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For the reaction `N_(2)H_(4) (g) rarr N_(2)H_(2)(g) +H_(2)(g)" "Delta_(r)H^(@)=109 KJ//mol` Calculate the bond enthalpy of `N=N`. Given : `B.E. (N-N) =163 KJ//mol, B.E. (N-H) =391 KJ//mol, B.E. (H-H)=436 KJ//mol` |
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Answer» The equation: `DeltaH=109 = in_(N-N)+4 in_(N-H)-in_(H-H)-2in_(N-H)-in_(N=N)` `in_(N equivN)=163+2xx391-436-109=400" kJ/mole"` |
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