Setup of Born-Haber cycle; calculate lattice energy of MgO (s) . The given that - enthalpy of formation of MgO (s) = –602, sublimation of Mg (s) = 148 ; 1 st & 2 nd ionization energy of Mg = 738 & 1450 respectively. For Oxygen bond dissociation energy = 498; 1 st & 2 nd electron gain enthalpy = –141 & 844 respectively (all unit in kJmole –1 ).
Text Solution
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(3890 kJmole –1 )
Required reaction Mg (s) + ½O 2 (g) ⎯→ MgO(s)
Mg (s) ⎯→ Mg (g) Δ H = 148
Mg (g) ⎯→ Mg (g) +1 + e – Δ H = 738
Mg +1 (g) ⎯→ Mg (g) +2 + e – Δ H = 1450
Adding above reactions
Mg (s) ⎯→ Mg (g) +2 + 2e – Δ H 1 = 2336 ...
½O 2(g) ⎯→ O (g) Δ H = 249
O (g) + e – ⎯→ O – (g) Δ H = –141
O – (g) + e – ⎯→ O –2 (g) Δ H = 844
adding above reactions
½O 2(g) + 2e – ⎯→ O –2 (g) Δ H 2 = 950 ...
Finally adding +
Mg (s) + ½O 2(g) ⎯→ MgO (s) Δ H rxn = Δ H f (MgO (s) ) = Δ H 1 + Δ H 2 – Δ H L.E
Δ H L.E = 2336 + 952 + 602 = 3890 kJmole –1
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