Fe\(_2\)O\(_{3(s)}\) + 2Al\(_{(s)}\) → Al\(_2\)O\(_3\) + 2Fe\(_{(s)}\). If the heats of formation of Al\(_2\)O\(_3\) and Fe\(_2\)O\(_3\) are - 1670 kJmol\(^{-1}\) and - 822 kJmol\(^{-1}\) respectively, the enthalpy change in kJ for the reaction is ?
Heat of formation = Heat of products - Heat of reactants
= - 1670 - (- 822)
= - 1670 + 822
= - 848 kJ
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