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)`?

Options

A. 391 kJ `mol^(-1)`
B. 160 kJ `mol^(-1)`
C. 1173 kJ `mol^(-1)`
D. 320 kJ `mol^(-1)`

Answer

B

Explanation

The structure of `N_(2)H_(4)` is as follows

Hence, in the reaction, `N_(2)H_(4)(g)to2N(g)+4H(g),`
`DeltaH=(N-N" bond energy")+(N-H" bond energy"xx4)`
`therefore 1724=N-N` bond energy+`4xx391`
or N-N bond energy=1724-1564=160kJ/mol.