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Calculate the resonance enegry of N(2)O ...

Calculate the resonance enegry of `N_(2)O` form the following data
`Delta_(f)H^(Theta) of N_(2)O = 82 kJ mol^(-1)`
Bond enegry of `N-=N, N=N, O=O,` and `N=O` bond is `946, 418, 498`, and `607 kJ mol^(-1)`, respectively.

A

`-88kJ mol^(-1)`

B

`-178 kJ mol^(-1)`

C

`-252 kJ mol^(-1)`

D

`-36 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2)O implies bar(N) + overset(+)(N) = O`
`N_(2) + 1/2 O_(2) rarr N_(2)O`
`implies (N -= N) + 1/2 (O -= O) - [(N = N) + (O = N)] implies 946 + 1/2(498) -418 - 607 = 170 kJ//mol`
`implies " "`Resonance Energy = 82 – 170 = –88 kJ/mol.
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