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Bond energy of (N-H) bond is yKJ mol^(-1...

Bond energy of `(N-H)` bond is `yKJ mol^(-1)` under standard state. Thus,change in internal energy in the following process is
`NH_(3)(g)rarrN(g)+3H(g)`

A

`-3yKJmol^(-1)`

B

`-yKJmol^(-1)`

C

`3yKJmol^(-1)`

D

`yKJmol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D

`H-underset(H)underset(|)(N)-H` (three `N-H`) bonds
Thus, `DeltaE=3yKJmol^(-1)`
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