Question

Given the bond energies N ≡ N, H – H and N – H bonds are 945, 436 and 391 kJ/mol respectively. The enthalpy of the reaction;

\[\ce{N2_{(g)} + 3H2_{(g)} -> 2NH3_{(g)}}\]

Options

• −93 kJ

• −102 kJ

• +93 kJ

• +102 kJ

Answer 1

−93 kJ

Explanation:

1 × N ≡ N = 945 kJ

3 × H – H = 1308 kJ

6 × N – H = 2346 kJ

∆H = sum of bond energies of reactants − sum of bond energies of products

∆H = 2253 − 2346 = −93 kJ