Calculate enthalpy of formation of NaCl2 from the following data using Born-Haber Cycle Heat of sublimation of Na = 109 kJ/mol 1st IE of Na = 494 kJ/mol 1st IE of Na+ = 4563 kJ/mol Bond Energy of Cl2 = 242 kJ/mol Electron gain enthalpy of Cl = –347 kJ/mol Lattice energy of NaCl2 = 2560 kJ/mol

Engineering

Chemistry

Born Haber Cycle

Question

Calculate enthalpy of formation of NaCl₂ from the following data using Born-Haber Cycle
Heat of sublimation of Na =   109 kJ/mol
1^st IE of Na =   494 kJ/mol
1^st IE of Na⁺    =   4563 kJ/mol
Bond Energy of Cl₂ =   242 kJ/mol
Electron gain enthalpy of Cl   =   –347 kJ/mol
Lattice energy of NaCl₂ =   2560 kJ/mol

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${ΔH}_{f}^{°} = s + ({IE}_{1} + {IE}_{2}) + D_{{Cl}_{2}} + 2 + ΔHe . g . + U$
= 109 + (494 + 4563) + 242 + [2 × (–347)] – 2560
= 2154 kJ/mol

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