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The bond dissociation enthalpies of H2(g),Cl2(g) and HCl(g) are 435, 243 and 431 kJ mol1 respectively. The enthalpy of formation of HCl(g) will be

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a
247 kJ mol1
b
770 kJ mol1
c
1109 kJ mol1
d
92 kJ mol1

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detailed solution

Correct option is D

H2(g)+Cl2(g)2HCl(g)ΔfH(HCl,g)=12[ε(HH)+ε(ClCl)2ε(HCl)]=12(435+2432×431)kJ mol1=92 kJ mol1

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