Consider the following data: Delta_(f) H^(2) (N_(2)H_(4), l) =50kJ//mol,, Delta_(f) H^(@) (NH_(3), g)=-46 kJ//mol, B.E. (N-H)= 393kJ//mol and B.E. (H-H)= 436 kJ/mol, also Delta_("vap") H (N_(2)H_(4),l)= 18kJ//mol. The N-N bond energy in N_(2)H_(4) is

Consider the following data: Delta_(f) H^(2) (N_(2)H_(4), l) =50kJ//m

1.	Consider the following data: Delta_(f) H^(2) (N_(2)H_(4), l) =50kJ//mol,, Delta_(f) H^(@) (NH_(3), g)=-46 kJ//mol, B.E. (N-H)= 393kJ//mol and B.E. (H-H)= 436 kJ/mol, also Delta_("vap") H (N_(2)H_(4),l)= 18kJ//mol. The N-N bond energy in N_(2)H_(4) is
Answer» 226 kJ/mol 154 kJ/mol 190 kJ/mol 45.45 K Cal/mole Solution :`N_(2) + 2H_(2) rarr N_(2)H_(4(l)), Delta H_(1) = 50` `(1)/(2) N_(2) + (3)/(2) H_(2) rarr NH_(3(g)), Delta H_(2) = - 46` `UL(N_(2)H_((l)) rarr N_(2)H_(4(g)), Delta H_(3) = 18)` `N_(2)H_(4(g)) + H_(2) rarr 2NH_(3) , Delta H` `Delta H = - Delta H_(3) + 2 Delta H_(2) - Delta H_(1) = - 160` Theoretically, `Delta H = ((N-N) + 4(N -H) + (H - H)) - (2 xx 3 xx (N-H))` `rArr (N-N) = 190` kJ/mol = 45.45 Kcal/mole