Al(s) and Fe(s) are oxidised by oxygen to form `Al_(2)O_(3)(s) and Fe_(2)O_(3)(s)` respectively. If the standard enthalpy of formation of `Al_(2)O_(3)(s) and Fe_(2)O_(3)(s)` are `-1669.8 " kJ mol"^(-1) and -824.2 " kJ mol"^(-1)` respectively, calculate the change in enthalpy of the given reaction :
`2Al(s)+Fe_(2)O_(3)(s) to 2Fe(s)+Al_(2)O_(3)(s)`

Explanation

`Delta H^(0)= 2Delta H_(f)^(0)[Fe(s)]+DeltaH_(f)^(0)[Al_(2)O_(3)(s)]-2DeltaH_(f)^(0)[Al(s)]-DeltaH_(f)^(0)[Fe_(2)O_(3)(s)]`
`= (0-1669.8-0+824.2) kJ = - 845.6 kJ`