Question

What is enthalpy of formation of $NH_3$ if bond enthalpies are as $(N \equiv N) = 941$ kJ, $(H-H) = 436$ kJ, $(N-H) = 389$ kJ?

• A. -84.5 kJ
• B. -21.25 kJ
• C. -42.5 kJ
• D. -63.45 kJ

Answer 1

Answer: (C)

-42.5 kJ/mol

Answer 2

For the formation of ammonia:

$$\text{Reaction: } N_2 + 3H_2 \rightarrow 2NH_3$$

1. Energy to break bonds:

   • Breaking 1 mole of N≡N: $941\, \text{kJ}$
   • Breaking 3 moles of H–H: $3 \times 436 = 1308\, \text{kJ}$

   Total energy absorbed $= 941 + 1308 = 2249\, \text{kJ}$.

2. Energy released forming bonds:

   • Forming 6 moles of N–H bonds: $6 \times 389 = 2334\, \text{kJ}$

3. Net Reaction Enthalpy ($\Delta H$):

   $$\Delta H_{\text{rxn}} = 2249 - 2334 = -85\, \text{kJ}$$

   Since this is for 2 moles of $NH_3$, the enthalpy of formation per mole is:

   $$\Delta H_f = \frac{-85}{2} = -42.5\, \text{kJ/mol}$$