Consider an ionic compound, MX_2, composed of generic metal M and generic, gaseous halogen X. The enthalpy of formation of MX_2 is delta H_t degree = -957 kj/mol. The enthalpy of sublimation of M is delta H_sub = 151 kJ/mol. The first and second ionization energies of M are IE_1 = 675 kj/mol and IE_2 = 1361 kJ/mol. The electron affinity of X is delta H_ea = -347 kj/mol. The bond energy of X_2 is BE = 187 kj/mol. Determine the lattice energy of MX_2.
consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X
The Correct Answer and Explanation is:
To determine the lattice energy (U) of the ionic compound MX₂, we apply the Born–Haber cycle, which relates the standard enthalpy of formation of an ionic compound to the energies involved in its formation from elements in their standard states.
Given Data:
- ΔH_f°(MX₂) = -957 kJ/mol (formation of solid MX₂ from M(s) and X₂(g))
- ΔH_sub(M) = +151 kJ/mol (sublimation of M from solid to gas)
- IE₁(M) = +675 kJ/mol (first ionization energy)
- IE₂(M) = +1361 kJ/mol (second ionization energy)
- BE(X₂) = +187 kJ/mol (bond dissociation energy of X₂ molecule to 2X atoms)
- EA(X) = -347 kJ/mol (electron affinity for one X atom, for two X atoms: 2 × EA = -694 kJ/mol)
Step-by-step Born–Haber Cycle:
We are forming MX₂(s) from its elements in their standard states:
M(s) + X₂(g) → MX₂(s)
We break this into steps:
- Sublimation of M:
M(s) → M(g) ΔH = +151 kJ/mol - Ionization of M to M²⁺:
M(g) → M⁺(g) ΔH = +675 kJ/mol
M⁺(g) → M²⁺(g) ΔH = +1361 kJ/mol
Total = +2036 kJ/mol - Dissociation of X₂:
X₂(g) → 2X(g) ΔH = +187 kJ/mol - Electron affinity of X atoms:
2X(g) + 2e⁻ → 2X⁻(g) ΔH = 2 × (-347) = -694 kJ/mol - Formation of MX₂ from gaseous ions (lattice energy):
M²⁺(g) + 2X⁻(g) → MX₂(s) ΔH = U
Now, apply Hess’s Law: ΔHf∘=ΔHsub+IE1+IE2+BE1+2×EA+U\Delta H_f^\circ = \Delta H_{sub} + IE_1 + IE_2 + \frac{BE}{1} + 2 \times EA + U −957=151+675+1361+187−694+U-957 = 151 + 675 + 1361 + 187 – 694 + U −957=1680+U-957 = 1680 + U U=−957−1680=−2637 kJ/molU = -957 – 1680 = -2637\ \text{kJ/mol}
Final Answer:
Lattice Energy (U) = -2637 kJ/mol
Explanation
The lattice energy of an ionic compound is the energy released when gaseous ions combine to form the solid crystal lattice. It reflects the strength of electrostatic attraction between ions. For MX₂, a compound consisting of a metal ion (M²⁺) and two halide ions (X⁻), we calculate the lattice energy using the Born–Haber cycle—a thermochemical pathway that breaks down the formation of the solid into several steps.
We begin by sublimating the metal M from its solid form to a gas. Next, we remove two electrons from M to form M²⁺; this requires the sum of the first and second ionization energies. The halogen X exists naturally as a diatomic gas (X₂), so we split one X₂ molecule into two atoms using its bond dissociation energy. Each X atom then gains an electron, releasing energy equal to its electron affinity. Finally, the M²⁺ and 2X⁻ gaseous ions come together to form the ionic crystal MX₂, releasing lattice energy.
By adding all the energy changes—sublimation, ionizations, bond dissociation, and electron affinities—we calculate the total energy input before lattice formation. Since the total enthalpy change of formation of MX₂ is known, we can rearrange the equation to solve for the lattice energy.
This method shows that lattice energy is crucial in stabilizing ionic compounds. A large negative lattice energy (in this case, -2637 kJ/mol) indicates a strong electrostatic interaction, which typically leads to high melting points, hardness, and stability of the compound.
