Calculate the standard reaction enthalpy for the reaction.
$\begin{array}{ll}\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l}) +\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \\\text { Given: }\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l}) +\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g}) +2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & \Delta H^{\circ}=-543 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\2 \mathrm{H}_{2}(\mathrm{~g}) +\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & \Delta H^{\circ}=-484 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\\mathrm{~N}_{2}(\mathrm{~g}) +3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) & \Delta H^{\circ}=-92.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\end{array}$

Question

Calculate the standard reaction enthalpy for the reaction.

\begin{array}{ll}\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l}) +\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \\\text { Given: }\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l}) +\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g}) +2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & \Delta H^{\circ}=-543 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\2 \mathrm{H}_{2}(\mathrm{~g}) +\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & \Delta H^{\circ}=-484 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\\mathrm{~N}_{2}(\mathrm{~g}) +3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) & \Delta H^{\circ}=-92.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\end{array}

Quizplus · Accepted Answer

To find the standard reaction enthalpy for the given reaction, we use Hess's Law, which allows us to add or subtract given reactions to obtain the target reaction. The steps are as follows:1. The first given reaction is used directly as it provides $N_2$ and $H_2O$ from $N_2H_4$ and $O_2$, with $\Delta H^{\circ} = -543 \, \text{kJ/mol}$.2. The second given reaction needs to be reversed to produce $H_2$ and $O_2$ from $H_2O$, which changes its $\Delta H^{\circ}$ to $+484 \, \text{kJ/mol}$ (since reversing a reaction changes the sign of $\Delta H^{\circ}$).3. The third reaction is used as given, providing $NH_3$ from $N_2$ and $H_2$, with $\Delta H^{\circ} = -92.2 \, \text{kJ/mol}$.Adding these together:$$\Delta H^{\circ}_{\text{total}} = (-543 \, \text{kJ}) + (484 \, \text{kJ}) + (-92.2 \, \text{kJ}) = -151.2 \, \text{kJ/mol}$$Therefore, the standard reaction enthalpy for the target reaction is $-151 \, \text{kJ/mol}$.

Calculate the Standard Reaction Enthalpy for the Reaction A) -935 KJ