Born-Haber Cycles

Use Born-Haber cycles to the determine the lattice energy for each salt.
Thermodynamic Data for Selected Atoms, Ions and Salts:
Sublimation Energy, ΔsubH
Vaporisation Energy, ΔvapH
Ionisation Energy, ΔieH
Dissociation Energy, ΔdisH
Electron Affinity (Gain) Energy, ΔeaH
Enthalpy (Heat) of Formation, ΔfH
1. KCl [Solution]
2. MgO [Solution]
3. Na2S [Solution]
4. Al2O3 [Solution]
5. CaF2 [Solution]

 

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Solutions

  
1.
      ΔfH  
K(s) + 1/2Cl2(g) → KCl
  ↓ ΔsubH     ↓ ΔdisH      
K(g)   Cl(g)      
  ↓ ΔieH     ↓ ΔeaH     U
K+(g)   Cl−(g)      
               


ΔfH = ΔsubH + ΔieH + 1/2×ΔdisH + ΔeaH + U

U = ΔfH − [ΔsubH + ΔieH + 1/2×ΔdisH + ΔeaH]

U = −436 − [(89) + (419) + 1/2(243) + (−349)]

U = −716 kJ mol−1

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2.
      ΔfH  
Mg(s) + 1/2O2(g) → MgO
  ↓ ΔsubH     ↓ ΔdisH      
Mg(g)   O(g)      
  ↓ ΔieH     ↓ ΔeaH     U
Mg2+(g)   O2−(g)      
               


ΔfH = ΔsubH + ΔieH + 1/2×ΔdisH + ΔeaH + U

U = ΔfH − [ΔsubH + ΔieH + 1/2×ΔdisH + ΔeaH]

U = −602 − [(147) + (738 + 1451) + 1/2(498) + (−141 + 844)]

U = −3890 kJ mol−1

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3.
      ΔfH  
2Na(s) + S(s) → Na2S
  ↓ ΔsubH     ↓ ΔsubH      
2Na(g)   S(g)      
  ↓ ΔieH     ↓ ΔeaH     U
2Na+(g)   S2−(g)      
               


ΔfH = 2×ΔsubH + 2×ΔieH + ΔsubH + ΔeaH + U

U = ΔfH − [2×ΔsubH + 2×ΔieH + ΔsubH + ΔeaH]

U = −365 − [2(108) + 2(496) + (277) + (−200 + 649)]

U = −2299 kJ mol−1

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4.
      ΔfH  
2Al(s) + 3/2O2(g) → Al2O3
  ↓ ΔsubH     ↓ ΔdisH      
2Al(g)   3O(g)      
  ↓ ΔieH     ↓ ΔeaH     U
2Al3+(g)   3O2−(g)      
               

ΔfH = 2×ΔsubH + 2×ΔieH + 3/2ΔdisH + 3×ΔeaH + U

U = ΔfH − [2×ΔsubH + 2×ΔieH + 3/2ΔdisH + 3×ΔeaH]

U = −1676 − [2(330) + 2(578 + 1817 + 2745) + 3/2(498) + 3(−141 + 844)]

U = −15,472 kJ mol−1

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5.
      ΔfH  
Ca(s) + F2(g) → CaF2
  ↓ ΔsubH     ↓ ΔdisH      
Ca(g)   2F(g)      
  ↓ ΔieH     ↓ ΔeaH     U
Ca2+(g)   2F−(g)      
               


ΔfH = ΔsubH + ΔieH + ΔdisH + 2×ΔeaH + U

U = ΔfH − [ΔsubH + ΔieH + ΔdisH + 2×ΔeaH]

U = −1228 − [(178) + (590 + 1145) + (159) + 2(−328)]

U = −2644 kJ mol−1

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