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Question Number 23694 by Tinkutara last updated on 04/Nov/17

Calculate the lattice enthalpy of MgBr_2 .  Given that  Enthalpy of formation of MgBr_2  = −524  kJ mol^(−1)   Sublimation energy of Mg = +2187  kJ mol^(−1)   Vaporisation energy of Br_2 (l) = +31  kJ mol^(−1)   Dissociation energy of Br_2 (g) = +193  kJ mol^(−1)   Electron gain enthalpy of Br = −331  kJ mol^(−1)

$${Calculate}\:{the}\:{lattice}\:{enthalpy}\:{of}\:{MgBr}_{\mathrm{2}} . \\ $$$${Given}\:{that} \\ $$$${Enthalpy}\:{of}\:{formation}\:{of}\:{MgBr}_{\mathrm{2}} \:=\:−\mathrm{524} \\ $$$${kJ}\:{mol}^{−\mathrm{1}} \\ $$$${Sublimation}\:{energy}\:{of}\:{Mg}\:=\:+\mathrm{2187} \\ $$$${kJ}\:{mol}^{−\mathrm{1}} \\ $$$${Vaporisation}\:{energy}\:{of}\:{Br}_{\mathrm{2}} \left({l}\right)\:=\:+\mathrm{31} \\ $$$${kJ}\:{mol}^{−\mathrm{1}} \\ $$$${Dissociation}\:{energy}\:{of}\:{Br}_{\mathrm{2}} \left({g}\right)\:=\:+\mathrm{193} \\ $$$${kJ}\:{mol}^{−\mathrm{1}} \\ $$$${Electron}\:{gain}\:{enthalpy}\:{of}\:{Br}\:=\:−\mathrm{331} \\ $$$${kJ}\:{mol}^{−\mathrm{1}} \\ $$

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