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)?
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)?
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Always reverse sign when reversing reaction and divide enthalpy when coefficients are scaled.
Updated On: Jun 10, 2026
- +1670
- -3340
- +3340
- -1670
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The Correct Option is D
Solution and Explanation
Concept:
The standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
Step 1: Write the formation reaction \[ 2Al(s) + \frac{3}{2}O_{2}(g) \rightarrow Al_{2}O_{3}(s) \]
Step 2: Relate given reaction The given reaction is the reverse of formation for 2 moles: \[ 2Al_{2}O_{3}(s) \rightarrow 4Al(s) + 3O_{2}(g), \quad \Delta H = +3340\,\text{kJ} \] Reversing gives: \[ 4Al(s) + 3O_{2}(g) \rightarrow 2Al_{2}O_{3}(s), \quad \Delta H = -3340\,\text{kJ} \]
Step 3: Per mole value Since 2 moles are formed: \[ \Delta H_f^\circ = \frac{-3340}{2} = -1670\,\text{kJ mol}^{-1} \]
Step 1: Write the formation reaction \[ 2Al(s) + \frac{3}{2}O_{2}(g) \rightarrow Al_{2}O_{3}(s) \]
Step 2: Relate given reaction The given reaction is the reverse of formation for 2 moles: \[ 2Al_{2}O_{3}(s) \rightarrow 4Al(s) + 3O_{2}(g), \quad \Delta H = +3340\,\text{kJ} \] Reversing gives: \[ 4Al(s) + 3O_{2}(g) \rightarrow 2Al_{2}O_{3}(s), \quad \Delta H = -3340\,\text{kJ} \]
Step 3: Per mole value Since 2 moles are formed: \[ \Delta H_f^\circ = \frac{-3340}{2} = -1670\,\text{kJ mol}^{-1} \]
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