Question

The electron affinity of oxygen is \(-141 \mathrm{~kJ} / \mathrm{mol}\), corresponding to the reaction $$ \mathrm{O}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(g) $$ The lattice energy of \(\mathrm{K}_{2} \mathrm{O}(s)\) is \(2238 \mathrm{~kJ} / \mathrm{mol}\). Use these data along with data in Appendix \(\mathrm{C}\) and Figure 7.9 to calculate the "second electron affinity" of oxygen, corresponding to the reaction $$ \mathrm{O}^{-}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(g) $$

Short answer

The second electron affinity of oxygen is approximately 4293.2 kJ/mol.

Step-by-step solution

Identify the Born-Haber cycle components

To find the second electron affinity of oxygen, we must first understand the Born-Haber cycle and identify the components involved in the formation of potassium oxide (K2O). The Born-Haber cycle consists of the following components: 1. The formation of K and O into gaseous atoms from their standard states, ΔH₁ (sublimation of K and dissociation of O₂) 2. Ionization of gaseous potassium atoms to produce K⁺ ions, ΔH₂ (first ionization energy of K) 3. Electron affinity of gaseous oxygen atoms to produce O⁻ ions, ΔH₃ (first electron affinity of oxygen) 4. Second ionization of gaseous potassium atoms to produce K⁺ ions, ΔH₄ (second ionization energy of K) 5. Second electron affinity of gaseous O⁻ ions to produce O²⁻ ions, ΔH₅ (second electron affinity of oxygen) 6. The lattice energy formation of K2O, ΔH₆ (lattice energy of K2O)

Writing the Born-Haber cycle equation

Using Hess's Law, we can write the equation for the Born-Haber cycle as: ΔH = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ + ΔH₅ + ΔH₆ We want to find ΔH₅, the second electron affinity of oxygen. Rearranging the equation: ΔH₅ = ΔH - (ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ + ΔH₆)

Gather the required data

We are given the following data: ΔH₃ (electron affinity of O) = -141 kJ/mol ΔH₆ (lattice energy of K₂O) = 2238 kJ/mol We also need data from Appendix C and Figure 7.9: ΔH₁ (sublimation of K) = 89 kJ/mol ΔH₂ (first ionization energy of K) = 419 kJ/mol ΔH₄ (second ionization energy of K) = 3052 kJ/mol Finally, we need the standard enthalpy of formation of K2O (ΔH): ΔH (Enthalpy of formation of K₂O) = -363.8 kJ/mol

Calculate the second electron affinity of oxygen

Now we can substitute the values into the equation and calculate the second electron affinity of oxygen: ΔH₅ = (-363.8 kJ/mol) - (89 kJ/mol + 419 kJ/mol - 141 kJ/mol + 3052 kJ/mol + 2238 kJ/mol) ΔH₅ = -363.8 - 4657 ΔH₅ = 4293.2 kJ/mol The second electron affinity of oxygen is approximately 4293.2 kJ/mol.

Key concepts

Electron Affinity

Electron affinity is the energy change that occurs when an electron is added to a neutral atom to form an anion. It is represented by the reaction:

• \( \text{O}(g) + e^- \rightarrow \text{O}^-(g) \)

In this reaction, energy is released and the process can have a negative value, indicating that it is exothermic. For oxygen, the first electron affinity is \(-141 \text{ kJ/mol}\). This means energy is released when a neutral oxygen atom gains an electron.

The second electron affinity, however, involves adding another electron to an already negatively charged ion \( \text{O}^- \), forming \( \text{O}^{2-} \). This process requires energy, making it endothermic and resulting in a positive value, as seen in this problem where it is calculated to be \(4293.2 \text{ kJ/mol}\). Understanding these calculations helps reveal how elements form various ionic charges, crucial for predicting compounds' stability and formation.

Lattice Energy

Lattice energy is the energy required to separate one mole of a solid ionic compound into gaseous ions. It's a measure of the strength of the bonds in an ionic lattice. For \(\text{K}_2\text{O}\), the lattice energy is provided as \(2238 \text{ kJ/mol}\). This signifies the strong forces keeping the ions together in the crystalline solid.

Lattice energy affects the melting point, solubility, and hardness of an ionic compound. A high lattice energy indicates a stable compound, meaning more energy is needed to break its bonds. It plays an essential role in the Born-Haber cycle, as it offsets the energy changes from processes, like ionization and electron affinity, helping to form stable compounds such as potassium oxide.

Enthalpy of Formation

The enthalpy of formation, \(\Delta H_f\), is the change in energy when one mole of a compound is formed from its constituent elements in their standard states. In this problem, the enthalpy of formation of \(\text{K}_2\text{O}\) is \(-363.8 \text{ kJ/mol}\). This negative value indicates that the overall formation of the compound from potassium and oxygen is exothermic, releasing energy.

Understanding \(\Delta H_f\) is crucial because it provides insights into how energetically favorable a compound's formation is. A negative enthalpy of formation generally suggests a stable compound capable of forming spontaneously under standard conditions. It is a critical component in the Born-Haber cycle, tying together all energy changes to reflect the formation of ionic compounds.

Hess's Law

Hess’s Law states that the total enthalpy change of a reaction is the same regardless of the number of stages or the particular path taken. In the Born-Haber cycle, it allows us to calculate unknown values by rearranging the sum of known energy changes.

By using Hess’s Law, we set up an equation involving the enthalpy of formation, ionization energies, electron affinities, and lattice energy. In this exercise, rearranging formulas and substituting known values helps find the second electron affinity of oxygen.

This demonstrates the flexibility and power of Hess's Law in thermochemistry. It is an invaluable tool for breaking down complex reactions into manageable steps and calculating unknown thermodynamic quantities. It ensures that the conservation of energy holds, affirming that all paths leading from reactants to products result in the same energy change.