Calculate the lattice energy of `CaCl_2` from the given data.
`Ca_((s)) + Cl_(2(g)) to CaCl_(2(s)) DeltaH_f^0 =-795 "kJ mol"^(-1)`
Sublimation : `Ca_((s)) to Ca_((g)) " " DeltaH_1^0=+121 "kJ mol"^(-1)`
Ionisation : `Ca_((g)) to Ca_((g))^(2+) +2e^(-) " " DeltaH_2^0=+2422 "kJ mol"^(-1)`
Dissociation : `Cl_(2(g)) to 2Cl_((g)) " " DeltaH_3^0=+242.8 "kJ mol"^(-1)`
Electron affinity : `Cl_((g)) + e^(-) to Cl_((g))^(-) " " DeltaH_4^0=-355 "kJ mol"^(-1)`

Explanation

`DeltaH_f=DeltaH_1+DeltaH_2+DeltaH_3+2DeltaH_4 + u`
-795=121+2422+242.8+(2 x -355) + u
-795 =2785.8-710 +u
-795=2075.8 + u
u=-795-2075.8
`u=-2870.8 "kJ mol"^(-1)`