The following reaction is performed at 298 K. 2NO(g) + O2(g) \rightleftharpoons 2NO2(g) The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K? (Kp = 1.6×1012)

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Question

The following reaction is performed at 298 K.

2NO(g) + O₂(g) $⇌$ 2NO₂(g)

The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO₂(g) at 298 K? (K_p = 1.6×10¹²)

0.5[2 × 86,600 – R(298) ℓn(1.6 × 10¹²)]

86600 + R(298) ℓn(1.6 × 10¹²)

$86600 - \frac{ℓn (1 .6 \times 10^{12})}{R (298)}$

R(298) ℓn(1.6 × 10¹²) – 86600

2NO(g) + O₂(g) $⇌$ 2NO₂(g)

$ΔG ° = 2 Δ_{f} G º ({NO}_{2}) - Δ_{f} G ° (O_{2}) - 2 Δ_{f} G ° (NO)$

= – RT ℓn K

$Δ_{f} G ° ({NO}_{2}) = \frac{1}{2} (2 \times 86600 - R \times 298 ℓn (1 .6 \times 10^{12}))$

This question involves calculating the standard free energy of formation of NO₂(g) using the relationship between the standard free energy change of a reaction (ΔG°), the equilibrium constant (K_p), and the standard free energies of formation (ΔG_f°) of the reactants and products.

Key Concepts:

The standard free energy change for a reaction (ΔG°) is related to the equilibrium constant (K) by the equation:

$Δ G^{°} = - R T \ln (K)$

where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant (K_p for gases when partial pressures are used).

ΔG° can also be calculated from the standard free energies of formation of the products and reactants:

$Δ G^{°} = [Σ n G_{f}^{°} (products)] - [Σ m G_{f}^{°} (reactants)]$

where n and m are the stoichiometric coefficients.

Step-by-Step Solution:

Step 1: Write the given reaction and known values.

Reaction: $2 NO (g) + O_{2} (g) ⇌ 2 N O_{2} (g)$

Given: ΔG_f°(NO) = 86.6 kJ/mol = 86600 J/mol (since R is in J/mol·K), K_p = 1.6 × 10¹², T = 298 K.

Note: The standard free energy of formation for O₂(g), being an element in its standard state, is 0.

Step 2: Express ΔG° for the reaction using both methods.

From the equilibrium constant: $Δ G^{°} = - R T \ln (K_{p})$

From formation energies: $Δ G^{°} = 2 \times Δ_{G}^{°} f (N O_{2}) - [2 \times Δ_{G}^{°} f (N O) + 1 \times Δ_{G}^{°} f (O_{2})]$

Since ΔG_f°(O₂) = 0, this simplifies to: $Δ G^{°} = 2 \times Δ_{G}^{°} f (N O_{2}) - 2 \times Δ_{G}^{°} f (N O)$

Step 3: Set the two expressions for ΔG° equal to each other.

$2 \times Δ_{G}^{°} f (N O_{2}) - 2 \times 86600 = - R T \ln (K_{p})$

Step 4: Solve for ΔG_f°(NO₂).

$2 \times Δ_{G}^{°} f (N O_{2}) = 2 \times 86600 - R T \ln (K_{p})$

$Δ_{G}^{°} f (N O_{2}) = \frac{2 \times 86600 - R T \ln (K_{p})}{2}$

$Δ_{G}^{°} f (N O_{2}) = 86600 - \frac{R T \ln (K_{p})}{2}$

Final Answer: Comparing this result with the given options, the correct expression is:

0.5[2 × 86,600 – R(298) ℓn(1.6 × 10¹²)]

This matches our derived formula, as 0.5[2×86600 - RT ln(K_p)] simplifies to 86600 - (RT ln(K_p))/2.

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