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

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