Thermodynamics: Born-Haber & Entropy

Define the standard lattice enthalpy of formation of an ionic compound.

• AThe enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions✓
• BThe enthalpy change when one mole of an ionic solid is separated into its gaseous ions
• CThe enthalpy change when one mole of a compound forms from its elements in their standard states
• DThe enthalpy change when one mole of gaseous atoms is formed from the element

Answer: The lattice enthalpy of formation (lattice enthalpy) is the enthalpy change when ONE mole of a solid ionic lattice is formed from its constituent gaseous ions under standard conditions, e.g. Na⁺(g) + Cl⁻(g) → NaCl(s). It is exothermic (negative). The reverse process (solid → gaseous ions) is the lattice enthalpy of dissociation (endothermic).

Which definition correctly describes the first electron affinity of chlorine?

• AThe enthalpy change to remove one electron from each atom in one mole of gaseous Cl atoms
• BThe enthalpy change when one mole of gaseous Cl⁻ ions is hydrated
• CThe enthalpy change when one mole of gaseous Cl atoms each gains one electron to form one mole of gaseous Cl⁻ ions✓
• DThe enthalpy change when one mole of Cl₂ is split into atoms

Answer: The first electron affinity is the enthalpy change when one mole of gaseous atoms each gains one electron to form one mole of gaseous 1− ions: Cl(g) + e⁻ → Cl⁻(g). For chlorine this first electron affinity is exothermic (about −349 kJ mol⁻¹).

Use the Born-Haber data for sodium chloride to calculate the lattice enthalpy of formation of NaCl(s). ΔHf(NaCl) = −411 kJ mol⁻¹; ΔHat(Na) = +107; first ionisation energy of Na = +496; ΔHat(Cl) = +122; first electron affinity of Cl = −349 (all kJ mol⁻¹).

• A−787 kJ mol⁻¹✓
• B−411 kJ mol⁻¹
• C−1163 kJ mol⁻¹
• D+787 kJ mol⁻¹

Answer: By Hess's law: ΔHlatt = ΔHf − (ΔHat(Na) + IE₁ + ΔHat(Cl) + EA₁) = −411 − (107 + 496 + 122 − 349) = −411 − 376 = −787 kJ mol⁻¹.

For magnesium chloride, MgCl₂, use the data to calculate the lattice enthalpy of formation. ΔHf(MgCl₂) = −642 kJ mol⁻¹; ΔHat(Mg) = +148; IE₁(Mg) = +738; IE₂(Mg) = +1451; ΔHat(Cl) = +122; first electron affinity of Cl = −349 (all kJ mol⁻¹).

• A−1795 kJ mol⁻¹
• B−2525 kJ mol⁻¹✓
• C−2174 kJ mol⁻¹
• D−3191 kJ mol⁻¹

Answer: Two moles of Cl atoms are needed: total atomisation of Cl = 2 × 122 = +244; total electron affinity = 2 × (−349) = −698. Sum of steps forming the gaseous ions = 148 + 738 + 1451 + 244 − 698 = +1883. ΔHlatt = ΔHf − 1883 = −642 − 1883 = −2525 kJ mol⁻¹.

The enthalpy of solution of an ionic solid can be calculated from its lattice enthalpy of dissociation and the enthalpies of hydration of its ions. Which expression is correct?

• AΔHsol = ΔHlatt(dissociation) − ΣΔHhydration
• BΔHsol = ΔHlatt(formation) + ΣΔHhydration
• CΔHsol = ΔHlatt(dissociation) + ΣΔHhydration✓
• DΔHsol = ΣΔHhydration − ΔHlatt(dissociation)

Answer: Dissolving = breaking the lattice into gaseous ions (lattice enthalpy of DISSOCIATION, endothermic, positive) then hydrating those ions (exothermic, negative). So ΔHsol = ΔHlatt(dissociation) + ΣΔHhydration. Equivalently, ΔHsol = ΣΔHhydration − ΔHlatt(formation), since the dissociation enthalpy is the negative of the formation lattice enthalpy.

Calculate the enthalpy of solution of sodium chloride. Lattice enthalpy of formation of NaCl = −787 kJ mol⁻¹; ΔHhyd(Na⁺) = −406; ΔHhyd(Cl⁻) = −364 (all kJ mol⁻¹).

• A−1557 kJ mol⁻¹
• B−17 kJ mol⁻¹
• C+1557 kJ mol⁻¹
• D+17 kJ mol⁻¹✓

Answer: ΔHsol = ΣΔHhydration − ΔHlatt(formation) = [(−406) + (−364)] − (−787) = −770 + 787 = +17 kJ mol⁻¹. The slightly endothermic value is consistent with NaCl dissolving readily because the process is entropically favourable.

On which factors does the magnitude of an ionic hydration enthalpy primarily depend, and how?

• AIt depends only on the mass of the ion
• BIt becomes more exothermic as ionic charge decreases and ionic radius increases
• CIt becomes more exothermic as ionic charge increases and ionic radius decreases✓
• DIt becomes more endothermic as charge density increases

Answer: Hydration enthalpy depends on the charge density of the ion (charge/size). A higher charge and a smaller radius give a greater charge density, stronger ion–dipole attraction to water, and therefore a more exothermic (more negative) hydration enthalpy. Lattice enthalpy depends on the same factors in the same way.

For silver chloride, the experimental (Born-Haber) lattice enthalpy is more exothermic than the value calculated from the perfect-ionic (electrostatic) model. What does this difference indicate?

• AThe compound is purely ionic with no polarisation
• BThe bonding has significant covalent character, so the lattice is more stable than a purely ionic model predicts✓
• CThe experimental value must be wrong
• DThe ions are larger than the model assumes

Answer: The perfect-ionic model assumes spherical, undistorted ions. When the experimental lattice enthalpy is appreciably MORE exothermic than the theoretical value, there is additional bonding from covalent character (the cation polarises the anion's electron cloud). Ag⁺ is polarising and Cl⁻ is polarisable, so AgCl shows this discrepancy.

Which change is expected to have the largest POSITIVE entropy change (ΔS)?

• A2SO₂(g) + O₂(g) → 2SO₃(g)
• BH₂O(l) → H₂O(s)
• CCaCO₃(s) → CaO(s) + CO₂(g)✓
• DN₂(g) + 3H₂(g) → 2NH₃(g)

Answer: Entropy increases most when the amount of gas (disorder) increases. CaCO₃(s) → CaO(s) + CO₂(g) creates a mole of gas from a solid (large +ΔS). Options A and D both decrease the number of gas moles (−ΔS), and freezing water (B) decreases disorder (−ΔS).

A reaction has ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Calculate the Gibbs free energy change ΔG at 298 K and state whether the reaction is feasible.

• AΔG = −130 kJ mol⁻¹; feasible at 298 K
• BΔG = +30 kJ mol⁻¹; feasible at 298 K
• CΔG = +178 kJ mol⁻¹; not feasible at 298 K
• DΔG = +130 kJ mol⁻¹; not feasible at 298 K✓

Answer: ΔG = ΔH − TΔS. Convert ΔS to kJ: 161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹. ΔG = 178 − (298 × 0.161) = 178 − 47.98 = +130 kJ mol⁻¹ (to 3 s.f.). Since ΔG is positive, the reaction is not feasible at 298 K. (These are the values for CaCO₃ decomposition.)

For the decomposition of calcium carbonate, ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Estimate the minimum temperature at which the reaction becomes feasible.

• AAbout 1106 K✓
• BAbout 298 K
• CAbout 1.1 K
• DAbout 5610 K

Answer: The reaction becomes feasible when ΔG ≤ 0, i.e. at the point where ΔH = TΔS. So T = ΔH / ΔS = 178 000 J mol⁻¹ ÷ 161 J K⁻¹ mol⁻¹ = 1106 K. Above this temperature TΔS exceeds ΔH and ΔG becomes negative.

Why does the entropy of a substance increase significantly when it melts and again, more so, when it boils?

• AThe particles lose energy and become more ordered
• BThe particles gain more freedom of movement and the number of ways of arranging their energy increases, so disorder rises✓
• CThe mass of the particles increases on heating
• DChemical bonds within the molecules break, increasing the number of particles

Answer: Entropy is a measure of the number of ways energy and particles can be arranged (disorder). Going solid → liquid → gas, particles gain progressively more freedom of movement and the number of accessible arrangements rises sharply, so ΔS is positive and largest for boiling (the liquid → gas change produces the greatest increase in disorder).