Given that bond energies of H - H and Cl - Cl are 430 ~kJ ~mol ⁻¹ and 240 ~kJ mol ⁻¹ respectively and H _ f for HCl is - kJ mol ⁻¹ , bond enthalpy of HCl is:

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Correct answer: B. 425 ~kJ ~mol ⁻¹

1 2 H ₂+ 1 2 Cl ₂ HCl H _ HCl = B.E. of reactant - B.E. of products -90= 1 2 430+ 1 2 240- B.E. of HCl B.E. of HCl =215+120+90=425 ~kJ ~mol ⁻¹

Given that bond energies of H - H and Cl - Cl are 430 ~kJ ~mol ⁻¹ and 240 ~kJ mol ⁻¹ respectively and H _ f for HCl is - kJ mol ⁻¹ , bond enthalpy of HCl is:

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