Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{array}{cr} \text { Equation } & \Delta H(\mathrm{k}\rfloor) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(I) & -1010 \\ \mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{H}_{2} \mathrm{O}(I) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{H}_{2} \mathrm{O}(I) & -143 \\ \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(I) & -286 \end{array} $$

The target reaction is: $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$

We need to manipulate the given reactions in such a way that, when summed, they reproduce the target reaction. Taking the third given reaction: $$ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \space\space\space\space\space\space\space (\Delta H = -143\space kJ) \\ $$ And the fourth given reaction, multiplying it by 2 to get 2 molecules of \(\mathrm{H}_{2} \mathrm{O}(I)\): $$ 2(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(I)) \space\space\space\space (\Delta H = 2\times-286\space kJ) \\ $$ Now, add these two reactions: $$ 2 \mathrm{NH}_{3}(g) + \mathrm{O}_{2}(g) + 2\mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l) + 2 \mathrm{H}_{2} \mathrm{O}(l) $$ Also, we can take the second given reaction: $$ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(I) \space\space\space\space\space\space\space (\Delta H = -317\space kJ) $$ And reverse it: $$ \mathrm{N}_{2} \mathrm{H}_{4}(l) + \mathrm{H}_{2} \mathrm{O}(I) \rightarrow \mathrm{N}_{2} \mathrm{O}(g) + 3\mathrm{H}_{2}(g) \space\space\space\space\space (\Delta H = 317\space kJ) $$ If we now subtract the reversed second reaction from the previous summed reaction, we will obtain our target reaction.

Combining the previous manipulations: Target Reaction: $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ Enthalpy Changes: $$ \Delta H_\text{target} = (-143 + 2\times-286) - 317 $$

Evaluate the expression for \(\Delta H_\text{target}\): $$ \Delta H_\text{target} = (-143 - 2\times286) + 317 \\ \Delta H_\text{target} = (-143 - 572) + 317 \\ \Delta H_\text{target} = -398\space kJ $$ The enthalpy change for the target reaction is -398 kJ.