Calculate lattice energy for the change,
`Li^(+)(g)+Cl^(-) (g) rarr LiCl(g)`
Given that
`{:(DeltaH_("sublimation")" of "Li=160.67 kJ mol^(-1)",",,DeltaH_("Dissociation")" of "Cl_(2) =244.34 kJ mol^(-1)","),(DeltaH_("ionisation")" of "Li(g)=520.07 kJ mol^(-1)",",,DeltaH_(E.A)" of "Cl(g)= -365.26 kJ mol^(-1)","),(DeltaH_(f)" of "LiCl(s)= -401.66 kJ mol^(-1)",",,):}`

Explanation

`DeltaH_(f)=DeltaH_(1)+DeltaH_(2)+DeltaH_(3)+2DeltaH_(4)+u`
`-795=121+2422+242.8+(2xx355)+u`
`-795=2785.8-710+u`
`-795=2075.8+u`
`u=-795-2075.8=-2870.8 kJ .mol^(-1)`