More practice questions

4. Q4. Using NCERT’s data for HCl, find the enthalpy of dilution of HCl·10 aq to HCl·40 aq.

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   HCl·10 aq + 30 aq → HCl·40 aq: ΔH = −72.79 − (−69.01) = −3.78 kJ mol⁻¹.

5. Q5. Why is the enthalpy of atomisation of sodium the same as its enthalpy of sublimation?

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   Sodium is a metal made of atoms, not molecules. Turning one mole of solid sodium into gaseous atoms, Na(s) → Na(g), is exactly sublimation: no bonds within molecules need breaking afterwards.

6. Q6. Use a Born–Haber cycle to find the lattice enthalpy of KCl. Data (kJ mol⁻¹): Δ_fH^⊖(KCl) = −436.75; sublimation of K = +89; ionisation enthalpy of K = +419; ½Δ_bondH(Cl₂) = +121; electron gain enthalpy of Cl = −348.6.

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   Δ_latticeH^⊖ = 436.75 + 89 + 121 + 419 − 348.6, about +717 kJ mol⁻¹, smaller than NaCl’s +788 because K⁺ is larger than Na⁺.

7. Q7. Many fluorides are less soluble in water than the corresponding chlorides (NCERT asks why on page 156). Suggest a reason.

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   F⁻ is much smaller than Cl⁻, so the ions in a fluoride lattice come closer and the lattice enthalpy is much larger. The larger hydration enthalpy of F⁻ often does not make up for this, so fluorides such as LiF, MgF₂ and CaF₂ dissolve far less than their chlorides; it is a tendency, not a rule (KF and AgF are exceptions).

8. Q8. For MgCl₂(s) → Mg²⁺(g) + 2Cl⁻(g), ΔH = +2526 kJ mol⁻¹. (a) Which Born–Haber steps differ from NaCl’s? (b) Find ΔU at 298 K.

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   (a) Both the first and the second ionisation enthalpies of Mg; the whole Cl–Cl bond enthalpy, not half; two electron gains, 2 × (−348.6). (b) Δn_g = 3: ΔU = 2526 − 3 × 8.314 × 10⁻³ × 298 = 2526 − 7.43, about +2519 kJ mol⁻¹.