Given the bond energies N-=N, H-H and N-H bonds are 945,436 and 391 kJ – InterviewSolution

1.	Given the bond energies N-=N, H-H and N-H bonds are 945,436 and 391 kJ "mole"^(-1) respectively, the enthalpy of the following reaction N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g) is
Answer» `-93` KJ 102 kJ 90 kJ 105 kJ Solution :`underset("ENERGY ABSORBED")ubrace(underset(945)(N-=)Nunderset(+3xx436)(+3H-H))rarrunderset("Energy released")undersetubrace(2xx(3xx391)=2346)(2underset(H)underset(\|)overset(H)overset(\|)N-H)` NET. Energy released = 2346-2253=93 kJ i.e. `DeltaH=-93 kJ`.

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