Given 2Al(s) + (3/2)O₂(g) $\rarr$Al₂O₃(s), $\Delta$H^$\circ$_f = -1,670 kJ/mol for Al₂O₃ (s).Determine $\Delta$H^$\circ$ for the reaction 2Al₂O₃(s) $\rarr$ 4Al(s) + 3O₂(g). A) -3,340 kJ/mol B) -1,670 kJ/mol C) -835 kJ/mol D) 1,670 kJ/mol E) 3,340 kJ/mol

Question

Quizplus · Accepted Answer

The given reaction is the reverse of the formation of Al₂O₃. The enthalpy change for the reverse reaction is the opposite sign of the formation enthalpy, so it is +1,670 kJ/mol for 1 mole of Al₂O₃. Since the reaction involves 2 moles of Al₂O₃, the enthalpy change is 2 x 1,670 kJ/mol = 3,340 kJ/mol.

Given 2Al(s) + (3/2)O2(g) →\rarr→ Al2O3(s) Δ\DeltaΔ H ∘\circ∘ f = -1,670 KJ/mol for Al2O3 (S)

Given 2Al(s) + (3/2)O₂(g) $\rarr$ Al₂O₃(s) $\Delta$ H^$\circ$_f = -1,670 KJ/mol for Al₂O₃ (S)