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)

Calculate the lattice energy of CaCl_2 from the given data. Ca_((s))

1.	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)
Answer» SOLUTION : `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)`