Certainly! Let’s determine the enthalpy change for the goal reaction [tex]\( \Delta H^{\circ} \)[/tex] by combining the given reactions.
The goal reaction is:
[tex]\[ N_2H_4(l) + H_2(g) \rightarrow 2 NH_3(g) \][/tex]
We have the following enthalpy changes for the provided reactions:
[tex]\[
\begin{array}{ll}
N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2 H_2O(g) & \Delta H^{\circ} \text{ unknown (to be determined)} \\
2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) & \Delta H^{\circ} = +543.0 \, \frac{kJ}{mol} \, \text{and} \, 484.0 \, \frac{kJ}{mol} \\
N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) & \Delta H^{\circ} = -92.2 \, \frac{kJ}{mol}
\end{array}
\][/tex]
First, let's consider the reverse reaction of
[tex]\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2 H_2O(g) \][/tex]
This reaction should be reversed to match part of the goal reaction:
[tex]\[ N_2(g) + 2 H_2O(g) \rightarrow N_2H_4(l) + O_2(g) \][/tex]
The enthalphy change for this reaction would be:
[tex]\[ \Delta H^{\circ} = -(543.0 + 484.0) / 2 = -513.5 \, \frac{kJ}{mol} \][/tex]
Now we combine the adjusted reaction with the third reaction:
[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]
Finally, we add the enthalpies of the acquired reaction and the third reaction to get the enthalpy change for the goal reaction:
[tex]\[ \Delta H^{\circ} = -513.5 + (-92.2) = -605.7 \, \frac{kJ}{mol} \][/tex]
Thus, the enthalpy change [tex]\( \Delta H^{\circ} \)[/tex] for the goal reaction [tex]\( N_2H_4(l) + H_2(g) \rightarrow 2 NH_3(g) \)[/tex] is:
[tex]\[ -605.7 \, \frac{kJ}{mol} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{-605.7} \][/tex]
