To solve this problem, we need to determine the enthalpy change of the reaction using the standard enthalpies of formation ([tex]\(\Delta H_f\)[/tex]) provided for the substances involved.
The given balanced chemical equation is:
[tex]\[
3 MnO_2(s) + 4 Al(s) \rightarrow 2 Al_2O_3(s) + 3 Mn(s)
\][/tex]
The standard enthalpies of formation for the reactants and products are:
[tex]\[
\Delta H_f(MnO_2(s)) = -520.0\, \text{kJ/mol}
\][/tex]
[tex]\[
\Delta H_f(Al_2O_3(s)) = -1699.8\, \text{kJ/mol}
\][/tex]
Step 1: Calculate the sum of the enthalpies of formation for the reactants:
- For [tex]\(MnO_2\)[/tex], there are 3 moles involved:
[tex]\[
3 \times (-520.0\, \text{kJ/mol}) = -1560.0\, \text{kJ}
\][/tex]
- For aluminum (Al), since it is in its elemental form, its standard enthalpy of formation is zero. Therefore, it does not contribute to the enthalpy change:
[tex]\[
4 \times 0\, \text{kJ/mol} = 0\, \text{kJ}
\][/tex]
Summing up the enthalpies of the reactants:
[tex]\[
\sum H_f(\text{reactants}) = -1560.0\, \text{kJ}
\][/tex]
Step 2: Calculate the sum of the enthalpies of formation for the products:
- For [tex]\(Al_2O_3\)[/tex], there are 2 moles involved:
[tex]\[
2 \times (-1699.8\, \text{kJ/mol}) = -3399.6\, \text{kJ}
\][/tex]
- For manganese (Mn), it is also in its elemental form, so its standard enthalpy of formation is zero:
[tex]\[
3 \times 0\, \text{kJ/mol} = 0\, \text{kJ}
\][/tex]
Summing up the enthalpies of the products:
[tex]\[
\sum H_f(\text{products}) = -3399.6\, \text{kJ}
\][/tex]
Step 3: Calculate the enthalpy change of the reaction:
The enthalpy change ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) can be found using the formula:
[tex]\[
\Delta H_{\text{reaction}} = \sum H_f(\text{products}) - \sum H_f(\text{reactants})
\][/tex]
Substituting in our values:
[tex]\[
\Delta H_{\text{reaction}} = -3399.6\, \text{kJ} - (-1560.0\, \text{kJ})
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = -3399.6 + 1560.0\, \text{kJ}
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = -1839.6\, \text{kJ}
\][/tex]
Therefore, the enthalpy change for the reaction is:
[tex]\[
-1839.6\, \text{kJ}
\][/tex]
Answer:
[tex]\(-1839.6\, \text{kJ}\)[/tex]
So the correct choice is:
[tex]\[
\boxed{-1839.6\, \text{kJ}}
\][/tex]
