To determine the enthalpy change for the altered reaction, we need to understand how changing the coefficients of a reaction affects its enthalpy change.
Given the original reaction:
[tex]\[ \text{Fe}_2\text{O}_3(s) + 3 \text{CO}(g) \rightarrow 2 \text{Fe}(s) + 3 \text{CO}_2(g) \][/tex]
with [tex]\(\Delta H = -23.44 \, \text{kJ}\)[/tex],
we see that this reaction has a specific enthalpy change associated with the transformation of 1 mole of [tex]\(\text{Fe}_2\text{O}_3\)[/tex].
Consider the altered reaction:
[tex]\[ 3 \text{Fe}_2\text{O}_3(s) + 9 \text{CO}(g) \rightarrow 6 \text{Fe}(s) + 9 \text{CO}_2(g) \][/tex]
We notice that the coefficients of this altered reaction are exactly 3 times the coefficients of the original reaction. When a reaction's coefficients are multiplied by a factor, the enthalpy change for the reaction is also multiplied by that same factor.
Thus, if the original enthalpy change [tex]\(\Delta H\)[/tex] is [tex]\(-23.44 \, \text{kJ}\)[/tex], for the altered reaction (with coefficients multiplied by 3), the enthalpy change [tex]\(\Delta H'\)[/tex] will be:
[tex]\[ \Delta H' = 3 \times (\Delta H_{\text{original}}) = 3 \times (-23.44 \, \text{kJ}) \][/tex]
Calculating this gives:
[tex]\[ \Delta H' = -70.32 \, \text{kJ} \][/tex]
Therefore, the enthalpy change for the altered reaction is:
[tex]\[
\Delta H = -70.32 \, \text{kJ}
\][/tex]
