Calculate the lattice energy of `CaCl_(2)` from the given data
`Ca(s) +Cl_(2)(g) rarr CaCl_(2)(s) Delta H_(f)^(0) =-795" kJ mol"^(-1)`
`{:("Atomisation ": Ca(s) rarr Ca(g)" "Delta_(1)^(0)=+121" kJ mol"^(-1)),("Ionisation ": Ca(g) to Ca^(2+)(g) +2e^(-)" "Delta H_(2)^(0) =+2422" kJ mol"^(-1)), ("Dissociation ": Cl_(2)(g) to 2Cl(g)" "Delta H_(3)^(0) =+242.8" kJ mol"^(-1)),("Electron affinity": Cl(g) +e^(-) to Cl^(-)(g)" "Delta H_(4)^(0)=-355" kJ mol"^(-1)):}`

Explanation

`Delta H_(f) = Delta H_(1) + Delta H_(2) + Delta H_(3)+ 2 Delta H_(4) + u`
`-795 =121+2422+242.8+(2xx-355) +u`
`-795=2785.8-710 +u`
`-795= 2075.8 +u`
`u =-795-2075.8`
`u = -2870.8" kJ mol"^(-1)`