To find the standard enthalpy change [tex]\(\Delta H^{\circ}_{\text{rxn}}\)[/tex] for the given reaction:
[tex]\[
6 \text{Cl}_2(g) + 2 \text{Fe}_2\text{O}_3(s) \rightarrow 4 \text{FeCl}_3(s) + 3 \text{O}_2(g),
\][/tex]
we need to use the provided standard enthalpy of formation ([tex]\(\Delta H_{f}^{\circ}\)[/tex]) values for each substance involved.
### Step-by-Step Solution:
1. Identify the enthalpy of formation ([tex]\(\Delta H_{f}^{\circ}\)[/tex]) for each substance:
[tex]\[
\begin{aligned}
&\text{Cl}_2(g) & : &\quad 0.0 \, \text{kJ/mol} \\
&\text{Fe}_2\text{O}_3(s) & : &\quad -824.2 \, \text{kJ/mol} \\
&\text{FeCl}_3(s) & : &\quad -399.5 \, \text{kJ/mol} \\
&\text{O}_2(g) & : &\quad 0.0 \, \text{kJ/mol} \\
\end{aligned}
\][/tex]
2. Use the stoichiometric coefficients from the balanced chemical equation:
[tex]\[
\begin{aligned}
&\text{Cl}_2(g) & : &\quad 6 \\
&\text{Fe}_2\text{O}_3(s) & : &\quad 2 \\
&\text{FeCl}_3(s) & : &\quad 4 \\
&\text{O}_2(g) & : &\quad 3 \\
\end{aligned}
\][/tex]
3. Calculate the total enthalpy of the reactants:
[tex]\[
\Delta H_{\text{reactants}} = (6 \times 0.0) + (2 \times -824.2)
\][/tex]
[tex]\[
\Delta H_{\text{reactants}} = 0 + (-1648.4) = -1648.4 \, \text{kJ}
\][/tex]
4. Calculate the total enthalpy of the products:
[tex]\[
\Delta H_{\text{products}} = (4 \times -399.5) + (3 \times 0.0)
\][/tex]
[tex]\[
\Delta H_{\text{products}} = -1598.0 + 0 = -1598.0 \, \text{kJ}
\][/tex]
5. Calculate the standard enthalpy change of the reaction ([tex]\(\Delta H^{\circ}_{\text{rxn}}\)[/tex]):
[tex]\[
\Delta H^{\circ}_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}}
\][/tex]
[tex]\[
\Delta H^{\circ}_{\text{rxn}} = -1598.0 - (-1648.4)
\][/tex]
[tex]\[
\Delta H^{\circ}_{\text{rxn}} = -1598.0 + 1648.4 = 50.4 \, \text{kJ}
\][/tex]
Thus, the standard enthalpy change for the reaction, [tex]\(\Delta H^{\circ}_{\text{rxn}}\)[/tex], is [tex]\(50.4 \, \text{kJ}\)[/tex].
