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Question
Write down the Born-Haber cycle for the formation of CaCl2
Solution
Born – Haber cycle for the formation of CaCl2
\[\ce{Ca_{(s)} + Cl2_{(l)} -> CaC2_{(S)}}\] = ∆Hf°
Sublimation: \[\ce{Ca_{(S)} -> Ca_{(S)}}\] = ∆H1°
Ionization: \[\ce{Ca_{(g)} -> Ca^2+_{(g)} + 2e^-}\] = ∆H2°
Vapourisation: \[\ce{Cl2_{(l)} -> Cl2_{(g)}}\] = ∆H3°
Dissociation: \[\ce{Cl2_{(g)} -> 2Cl2_{(g)}}\] = ∆H4°
Electron affinity: \[\ce{2Cl2_{(g)} + 2e^- -> 2Cl^{-2}_{(g)}}\] = ∆H5°
Lattice enthalpy: \[\ce{Ca^2+_{(g)} + 2Cl^-_{(g)} -> CaCl2_{(S)}}\] = ∆H6°
∆Hf° = ∆H1° + ∆H2° + ∆H3° + ∆H4° + ∆H5° + ∆H6°
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\[\ce{Ca_{(s)} + Cl2_{(g)} -> CaCl2_{(s)}}\] ∆`"H"_"f"^0` = − 795 kJ mol−1
Sublimation: \[\ce{Ca_{(s)} -> Ca-{(g)}}\] ∆`"H"_1^0` = + 121 kJ mol−1
Ionisation: \[\ce{Ca_{(g)} -> Ca^2+_{(g)} + 2e^-}\] ∆`"H"_2^0` = + 2422 kJ mol−1
Dissociation: \[\ce{Cl2_{(g)} -> 2Cl_{(g)}}\] ∆`"H"_3^0` = + 242.8 kJ mol−1
Electron affinity: \[\ce{Cl_{(g)} + e^- -> Cl^-_{(g)}}\] ∆`"H"_4^0` = −355 kJ mol−1
