Answer:
ΔG° = -133,1 kJ
Explanation:
For the reactions:
<em>(1) </em>V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K
<em>(2) </em>V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234,2 kJ; ΔS° = 0,2 J/K
By Hess's law it is possible to obtain the ΔH° and ΔS° of:
2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)
Substracting -(1)-(2), that means:
ΔH° = -369,8 kJ - (-234,2 kJ) = <em>-135,6 kJ</em>
ΔS° = - 8,3 J/K - 0,2 J/K =<em> -8,5 J/K</em>
Using: ΔG° = ΔH° - TΔS° at 298K
ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K
<em>ΔG° = -133,1 kJ</em>
I hope it helps!
