Question

Calculate:

1. ∆G^Θ and 
2. the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K

\(NO (g) + \frac 12O_2 (g) ⇋ NO_2 (g)\)
where,
 ∆_fG^Θ (NO₂) = 52.0 kJ/mol
∆_fG^Θ (NO) = 87.0 kJ/mol
∆_fG^Θ (O₂) = 0 kJ/mol.

Answer 1

Given:

Reaction:
NO(g) + 1/2 O₂(g) ⇋ NO₂(g)

Standard Gibbs free energies of formation:
Δ_fG°(NO₂) = 52.0 kJ mol⁻¹
Δ_fG°(NO) = 87.0 kJ mol⁻¹
Δ_fG°(O₂) = 0 kJ mol⁻¹

Temperature, T = 298 K

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(a) Calculation of ΔG° for the reaction

ΔG° = Σ Δ_fG°(products) − Σ Δ_fG°(reactants)

ΔG° = Δ_fG°(NO₂) − [Δ_fG°(NO) + 1/2 Δ_fG°(O₂)]

ΔG° = 52.0 − [87.0 + 1/2 (0)]

ΔG° = 52.0 − 87.0

ΔG° = −35.0 kJ mol⁻¹

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(b) Calculation of equilibrium constant (K)

Relationship between ΔG° and equilibrium constant:

ΔG° = −RT ln K

Here,
R = 8.314 J mol⁻¹ K⁻¹
T = 298 K
ΔG° = −35.0 kJ mol⁻¹ = −35000 J mol⁻¹

−35000 = −(8.314 × 298) ln K

ln K = 35000 / (8.314 × 298)

ln K = 14.13

K = e^14.13

K = 1.36 × 10⁶

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Final Answer:

Standard Gibbs free energy change for the reaction:
ΔG° = −35.0 kJ mol⁻¹

Equilibrium constant for the formation of NO₂ at 298 K:
K = 1.36 × 10⁶