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Given that bond energies of H−H and Cl−Cl are 430 kJ mol⁻¹ and 240 kJ mol⁻¹ respectively and ΔHf for HCl is −90 kJ mol⁻¹, bond enthalpy of HCl is:

1. 380 kJ mol⁻¹
2. 425 kJ mol⁻¹
3. 245 kJ mol⁻¹
4. 290 kJ mol⁻¹

Correct answer: 380 kJ mol⁻¹

Solution

The bond enthalpy of HCl can be calculated using the equation: ΔHf = Bond energies of reactants - Bond energies of products. Substituting the values: -90 = 430 + 240 - 2 × Bond energy of HCl. Solving gives Bond energy of HCl = 380 kJ mol⁻¹.

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