The enthalpies of formation of `Al_(2)O_(3)` and `Cr_(2)O_(3)` are `-1596KJ` and `-1134KJ` respectively. `DeltaH` for the reaction
`2Al+Cr_(2)O_(3)rarr2Cr+Al_(2)O_(3)` is

Options

A. `-2730 kJ`
B. `-462 kJ`
C. `-1365 kJ`
D. `+2730 kJ`

Explanation

To find the enthalpy change (ΔH) for the reaction:

\[ 2Al + Cr_2O_3 \rightarrow 2Cr + Al_2O_3 \]

we will use the enthalpies of formation of the products and reactants.

### Step 1: Write down the enthalpy of formation values
- The enthalpy of formation of \( Al_2O_3 \) is \( -1596 \, \text{kJ/mol} \).
- The enthalpy of formation of \( Cr_2O_3 \) is \( -1134 \, \text{kJ/mol} \).
- The enthalpy of formation for elements in their standard state (like \( Al \) and \( Cr \)) is \( 0 \, \text{kJ/mol} \).

### Step 2: Write the formula for ΔH
The formula for calculating the enthalpy change of a reaction is:

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

### Step 3: Identify products and reactants
In our reaction:
- Products: \( 2Cr \) and \( Al_2O_3 \)
- Reactants: \( 2Al \) and \( Cr_2O_3 \)

### Step 4: Substitute the enthalpy of formation values into the formula
Using the values we have:

\[ \Delta H = [\Delta H_f(Al_2O_3) + 2 \times \Delta H_f(Cr)] - [2 \times \Delta H_f(Al) + \Delta H_f(Cr_2O_3)] \]

Since the enthalpies of formation for \( Al \) and \( Cr \) are \( 0 \):

\[ \Delta H = [\Delta H_f(Al_2O_3) + 0] - [0 + \Delta H_f(Cr_2O_3)] \]

Substituting the values:

\[ \Delta H = [-1596 \, \text{kJ/mol}] - [-1134 \, \text{kJ/mol}] \]

### Step 5: Calculate ΔH
Now, perform the calculation:

\[ \Delta H = -1596 + 1134 = -462 \, \text{kJ} \]

### Final Answer
Thus, the enthalpy change (ΔH) for the reaction is:

\[ \Delta H = -462 \, \text{kJ} \]

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