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)

Based on the values of B.E. given, Delta_(f)H^(0)" of " N_(2)H_(4(g))

1.	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)
Answer» `711 kJ MOL^(-1)` `62 kJ mol^(-1)` `-98 kJ mol^(-1)` `-711 kJ mol^(-1)` Solution :`N_(2) + 2H_(2) rarr N_(2)H_(4)` `Delta H_(f) = ((N -= N ) + 2 XX (H - H))- ((N - N) + 4(N - H))` `=((941) + 2 xx (436)) - (159 + 4 xx (398))` = 62 kJ