When 2.500 g of hydrogen gas, H2(g), was completely reacted with gaseous nitroge

Question

When 2.500 g of hydrogen gas, H2(g), was completely reacted with gaseous nitrogen, N2(g), at
constant pressure to produce gaseous ammonia, NH3(g), i.e., in the reaction:
3 H2(g) + N2(g) 2 NH3(g)
the heat (energy) released was 38.19 kJ. If the bond energy of the H—H bond in H2(g) is 436 kJ/mol
and that of the NN bond in N2(g) is 941 kJ/mol, calculate the average bond energy of the N—H
bond in NH3(g).

What I am doing is 3(436) + 941 - 2(3)(N-H) = 38.19 and I get the answer as 368KJ. The correct answer is 390kJ/mol but I can not figure out the proper moles.

Explanation / Answer

Moles of H2 = 2.5/2 = 1.25

Heat released per 1.25 moles H2 = - 38.19 KJ

Heat released per mole = 38.19/1.25 = - 30.55 KJ /mol

Now since reaction involves 3H2 , hence heat of reaction per 3mol H2 = 3 x - 30.55 = - 91.65 KJ

Now enthalphy of reaction = bond energies of bodn broken -bond enegies of bonds formed

3 BE( H-H) + BE( NN) - 6BE ( N-H)   = - 91.65

3(436) + 941 - 6BE ( N-H) = - 91.65

BE ( N-H) = 390 KJ/mol

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