In a closed flask at 600 K, one mole of X2Y4(g) attains equilibrium as given below : X2Y4(g) 2XY2(g) At equilibrium, 75% X2Y4(g) was dissociated and the total pressure is 1 atm. The magnitude of rG (in kJ mol-1) at this temperature is . (Nearest Integer) (Given : R = 8.3 J mol-1 K-1; ln 10 = 2.3, log 2 = 0.3, log 3 = 0.48, log 5 = 0.69, log 7 = 0.84)
Answer
8
Solution
X2Y4(g) 2XY2(g)
Equilibrium moles: X2Y4 = 1 − 0.75 = 0.25; XY2 = 2 × 0.75 = 1.5; total = 1.75
P = 1 atm ⇒ pXY2 = 1.51.75 = 67 atm, pX2Y4 = 0.251.75 = 17 atm
Kp = (6/7)21/7 = 367
log Kp = 2 log 6 − log 7 = 2(0.3 + 0.48) − 0.84 = 0.72
rG = -2.3RT Kp
= −2.3 × 8.3 × 600 × 0.72
= −8246.9 J mol-1 ≈ −8.25 kJ mol-1
∴ Magnitude ≈ 8 kJ mol-1
Answer: 8