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calculateth resonance energy of N(2)O ...

calculateth resonance energy of `N_(2)O`
`Delta_(f)H^(-) of N_(2)O = 82 kJ mol ^(-1)`
bond energy of N=O = `607 kJmol ^(-1)`
bond energy of O=O = `498 kJ mol ^(-1)`
nond energy of N = N = `418 kJ mol^(-1)`
bond energy of N= N = `946 kJ mol ^(-1)`

A

`82 kJ mol ^(-1)`

B

`-88 kJ mol ^(-1)`

C

`- 82 kJ mol ^(-1)`

D

`+ 88 kJ mol ^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
b
`N equivN(g) + 1/2 O_(2)(g)to N=N=0`
calculated value of `Delta_(f)H^(Theta)=`
`Delta_(f)H^(-)=[BE( NequivN)+1/2BE(=O)]`
`[BE(N=N)+BE(N=O)]`
`[946+1/2(498)]- [418+607]=170 kJ mol^(-1)`
Respmace energy = observed `Delta_(f)H^(Theta)` - calculated `Delta_(f)H^(Theta)`
= 82 -170 =-88 kJ `mol^(-1)`
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