Copper(I) oxide can be oxidized to copper(II) oxide: Cu2O(s) + 1/2 O2(g) → 2 CuO(s).....

Question

ΔH⁰_rxn = -146.0 kJ

Given ΔH⁰_f of Cu₂O(s) = −168.6 kJ/mol, find ΔH⁰_f of CuO(s)

Answer 1

Cu₂O(s) + 1/2O₂(g) ==> 2CuO(s) ... ∆Hºrxn = -146.0 kJ

∆Hºrxn = 2 x ∆Hf CuO(s) - [1/2 ∆Hºf O₂(g) + ∆Hºf Cu₂O]

-146 = 2 x ∆Hf CuO(s) - (0 + -168.6)

-146 - 168.6 = 2 x ∆Hf CuO(s)

314.6 kJ = 2 x ∆Hf CuO(s)

∆Hf CuO(s) = 157.3 kJ/mol