Given the bond energies `N-N` , `N-H` and `H-H` bond are `945, 436` and `391KJmol^(-1)` respectively, the enthalpy change of the reaction
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` is

Options

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

Answer

A

Explanation

`N=- N +3H-H rarr 3 underset(H)underset(|)overset(H)overset(|)N- H`
`Delta H = [Delta H_("Reactants") - Delta H _("Products ")]`
`Delta H = [ 945 +3 xx 436 ]-[6 xx 391] =-93 kJ`