Question

Consider the equilibrium, N2(g) + 2O2(g) <----> N2O4(g). Calculate the equilibrium constant, Kp if the equilibrium partial pressures are 0.431 atm for N2O4, 0.0867 atm for N2 and 0.00868 atm for O2

6.60 x 104
1.75 x 10-3
573
1.52 x 10-5

Answer 1

The reaction given is

N2(g) + 2O2(g) <----> N2O4(g)

Kp of reaction will be

Kp = pN2O4 / pN2 X p^2O2

Given

pN2O4 = 0.431

pN2 = 0.0867

pO2 = 0.00868

Putting values

Kp = 0.431 / 0.0867 X (0.00868)^2 = 6.60 X 10^4