What is the value of ∆H ̊
(in kJ) for this reaction? 2CuO(s) r Cu2O(s) + 1⁄2O2(g)?

∆Hf ̊ / kJ·mol–1
CuO –156.1
Cu2O –170.7

Question

What is the value of ∆H ̊
(in kJ) for this reaction? 2CuO(s) r Cu2O(s) + 1⁄2O2(g)?

∆Hf ̊ / kJ·mol–1
￼￼￼CuO –156.1 
Cu2O –170.7

abb0t · Answer

To solve for $\sf \Delta H$ you would use $\sf \color{blue}{\Delta H_{reaction}^º= \Sigma \Delta _f^º(products)- \Sigma \Delta H_f^º(reactants)}$

abb0t · Answer

it essentially is saying that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products $\sf \color{red}{subtracted}$ by the sum of the standard enthalpies of formation of the reactants.

anonymous · Answer

but how would you get the 1/2O2, since the rest are given?

abb0t · Answer

Oxygen gas = 0

anonymous · Answer

so basically you use oxidation state since that says free elements are 0?

anonymous · Answer

haha got it. Thank you!!