Doc Brown's A Level Chemistry - Advanced Level Theoretical Physical Chemistry Revision Notes - Basic Thermodynamics

GCE Thermodynamics–thermochemistry sub–index links below

Part 2 ΔH Enthalpy Changes contd. – Lattice Enthalpy, Born Haber Cycle, Enthalpies of Solution, Enthalpies of Ion Hydration

2.2a Born Haber Cycle Calculations

How to construct and do calculations with Born–Haber cycles from enthalpy level diagrams

Energetics index: GCSE Notes on the basics of chemical energy changes – important to study and know before tackling any of the three Advanced Level Chemistry pages Parts 1–3 here * Part 1a–b ΔH Enthalpy Changes 1.1 Advanced Introduction to enthalpy changes – reaction, formation, combustion : 1.2a & 1.2b(i)–(iii) Thermochemistry – Hess's Law and Enthalpy Calculations – reaction, combustion, formation etc. : 1.2b(iv) Bond Enthalpy Calculations  : 1.3a–b Experimental methods for determining enthalpy changes and treatment of results : 1.4 Some enthalpy data patterns : 1.4a The combustion of linear alkanes and linear aliphatic alcohols : 1.4b Some patterns in Bond Enthalpies and Bond Length : 1.4c Enthalpies of Neutralisation : 1.4d Enthalpies of Hydrogenation of unsaturated hydrocarbons and evidence of aromatic ring structure in benzene : Extra Q page A set of practice enthalpy calculations with worked out answers ** Part 2 ΔH Enthalpies of ion hydration, solution, atomisation, lattice energy, electron affinity and the Born–Haber cycle : 2.1a–d What happens when a salt dissolves in water and why? : 2.1d–e Enthalpy cycles involving a salt dissolving : 2.2a–c The Born–Haber Cycle *** Part 3 ΔS Entropy and ΔG Free Energy Changes : 3.1a–g Introduction to Entropy : 3.2 Examples of entropy values and comments * 3.3a ΔS, Entropy and change of state : 3.3b ΔS, Entropy changes and the feasibility of a chemical change : 3.4a–d More on ΔG, Free energy changes, feasibility and applications : 3.5 Calculating Equilibrium Constants : 3.6 Kinetic stability versus thermodynamic feasibility

PLEASE note that delta H/S/G values vary slightly from source to source, so I apologise in advance for any inconsistencies that may arise as I've researched and developed each section. However, any discrepancy should not detract from learning how to construct and calculate using Born–Haber Cycles.

 

2.2a The Born Haber Cycle

This section looks at the application of Hess's Law to the theoretical formation of an ionic lattice from its constituent elements in their standard states i.e. most stable state at 298K (25oC), 101kP (1 atm).

Brown arrows = exothermic

Blue arrows = endothermic

Fig 1. The simplest way, and the easiest way to learn to construct and present a Born–Haber cycle – in this case for sodium chloride NaCl.

Right Fig 2. The enthalpy level diagram for the Born–Haber cycle for sodium chloride.

Left Fig 3. A way of thinking to solve the cycle for one unknown enthalpy component in a Born–Haber cycle presented in the style of Fig 2.

All the delta H values are defined and their use explained below. You need to be able to recognise any of ΔH 1–6 and be familiar with the 'styles' of figs 1–2 and be able to complete/construct a cycle and solve a problem to obtain an unknown value.

  • In any enthalpy value with a superscript theta, (θ), i.e. ΔHθx = ??, the θ denotes a standard enthalpy value, which usually means 1 atm. pressure (101 kPa) and a 298 K temperature (25oC). Where relevant, any such criteria need to be added to the definitions below if quoted as standard values.
  • ΔH1 = ΔHθf(NaCl) the enthalpy of formation of sodium chloride:
    • Na(s) + 1/2Cl2(g) ==> NaCl(s)  ΔHf(NaCl) = –411 kJ mol–1
    • The standard enthalpy of formation is defined as the energy released or absorbed when 1 mole of a compound is formed from its constituent elements in their normal stable states at 298K and 1atm.
  • ΔH2 = ΔHθatom(Na)  the enthalpy of atomisation of sodium:
    • Na(s) ==> Na(g)  ΔHatom(Na) = +107 kJ mol–1
    • The standard enthalpy of atomisation is defined as the energy absorbed when 1 mole of gaseous atoms is formed from the element in its normal stable state at 298K and 1atm.
      • other examples, but not needed here are ...
        • 1/2O2(g) ==> O(g)  ΔHatom(O) = +249 kJ mol–1
        • 1/2Br2(l) ==> Br(g)  ΔHatom(Br) = +112 kJ mol–1
        • 1/2I2(l) ==> I(g)  ΔHatom(I) = +107 kJ mol–1
        • Note that the enthalpy of atomisation for gaseous species is related to the 'gaseous' bond enthalpy value
        • e.g. the enthalpy of atomisation of oxygen is half the bond enthalpy of the O=O double bond,
  • ΔH3 = ΔHθatom(Cl2) the enthalpy of atomisation of chlorine:
    • 1/2Cl2(g) ==> Cl(g)  ΔHatom(Cl) = +121 kJ mol–1
    • Already defined above and is also half the bond enthalpy for chlorine molecules
      • Cl2(g) ==> 2Cl(g)  ΔHBE(Cl2) = +242 kJ mol–1
      • so take care with how the data is presented.
  • ΔH4 = ΔHθel.affin.(Cl)  the 1st electron affinity of chlorine:
    • Cl(g) + e ==> Cl(g)  ΔHelec.affin.(Cl) = –355 kJ mol–1
    • The standard enthalpy change for the first electron affinity is defined as the energy released or absorbed when one mole of gaseous neutral atoms gain one electron each to form one mole of singly charged negative gaseous ions at 298K and 1atm.
      • I wouldn't worry about the 2nd of chlorine, the Cl2– ion will be too endothermic to be form in a chemical change and is electronically very unstable.
        • The 2nd electron affinity would be defined as the energy absorbed or released when 1 mole of singly charged negative ions gain one electron each to form 1 mole of doubly charged negative ions.
      • However in a Born–Haber Cycle for the formation of an ionic metal oxide e.g. MgO (shown below) you need two electron affinities
        • 1st electron affinity of oxygen is exothermic
          • O(g) ==> O(g)  ΔH1st elec.affin.(O) = –142 kJ mol–1
        • but the 2nd is very endothermic
          • O(g) + e ==> O2–(g)  ΔH2nd elec.affin.(O) = +844 kJ mol–1
          • because of the O... e repulsion, but must be considered in the cycle for MgO etc. (see further down the page
    • The 2nd electron affinity (not needed here) would be defined as:
      • The energy released or absorbed when one mole of gaseous singly charged negative ions gain one electron each to form one mole of doubly charged negative gaseous ions at 298K and 1atm.
      • This would be needed for the Born–Haber cycle for magnesium oxide (Mg2+O2–).
  • ΔH5 = ΔHθ1st IE(Na)  the 1st ionization enthalpy of sodium:
    • Na(g) ==> Na+(g) + e  ΔH1st IE(Na) = +502 kJ mol–1
    • The standard first ionisation energy is defined as the energy absorbed when the most loosely bond electron is removed from one mole of neutral gaseous atoms to form one mole of singly positively charged gaseous ions at 298K and 1atm.
    • Again, in a B–H cycle for MgO etc. the 2nd ionisation energy must be taken into consideration and would be defined as:
      • The energy absorbed when the most loosely bond electron is removed from one mole of singly positively charged gaseous ions to form one mole of doubly positively charged gaseous ions at 298K and 1atm.
  • ΔH6 = ΔHθLE(NaCl)  the lattice enthalpy of sodium chloride:
    • Na+(g) + Cl(g) ==> NaCl(s)  ΔHLE(NaCl) = –786 kJ mol–1
    • The standard enthalpy change, known as the lattice enthalpy (lattice energy) is defined as the energy released when 1 mole of an ionic compound is formed from its constituent gaseous positive and negative ions at 298K and 1atm.
    • Factors affecting the value of lattice enthalpylattice enthalpy previously discussed – brief summary below.
      • The greater the force of attraction between the ions, the greater the energy release, in coming together to the point of minimum potential energy in forming the most stable ionic crystal lattice.
      • From the laws of electrostatics, the force of attraction is proportional to ...
        • +ve charge x –ve charge/(distance between the centres of the charges)2,
        • translating this into the ionic lattice situation, the force of attraction is proportional to the ..
        • charge on cation x charge on anion/(radius of cation + radius of anion)2,
        • so quite simply, the smaller the radius of an ion or the greater the charge on an ion, the greater will be the lattice enthalpy because the electric field intensity is increases, hence the force of attraction is stronger as the ions are closer together, hence the greater amount of energy released when the ions come together.
          • the sum of the cation and anion radius = the distance between the positive and negative centres of attraction
        • From this simple trends can be noted e.g. in terms of LE (in kJmol–1) ...
          • LiF (1022) > LiCl (846) > LiBr (800) > LiI (744), from L to R, the anion radius becomes larger,
          • Al2O3 (15916) > MgO (3889) > Na2O (2478), from L to R, the cation charge decreases (and the cation radius increases),
          • MgO (3889) is over 4 x NaCl (780), charges 2+ x 2– versus 1+ x 1–, ignoring ionic radii differences.

ENERGETICS INDEX 


2.2b Problem solving using a Born–Haber Cycle as presented above

  • In Fig. 1 above for sodium chloride ΔH1 = ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6 because all the arrows in the cycle point from the start to the finish
    • This shows how to present and solve a Hess's Law cycle for the formation of an ionic metal chloride and is an example of the so–called Born–Haber Cycle.
    • ΔH1 = ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6
    • ΔHf(NaCl) = ΔHatom(Na) + ΔHatom(Cl) + ΔHelecaffin(Cl) + ΔH1stIE(Na) + ΔHLE(NaCl)
    • –411 = (+107) + (+121) + (–355) + (+502) + (–786)
      • I often work in () at first retaining the original delta H sign, it reduces the risk of 'sign error'!
    • –411 = +107 +121 –355 +502 –786 = –411
    • If you know five of the six values you can theoretically calculate the 6th enthalpy value which is the most important application of a Hess's Law cycle.
  • Fig 2. shows one how to present a Born–Haber cycle diagram for the formation of an ionic metal chloride by using an enthalpy level diagram.
    • However, in solving problems via enthalpy diagrams you need to define your start and end points and watch the signs!
    • Fig. 3 shows one possible start and end point.
    • The sum of the 'upper loop' enthalpy changes must equal the sum of the 'lower loop' changes, but the signs of the numerical enthalpy values must be appropriate,
    • so ΔH2 + ΔH3 + ΔH5 + ΔH4 equals the upper loop, and no change in signs required, the arrows point towards the end products,
    • but the sum of the lower loop enthalpies = ΔH1 + (–ΔH6) because the arrow must point in the opposite direction to Fig. 2 and the sign of ΔH6 must therefore be reversed,
    • so ΔH2 + ΔH3 + ΔH5 + ΔH4 =  ΔH1 –ΔH6 to solve a problem.
    • +107 +121 + 502 –355 = –411 – –786
    • so, +375 = +375, so, if 5 values known, the 6th can be theoretically calculated.

ENERGETICS INDEX 


2.2c Problem solving using other cycles (not using enthalpy level diagrams – see 2.2d)

  • Δ abbreviations used (all are defined further up the page)

    • f = enthalpy of formation

    • atom = atomisation energy

      • note that for Cl2and O2 the enthalpy of atomisation = half the bond dissociation enthalpy for Cl–Cl and O=O.

    • BE = bond enthalpy

    • IE = ionisation energy (1st or 2nd etc.)

    • LE = lattice enthalpy expressed endothermically i.e. from ionic crystals to free gaseous ions.

    • elec.affin = electron affinity (1st or 2nd)

  • Each cycle involves 6–8 enthalpy values, of which you must know all of them except one!

  • You can then calculate the unknown enthalpy value by substitution and simple algebraic rearrangement.

  • No numerical values are shown on these Born–Haber cycle diagrams, but they are in section 2.2d enthalpy level diagrams.

(a)   K(s)

+ 1/2I2(s) (c) doc b ΔHθf(KI) (c) doc b  K+I(s)
ΔHθatom(K)(c) doc b  

 

(c) doc bΔHθatom(I)

ΔHθLE(KI)

(c) doc b

+ I(g) (c) doc b 2 x ΔHelec.affin(I) (c) doc b I(g) +

K(g)

 (c) doc b ΔHθ1st IE(K) (c) doc b

 K+(g)

The Born–Haber Cycle for the formation of a potassium iodide, an ionic halide salt
  • (a) ΔHθf(KI) = ΔHθatom(K) + ΔHθatom(I2) + ΔHθ1st IE(K) + ΔHθel.affin.(I)  =  + ΔHθLE(KI)

 

(b)   M(s)

+ X2(g/l/s) (c) doc b ΔHθf(MX2) (c) doc b  M2+(X)2(s)
ΔHθatom(M)(c) doc b  

 

(c) doc b2xΔHθatom(X)

ΔHθLE(MX2)

(c) doc b

+ 2X(g) (c) doc b 2 x ΔHelec.affin.(X) (c) doc b 2X(g) +

M(g)

 (c) doc b ΔHθ1st+2nd IE(M) (c) doc b

 M2+(g)

The Born–Haber Cycle for the formation of an MX2 ionic halide salt
  • (b) ΔHθatom(M) + {2 x ΔHθatom(X2)} + ΔHθ1st IE(M) + ΔHθ2nd IE(M) + {2 x ΔHθel.affin.(X)}  = ΔHθf(MX2) + ΔHθLE(MX2)

 

(c)   M(s)

+ 1/2O2(g) (c) doc b ΔHθf(MO) (c) doc b  MO(s)
ΔHθatom(M)(c) doc b  

 

(c) doc bΔHθatom(O2)

ΔHθLE(MO)

(c) doc b

+ O(g) (c) doc b ΔH1st+2nd elec.affin.(O) (c) doc b O2–(g) +

M(g)

 (c) doc b ΔHθ1st+2nd IE(O) (c) doc b

 M2+(g)

The Born–Haber Cycle for the formation of an MO ionic oxide
  • (c) ΔHθf(MO) = ΔHθatom(M) + ΔHθatom(O2) + ΔHθ1st IE(M) + ΔHθ2nd IE(M) + ΔHθ1st elec.affin.(O) + ΔHθ2nd elec.affin.(O) + ΔHθLE(MgO)

 


2.2d – more Born–Haber Cycle Enthalpy Level Diagrams

  • Δ abbreviations used (all are defined further up the page)

    • f = enthalpy of formation

    • at = atomisation energy

      • note that for Cl2and O2 the enthalpy of atomisation = half the bond dissociation enthalpy for Cl–Cl and O=O.

    • IE = ionisation energy (1st or 2nd etc.)

    • LE = lattice enthalpy expressed endothermically i.e. from ionic crystals to free gaseous ions.

    • ea = electron affinity (1st or 2nd)

  • Each cycle involves 6–8 enthalpy values, of which you must know all of them except one!

  • You can then calculate the unknown enthalpy value by substitution and simple algebraic rearrangement.

  • All the numerical values on the Born–Haber cycle diagrams are in kJ mol–1.


Born–Haber Cycle for Sodium Chloride

  • route A = route B = +377 kJ mol–1 (but watch the signs!)

  • ΔHθatom(Na) + ΔHθatom(Cl2) + ΔHθ1st IE(Na) + ΔHθel.affin.(Cl)  = ΔHθf(NaCl) + ΔHθLE(NaCl)

  • This Born–Haber cycle can be adapted for any Group 1 Alkali Metal Halide MX eg LiF, NaBr, KCl, KBr etc.


Born–Haber Cycle for Magnesium Chloride

  • route A = route B = +1883 kJ mol–1  (but watch the signs!)

  • ΔHθatom(Mg) + {2 x ΔHθatom(Cl2)} + ΔHθ1st IE(Mg) + ΔHθ2nd IE(Mg) + {2 x ΔHθel.affin.(Cl)}  = ΔHθf(MgCl2) + ΔHθLE(MgCl2)

  • This Born–Haber cycle can be adapted for any Group 2 Alkaline Earth Halide MX2 eg MgF2, CaCl2, CaBr2 etc.

  • Why not MgCl?

    • route A = route B = +659 kJ mol–1  (but watch the signs!)

    • ΔHθatom(Mg) + ΔHθatom(Cl2) + ΔHθ1st IE(Mg) + ΔHθel.affin.(Cl)  = ΔHθf(MgCl) + ΔHθLE(MgCl)

    • The lattice enthalpy for MgCl can be theoretically calculated (its similar to NaCl, and so with known values for the rest of the cycle, you can then calculate the enthalpy of formation for MgCl. This turns out to –94 kJ mol–1, which is much less exothermic than the energy released when the favoured MgCl2 is formed.

  • Why not MgCl3?

    • I've found one quoted value of +3949 kJ mol–1 for the enthalpy of formation of MgCl3.

    • I think one could reasonably deduce that this highly endothermic compound is hardly likely to exist!

    • Although the more highly charged Mg3+ ion would increase the lattice enthalpy, favouring MgCl3 formation, the formation of Mg3+ requires far more energy because the 3rd ionisation requires the removal of an electron from an inner less shielded shell.

      • The 3rd ionisation energy of magnesium is +7740 kJ mol–1, which completely outweighs any increase in lattice energy and accounts for such an endothermic enthalpy of formation of MgCl3.


Born–Haber Cycle for Sodium Oxide

  • route A = route B = +2114 kJ mol–1  (but watch the signs!)

  • {2 x ΔHθatom(Na)} + ΔHθatom(O2) + {2 x ΔHθ1st IE(Na)} + ΔHθ1st elec.affin.(O) + ΔHθ2nd elec.affin.(O)  = ΔHθf(Na2O) + ΔHθLE(Na2O)

  • 2nd electron affinity of oxygen is the same as the 1st electron affinity of the O ion, but it seems a lot safer to think of these as the 1st and 2nd electron affinities of oxygen atoms themselves, in the same way that you refer to the 1st and 2nd ionisation energies of gaseous sodium atoms.

  • This Born–Haber cycle can be adapted for any Group 1 Alkali Metal oxide M2O eg Li2O, K2O etc.


Born–Haber Cycle for Magnesium Oxide

  • route A = route B = +3243 kJ mol–1  (but watch the signs!)

  • ΔHθatom(Mg) + ΔHθatom(O2) + ΔHθ1st IE(Mg) + ΔHθ2nd IE(Mg) + ΔHθ1st elec.affin.(O) + ΔHθ2nd elec.affin.(O)  = ΔHθf(MgO) + ΔHθLE(MgO)

  • Using this data gives a lattice energy of kJ mol–1, but one of my data books quotes 3889.

  • This Born–Haber cycle can be adapted for any Group 2 Alkaline Earth oxide MO eg CaO, BaO etc.


ENERGETICS INDEX

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