Calculate \Delta H_{rxn}\circ for 2Al_{2}O_{3} to 4Al+3O_{2}

[[V]]

\Delta H_{f}\circ=-1670kJ/mol for Al_{2}O_{3}

What is \Delta H_{rxn}\circ for
2Al_{2}O_{3}(s)\rightarrow4Al+3O_{2}(g)

So clearly we simply multiply -1670*2=-3340kJ/mol.

The answer is actually POSITIVE 3340kJ/mol!

Can someone please explain to me how this reasoning works?

Thank you!

 

[Borek]

You reverse the reaction direction, so you need to reverse the sign. If reaction is exothermic when going one side, it must be endothermic when going back.