Given that bond energies of N=N,H-H and N-H bonds as 945, 436 and 391 kJ/ mol respectively ,the enthalpy of the reaction `N_(2)(g)+3H_(2)(g)+to2NH_(3)(g),`is

Options

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

Answer

A

Explanation

`N_(2)(g)+3H_(2)(g)to2NH_(3)(g)`
Bond energies of N=N=954 kJ
H-H=436 kJ
N-H=391 KJ
(i) To break N=H ,energy required is 945 kJ
(ii) To break 3 (H-N) eneegy required is = 436`xx`3 =1308 kJ
`therefore` Total energy required =945 kJ + 1308 kJ =2253 kJ
(iii) Energy liberated while forming 2`xx`3=6(N-H) bonds is 391`xx`6=236 kJ
`therefore`Enthalphy of the reaction = Energy required - Energy liberated =2253-2346=-93 kJ