Question

Use the Born-Haber Cycle and information given below to determine the lattice energy of LiF.

            Li(s) → Li(g) +159.3 kJ      

            Li(g) → Li⁺(g) + e^–                   +500.9 kJ      

            F₂(g) → 2 F(g) +158.8 kJ

            F(g) + e^– → F^–(g)                     –332.6 kJ      

            Li⁺(g) + F^–(g) → LiF(s)                 ?                          

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            Li(s) + ½ F₂(g) → LiF(s)          – 616.0 kJ

Group of answer choices

+209.0 kJ

–1023 kJ

+1023 kJ

–209.0 kJ

–1102.4 kJ

Answer 1

--787 0 Lattice formation enthalpy: The lattice formation enthalpy Hilford is the enthalpy Change when I mole of an ionic solidis formed form its gaseous ions. o sodium chloride (nach) - Mat & gt ce ig n ace (s) «Histown 0 Energy is released when ionic bonds form. Therefore it is an exothermic process. . same value Lattice enthalpy of formation has the as the lattice I enthalpy of dissociation but with the reverse sign calculation of lattice energy formation of Lif Enthalpy of formation of Lif; Li(s) + 1/2 F2 9 → Lif(s) THE = -6160k Enthalpy of Sublimation of Li(s); Lics) - Licg c) Has = +159.3 kJ Enthalpy of ionisation of Li (9) d) Lig Lit (g) te Enthalpy of Bond dissociation of te ohie= +550.9 kJ P21f2 ) +26 ) 10H 80 = (1/2*458-87kJ Lattice ã on ħ e) Enthalpy of electron affinity of F (92: FG) te- F-8) , SHE A=-332.6 kJ - Inthalpy of formation of Lif(s); li+c8) + FB → Lif(s), s Hictori? Born-Haber cycle Li (s) + 1/2 Fy @ th Lif (s) AHLctom Li (3) + 1/2 62 & SHE Li (9) + 112 fz69) According to Born-Haber cycle T meer que 1 120H&D otes P lita)+ * Fcg) — SHE = SHS + Hie + y2 +HBD + SHA+4+ 4 HCA Li+3)+fc) (1588); - 332.6 + & Hi (form) -6160 = +159-3 +500.9 + 1/2(1588), -3 4 Hi (form, & – 1023 kJ & Lattice enthalpy of - 1028 kJ formation of Lifrs)