Question

Calculate the lattice enegry of CaCl₂ from the given data

Ca_(s) + Cl_2(g) → CaCl_2(s) ∆H_f⁰ = -795 KJ mol^-1

Atomisation : Ca_(s) → Ca_(g)∆H₁° = + 121 KJ mol^-1 

Ionization : Ca_(g) → Ca²⁺_(g) + 2e^-∆H₂° = + 242.8 KJ mol^-1 

Dissociation : Cl_2(g) → 2Cl_(g) ∆H₃° = +242.8 KJ mol^-1 

Electron affinity : Cl_(g) + e^- → Cl^-_(g) ∆H₃° = -355 KJ mol^-1

Answer 1

∆H_f = ∆H₁ + ∆H₂ + ∆H₃ + 2∆H₄+ 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