Sure, let's go through the problem step-by-step.
1. First, we need to understand the reaction provided:
[tex]\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g) \][/tex]
2. The standard enthalpy change ([tex]\(\Delta H^{\circ}\)[/tex]) for this reaction is given as:
[tex]\[ \Delta H^{\circ} = -543.0 \ \frac{kJ}{mol} \][/tex]
This means that for every mole of [tex]\(N_2H_4\)[/tex] reacting with [tex]\(O_2\)[/tex] under standard conditions, 543.0 kJ of energy is released (hence the negative sign).
3. Since the given problem states that the reaction remains unchanged to connect with the goal reaction, the enthalpy change for the modified reaction is the same as the standard enthalpy change provided.
4. Therefore, the enthalpy for the modified reaction is:
[tex]\[ \Delta H^{\circ} = -543.0 \ \frac{kJ}{mol} \][/tex]
This concludes the detailed explanation. The final answer should include the sign and the magnitude of the enthalpy change with an appropriate unit and significant figures.
[tex]\[ \boxed{-543.0 \ \frac{kJ}{mol}} \][/tex]
