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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

160kJ `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(overset(H)(|))(N)- overset(overset(H)(|))(N)- H` (So, 4N- H bond present) means their energy `=391 xx 4= 1564`
so the bond energy of N-N in `N_(2)H_(4)` = 1724-1564= 160kJ/mol
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