Born-Haber Cycles and Enthalpy of Solution – Comprehensive Study Notes

Born-Haber Cycles and Enthalpy of Solution – Study Notes

• Key concept: Born-Haber cycle is a specific Hess’ Law cycle for ionic compounds used to calculate lattice enthalpy (ΔH_lat) when it cannot be measured directly.

• Lattice energy (ΔH_lat): the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions (standard conditions). It is exothermic and thus negative because forming the solid from gaseous ions releases a large amount of energy.

  • Larger magnitude (more negative) ΔH_lat indicates stronger ionic bonds in the lattice and greater stability of the ionic solid relative to gaseous ions.
  • Lattice energy cannot be measured directly from a single experiment; it is found via an energy cycle combining several steps.
  • Example (formation from gaseous ions):
    ext{Na}(g) + ext{Cl}(g)
    ightarrow ext{NaCl}(s)
    \Delta H = -776\ \text{kJ mol}^{-1}

• Formation enthalpy vs dissociation enthalpy: lattice energy can be viewed as a formation process (formation of the solid from gaseous ions) or as a dissociation process (breaking the solid into gaseous ions).

• Enthalpy change of atomisation (ΔH_atom): the enthalpy change when 1 mole of gaseous atoms is formed from its element in its standard state. It is always endothermic (positive) because bond-breaking requires energy.

  • Example conventions:
  • Na(s) → Na(g) ΔH = +108 kJ mol⁻¹
  • For Cl₂, ½Cl₂(g) → Cl(g) ΔH = +121 kJ mol⁻¹
  • For mercury (Hg): elemental Hg is liquid; atomisation means forming Hg(g) from Hg(l) (ΔH_atom(Hg) is the energy required to produce Hg(g) from the liquid phase).

• Electron affinity (ΔH_ea): the energy change when one mole of electrons is gained by one mole of gaseous atoms to form gaseous ions.

  • First electron affinity (e⁻ added to a neutral atom): exothermic (negative) for most elements (energy released when a neutral atom attracts an electron).
  • Example: chlorine:
    ext{Cl}(g) + e^-
    ightarrow ext{Cl}^-(g)\Delta H = -364\ \text{kJ mol}^{-1}
  • Second electron affinity (adding a second electron) can be endothermic due to repulsion between the added electron and the existing negative ion (energetic penalty).
  • Example: oxygen:
    ext{O}(g) + e^-
    ightarrow ext{O}^-(g)\Delta H = +844\ \text{kJ mol}^{-1}
  • Note: values given here are illustrative from the transcript; be aware that actual data have specific tabulated values.

• Born-Haber cycle: construction principles

  • Start with elements in their standard states on the left (as atoms or diatomic molecules, depending on the element).
  • Move upward to form gaseous atoms (atomisation): energy goes up (endothermic).
  • Move to ions by ionisation (for cations) and electron addition (for anions): ionisation energies are positive (upward), electron affinity for halogens is negative (downward) when a second step is needed.
  • Form the ionic lattice from gaseous ions: lattice enthalpy is shown as a downward step (formation of solid).

- The alternative route starts from the elements in their standard states and goes directly to the ionic solid via enthalpy of formation of the solid; both routes have the same net ΔH (Hess’ Law).

• Example: drawing a Born-Haber cycle for NaCl (conceptual steps)
  1) Start with Na(s) and 1/2 Cl₂(g) at a lower energy level.
  2) Atomisation: Na(s) → Na(g) (ΔH = +108 kJ mol⁻¹); ½Cl₂(g) → Cl(g) (ΔH = +121 kJ mol⁻¹).
  3) Create gaseous ions: Na(g) → Na⁺(g) + e⁻ (IE₁ ≈ +500 kJ mol⁻¹); Cl(g) + e⁻ → Cl⁻(g) (EA ≈ -364 kJ mol⁻¹).
  4) Lattice formation: Na⁺(g) + Cl⁻(g) → NaCl(s) (ΔHlat, negative). 5) Enthalpy of formation of NaCl(s): Na(s) + ½Cl₂(g) → NaCl(s) (ΔHf° ≈ -411 kJ mol⁻¹).

• The cycle is completed by linking the left-hand starting state to the right-hand ionic lattice using Hess’ Law to determine ΔH_lat from known values.

• Born-Haber cycle – Calculations (general approach)

  • After constructing the cycle, lattice energy can be computed from the known enthalpies using Hess’s law.
  • General equation (in symbolic form):
    ext{ΔH}{ ext{latt}} = \text{ΔH}f^\uparrow{}^\
    (lattice) - [\Delta H{ ext{atom}}(M) + \tfrac{1}{2}\Delta H{ ext{atom}}(X2) + \, IEM + EA_X]\n
    where M is the metal, X is the non-metal forming the halide/oxide/etc., IE is the ionisation energy of the metal, and EA is the electron affinity of the non-metal.
  • Note: when multiple ions are involved (e.g., MgO, MgCl₂), include the appropriate multiple ionisations (e.g., IE for Mg twice for Mg²⁺) and multiple electron affinities if required.

• Worked examples (conceptual outcomes extracted from the transcript)

  • Potassium chloride (KCl): lattice energy can be calculated from the cycle once ΔH_f(KCl), atomisation enthalpies, ionisation energy of K, and electron affinity of Cl are known.
  • Magnesium oxide (MgO): similar calculation with two ionisation steps for Mg and two steps for O where needed, using the appropriate (two) ionisation energies for Mg and two hydration/affinity steps for O depending on the cycle used.
  • Typical results shown in the transcript include calculated lattice energies such as around -718 kJ mol⁻¹ for KCl and -3812 kJ mol⁻¹ for MgO (values provided in worked steps in the transcript).

• Polarisation (covalent character in ionic solids)

  • Real lattice energies often deviate from purely theoretical (purely ionic) values because anions can be polarised by cations, introducing covalent character.
  • Polarisation mechanism: a cation with high polarising power distorts the electron cloud of the anion, creating partial covalent character in the bond.
  • Factors affecting polarisation:
  • Cations: higher charge density (smaller radius and higher charge) → greater polarising power. E.g., Li⁺ > Na⁺ > K⁺ in terms of polarising power for a given anion.
  • Anions: larger polarisability with larger ionic radius; larger anions are more easily distorted (e.g., Br⁻ more polarisable than Cl⁻).
  • Rough charge-density estimate: charge density ∝ charge / r², used to gauge covalent character tendency.
  • Consequence: polarisation reduces the observed lattice energy relative to the purely ionic model and explains deviations between theoretical and measured lattice energies.

• Enthalpy of Solution – Key Terms

  • Enthalpy of solution (ΔHsol): the enthalpy change when 1 mole of an ionic substance dissolves in sufficient water to form an infinitely dilute solution. For example: ext{KCl(s)} + aq ightarrow ext{KCl(aq)} or ext{KCl(s)} + aq ightarrow ext{K}^+(aq) + ext{Cl}^-(aq) ΔHsol can be exothermic (negative) or endothermic (positive).
  • Enthalpy of hydration (ΔH_hyd): the enthalpy change when 1 mole of a gaseous ion dissolves in water to form an infinitely dilute solution. Example for Mg²⁺:
    ext{Mg}^{2+}(g) + aq
    ightarrow ext{Mg}^{2+}(aq)
    Hydration enthalpies are typically exothermic (negative) because energy is released when ions are solvated by water molecules.
  • Water as a solvent: water is polar; the δ- oxygen attracts cations and δ+ hydrogens attract anions; ion–dipole interactions lead to hydration shells around ions.
  • Hydration enthalpies measure the energy released upon ion–water interactions; smaller ions or ions with higher charge have more negative hydration enthalpies (greater hydration).
  • The enthalpy of solution is the sum of lattice enthalpy and hydration enthalpies of the ions:
    \Delta H{ ext{sol}} = \Delta H{ ext{latt}} + \Delta H{ ext{hyd}}(\text{cation}) + \Delta H{ ext{hyd}}(\text{anion})
  • The two-step view: there are two routes from gaseous ions to ions in aqueous solution:
  • Indirect route: gaseous ions → ionic solid (lattice formation) → ions in solution (hydration of ions).
  • Direct route: gaseous ions → ions in solution (hydration) directly.
    Equivalently, both routes have the same ΔH_sol by Hess’s Law when all terms are included.

• Enthalpy of Solution – Calculations (energy cycles and Hess’ Law)

  • To compute a missing hydration enthalpy, build the energy cycle for dissolution and rearrange to isolate the unknown.
  • For a generic salt MX, the cycle includes lattice enthalpy (formation of MX(s) from MX(g) or from ions), and hydration enthalpies of M⁺ and X⁻.
  • Example approach (KCl):
  • Known data: ΔHsol(KCl), lattice energy, ΔHhyd(K⁺), etc.
  • Solve for ΔHhyd(Cl⁻) using: ΔHsol = ΔHlatt + ΔHhyd(K⁺) + ΔH_hyd(Cl⁻).
  • Therefore: ΔHhyd(Cl⁻) = ΔHsol − ΔHlatt − ΔHhyd(K⁺).
  • MgCl₂ example (Mg²⁺):
  • By applying the cycle with two chloride ions and Mg²⁺, one can derive the hydration enthalpy of Mg²⁺ or Cl⁻ from the given data. A representative calculation in the transcript yields (for Mg²⁺ example):
    \Delta H_{[\mathrm{Mg^{2+}}]} = (-2592) + (-55) - (2 \times -363) = -1921\ \text{kJ mol}^{-1}
  • This demonstrates how multiple ion hydration terms combine in the cycle.
  • Important: in these calculations you may need to double or halve certain terms depending on the number of ions involved in the formula unit (e.g., two Cl⁻ in MgCl₂).

• Enthalpy of Solution – Predictions

  • Dissolution involves lattice breakdown (endothermic) and hydration (exothermic). Lattice breakdown increases the number of particles, increasing entropy; hydration can decrease entropy due to solvent structuring around ions.
  • Overall spontaneity in solution also depends on entropy change (ΔS) and temperature T via the relation ΔG = ΔH − TΔS (noted in the transcript as a discussion of ΔS for solutions).
  • Entropy changes: total ΔS = ΔSsolvating (solute in solution) + ΔSsurr (surroundings). For dissolution at constant T, ΔSsurr = −ΔHsol / T.
  • Example (ammonium nitrate, NH₄NO₃, at 298 K):
  • ΔH_sol = +25.8 kJ mol⁻¹; T = 298 K
  • ΔStotal ≈ +22.1 J K⁻¹ mol⁻¹ (positive), implying the dissolution is thermodynamically spontaneous at 298 K despite ΔHsol being positive, due to a favorable entropy term.
  • A full approach requires careful accounting of the two changes: lattice (endothermic, ΔS_solv increases) and hydration (exothermic, but water becomes more ordered). The net ΔS determines spontaneity at a given temperature.

• Enthalpy of Solution – Ionic Charge & Radius (factors affecting lattice enthalpy and hydration)

  • Lattice energy is strongly affected by ion size and charge.
  • Ionic radius: larger ions lead to less exothermic lattice energies because charges are spread over a larger volume and ions are further apart in the lattice; e.g., CsF has a less exothermic lattice energy than KF.
  • Ionic charge: higher charges increase lattice energy magnitude (more exothermic) due to stronger electrostatic attraction; e.g., CaO (Ca²⁺ and O²⁻) has a more exothermic lattice energy than KCl (K⁺ and Cl⁻).
  • Covalent contributions (polarisation) can modify lattice energy: smaller, highly charged cations polarise anions more, increasing covalent character and reducing measured lattice energy relative to the purely ionic model.

• Hydration enthalpies (factors affecting hydration strength)

  • Hydration enthalpies are exothermic (negative); water molecules stabilise ions via ion–dipole interactions.
  • Higher charge density (smaller ions with higher charges) yields more negative hydration enthalpies (stronger hydration):
  • F⁻ has more negative hydration enthalpy than Cl⁻.
  • Mg²⁺ has a more negative hydration enthalpy than Ba²⁺.
  • Therefore, smaller ions or ions with higher charges tend to hydrate more strongly, releasing more energy when solvated by water.

• Practical notes and exam-style tips

  • Build the Born-Haber cycle step by step and label directions of arrows clearly (up for endothermic steps, down for exothermic steps).
  • When solving lattice energy problems, use the equation
    \Delta H{\text{latt}} = \Delta Hf^{\circ} (\text{ionic solid}) - [\Delta H{\text{atom}}(M) + \tfrac{1}{2}\Delta H{\text{atom}}(X2) + IEM + EA_X]
    with the appropriate factors for your salt (e.g., two ionisations for Mg²⁺, two electron affinities if forming two anions, etc.).
  • In dissolution problems, remember the two-route concept: indirect route via lattice + hydration vs direct hydration of gaseous ions; both routes must give the same ΔH_sol when all terms are included.
  • For entropy considerations, calculate ΔStotal by combining the solute/solvation entropy and the surroundings entropy: \Delta S{\text{total}} = \Delta S{\text{solv}} + \Delta S{\text{surr}} with \Delta S{\text{surr}} = -\frac{\Delta H{\text{sol}}}{T}. A positive ΔS_total at the operation temperature indicates thermodynamic spontaneity.

• Summary of sign conventions (quick reference)

  • Lattice formation from gaseous ions: ΔH_lat is negative (exothermic).
  • Atomisation: ΔH_atom is positive (endothermic).
  • Ionisation energy: ΔH_IE is positive for forming cations.
  • Electron affinity: ΔH_EA is negative for the first electron added; may be positive for subsequent electrons (second EA can be endothermic).
  • Enthalpy of dissolution: ΔH_sol can be negative or positive depending on balance of lattice energy and hydration enthalpies.

• Quick worked exemplars (conceptual results)

  • KCl: Using a Born-Haber cycle with given data yields a lattice energy on the order of several hundred kJ mol⁻¹ (example in transcript gives -718 kJ/mol for a similar setup).
  • MgO: A cycle-based calculation yields a large negative lattice energy in the thousands of kJ/mol range (example in transcript gives about -3812 kJ/mol for MgO) due to high charges and strong electrostatic attraction.
  • Hydration enthalpy derivation: Given ΔH_sol, lattice enthalpy, and one ion’s hydration enthalpy, solve for the other ion’s hydration enthalpy via a simple Hess’ law rearrangement.

• Final takeaway

  • Born-Haber cycles tie together atomisation, ionisation, electron affinity, lattice formation, and dissolution to enable calculation of lattice energies and related thermochemical quantities that are not directly measurable.
  • Polarisation and ion size/charge critically influence both lattice and hydration enthalpies, affecting solubility and the stability of ionic compounds in solution.