The bond dissociation enthalpies of H2(g) and N2 (g) are + 435 95 kJmol-1 and + 941 8 kJ mol-1 - Chemistry - Thermodynamics

Question

T h e   b o n d   d i s s o c i a t i o n   e n t h
a l p i e s   o f   H 2 ( g )   a n d   N 2
  ( g )   a r e   +   435 95   k J m o l -
1   a n d   +   941 8   k J   m o l - 1
    a n d   e n t h a l p y   o f   f o r
m a t i o n   o f   N H 3 ( g )   i s   —
46 024   k J   m o l - 1     ( a )   W h a
t   i s   t h e   e n t h a l p y   o f  
a t o m i z a t i o n   o f   N H 3 ( g )   ?  
  ( b )   W h a t   i s   t h e   b o n d
  e n t h a l p y   o f   N   —   H
  b o n d   ?

Raveena Sharma · Accepted Answer

The solution is as follows:

(a) Enthalpy of formation of NH3 = Bond dissociation in
N2 + Bond dissociation in H2 - Enthalpy of
atomisation of NH3

Enthalpy of atomisation of NH3 = 941 82+435
95×32-(-46 024)=470 9 + 653 92 + 46
024= 1170 84 kJ/mol

(b) Bond enthalpy for N-H bond = 1170 843=90 28 kJ/mol