If enthaopy of hydrogenation of `C_(6)H_(6)(l)` into `C_(6)H_(12)(l)` is `-205kJ//mol` and resonance energy of `C_(6)H_(6)(l)` is `-152kJ//mol` then enthaopy of hydrogenation of Assume `DeltaH_("vap")` of `C_(6)H_(6)(l),C_(6)H_(8)(l)` all equal :
A
`-535.5kJ//mol`
B
`-238kJ//mol`
C
`-357kJ//mol`
D
`-119` kJ/mol
Text Solution
Verified by Experts
The correct Answer is:
D
Theoretical heat of hydrogenation of benzene = (Actual heat of hydrogenation) + (Resonance energy) =-205 - 152 =- 357 Enthalpy of hydrogenation of , `=(-357)/(3) = - 119` kJ/mol
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