Calculate the `DeltaH^(Theta)` for the reduction of `Fe_(2)O_(3)(s)` by `AI(s)` at `25^(@)C`. The enthalpies of formation of `Fe_(2)O_(3)` and `AI_(2)O_(3)` are `-825.5` and `-1675.7 kJ mol^(-1)` respectively.

Explanation

`Fe_(2)O_(3)(s) +2AI(s) rarr 2Fe(s) +AI_(2)O_(3), DeltaH = ?`
`3Fe(s) +(3)/(2) O_(2)(g) rarr Fe_(2)O_(3), DeltaH_(1) =- 825.5 kJ mol^(-1)`
`2AI(s) +(3)/(2)O_(2)(g) rarr AI_(2)O_(3), DeltaH_(2) =- 1675.7 kJ mol^(-1)`
Operate:
`DeltaH = DeltaH_(1) - DeltaH_(2)`
`=- 1675.7 -(-825.5) =- 850.2 kJ mol^(-1)`