Question

If 50% of N2o4 is dissociated at 2atm pressure then calculate Kp

Answer 1

N2O4 --- 2NO2

50% consumed = 0.5

__________N2O4. --2NO2
initial. === 1 mole. 0
change ==.5 ----------–-.0.25
equilibrium= =.5----------0.25
Xno2 = .25 mole/.75 mole = 0.33

Xn2o4 = .5/.75 = 0.67

Since the pressure is 2 atm

Pno2= Xno2× P = 0.33×2=0.66 atm
Pn2o4 = Xn2o4× P= 0.67×2= 1.34 atm

Kp =( Pno2)^2/Pn2o4= (.66)×(.66) /1.34

Kp= 0.325

mark it B