Question

Born-Haber cycle of BaO crystal.

Answer 1

Born hyber cycle is based on the fact that formation of an ionic Crystal may occur either by direct combination of the elements or, by an alternative process .
here BaO crystal given
Ba (s) +1/2O2(g)====>BaO(s) ∆H1=Q
now,
alternatively ,,,
(1) Ba(s)---------->Ba(g) ∆H=S
(2)Ba(g)--------->Ba+ +e^- ∆H=I
(3)1/2O2(g)-------->O(g) ∆H=D/2
(4)O(g) +e^- ------>O-(g) ∆H=-E
(5)Ba+ +O- ----->BaO(s) ∆H=-U

∆H2=S+I+D/2-E-U

where
S is enthalpy of sublimation
I is enthalpy of ionisation
D is dissociation enthalpy
E is eletron gain enthalpy
U is lattice enthalpy

now,
∆H1=∆H2
because path are different but reaction is same so , ∆H1=∆H2