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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.

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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!

The study of chemicals and bonds is called chemistry.  According to the law of thermodynamics, heat is a transfer from the high temperature to the low temperature.

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

What is Gibbs free energy?

  • The Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure

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

Where:

  • Keq is equilibrium constant
  • ΔG° is standard state free energy change
  • R is gas constant
  • T is temperature.

It is possible to know that the large negative ΔG° is the largest equilibrium constant. That is because R and T are always positive and to cancel the negative of the equation it is necessary that ΔG° be negative. Knowing this is the oxidation of Hg the reaction that has the largest negative ΔG°. So, this reaction will have the largest equilibrium constant.

[tex]2 Hg(g) + O_2(g) ---> 2 HgO(s) G = -180.8 kJ\\ CaCO_3(s) ----> CaO(s) + CO_2(g) G =+131.1 kJ\\ 3O_2(g) ---> 2O_3(g) G = +326 kJ \\ Fe_2O_3(s) + 3CO(g) ---> 2 Fe(s) + 3CO_2(g) G = -28.0 kJ [/tex]

Hence, the correct answer is mentioned above.

For more information about Gibbs free energy, refer to the link:-

https://brainly.com/question/1673533

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