Calculate \(\Delta H\) for the reaction: $$ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-1010 . \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-317 \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-623 \mathrm{kJ}\) $$ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-286 \mathrm{kJ}\)

We need to manipulate the given reactions in such a way that they can be added together to equal the target reaction. The following adjustments are made: a. In the first reaction, multiply by -2/3 to have -2 NH3(g) on the left side: $$ \frac{-2}{3}(2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)) $$ b. In the second reaction, reverse it to have the N2H4 (l) on the right side: $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longleftrightarrow \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) $$ c. In the third reaction, multiply it by 1/2 to have 1/2 O2(g) on the left side: $$ \frac{1}{2}(\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)) $$ d. In the fourth reaction, no need to manipulate as it fits with the target reaction. #Step 3: Calculate the corresponding enthalpy changes for the adjusted reactions#

Now add the adjusted reactions and their respective enthalpies: Reactions: $$ \frac{-2}{3}(2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g))+( \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2}(\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g))+ (\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)) = 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ Enthalpies: $$ \Delta H = \Delta H_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} $$ $$ \Delta H = 674\,\text{kJ} + 317\,\text{kJ} - 311.5\,\text{kJ} - 286\,\text{kJ} = 393.5\,\text{kJ} $$ So, the enthalpy change for the target reaction is: $$ \Delta H = 393.5\,\text{kJ} $$