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 the compound is formed from its constituent elements in their standard states under standard conditions
• DThe enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions

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 when one electron is removed from each atom in one mole of gaseous Cl atoms to form one mole of gaseous Cl⁺ ions
• BThe enthalpy change when one mole of gaseous Cl⁻ ions is dissolved in enough water to give an infinitely dilute aqueous solution under standard conditions
• 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 gaseous Cl₂ molecules is split apart to form two moles of separate gaseous chlorine atoms under standard conditions

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 of the chloride ion
• BThe bonding has significant covalent character, so the lattice is more stable than a purely ionic model predicts✓
• CThe experimental Born-Haber value must be wrong
• DThe ions are larger and softer 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). Both 2SO₂ + O₂ → 2SO₃ and N₂ + 3H₂ → 2NH₃ decrease the number of gas moles (−ΔS), and freezing water 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
• DThe covalent bonds within the molecules themselves break as the substance melts and boils, so the number of separate particles present increases sharply

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).

What is lattice enthalpy of formation?

• AEnthalpy change forming one mole of ionic lattice from gaseous ions✓
• BEnthalpy change forming one mole of ionic lattice from its elements
• CEnthalpy change separating one mole of lattice into gaseous ions
• DEnthalpy change dissolving one mole of lattice in excess water

Answer: Formation of the lattice from gaseous ions is strongly exothermic. The reverse, lattice enthalpy of dissociation, is equally endothermic — always check which convention a question uses, since the sign flips.

Why do experimental and theoretical lattice enthalpies differ for silver chloride?

• AThe bonding is purely ionic in every respect
• BThe bonding has significant covalent character✓
• CThe experimental value is measured inaccurately
• DSilver chloride does not form a lattice at all

Answer: Theoretical values assume perfectly spherical ions with no electron sharing. A small, highly charged cation polarises the anion, giving covalent character and a stronger lattice than predicted — the gap is a measure of that polarisation.

What is entropy a measure of?

• AThe total energy content stored within a system
• BThe rate at which a reaction reaches equilibrium
• CThe dispersal of energy and disorder in a system✓
• DThe energy needed to start a chemical reaction

Answer: Entropy rises with the number of ways energy can be arranged, so gases have far higher entropy than liquids or solids. Reactions producing more gas moles show a large positive entropy change.

What is the Gibbs free energy equation?

• AΔG = ΔH + TΔS, with T in kelvin
• BΔG = ΔS − TΔH, with T in kelvin
• CΔG = ΔH − TΔS, with T in celsius
• DΔG = ΔH − TΔS, with T in kelvin✓

Answer: A reaction is feasible when ΔG is negative or zero. Because the entropy term is multiplied by temperature, an endothermic reaction with a positive entropy change becomes feasible once it is hot enough.

What sign of ΔG indicates a feasible reaction?

• APositive or zero
• BPositive only
• CNegative or zero✓
• DZero only

Answer: ΔG ≤ 0 means the reaction is thermodynamically feasible; at exactly zero the system is at equilibrium. Feasibility says nothing about rate — a feasible reaction may still be far too slow to observe.

Why does dissolving ammonium nitrate feel cold yet happen spontaneously?

• AThe large positive entropy change outweighs the endothermic enthalpy✓
• BThe large negative entropy change outweighs the endothermic enthalpy
• CThe process is actually exothermic despite feeling cold
• DThe enthalpy change is zero so entropy alone decides

Answer: Breaking the ordered lattice and dispersing ions through the solvent raises entropy substantially. TΔS exceeds the positive ΔH, so ΔG is negative and the process proceeds while absorbing heat.

How does entropy change when a solid melts?

• AIt increases as the particles become less ordered✓
• BIt decreases as the particles become less ordered
• CIt increases as the particles become more ordered
• DIt stays the same because the substance is unchanged

Answer: Liquid particles can move past one another, so there are far more possible arrangements than in a fixed lattice. Boiling produces a much larger increase again, because gas particles are essentially free.

What is electron affinity?

• AEnthalpy change when gaseous atoms each gain one electron✓
• BEnthalpy change when gaseous atoms each lose one electron
• CEnthalpy change when gaseous ions each gain one proton
• DEnthalpy change when solid atoms each gain one electron

Answer: The first electron affinity is exothermic because the nucleus attracts the incoming electron. The second is endothermic, since a negative ion repels a further electron — which matters in Born-Haber cycles for oxides.

Why is atomisation enthalpy always endothermic?

• ABonds are formed when separate gaseous atoms are produced
• BBonds must be broken to produce separate gaseous atoms✓
• CThe atoms release energy as they enter the gaseous state
• DThe process involves electrons being gained by the atoms

Answer: Producing one mole of gaseous atoms means overcoming whatever holds the element together — metallic, covalent or intermolecular forces. Breaking bonds always costs energy, so the value is positive without exception.