For ionic compound A²⁺B₂^2- and C₂²⁺D^2-

(a) A_4(S) + B_2(g) → AB₂ ΔH_f = -500

Find the lattice energy of AB₂.

(b)Find ΔH_f for according D_3(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) → B^2-_(g) ;ΔH=250

D_(g) → D^-_(g) ; ΔH=-50

D^-_(g) → D^2-_(g) ; ΔH=100

C⁺_(g) → C_(g) ;ΔH=-150

C₂D_(s) → 2C⁺_(g) + D^2-_(g) ; ΔH=1000 [All the values are in kJ/mol]

ans: (a)= -780 kJ/mol , (b) = -420 kJ/mol

i need explanation...

suchita undare , 14 Years ago

Grade 12th Pass