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Heat of hydrogenation of ethene is x(1) ...

Heat of hydrogenation of ethene is `x_(1)` and that of benzene is `x_(2)` . Hence resonance energy of benezene is

A

`x_(2)-3x_(1)`

B

`x_(1)+x_(2)`

C

`3x_(1)-x_(2)`

D

`x_(1)-3x_(2)`

Text Solution

Verified by Experts

The correct Answer is:
B

`+3H_(2)rarrC_(6)H_(12)(Delta_(r)H)_(cal)=3x_(1)`
and `(Delta_(r)H)_(obs)=3x_(1)-x_(2)`
`therefore RE_(=) (Delta_(r)H)_("cal")-(Delta_(r)H)_("obs") = 3x_(1) -x_(2)`
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