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# For ionic compound A2+B22- and C22+D2-(a) A4(S) + B2(g) → AB2     ΔHf = -500Find the lattice energy of AB2.(b)Find ΔHf for according D3(g) + C(S) → C2D(s)Given on sublimation of above metal (A4) it dissociates into individual atom.Given: Sublimation energy of A4 is 1600 ; Sublimation energy of C is 100Dissociation energy of B2 IS 200            ; Dissociation energy of D3 is 90A(g)  → A+(g) ; ΔH = 50A+(g) → A2+(g) ; ΔH=150B-(g) → B(g) ;ΔH=260B-(g) → B2-(g) ;ΔH=250D(g) → D-(g) ; ΔH=-50D-(g) → D2-(g) ; ΔH=100C+(g) → C(g) ;ΔH=-150C2D(s) → 2C+(g) + D2-(g) ; ΔH=1000  [All the values are in kJ/mol]ans: (a)= -780 kJ/mol , (b) = -420 kJ/moli need explanation... 6 years ago
Hi student ,
Here is the explaination for calculation.

Accodring to born Habber cycle,
Enthalphy of foramtion of AB2 = Entyalpy sublimation of A + 1stionization enthaly pf A + 2ndionization enthalpy of A+ Enthalpy of atomization of B +2(1st ionization energy of B)+ Lattice energy

Substitution the values and calculating for the value of lattice enthalpy , we get the answer.