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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`

B

`-66kJ`

C

`-62kJ`

D

`-44kJ`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2)(g)+1/2O_(2)toN_(2)O(g)`
`DeltaH_(f)^(@)=(DeltaH_(N-=N)+1/2DeltaH_(O=O))-(DeltaH_(N=N)+DeltaH_(N=O))`
`=(946+1/2xx498)-(418+607)=170kJ"mol"^(-1)`
Resonance energy `=170-82=88kJ "mol"^(-1)`
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Calculate the resonance energy of N_(2)O from the following data : DeltaH_(f)^(@) of N_(2)O = 82 kJ mol^(-1) , Bond energies of N equivN , N = N , O =O and N =O bonds are 946, 418, 498 and 607 kJ mol^(-1) respectively.

Calculate the resonance energy of N_(2)O from the following data: DeltaH_(1)^(@) of N_(2)O = 82 kJ "mole"^(-1) . Bond energies N=N = 946 kJ "mole"^(-1) N=N = 418 kJ "mole"^(-1) O=O = 498 kJ "mole"^(-1) N=O = 607 kJ "mol"^(-1)

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