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

K = 1.15 x 10⁻⁹.

Explanation:

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.

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.

b is the correct answer

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a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.