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Ciclo de Born-Haber

Ciclo de Born-Haber. Una aplicación de la Ley de Hess. Li (s) + ½ F 2 (g) Li + F - (s). Δ H f 0 = -594,1 kJ. Por partes:. Li (s) Li (g) Δ H(1) = 155,2 kJ Li (g) Li + (g) + 1e - Δ H(2) = 520 kJ. 1/2F 2 (g) F (g) Δ H(3) = ½.150,6 kJ

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Ciclo de Born-Haber

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  1. Ciclo de Born-Haber Una aplicación de la Ley de Hess

  2. Li (s) + ½ F2(g) Li+F- (s) ΔHf0 = -594,1 kJ

  3. Por partes: Li (s) Li (g)ΔH(1) = 155,2 kJ Li (g) Li+ (g) + 1e-ΔH(2) = 520 kJ 1/2F2 (g) F (g) ΔH(3) = ½.150,6 kJ F (g)+ 1e- F- (g) ΔH(4) = - 333 kJ Li+(g) + F-(g) Li+ F-(s) ΔU ΔHf0 =ΔH(1) + ΔH(2) + ΔH(3) + ΔH(4) + ΔU

  4. Video 1 Video 2

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