Use the Born-Haber cycles to determine the lattice energies of the two ionic compounds, then use these answers to justify the large energy release from this reaction

Use the Born-Haber cycles to determine the lattice energies of the two ionic compounds, then use these answers to justify the large energy release from this reaction. Al(s) + Fe2O3(s) –> Fe(I) + Al2O3 (s)

The Correct Answer and Explanation is:

To determine the lattice energies of the ionic compounds involved in the reaction Al(s)+Fe2O3(s)→Fe(l)+Al2O3(s)\text{Al}(s) + \text{Fe}_2\text{O}_3(s) \rightarrow \text{Fe}(l) + \text{Al}_2\text{O}_3(s)Al(s)+Fe2​O3​(s)→Fe(l)+Al2​O3​(s), we will first use the Born-Haber cycle, which is a thermodynamic cycle used to calculate the lattice energy of an ionic compound.

Step 1: Ionization of Aluminum (Al) and Iron (Fe)

• Aluminum (Al) is first ionized to Al3+\text{Al}^3+Al3+, which requires three ionization energies:
  Al(s)→Al3+(g)+3e−\text{Al}(s) \rightarrow \text{Al}^{3+}(g) + 3e^-Al(s)→Al3+(g)+3e−
• Iron (Fe) in Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​ is present as Fe3+\text{Fe}^{3+}Fe3+, so for Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​, each iron atom is ionized into Fe3+\text{Fe}^{3+}Fe3+:
  Fe(s)→Fe3+(g)+3e−\text{Fe}(s) \rightarrow \text{Fe}^{3+}(g) + 3e^-Fe(s)→Fe3+(g)+3e−

Step 2: Formation of Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​ and Al2O3\text{Al}_2\text{O}_3Al2​O3​

• Oxygen atoms in the Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​ and Al2O3\text{Al}_2\text{O}_3Al2​O3​ structures are ionized, forming O2−\text{O}^{2-}O2− ions:
  O2(g)→2O2−(g)+2e−\text{O}_2(g) \rightarrow 2\text{O}^{2-}(g) + 2e^-O2​(g)→2O2−(g)+2e−

Step 3: Lattice Formation Energy

• The lattice energy is the energy required to assemble the ions into a solid ionic lattice from the individual gaseous ions.
• For Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​, the ions Fe3+\text{Fe}^{3+}Fe3+ and O2−\text{O}^{2-}O2− attract each other to form a highly stable lattice structure. The same holds for Al2O3\text{Al}_2\text{O}_3Al2​O3​, where the lattice energy can be calculated based on the electrostatic forces between Al3+\text{Al}^{3+}Al3+ and O2−\text{O}^{2-}O2−.

Step 4: Calculation of Lattice Energy

The lattice energy UUU can be estimated using the formula derived from the Coulomb’s Law: U=k⋅Q1⋅Q2rU = \dfrac{k \cdot Q_1 \cdot Q_2}{r}U=rk⋅Q1​⋅Q2​​

where:

• kkk is Coulomb’s constant,
• Q1Q_1Q1​ and Q2Q_2Q2​ are the charges of the ions,
• rrr is the sum of the ionic radii.

The lattice energies for both Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​ and Al2O3\text{Al}_2\text{O}_3Al2​O3​ are large due to the high charges on the ions and the small ionic radii of Fe3+\text{Fe}^{3+}Fe3+ and Al3+\text{Al}^{3+}Al3+. This leads to a very strong electrostatic attraction, resulting in high lattice energy values.

Step 5: Justifying the Large Energy Release from the Reaction

The large energy release from the reaction: Al(s)+Fe2O3(s)→Fe(l)+Al2O3(s)\text{Al}(s) + \text{Fe}_2\text{O}_3(s) \rightarrow \text{Fe}(l) + \text{Al}_2\text{O}_3(s)Al(s)+Fe2​O3​(s)→Fe(l)+Al2​O3​(s)

can be attributed to the combination of several factors:

1. High Lattice Energy: Both Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​ and Al2O3\text{Al}_2\text{O}_3Al2​O3​ have extremely high lattice energies due to the high charges on the metal ions and small ionic sizes.
2. Reduction of Fe3+\text{Fe}^{3+}Fe3+: The iron ion Fe3+\text{Fe}^{3+}Fe3+ is reduced to elemental iron (Fe), which is a significant reduction in energy, making the reaction highly exothermic.
3. Oxidation of Aluminum: Aluminum metal undergoes oxidation to form Al2O3\text{Al}_2\text{O}_3Al2​O3​, releasing a considerable amount of energy.

Thus, the total energy released in the reaction is a result of the large negative lattice energies of the products, combined with the reduction and oxidation processes.