Q. Bond dissociation enthalpy of H₂, Cl₂ and HCl are 434, 242 and 431 kJ mol⁻¹ respectively. Enthalpy of formation of HCl is
A) 245 kJ mol⁻¹
B) 93 kJ mol⁻¹
C) −245 kJ mol⁻¹
D) −93 kJ mol⁻¹
Explanation
[Solution from source] For ½H₂ + ½Cl₂ → HCl, ΔHf = ½(434) + ½(242) − 431 = 338 − 431 = −93 kJ mol⁻¹. The ±245 values come from using the full bond enthalpies of H₂ and Cl₂ with only one H–Cl bond (434 + 242 − 431 = 245); that equation actually forms 2 mol HCl. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.