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COMEDK202510 May 2025Evening ShiftChemistryThermodynamics (C)Actual

For the reaction A ₂ + B ₂ 2 AB , H_f (AB) = -400 kJ/mol . The bond dissociation enthalpies of A ₂ , B ₂ and AB are in the ratio 1 : 0.75 : 1. What is the bond dissociation enthalpy of B ₂ in kJ/mol?

Options

  1. A800
  2. B1600
  3. C2400
  4. D3200

Correct answer

C. 2400

Step-by-step solution

H_ rxn = 2 H_f(AB) = 2 (-400) = -800 kJ/mol Let BE_ A₂ = x , BE_ B₂ = 0.75x , BE_ AB = x H_ rxn = BE( reactants ) - BE( products ) -800 = (x + 0.75x) - 2x = 1.75x - 2x = -0.25x x = -800 -0.25 = 3200 kJ/mol BE_ B₂ = 0.75 3200 = 2400 kJ/mol

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