2[H]+2[Cl]⟶2HCl,ΔH=−862 kJ mol−1H2⟶2[H],ΔH=434 kJCl2⟶2[Cl],ΔH=242 kJH2+Cl2⟶2HCl,ΔH=−186 kJ∴ΔH4 of HCl=2−186=−93 kJ mol−1
NEET UG 2008 — Chemistry Physical Chemistry
Bond dissociation enthalpy of H2,Cl2 and HCl are 434,242 and 431kJmol− 1 respectively. Enthalpy of formation of HCl is
Held on 30 Apr 2008 · Verified 9 Jul 2026.
245kJmol−1
93kJmol−1
−245kJmol−1
−93kJmol−1
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