Question

At 60°C, dinitrogen tetroxide dissociates to 50%. Calculate standard free energy change at 1 atm and same temperature.

Answer 1

Dissociation of N₂O₄ proceed as
N₂O₄ ➝ 2NO₂
Mole fraction, dissociation is 50% is 0.5
= (1- 0.5)/(1 + 0.5) = 0.5/1.5 = 0.33
Therefore, P = 0.33 × 1 atm. = 0.33
Similarly for NO₂ = 2 × 0.5/(1.0.5) = 1/1.5 = p = 0.66
Equilibrium constant Kp = P(NO₂)₂/P(N₂O₄)
= (0.66)²/(0.33) = 1.33
G = -RT In Kp
= – 8.314 × 333 × 2.303 × (0.1239)
= -768.3 kJ/mol