Question

Bond energy of n triple bond n ,h-h ,and n-h are 945.6,435.1,389.1 kj per mole. Calculate heat of formation of nh3.

Answer 1

hey! the heat of formation of NH3 will
(BE) of N2 +3(BE)of H2 -2(BE)of NH3

Answer 2

Explanation:

The chemical equation for given reaction will be as follows.

              $N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)$

Therefore, calculate the heat of formation of $NH_{3}$ as follows.

  $\Delta H$ = Bond energies of reactants - Bond energies of products

         $\Delta H_{NH_{3}}$ = Bond energy of $N_{2}$ + bond energy of $H_{2}$ - 2 × Bond energy of $NH_{3}$  

                       = $945.6 kj/mol + (3 \times 435.1 kj/mol) - (6 \times 389.1)kj/mol$

                        = (945.6 + 1305.3 - 2334.6) kJ/mol     (There are 6 N-H bonds in $2NH_{3}$)

                        = -83.7 kJ/mol

Thus, we can conclude that heat of formation of $NH_{3}$ is -83.7 kJ/mol.