Question

What is enthalpy of formation of NH₃ if bond enthalpies as (N ≡ N) = - 941 kJ/mol.

\[\ce{(H - H)}\] = 436 kJ/mol and \[\ce{(N - H)}\] = 389 kJ/mol?

Options

• - 84.5 kJ/mol

• - 63.45 kJ/mol

• - 21.25 kJ/mol

• - 42.5 kJ/mol

Answer 1

- 42.5 kJ/mol

Explanation:

\[\ce{N2 + 3H2 -> 2NH3}\]

`Delta_"f""H"("NH"_3)` = Bond energy of reactant - Bond energy of product

`= (1/2 xx "BE of N" ≡ "N bond" + 3/2 "BE of H - H bond") - ("3 BE of N - H bond")`

`= (1/2 xx 941 + 3/2 xx 436) - (3 xx 389)`

= 470.5 + 654 - 1167

= - 42.5 kJ/mol