Question #2632
Write down the Born-Haber cycle for SrBr[sub]2[/sub] formation.
Expert's answer
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)
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)
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#339914 on Dec 2023
#339914 on Dec 2023