Question

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
A. `-93` kJ
B. 102 kJ
C. 90 kJ
D. 105 kJ

Answer 1

Correct Answer - A
`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`.