Consider the following data : `Delta_(f)H^(@)(N_(2)H_(4),l)=50kJ//mol,Delta_(f)H^(@)(NH_(3),g)=-46kJ//mol` `B.E(N-H)=393" kJ//mol and B.E."(H-H)=436kJ//mol` `Delta_("vap")H(N_(2)H_(4),l)=18kJ//mol` The N-N bond energy in `N_(2)H_(4)` is :
A
226 kJ/mol
B
154 kJ/mol
C
190 kJ/mol
D
None of these
Text Solution
Verified by Experts
The correct Answer is:
C
`(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)rarrNH_(3)(g)` Let B.E. of `N-=N` is x `-46 = (x)/(2)+(3)/(2)xx436 - 3xx393 implies x = 958` `N_(2)H_(4)(l)
rarrN_(2)(g)+2H_(2)(g),` `Delta_(r )H=-50 " kJ"//"mol"` `Delta_(r)H=[(Delta_("vap")H(N_(2)H_(4),l)),(+4xxB.E.(N-H)),(+B.E.(N-N))]` `-((B.E.(N-=N)),(+2B.E.(H-H)))` `-50 =(18+4xx393+y)-(958 + 2xx436)` `-50=(1590+y)-(1830)` `B.E. (N-N) or y = 190 kJ/mol
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