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The enthalpy change (DeltaH) for the pro...

The enthalpy change `(DeltaH)` for the process, `N_(2)H_(4)(g)to 2N(g)+4H(g)` is
is 1724 kJ `mol^(-1)`. If the bond energy of N-H bond in ammonia is 391 kJ `mol^(-1)`, what is the bond energy for N-N bond in `N_(2)H_(4)`?

A

`160 kJ mol^(-1)`

B

`391 kJ mol^(-1)`

C

`1173 kJ mol^(-1)`

D

`320 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`H-overset(H)overset(|)(N)-overset(H)overset(|)(N)-H` (So, 4N-H bond present) means their energy `= 391xx4=1564` so the bond energy of N-N in `N_(2)H_(4)`
`= 1724-1564=160 KJ//mol`.
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