Consider the following reaction and use bond energies to cal

Consider the following reaction and use bond energies to calculate the Delta H_rxn N_2(g) rightarrow 3H_2(g) rightarrow 2NH_3(g)

Solution

H° = sum of bond energies of bonds being broken - sum of bond energies of bonds being formed.

Since, bond energies we require are N2 - 945 kJ/mole (triple bond)

H-H - 436 kJ/mole

N-H - 391 kJ/mole

Hence, H° = [ 945 + (3 x 436) - (6 x 391)]

= [ 945 + 1308 - 2346]

= - 93 kJ/mole