Answer to Question #242191 in General Chemistry for KING

Question

Copper(I) oxide can be oxidized to copper(II) oxide: Cu2O(s) + 1⁄2 O2(g) →2CuO(s) ∆H°rxn = - 146.0 kJ

Given that ∆H°f of Cu2O(s) = -168.6 kJ/mol, what is ∆H°f of CuO(s)?

Answer 1

$Cu_2O(s) +\frac{ 1}{2}O_2(g) \to 2CuO(s)$

$∆Hºrxn = -146.0 kJ$

$∆Hºrxn = 2 ×∆Hf CuO(s) - [\frac{1}{2 }∆Hºf O_ 2(g) + ∆Hºf Cu_2O]$

-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