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 ?
- +2492
- +848
-
-848
- -2492
The correct answer is: C
Explanation
Heat of formation = Heat of products - Heat of reactants
= - 1670 - (- 822)
= - 1670 + 822
= - 848 kJ