Question

For the reaction, \(2Al_{2}O_{3}(s) \rightarrow 4Al(s) + 3O_{2}(g)\), \(\Delta H = +3340\,\text{kJ}\). What is the enthalpy of formation of \(Al_{2}O_{3}(s)\) (in kJ)?

• A. +1670
• B. -3340
• C. +3340
• D. -1670

Hint:

Always reverse sign when reversing reaction and divide enthalpy when coefficients are scaled.

Answer 1

The Correct Option is D

Step 1: Recall what enthalpy of formation means.
The enthalpy of formation is the heat change when one mole of a compound is made from its elements in their normal states. We must end with exactly one mole of $Al_2O_3$.

Step 2: Write the given reaction.
The problem gives \[ 2Al_2O_3(s) \rightarrow 4Al(s) + 3O_2(g), \quad \Delta H = +3340 \text{ kJ} \] This is decomposition, the opposite of formation, and it involves two moles of the oxide.

Step 3: Reverse the reaction.
Formation is the reverse process, so we flip the equation: \[ 4Al(s) + 3O_2(g) \rightarrow 2Al_2O_3(s) \] When a reaction is reversed, the sign of its enthalpy flips, giving $\Delta H = -3340$ kJ.

Step 4: Notice the amount is two moles.
This reversed reaction makes $2$ moles of $Al_2O_3$ and releases $3340$ kJ. Formation enthalpy is defined for just one mole.

Step 5: Divide by two.
For one mole, we halve the enthalpy: \[ \Delta H_f = \frac{-3340}{2} = -1670 \text{ kJ} \]

Step 6: State the answer.
So forming one mole of aluminium oxide releases $1670$ kJ, and the formation enthalpy is negative.
\[ \boxed{-1670 \text{ kJ}} \]