Lattice Enthalpy and Born–Haber Cycles: Quiz

Lattice Enthalpy and Born–Haber Cycles: Quiz

7 questions

1. Question 1EasyWhich equation represents the lattice enthalpy of potassium chloride (formation convention)?

   AK(s) + ½Cl₂(g) → KCl(s)BK⁺(g) + Cl⁻(g) → KCl(s)CKCl(s) → K⁺(g) + Cl⁻(g)DK⁺(aq) + Cl⁻(aq) → KCl(s)
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   Answer: K⁺(g) + Cl⁻(g) → KCl(s)

   Lattice enthalpy starts from gaseous ions and forms 1 mol of solid. The first equation is the enthalpy of formation; the third is the dissociation convention.

2. Question 2EasyWhich step in a Born–Haber cycle is usually exothermic?

   ASublimation of the metalBFirst ionization energy of the metalCFirst electron affinity of the non-metalDSecond electron affinity of oxygen
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   Answer: First electron affinity of the non-metal

   The nucleus attracts the added electron, releasing energy. Sublimation, atomization and ionization need energy, and adding a second electron to O⁻ is endothermic.

3. Question 3EasyThe Cl–Cl bond enthalpy is 242.6 kJ/mol. What value is used for the atomization of chlorine in the cycle for NaCl?

   A+121.3 kJ/molB+242.6 kJ/molC−121.3 kJ/molD+485.2 kJ/mol
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   Answer: +121.3 kJ/mol

   NaCl needs one Cl atom, so half a mole of Cl₂ is atomized: 242.6 kJ/mol ÷ 2 = +121.3 kJ/mol.

4. Question 4MediumWhich compound has the most negative lattice enthalpy?

   ANaClBNaICKClDMgO
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   Answer: MgO

   Mg²⁺ and O²⁻ have double charges and are small, so the attraction is far stronger (about −3800 kJ/mol, compared with −787 kJ/mol for NaCl).

5. Question 5MediumΔHf(NaCl) = −411.2 kJ/mol and the steps to form Na⁺(g) + Cl⁻(g) from the elements total +376.0 kJ/mol. What is the lattice enthalpy?

   A−35.2 kJ/molB+787.2 kJ/molC−787.2 kJ/molD−411.2 kJ/mol
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   Answer: −787.2 kJ/mol

   ΔH(latt) = −411.2 kJ/mol − 376.0 kJ/mol = −787.2 kJ/mol.

6. Question 6HardIn the Born–Haber cycle for MgCl₂, which set of steps is correct?

   AIE₁(Mg), 1 × atomization of Cl, 1 × EA(Cl)BIE₁ + IE₂ (Mg), 2 × atomization of Cl, 2 × EA(Cl)CIE₁ + IE₂ (Mg), 1 × atomization of Cl, 2 × EA(Cl)DIE₂(Mg) only, 2 × atomization of Cl, 2 × EA(Cl)
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   Answer: IE₁ + IE₂ (Mg), 2 × atomization of Cl, 2 × EA(Cl)

   Mg must lose two electrons (IE₁ then IE₂), and two Cl atoms must each form and each gain one electron.

7. Question 7HardThe Born–Haber lattice enthalpy of AgI is more negative than the value from a purely ionic model. What does this suggest?

   AThe data are wrongBAgI has some covalent characterCThe ions in AgI are larger than expectedDAgI is a metal
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   Answer: AgI has some covalent character

   The large, polarizable I⁻ ion shares some electron density with Ag⁺, adding covalent bonding to the ionic attraction.