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

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$$\Delta H_{f}\circ$$=-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.