madey21
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[tex]\[
\begin{aligned}
N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g) \\
\Delta H^{\circ} = -543.0 \ \frac{kJ}{mol}
\end{aligned}
\][/tex]

This reaction remains unchanged to connect with the target reaction. What is the enthalpy for the modified reaction?

Enter either a + or - sign and the magnitude. Use significant figures.



Answer :

To determine the enthalpy change for the modified reaction, we can follow a clear, logical process based on known thermodynamic principles.

### Step-by-Step Solution:

1. Identify the Enthalpy Change for the Given Reaction:
The provided reaction is:
[tex]\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g) \][/tex]
The standard enthalpy change (ΔH°) for this reaction is given as:
[tex]\[ \Delta H^\circ = -543.0 \ \frac{\text{kJ}}{\text{mol}} \][/tex]

2. Understanding the Modifications:
The problem statement indicates that we are to find the enthalpy change for a modified reaction, which stays the same as the initial reaction and simply connects with the goal reaction. This implies that no changes have been made in the stoichiometry or the species involved in the reaction.

3. Consistency of Enthalpy Changes:
Since the modified reaction essentially remains the same as the original given reaction, the enthalpy change will also remain the same.

4. Conclusion:
Therefore, the enthalpy change for the modified reaction is consistent with the given reaction's enthalpy change.

The enthalpy change for the modified reaction is:
[tex]\[ -543.0 \ \frac{\text{kJ}}{\text{mol}} \][/tex]

Thus, the final answer with the appropriate sign and magnitude, considering significant figures, is:
[tex]\[ -543.0 \ \text{kJ/mol} \][/tex]

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