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
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)`
8 views
1 Answers
Correct Answer - 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)`
8 views
Answered