To determine [tex]\(\Delta H^\circ\)[/tex] for the goal reaction:
[tex]\[N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g),\][/tex]
we use Hess's Law, which states that the total enthalpy change for a given reaction is the sum of the enthalpy changes for the intermediate steps that lead to that reaction.
Given the adjusted reactions and their respective enthalpies, we can derive the goal reaction.
1. Input adjusted reaction:
[tex]\[N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g)\][/tex]
2. Reverse the enthalpy of the second reaction (multiply by -1):
[tex]\[2H_2O(g) \rightarrow 2H_2(g) + O_2(g), \quad \Delta H^\circ = +543.0 \, \text{kJ/mol}\][/tex]
3. Retain the third reaction as it is:
[tex]\[N_2(g) + 3H_2(g) \rightarrow 2NH_3(g), \quad \Delta H^\circ = -984.0 \, \text{kJ/mol}\][/tex]
Now, sum these reactions to obtain the goal reaction:
[tex]\[
\begin{align*}
&N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g)\\
&2H_2O(g) \rightarrow 2H_2(g) + O_2(g), \quad \Delta H^\circ = +543.0 \, \text{kJ/mol}\\
&N_2(g) + 3H_2(g) \rightarrow 2NH_3(g), \quad \Delta H^\circ = -984.0 \, \text{kJ/mol}
\end{align*}
\][/tex]
Adding them together:
[tex]\[
N_2H_4(l) + O_2(g) + 2H_2O(g) + N_2(g) + 3H_2(g) \rightarrow N_2(g) + 2H_2O(g) + 2H_2(g) + 2NH_3(g)
\][/tex]
On simplifying:
[tex]\[N_2H_4(l) + H_2(g) \rightarrow 2 NH_3(g)\][/tex]
Sum the enthalpies for the reactions:
[tex]\[
\Delta H^\circ = +543.0 + (-984.0) = -441.0 \, \text{kJ/mol}
\][/tex]
Thus,
[tex]\[
\Delta H^\circ = -1527.0 \, \text{kJ/mol}
\][/tex]
Hence, the enthalpy change [tex]\(\Delta H^\circ\)[/tex] for the goal reaction [tex]\(N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g)\)[/tex] is:
[tex]\[
\boxed{-1527.0 \, \text{kJ/mol}}
\][/tex]
