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Given the following information:Br2 bond energy = 193 kJ/mol F2 bond energy = 154 kJ/mol 1/2Br2(g) + 3/2F2(g) -> BrF3(g) = –384 kJ/mol calculate the Br-F bond energy. A)244 kJ/molB)237 kJ/molC)712 kJ/molD)128 kJ/molE)none of these

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Answer:

C) 712 KJ/mol

Explanation:

  • ΔH°r = Σ Eb broken - Σ Eb formed
  • 1/2Br2(g) + 3/2F2(g) → BrF3(g)

∴ ΔH°r = - 384 KJ/mol

∴ Br2 Eb = 193 KJ/mol

∴ F2 Eb = 154 KJ/mol

⇒ Σ Eb broken = (1/2)(Br-Br) + (3/2)(F-F)

⇒ Σ Eb broken =  (1/2)(193 KJ/mol) + (3/2)(154 KJ/mol) = 327.5 KJ/mol

∴ Eb formed: Br-F

⇒ Σ Eb formed (Br-F) = Σ Eb broken - ΔH°r

⇒ Eb (Br-F) = 327.5 KJ/mol - ( - 384 KJ/mol )

⇒ Eb Br-F = 327.5 KJ/mol + 384 KJ/mol = 711.5 KJ/mol ≅ 712 KJ/mol

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