Question

2) Suppose the initial pressure Po= 0.600 atm at 25 oC for N2O4 before dissociation. Calculate the partial pressures PN2O4 and PNO2 at equilibrium and the fraction of N2O4 molecules that have dissociated. Keq = 8.8 at 25 oC.

Answer 1

At 25 oC, Po (N2O4) = 0.600 atm

Keq = 8.8

N2O4 2NO2
Initially 0.600 0
Finally 0.600 - P 2P

Keq = [P (NO2)]² / P(N2O4)

8.8 = (2P)² / (0.600 - P)

8.8 * (0.600 - P) = (2P)²

5.28 - 8.8P = 4P²  

4P² + 8.8P - 5.28 = 0

Solving above equation:

P = 0.49 atm

Hence,

Partial pressure of NO2 = 2P = 0.98 atm

Partial pressure of N2O4 = 0.600 - P = 0.11 atm