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Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K = ? ΔG°f (kJ/mol) -110.9 87.6 51.3 -237.1 Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K = ? ΔG°f (kJ/mol) -110.9 87.6 51.3 -237.1 0.980 1.02 1.15 × 10-9 5.11 × 10-4 8.71 × 108

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superman1987

Answer:

K = 1.15 x 10⁻⁹.

Explanation:

  • We have the relation:

ΔG°rxn = - RTlnK.

where, ΔG°rxn is the standard free energy change of the reaction.

R is the general gas constant (R = 8.314 J/mol.K).

T is the temperature of the reaction (K) (T = 298.0 K).

K is the equilibrium constant.

  • We can get ΔG°rxn using the relation:

ΔG°rxn = ΣG°f, products – ΣG°f, reactants

ΔG°rxn = [3(G°f, NO₂) + (G°f, H₂O)] - [2(G°f, HNO₃) + (G°f, NO)]

ΔG°rxn = [3(51.3 kJ/mol) + (- 237.1 kJ/mol)] - [2(- 110.9 kJ/mol) + (87.6 kJ/mol)

ΔG°rxn = (- 83.2 kJ/mol) - (- 134.2 kJ/mol) = 51.0 kJ/mol.

∵ ΔG°rxn = - RTlnK.

∴ lnK = - (ΔG°rxn)/(RT) = - (51000 J/mol)/(8.314 J/mol.K)(298.0 K) = - 20.58.

∴ K = 1.15 x 10⁻⁹.

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