To solve for the enthalpy change ([tex]\(\Delta H\)[/tex]) of the altered reaction, we need to understand how the enthalpy change scales with the reaction. Here is the step-by-step breakdown:
1. Original Reaction Given:
[tex]\[
Fe_2O_3(s) + 3 CO(g) \rightarrow 2 Fe(s) + 3 CO_2(g)
\][/tex]
With an enthalpy change of:
[tex]\[
\Delta H = -23.44 \text{ kJ}
\][/tex]
2. Altered Reaction:
[tex]\[
3 Fe_2O_3(s) + 9 CO(g) \rightarrow 6 Fe(s) + 9 CO_2(g)
\][/tex]
3. Comparison Between Reactions:
The altered reaction is exactly three times the original reaction. This can be verified by comparing the coefficients of each substance in both reactions:
- Original reaction: [tex]\(Fe_2O_3\)[/tex] [tex]\(\times 1\)[/tex], [tex]\(CO\)[/tex] [tex]\(\times 3\)[/tex], [tex]\(Fe\)[/tex] [tex]\(\times 2\)[/tex], [tex]\(CO_2\)[/tex] [tex]\(\times 3\)[/tex]
- Altered reaction: [tex]\(Fe_2O_3\)[/tex] [tex]\(\times 3\)[/tex], [tex]\(CO\)[/tex] [tex]\(\times 9\)[/tex], [tex]\(Fe\)[/tex] [tex]\(\times 6\)[/tex], [tex]\(CO_2\)[/tex] [tex]\(\times 9\)[/tex]
4. Determination of Enthalpy Change for the Altered Reaction:
Since the altered reaction is three times the original reaction, the enthalpy change will also be three times the enthalpy change of the original reaction.
5. Calculation:
Multiply the original enthalpy change by 3:
[tex]\[
\Delta H_{\text{altered}} = 3 \times \Delta H_{\text{original}}
\][/tex]
[tex]\[
\Delta H_{\text{altered}} = 3 \times (-23.44 \text{ kJ})
\][/tex]
[tex]\[
\Delta H_{\text{altered}} = -70.32 \text{ kJ}
\][/tex]
6. Result with Significant Figures:
The given original enthalpy change (-23.44 kJ) has four significant figures, so our final answer should also be presented with the same precision.
Thus, the enthalpy change ([tex]\(\Delta H\)[/tex]) for the altered reaction is:
[tex]\[
\Delta H = -70.32 \text { kJ}
\][/tex]
