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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. 190kJ/molB. `-190kJ//"mol"`C. 95 kJ/molD. `-95 kJ//"mol"` |
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Answer» Correct Answer - A `1/2N_2(g)+3/2H_2(g)toNH_3(g)` Let B.E. of N = N is x `-46=x/2+3/2xx436-3xx393impliesx=958` `N_2H_4(l) to N_2(g)+2H_2(g)` `Delta_rH=-50` kJ/mole `implies Delta_rH=[underset(+B.E.(N-N))overset(Delta_(vap)H(N_2H_4,l))(+4xxB.E.(N-H))]-[overset(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/mole |
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