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).

Question

Quizplus · Accepted Answer

The given reaction is the reverse of the formation of Al₂O₃. The enthalpy change for the reverse reaction is the negative of the formation enthalpy. Since the formation of 1 mole of Al₂O₃ releases 1,670 kJ, reversing 2 moles will require 2 x 1,670 kJ = 3,340 kJ.

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)