First, let's analyze the given reaction:
[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \][/tex]
The enthalpy change ([tex]\(\Delta H^\circ\)[/tex]) for this reaction is given as:
[tex]\[ \Delta H^\circ = -92.2\, \frac{kJ}{mol} \][/tex]
Now, we are asked to determine the enthalpy change for a modified reaction. Since it is specified that the reaction remains the same in order to connect with the goal reaction, we can infer that there are no changes in stoichiometry or the chemical equation itself.
Given that the reaction and conditions have not changed, the enthalpy change will remain the same. Hence, the enthalpy change for the modified reaction will be:
[tex]\[ \Delta H^\circ_{modified} = -92.2\, \frac{kJ}{mol} \][/tex]
The magnitude of the enthalpy change, which is the absolute value, is:
[tex]\[ |\Delta H^\circ_{modified}| = 92.2\, \frac{kJ}{mol} \][/tex]
Therefore, the enthalpy for the modified reaction is:
[tex]\[ -92.2\, \frac{kJ}{mol} \][/tex]
And the magnitude of this enthalpy change is:
[tex]\[ 92.2\, \frac{kJ}{mol} \][/tex]
When expressing our final answer with significant figures, we note that the value given ([tex]\(-92.2\)[/tex]) has three significant figures, and our final answer should preserve this precision. Therefore, the enthalpy change for the modified reaction, with both the sign and magnitude, is:
[tex]\[ -92.2 \, \frac{kJ}{mol} \][/tex]
And the magnitude alone is:
[tex]\[ 92.2 \, \frac{kJ}{mol} \][/tex]
