Given 2Al(s)+ (3/2)O₂(g)$\rarr$Al₂O₃(s), $\Delta$H°_f = -1,670 kJ/mol for Al₂O₃ (s). Determine $\Delta$H° 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 reaction, so $\Delta$H° = +3,340 kJ/mol.

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

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