Question

Which of the following reactions will have the largest equilibrium constant (K) at 298 K? Which of the following reactions will have the largest equilibrium constant (K) at 298 K? 2 Hg(g) + O2(g) → 2 HgO(s) ΔG° = -180.8 kJ CaCO3(s) → CaO(s) + CO2(g) ΔG° =+131.1 kJ 3 O2(g) → 2 O3(g) ΔG° = +326 kJ Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔG° = -28.0 kJ It is not possible to determine without more information.

Answer 1

Answer:

2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ

Explanation:

If we know the ΔG° of a chemical reaction it is possible to calculate the equilibrium constant (k) of this procedure with the next equation:

ln Keq = -ΔG° / RT (1)

Where: Keq is equilibrium contant, ΔG° is standard state free energy change, R is gas constant and T is temperature.

Watching (1), it is possible to know that the large negative ΔG° the largest equilibrium constant. That is because R and T are always positive and to cancel the negative of equation it is necessary that ΔG° be negative.

Knowing this, is the oxidation of Hg the reaction that has the largely negative ΔG°. So, this reaction will have the largest equilibrium constant.

2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ

CaCO₃(s) → CaO(s) + CO₂(g) ΔG° =+131.1 kJ

3 O₂(g) → 2 O₃(g) ΔG° = +326 kJ

Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g) ΔG° = -28.0 kJ

I hope it helps!