Question

From these two reactions at 298 K, V2O3(s) + 3CO(g) → 2V(s) + 3CO2(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K V2O5(s) + 2CO(g) → V2O3(s) + 2CO2(g); ΔH° = –234.2 kJ; ΔS° = 0.2 J/K calculate ΔG° for the following at 298 K: 2V(s) + 5CO2(g) → V2O5(s) + 5CO(g

Answer 1

ΔG° = -133,1 kJ

Step-by-step explanation:

For the reactions:

(1) V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K

(2) 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) = -135,6 kJ

ΔS° = - 8,3 J/K - 0,2 J/K = -8,5 J/K

Using: ΔG° = ΔH° - TΔS° at 298K

ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K

ΔG° = -133,1 kJ

I hope it helps!