For the following reaction at 50 °C and at 2 atm pressure, 2N2O5(g) 2N2O4(g) + O2(g) N2O5 is 50% dissociated. The magnitude of standard free energy change at this temperature is x. x = J mol-1 [Nearest integer]. Given : R = 8.314 J mol-1 K-1, log 2 = 0.30, log 3 = 0.48, ln 10 = 2.303, °C + 273 = K
Answer
2474
Solution
2N2O5 2N2O4 + O2
Initial (mol): 2, 0, 0
At equilibrium (50% of 2 mol, i.e. 1 mol, dissociates): 1, 1, 0.5 ⇒ total = 2.5 mol
Partial pressures at P = 2 atm:
pN2O5 = 12.5 2 = 0.8 atm; pN2O4 = 0.8 atm; pO2 = 0.52.5 2 = 0.4 atm
Kp = (pN2O4)2 pO2(pN2O5)2 = (0.8)2(0.4)(0.8)2 = 0.4
T = 50 + 273 = 323 K; log 0.4 = log 4 − 1 = 2(0.30) − 1 = −0.40
G = -2.303RT Kp
= −2.303 × 8.314 × 323 × (−0.40)
= +2473.8 J mol-1
∴ x ≈ 2474
Answer: 2474