Find the Bond enthalpy (in kJ/mol) of one "three centre two electron bond" in B2H6 {B–H–B → 2B(g) + H(g)} from the given data. ΔHf° [BH3(g)] = 100 kJ/mol ΔHf° [B2H6(g)] = 36 kJ/mol ΔHf° ΔHatm [B(s)] = 565 kJ/mol ΔHatm = [H2(g)] = 218 kJ/mol

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Chemistry

Ionisation Electron Gain Atomisation enthalpy

Question

Find the Bond enthalpy (in kJ/mol) of one "three centre two electron bond" in B₂H₆
{B–H–B → 2B(g) + H(g)} from the given data.
ΔH_f^° [BH₃(g)] = 100 kJ/mol ΔH_f^° [B₂H₆(g)] = 36 kJ/mol ΔH_f^°
ΔH_atm [B(s)] = 565 kJ/mol ΔH_atm = [H₂(g)] = 218 kJ/mol

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2E_B–H–B = 2 × 565 + 6 × 218 – 36 – 4 × 373
E_B–H–B = 455 kJ/mol = 4.55 × 102 kJ/mol