problem 8.141QP

Use your knowledge of thermochemistry to calculate the ΔH for the following processes: (a) Cl⁻(g) → Cl⁺(g) + 2e⁻; (b) K⁺(g) + 2e⁻ → K⁻(g).

Interpretation Introduction

Interpretation: The value of $\text{ΔH}$ for the reaction $C{l}^{-}{}_{\left(g\right)}\underset{}{\overset{}{\to }}C{l}^{+}{}_{\left(g\right)}+\text{2}\stackrel{-}{e}$ has to be determined.

Concept Introduction:

Second ionization energy: The minimum energy needed to remove an electron from a unipositive gaseous ion to form a dipositive ion in the ground state is known as second ionization energy.

Electron Affinity: It is energy change taking place when one electron is added to an atom which is neutral in its gaseous state forming an anion.

Answer to Problem 8.141QP

The value of $\text{ΔH}$ for the given reaction is $1600\text{kJ/mol}$

Explanation of Solution

The reaction is given below:

$C{l}^{-}{}_{\left(g\right)}\underset{}{\overset{}{\to }}C{l}^{+}{}_{\left(g\right)}+\text{2}\stackrel{-}{e}$

Here, one electron is removed from $C{l}^{-}$ which is the opposite of electron affinity of $Cl$ atom. Then one electron is accepted by $Cl$ to form $C{l}^{+}$

Hence the change in energy for the given reaction is the sum of first ionization energy of $Cl$ and the electron affinity of $Cl$

Therefore,

The $\text{ΔH}$ can be given as:

$\begin{array}{l}ElectronaffinityofCl,\\ C{l}^{-}{}_{\left(g\right)}\underset{}{\overset{}{\to }}C{l}_{\left(g\right)}+\stackrel{-}{e}\text{ΔH}\text{=}349\text{kJ/mol}\\ \\ \text{First}\text{i}onizationofCl,\\ C{l}_{\left(g\right)}\underset{}{\overset{}{\to }}C{l}^{+}{}_{\left(g\right)}+\stackrel{-}{e}\text{ΔH}\text{=}\text{1251}\text{kJ/mol}\\ \underset{\_}{}\\ C{l}^{-}{}_{\left(g\right)}\underset{}{\overset{}{\to }}C{l}^{+}{}_{\left(g\right)}+\text{2}\stackrel{-}{e}\text{ΔH}\text{=}1600\text{kJ/mol}\end{array}$

The value of $\text{ΔH}$ for the given reaction is $1600\text{kJ/mol}$

Interpretation Introduction

Interpretation: The value of $\text{ΔH}$ for the reaction $K^{+}{}_{\left(g\right)}+\text{2}\stackrel{-}{e}\underset{}{\overset{}{\to }}{K}^{-}{}_{\left(g\right)}$ has to be determined.

Concept Introduction:

Second ionization energy: The minimum energy needed to remove an electron from a unipositive gaseous ion to form a dipositive ion in the ground state is known as second ionization energy.

Electron Affinity: It is energy change taking place when one electron is added to an atom which is neutral in its gaseous state forming an anion.

Answer to Problem 8.141QP

The value of $\text{ΔH}$ for the given reaction is $-\text{467}\text{kJ/mol}$

Explanation of Solution

The reaction is given below:

$K^{+}{}_{\left(g\right)}+\text{2}\stackrel{-}{e}\underset{}{\overset{}{\to }}{K}^{-}{}_{\left(g\right)}$

Here, one electron is added to $K^{+}$ which is the opposite of first ionization energy of $K$ atom. Then one electron is removed from by $K$ to form $K^{+}$

Hence the change in energy for the given reaction is the sum of the electron affinity of $K$ and first ionization energy of $K$

Therefore,

The $\text{ΔH}$ can be given as:

$\begin{array}{l}\text{First}\text{i}onizationofK,\\ K^{+}{}_{\left(g\right)}+\stackrel{-}{e}\underset{}{\overset{}{\to }}{K}_{\left(g\right)}\text{ΔH}\text{=}-418.7\text{kJ/mol}\\ \\ ElectronaffinityofK,\\ K_{\left(g\right)}+\stackrel{-}{e}\underset{}{\overset{}{\to }}{K}^{-}{}_{\left(g\right)}\text{ΔH}\text{=}-48\text{kJ/mol}\\ \underset{\_}{}\\ K^{+}{}_{\left(g\right)}+\text{2}\stackrel{-}{e}\underset{}{\overset{}{\to }}{K}^{-}{}_{\left(g\right)}\text{ΔH}\text{=}-\text{467}\text{kJ/mol}\end{array}$

The value of $\text{ΔH}$ for the given reaction is $-\text{467}\text{kJ/mol}$