To calculate the enthalpy change, [tex]\(\Delta H\)[/tex], for the target reaction:
[tex]\[ 4 \text{Al} + 3 \text{MnO}_2 \rightarrow 2 \text{Al}_2\text{O}_3 + 3 \text{Mn} \][/tex]
We can break the process into several steps using the provided reactions and their associated enthalpy changes.
### Given Reactions:
1. [tex]\(4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \quad \Delta H = -3352 \text{kJ}\)[/tex]
2. [tex]\( \text{Mn} + \text{O}_2 \rightarrow \text{MnO}_2 \quad \Delta H = -521 \text{kJ}\)[/tex]
### Step-by-Step Solution:
#### Step 1: Adjust the [tex]\(\text{MnO}_2\)[/tex] formation reaction
For the target reaction, [tex]\(\text{MnO}_2\)[/tex] appears on the reactant side, whereas it is formed on the product side in the given second reaction. Therefore, we can reverse the second reaction to align with our target reaction:
[tex]\[ \text{MnO}_2 \rightarrow \text{Mn} + \text{O}_2 \][/tex]
Reversing the reaction changes the sign of [tex]\(\Delta H\)[/tex]:
[tex]\[ \Delta H = +521 \text{kJ} \][/tex]
#### Step 2: Scale the reversed reaction
Since the target reaction has 3 [tex]\(\text{MnO}_2\)[/tex], we multiply the reversed reaction by 3:
[tex]\[ 3 \text{MnO}_2 \rightarrow 3 \text{Mn} + 3 \text{O}_2 \][/tex]
[tex]\[ \Delta H = 3 \times 521 \text{kJ} = 1563 \text{kJ} \][/tex]
#### Step 3: Combine with the first reaction
Now we combine the modified reactions to achieve the target reaction. The first reaction does not need to be adjusted:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \quad \Delta H = -3352 \text{kJ} \][/tex]
### Combining Reactions
Add the enthalpies of the reactions to find the total enthalpy change for the target reaction:
[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_{\text{MnO}_2} \][/tex]
[tex]\[ \Delta H_{\text{total}} = -3352 \text{kJ} + 1563 \text{kJ} \][/tex]
#### Final Calculation:
[tex]\[ \Delta H_{\text{total}} = -1789 \text{kJ} \][/tex]
Therefore, the enthalpy change [tex]\(\Delta H\)[/tex] for the reaction [tex]\(4 \text{Al} + 3 \text{MnO}_2 \rightarrow 2 \text{Al}_2\text{O}_3 + 3 \text{Mn}\)[/tex] is [tex]\(-1789 \text{kJ}\)[/tex].
