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At 298 K, consider the equilibrium reaction: 2NO(g) + O2(g) ⇌ 2NO2(g). If the standard free energy of formation of NO(g) is 86.6 kJ mol⁻¹ at 298 K and the equilibrium constant Kp is 1.6 × 10¹², what is the standard free energy of formation of NO2(g) at 298 K?

1. 86600 − R(298)
2. R(298) ln(1.6 × 10¹²) − 86600
3. 0.5[2 × 86600 − R(298) ln(1.6 × 10¹²)]
4. 86600 + R(298) ln(1.6 × 10¹²)

Correct answer: 0.5[2 × 86600 − R(298) ln(1.6 × 10¹²)]

Solution

This option correctly applies the relationship between the standard free energies of formation and the equilibrium constant, taking into account the stoichiometry of the reaction. It averages the contributions from both reactants and products, leading to the correct calculation for the standard free energy of formation of NO2.

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