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

Question

Dr. Abubacker · Accepted Answer

[Solution from source] ½H₂ + ½Cl₂ → HCl: ΔHf = ½(430) + ½(240) − x = −90, so x = 335 + 90 = 425 kJ mol⁻¹. 245 kJ mol⁻¹ subtracts the 90 instead of adding it. Answer (2).

Q. 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

Explanation

Reviewed by Dr. Abubacker

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