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slavikrds [6]
3 years ago
5

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
Chemistry
1 answer:
gregori [183]3 years ago
7 0

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!

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