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COMEDK2024Morning ShiftChemistryThermodynamics (C)Actual

If the enthalpy of formation of a diatomic molecule AB is -400 ~kJ / mol and the bond dissociation energies of A ₂ and B ₂ and AB are in the ratio 2: 1: 2 , what is the bond dissociation enthalpy of B ₂ ?

Options

  1. A600 kJ/mol
  2. B1600 kJ/mol
  3. C800 kJ/mol
  4. D400 kJ/mol

Correct answer

C. 800 kJ/mol

Step-by-step solution

The formation reaction for the diatomic molecule AB is given by: 1 2 A ₂(g) + 1 2 B ₂(g) AB (g) . The enthalpy of formation H_f^ is related to the bond dissociation energies (BDE) by the equation: H_f^ = BDE ( reactants ) - BDE ( products ) . Let the bond dissociation energies of A ₂ , B ₂ , and AB be 2x , x , and 2x respectively, based on the given ratio 2:1:2 . Substituting these into the enthalpy equation: -400 = ( 1 2 (2x) + 1 2 (x) ) - (2x) . Simplifying the expression: -400 = x + 0.5x - 2x . -400 = -0.5x . So

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