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Based on the values of B.E. given, Delta...

Based on the values of B.E. given, `Delta_(f)H^(0) " of " N_(2)H_(4(g))` is : Given BE of : `N- N` is 159 kJ `mol^(-1)`, H-H is 436 kJ `mol^(-1), N =- N` is 941 kJ `mol^(-1), N-H` is 398 kJ `mol^(-1)`

A

711 kJ `" mol"^(-1)`

B

62 kJ `" mol"^(-1)`

C

`-98 " kJ mol"^(-1)`

D

`-711 " kJ mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`N_(2)(g)+2H_(2)(g)rarrN_(2)H_(4)(g)`
`Delta_(f)H(N_(2)H_(4),g)`
`=(941+2xx436)-(159+4xx398)`
`=1813 - 1751 = 62 " kJ mol"^(-1)`
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