To determine the enthalpy change for the reaction, we start with the given balanced chemical reaction and its associated enthalpy change:
[tex]\[
\begin{aligned}
N_2(g) + 3H_2(g) \rightarrow 2 NH_3(g) \\
\Delta H^\circ = -92.2 \text{ kJ/mol}
\end{aligned}
\][/tex]
This enthalpy change, [tex]\(\Delta H^\circ\)[/tex], is given as [tex]\(-92.2 \text{ kJ/mol}\)[/tex]. It corresponds to the enthalpy change when 1 mole of nitrogen gas ([tex]\(N_2\)[/tex]) reacts with 3 moles of hydrogen gas ([tex]\(H_2\)[/tex]) to produce 2 moles of ammonia gas ([tex]\(NH_3\)[/tex]) under standard conditions.
Since the reaction itself and its enthalpy change are given directly and no modifications to the reaction have been indicated, the enthalpy change for the modified reaction—if it remains the same as the given reaction—also stays the same.
Thus, the enthalpy for the modified reaction is:
[tex]\[
\Delta H^\circ = -92.2 \text{ kJ/mol}
\][/tex]
Remember to take careful note of the significant figures in the given data. In this case, [tex]\(-92.2\)[/tex] has three significant figures.
So, the enthalpy for the modified reaction is:
[tex]\[
-92.2 \text{ kJ/mol}
\][/tex]
