Given the bond dissociation enthalpies: $ ext{N}\equiv ext{N} = 946 ext{ kJ mol}^{-1}$, $ ext{H

Question

Given the bond dissociation enthalpies: $	ext{N}\equiv	ext{N} = 946 	ext{ kJ mol}^{-1}$, $	ext{H-H} = 436 	ext{ kJ mol}^{-1}$, and $	ext{N-H} = 389 	ext{ kJ mol}^{-1}$. What is the standard enthalpy of formation ($\Delta_f H^\circ$) of ammonia gas ($	ext{NH}_3$)?

Professor PC Thomas Classes · Answer

Equation: $\frac{1}{2}	ext{N}_{2(g)} + \frac{3}{2}	ext{H}_{2(g)} ightarrow 	ext{NH}_{3(g)}$
(Note: Standard enthalpy of formation is for exactly 1 mole of product.)
Formula: $\Delta_f H^\circ = \sum 	ext{Bonds Broken} - \sum 	ext{Bonds Formed}$
Bonds Broken (Reactants): $\frac{1}{2}(	ext{N}\equiv	ext{N}) + \frac{3}{2}(	ext{H-H})$
$\frac{1}{2}(946) + \frac{3}{2}(436) = 473 + 654 = \mathbf{1127 	ext{ kJ}}$
Bonds Formed (Products): $3(	ext{N-H})$
$3(389) = \mathbf{1167 	ext{ kJ}}$
$\Delta_f H^\circ$ $= 1127 - 1167 = \mathbf{-40 	ext{ kJ mol}^{-1}}$