The enthalpies of formation of ${Al}_{2} O_{3} (s)$ and ${Fe}_{2} O_{3} (s)$ are $- 1670 kJ {mol}^{- 1}$ and $- 834 kJ {mol}^{- 1}$ respectively. The enthalpy change of the reaction ${Fe}_{2} O_{3} (s) + Al (s) ⟶ {Al}_{2} O_{3} (s) + 2 Fe (s)$ would be

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$836 kJ {mol}^{- 1}$

$2504 kJ {mol}^{- 1}$

$- 2504 kJ {mol}^{- 1}$

$- 836 kJ {mol}^{- 1}$

answer is D.

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Detailed Solution

$Δ_{r} H = Δ_{f} H ({Al}_{2} O_{3}, s) - Δ_{f} H ({Fe}_{2} O_{3}, s) = (- 1670 + 834) kJ {mol}^{- 1} = - 836 kJ {mol}^{- 1}$

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