Born – Haber cycle numerical

Born – Haber cycle

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Born – Haber cycle-

Question 1) – Calculate the lattice enthalpy of KCl from the following data at standard states-

Enthalpy of sublimation of K = 89 KJ/mole

Enthalpy of dissociation of chlorine = 244 KJ/ mole

Ionisation energy of K = 425 KJ/mole

Electron gain enthalpy of chlorine = -355 KJ/mole

Enthalpy of formation of KCl = -438 KJ/mole

Solution )

K(s) + ΔH _sub ———> K(g)  ;  ΔH _sub = 89 KJ/mole

1/2 Cl₂ (g) + D/2 ——-> Cl(g)  ;  D/2 = 244/2 = 122 KJ/mole

K(g) + IE ——–> K⁺ (g) + e^–  ;  IE= 425 KJ/mole

Cl(g) + e ———–> Cl^– (g) +E_A    ;  E_A = -355 KJ/mole

K⁺ (g)  + Cl^– (g) ———> KCl (s)  ;  Δ H _f ⁰ = -438 KJ/mole

According to Born – Haber cycle-

 Δ H _f ⁰ = ΔH _sub + D/2 + IE + E_A +U

-438 = 89 + 122 + 425 – 355 + U

U = – 719 KJ/mole

Lattice energy of KCl = 719 KJ/mole Ans.

Question 2) – Calculate the lattice enthalpy of CaCl₂ , given that the enthalpy of –

Enthalpy of sublimation for  Ca (s) —-> Ca(g) = 121 KJ/mole

Enthalpy of dissociation of Cl₂ (g) —-> 2Cl(g) = 242.8 KJ/ mole

Ionisation energy of Ca(g) —–>Ca⁺⁺ = 2422 KJ/mole

Electron gain enthalpy of  2Cl —–> 2 Cl^– = 2 x -355 = -710 KJ/mole

Enthalpy of formation of CaCl2= -795 KJ/mole

Solution )

Ca (s) + ΔH _sub ———> Ca (g)  ;  ΔH _sub = 121 KJ/mole

 Cl₂ (g) + D ——-> 2Cl(g)  ;  D = 242.8KJ/mol

Ca(g) + IE ——–> Ca⁺⁺ (g) + 2e^–  ;  IE= 2422 KJ/mole

2Cl(g) +2 e ———–> 2Cl^– (g) +E_A    ;  E_A = – 2x 355  = 710 KJ/mole

Δ H _f ⁰ = -795KJ/mole

Ca⁺⁺ (g)  + 2Cl^– (g) ———> CaCl₂ (s)   

According to Born – Haber cycle-

 Δ H _f ⁰ = ΔH _sub + D + IE + E_A +U

-795 = 121 + 242.8 + 2422 – 710 + U

U = –  2870.8 KJ/mole

Lattice energy of CaCl₂ = 2870.8 KJ/mole Ans.

Question 3) Calculate the enthalpy of formation of MgF₂ from the following data-

Enthalpy of sublimation of Mg  = 146.4 KJ/mole

Enthalpy of dissociation of Fluorine = 155.8 KJ/ mole

Ionisation energy of Mg (IE₂ )   = 2186.0 KJ/mole

Electron gain enthalpy of Fluorine = -322.6 KJ/mole

Lattice Enthalpy of  MgF₂ = – 2922.5 KJ/mole

Solution )

The heat of formation , Δ H _f ⁰     may be expressed as,

Δ H _f ⁰ = ΔH _sub + D + IE + E_A + U

    Δ H _f ⁰     = 146.4 + 155.8 + 2186 –  2 x 332.6  – 2922.5

Δ H _f ⁰= 1099.5 KJ/mole Ans.