For the given reaction,

NO (g) + 1/2 O₂ (g) ⇌ NO₂ (g)

Given:

Standard free energy change of the formation of the product (NO₂)

⇒Δ_fG^° (NO₂) = 52.0 kJ/mol

Standard free energy change of the formation of one reactant (NO)

⇒Δ_fG^° (NO) = 87.0 kJ/mol

Standard free energy change of the formation of second reactant(O₂)

⇒Δ_fG^° (O₂) = 0 kJ/mol

a) Calculation of ΔG^°

ΔG^° = Difference in free energy of the reaction when all the reactants and products are in the standard state (1 atmospheric pressure and 298K) Hence,

ΔG^° = ∑ Δ_fG^° (Products) - ∑ Δ_fG^° (Reactants)

⇒ΔG^° = Δ_fG^° (NO₂) – [Δ_fG^° (NO) + 1/2 Δ_fG^° (O₂)]

⇒ΔG^° = 52.0 – [87.0 + 1/2 x 0]

⇒ΔG^° = -35.0 kJ mol^-1

b) Equilibrium constant for the formation of NO₂ from NO and O₂ at 298K

From standard free change of a reaction, we know that,

ΔG^° = -2.303 RT log K_c

Where, ΔG^° is the standard free change of a reaction, value of ΔG^° is -35000 (calculated in part(a))

R = gas constant, value of R is 8.314 J/mol-K

T= Absolute temperature, value of T is 298K (given)

K_c= Equilibrium constant

Using the formula, we write,

log K_c =

⇒log K_c =

⇒log K_c = 6.134

∴ K_c = antilog (6.134)

⇒K_c = 1.361 x 10⁶