Determine the heat of reaction for the combustion of ammonia,
4NH3(g) + 7O2(g) ? 4NO2(g) + 6H2O(l) using molar enthalpies of formation.
$\begin{array}{ll}
\text { molecule } & \Delta f H^{\circ}(\mathrm{kJ} / \mathrm{mol}) \
\hline \mathrm{NH}_{3}(g) & -45.9 \
\mathrm{NO}_{2}(g) & +33.1 \
\mathrm{H}_{2} \mathrm{O}(l) & -285.8
\end{array}$

Question

Determine the heat of reaction for the combustion of ammonia, 
4NH3(g) + 7O2(g) ? 4NO2(g) + 6H2O(l) using molar enthalpies of formation.
$\begin{array}{ll}
\text { molecule } & \Delta f H^{\circ}(\mathrm{kJ} / \mathrm{mol}) \
\hline \mathrm{NH}_{3}(g) & -45.9 \
\mathrm{NO}_{2}(g) & +33.1 \
\mathrm{H}_{2} \mathrm{O}(l) & -285.8
\end{array}$

Quizplus · Accepted Answer

The heat of reaction ($\Delta H_{rxn}^{\circ}$) can be calculated using the formula: $\Delta H_{rxn}^{\circ} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants})$. For the given reaction, this becomes: $[4(\Delta H_f^{\circ}(\mathrm{NO}_2)) + 6(\Delta H_f^{\circ}(\mathrm{H}_2\mathrm{O}))] - [4(\Delta H_f^{\circ}(\mathrm{NH}_3))] = [4(33.1) + 6(-285.8)] - [4(-45.9)] = [132.4 - 1714.8] - [-183.6] = -1582.4 + 183.6 = -1398.8 \, \mathrm{kJ}$.

Determine the Heat of Reaction for the Combustion of Ammonia