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

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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 Δ H f ​ for HCl is − 90 kJ mol − 1 , bond enthalpy of HCl is

TestHub · Accepted Answer

2 1 ​ H 2 ​ + 2 1 ​ Cl 2 ​ ⟶ HCl ​ Δ H HCl ​ = ∑ B.E. of reactant − ∑ B.E. of products − 90 = 2 1 ​ × 430 + 2 1 ​ × 240 − B.E. of HCl ∴ B.E. of HCl = 215 + 120 + 90 = 425 kJ mol − 1 ​

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