Question

Calculate ΔH°298 for the process Co₃O₄(s) → 3Co(s) + 2O₂(g) from the following information:

Co(s) + 1/2O₂(g) → CoO(s) ΔH° = −237.9kJ
3CoO(s) + 1/2O₂(g) → Co₃O₄(s) ΔH° = −177.5kJ

Answer 1

The standard enthalpy change (ΔH°) for the process Co₃O₄(s) → 3Co(s) + 2O₂(g) is calculated to be − 536.2 kJ using Hess's Law.

Step-by-step explanation:

To calculate the standard enthalpy change (ΔH°) for the reaction Co₃O₄(s) → 3Co(s) + 2O₂(g), we can use Hess's Law and the given reactions. The objective is to manipulate and combine the given reactions to resemble the target reaction.

The two reactions provided are:

1. Co(s) + ½O₂(g) → CoO(s) with ΔH° = − 237.9 kJ
2. 3CoO(s) + ½O₂(g) → Co₃O₄(s) with ΔH° = − 177.5 kJ

Firstly, we need to reverse reaction 2 to match the direction of the target reaction, which also inverts the sign of ΔH°:

Co₃O₄(s) → 3CoO(s) + ½O₂(g) with ΔH° = +177.5 kJ

Now we multiply reaction 1 by 3 to get the correct stoichiometry:

3Co(s) + 3(½O₂(g)) → 3CoO(s) with ΔH° = 3(− 237.9 kJ) = − 713.7 kJ

Finally, we add the modified reaction 1 to the reversed reaction 2 to obtain the target reaction:

Co₃O₄(s) + 3Co(s) + ¾O₂(g) → 3CoO(s) + 3CoO(s) + ½O₂(g),

with the total ΔH° = +177.5 kJ (± from reversion of reaction 2) + − 713.7 kJ (± from three times reaction 1) = − 536.2 kJ.