At 60°C, dinitrogen tetroxide (N₂O₄) is 50% dissociated into NO₂. Calculate the standard free energy

Q: What is the correct answer for CBSE11?

The correct answer is C. This is Problem 5.14 from the document. Reaction: N₂O₄(g) ⇌ 2NO₂(g). T = 60°C = 333 K. If N₂O₄ is 50% dissociated, initial moles = 1. At equilibrium, N₂O₄ = 0.5, NO₂ = 1. Total moles = 1.5. Mole fraction of N₂O₄ = 0.5/1.5, NO₂ = 1/1.5. Partial pressures (at 1 atm total pressure): p_N₂O₄ = (0.5/1.5) * 1 atm = 1/3 atm. p_NO₂ = (1/1.5) * 1 atm = 2/3 atm. K_p = (p_NO₂)² / p_N₂O₄ = (2/3)² / (1/3) = (4/

Q: At 60°C, dinitrogen tetroxide (N₂O₄) is 50% dissociated into NO₂. Calculate the standard free energy change (Δ_r G°) at this temperature and at one atmosphere. (R = 8.314 J K⁻¹ mol⁻¹)

The correct answer is C. This is Problem 5.14 from the document. Reaction: N₂O₄(g) ⇌ 2NO₂(g). T = 60°C = 333 K. If N₂O₄ is 50% dissociated, initial moles = 1. At equilibrium, N₂O₄ = 0.5, NO₂ = 1. Total moles = 1.5. Mole fraction of N₂O₄ = 0.5/1.5, NO₂ = 1/1.5. Partial pressures (at 1 atm total pressure): p_N₂O₄ = (0.5/1.5) * 1 atm = 1/3 atm. p_NO₂ = (1/1.5) * 1 atm = 2/3 atm. K_p = (p_NO₂)² / p_N₂O₄ = (2/3)² / (1/3) = (4/

Q: What are the options for this CBSE11 question?

A. -763.8 kJ mol⁻¹ · B. +763.8 kJ mol⁻¹ · C. -763.8 J mol⁻¹ · D. +763.8 J mol⁻¹

Question

At 60°C, dinitrogen tetroxide (N₂O₄) is 50% dissociated into NO₂. Calculate the standard free energy change (Δ_r G°) at this temperature and at one atmosphere. (R = 8.314 J K⁻¹ mol⁻¹)

FirstInTest · Accepted Answer

This is Problem 5.14 from the document. Reaction: N₂O₄(g) ⇌ 2NO₂(g). T = 60°C = 333 K. If N₂O₄ is 50% dissociated, initial moles = 1. At equilibrium, N₂O₄ = 0.5, NO₂ = 1. Total moles = 1.5. Mole fraction of N₂O₄ = 0.5/1.5, NO₂ = 1/1.5. Partial pressures (at 1 atm total pressure): p_N₂O₄ = (0.5/1.5) * 1 atm = 1/3 atm. p_NO₂ = (1/1.5) * 1 atm = 2/3 atm. K_p = (p_NO₂)² / p_N₂O₄ = (2/3)² / (1/3) = (4/9) / (1/3) = 4/3 = 1.333. Δ_r G° = -RT ln K_p = -8.314 J K⁻¹ mol⁻¹ * 333 K * ln(1.333). ln(1.333) ≈ 0.287. Δ_r G° = -8.314 * 333 * 0.287 ≈ -763.8 J mol⁻¹.

Answer

-763.8 kJ mol⁻¹

Answer

+763.8 kJ mol⁻¹

Answer

-763.8 J mol⁻¹

Answer

+763.8 J mol⁻¹

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