Question

Given the thermochemical equation Al2O3(s) → 2Al(s) + (3/2)O2(g) ∆Hº = 1676 kJ/mol what is ∆Hº for the reaction? 4Al(s) + 3O2(g) → 2Al2O3(s)

Answer 1

So you gots....

`Al_2O_3(s) rarr 2Al(s) + 3/2O_2(g)` `;DeltaH_"rxn"^@=1676*kJ*mol^-1`

Explanation:

Of course the enthalpy is quoted per moles of reaction of written. And another way I could represent this is as......

`Al_2O_3(s) +1676*kJ rarr 2Al(s) + 3/2O_2(g)`

The positive sign of the quoted enthalpy indicates that the reaction as written is ENDOTHERMIC, which is reasonable in that we have to break strong `Al-O` bonds....

But because mass, charge, and enthalpy are always are conserved, we could write....

`2Al(s) + 3/2O_2(g)rarrAl_2O_3(s) +1676*kJ`

And thus the REVERSE reaction gives an EXOTHERM. An equivalent representation would be.....

`2Al(s) + 3/2O_2(g)rarrAl_2O_3(s)` `;DeltaH_"rxn"^@=-1676*kJ`

And the negative enthalpy term indicates that energy is released from the reaction. Claro?