Heat of Formation from Heat of Reaction?

Question

Copper(I) oxide is oxidized to copper(II) oxide according to the following equation:
Cu2O(s) + 1/2O2(g) → 2CuO(s) [ΔHorxn = -146.0 kJ]
Given that ΔHf of Cu2O(s) = -168.6 kJ/mol.
What is ΔHf of CuO(s)?

Answer 1

∆H˚rxn = ∑∆H˚f(products) - ∑∆H˚f(reactants)

∆H˚rxn = 2 ∆H˚f(CuO (s)) - [ ∆H˚f(Cu2O (s)) + 1/2 ∆H˚f(O2 (g))]

but, ∆H˚f (O2 (g)) = 0 (by definition, for any element in its most stable form, the value is always zero.

-146 kJ = 2∆H˚f(CuO (s)) - (-168.6 kJ)

-146 kJ - 168.6 kJ = 2∆H˚f(CuO (s))
-314.6 kJ/2 = ∆H˚f(CuO (s))
-157.3 kJ = ∆H˚f(CuO (s))

Answer 2

William Q is absolutely correct, and you should award him best answer. The way I worked it out:

Cu2O + 1/2O2 ===> 2CuO -146.0kJ

2Cu + 1/2O2 ===> Cu2O -168.6kJ (heat of formation of Cu2O is given)

Add them together and the Cu2O's cancel out:

2Cu + O2 ===> 2CuO -314.6kJ

Divide through by 2:

Cu + 1/2O2 ===> CuO -157.3kJ