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 sturcture of `N_(2)H_(4)` is as follows

Hence, in the reaction,
`N_(2)H_(4)(g) to 2N(g) + 4H(g)`,
`Delta H =` (N - N bond energy ) + (N - N bond energy `xx 4)`
`:. " "` bond energy = `1724 - 1564 = 160 kJ//mol`.