Question

plzz solve this question fast i need a perfect answer​

Answer 1

N2O4 ↔ 2NO2

1 0

1−x 2x

Total moles = 1−x + 2x = 1 + x

pN2O4 = (1−x) /(1+x) × P

It is given that x=0.25 and P=1 atm

So,

pN2O4 =1−0.251+ 0.25 × 1 = 0.6 atm

pNO2 = 2x / (1+ x) * P

pNO2 = 2 × 0.251 + 0.25 ×1=0.4 atm

So,

Kp = (pNO2)²/(pN2O4) = (0.4)²/(0.6) = 0.267 atm

Now, let the degree of dissociation at 0.1 atm be y

pN2O4 = (1− y)/ (1+y) × P

. . . . . . .= (1−y) /(1+y) × 0.1

pNO2 = 2y/1+y × P = 2y/1+y × 0.1

Kp= (pNO2)²/ (pN2O4)

=(2y/(1+y) × 0.1)²/((1−y) /(1 + y) × 0.1)

= 0.267 atm

or, 0.667y² = 0.276 atm

or, y = 0.632 %

or, y=63.2%

Hope it will help you...