What is the enthalpy change for the given reaction, if enthalpies of formation of `Al_(2)O_(3)` and `Fe_(2)O_(3)` are `-"1670 kJ mol"^(-1) and -"834 kJ mol"^(-1)` respectively?
`Fe_(2)O_(3) +2Al rarr Al_(2)O_(3)+2Fe`

Options

A. `-"836 kJ mol"^(-1)`
B. `+"836 kJ mol"^(-1)`
C. `-"424 kJ mol"^(-1)`
D. `+"424 kJ mol"^(-1)`

Explanation

To calculate the enthalpy change for the given reaction, we can use the enthalpy of formation values provided. The reaction is:

\[ \text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe} \]

### Step 1: Identify the enthalpy of formation values
We are given:
- Enthalpy of formation of \( \text{Al}_2\text{O}_3 = -1670 \, \text{kJ/mol} \)
- Enthalpy of formation of \( \text{Fe}_2\text{O}_3 = -834 \, \text{kJ/mol} \)
- Enthalpy of formation of elements (Al and Fe) = 0 (since they are in their standard states)

### Step 2: Write the formula for enthalpy change
The enthalpy change (\( \Delta H \)) for the reaction can be calculated using the following formula:

\[ \Delta H = \sum (\text{Enthalpy of formation of products}) - \sum (\text{Enthalpy of formation of reactants}) \]

### Step 3: Calculate the enthalpy of formation for products and reactants
- **Products**:
- Enthalpy of formation of \( \text{Al}_2\text{O}_3 \) = \( -1670 \, \text{kJ/mol} \)
- Enthalpy of formation of \( 2\text{Fe} \) = \( 2 \times 0 = 0 \, \text{kJ/mol} \)

Therefore, the total enthalpy of formation for products:
\[ \text{Total for products} = -1670 + 0 = -1670 \, \text{kJ/mol} \]

- **Reactants**:
- Enthalpy of formation of \( \text{Fe}_2\text{O}_3 \) = \( -834 \, \text{kJ/mol} \)
- Enthalpy of formation of \( 2\text{Al} \) = \( 2 \times 0 = 0 \, \text{kJ/mol} \)

Therefore, the total enthalpy of formation for reactants:
\[ \text{Total for reactants} = -834 + 0 = -834 \, \text{kJ/mol} \]

### Step 4: Substitute the values into the formula
Now we can substitute the values into the enthalpy change formula:

\[ \Delta H = (-1670) - (-834) \]

### Step 5: Calculate the final result
\[ \Delta H = -1670 + 834 = -836 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy change for the reaction is \( -836 \, \text{kJ/mol} \).

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