Calculate the lattice enthalpy of CaCl2 , given that : Enthalpy of sublimation for Ca(s) → Ca(g) = 121 kJ mol−1 Enthalpy of dissociation of Cl2(g) → 2Cl(g) =242.8 kJ mol−1 Ionisation energy of Ca(g)→ Ca++(g) =2422 kJ mol−1 Electron gain enthalpy of 2Cl→ 2Cl−1=(2×−355) kJ mol−1=−710 kJ mol−1 Enthalpy of formation of CaCl2=−795 kJ mol−1

Calculate the lattice enthalpy of CaCl2 , given that : Enthalpy of sub

1.	Calculate the lattice enthalpy of CaCl2 , given that : Enthalpy of sublimation for Ca(s) → Ca(g) = 121 kJ mol−1 Enthalpy of dissociation of Cl2(g) → 2Cl(g) =242.8 kJ mol−1 Ionisation energy of Ca(g)→ Ca++(g) =2422 kJ mol−1 Electron gain enthalpy of 2Cl→ 2Cl−1=(2×−355) kJ mol−1=−710 kJ mol−1 Enthalpy of formation of CaCl2=−795 kJ mol−1
Answer» Calculate the lattice enthalpy of $C a C l_{2}$ , given that : Enthalpy of sublimation for $C a (s) \to C a (g) = 121 k J m o l^{- 1}$ Enthalpy of dissociation of $C l_{2} (g) \to 2 C l (g) = 242.8 k J m o l^{- 1}$ Ionisation energy of $C a (g) \to C a^{+ +} (g) = 2422 k J m o l^{- 1}$ Electron gain enthalpy of $2 C l \to 2 C l^{- 1} = (2 \times - 355) k J m o l^{- 1} = - 710 k J m o l^{- 1}$ Enthalpy of formation of $C a C l_{2} = - 795 k J m o l^{- 1}$