Answer to Question #2632 in Inorganic Chemistry for Mandy
Write down the Born-Haber cycle for SrBr[sub]2[/sub] formation.
1. Sr(s) → Sr(g) + VBr2(g) → 2Br(g) + B 2. Sr(g) → Sr(g)+ + e- + IEM1 Sr+(g) → Sr(g)2++ e- + IEM2 Br + e → Br- - EAx 3. Sr2+ + 2Br-→ SrBr2 + UL 4. Br2(g) + Sr(s) → SrBr2 + ∆Hf Therefore, ∆Hf = V + B + IEM1 + IEM2 - EAx + UL V is the enthalpy of sublimation for metal atoms B is the bond energy of halogen atom IEM is the ionization energy of the metal atom EAX is the electron affinity of halogen atom UL is the lattice energy (defined as exothermic here)