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For ionic compound A2+B22- and C22+D2-
(a) A4(S) + B2(g) → AB2 ΔHf = -500
Find 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 100
Dissociation energy of B2 IS 200 ; Dissociation energy of D3 is 90
A(g) → A+(g) ; ΔH = 50
A+(g) → A2+(g) ; ΔH=150
B-(g) → B(g) ;ΔH=260
B-(g) → B2-(g) ;ΔH=250
D(g) → D-(g) ; ΔH=-50
D-(g) → D2-(g) ; ΔH=100
C+(g) → C(g) ;ΔH=-150
C2D(s) → 2C+(g) + D2-(g) ; ΔH=1000 [All the values are in kJ/mol]
ans: (a)= -780 kJ/mol , (b) = -420 kJ/mol
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